TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(b)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(b)

Resolve the following fractions into partial fractions.

Question 1.
\(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}\)
Solution:
Let \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}\) = \(\frac{A}{(x-1)}+\frac{B x+C}{\left(x^2+2\right)}\)
⇒ A (x2 + 2) + (Bx + C) (x – 1)
= 2x2 + 3x + 4 …………(1)
Substituting x = 1 in (1),
we get 3A = 9
⇒ A = 3
Substituting x = 0 in (1),
we get 2A – C = 4
⇒ C = 2A – 4
= 2(3) – 4
⇒ C = 2
Equating coefficient of x2 on both sides in (1)
∴ A + B = 2
= 3 + B = 2
⇒ B = – 1
∴ \(\frac{2 x^2+3 x+4}{(x-1)\left(x^2+2\right)}=\frac{3}{x-1}+\frac{-x+2}{x^2+2}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(b)

Question 2.
\(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}\)
Solution:
Let \(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}\) = \(\frac{A}{x+2}+\frac{B x+C}{\left(1-x+x^2\right)}\)
⇒ A (1 – x + x2) + (Bx + C) (x + 2) = 3x – 1 ………..(1)
Substituting x = – 2 in (1), we get
7A = – 7
⇒ A = – 1
Substituting x = 0 in (1), we get
A + 2C = – 1
⇒ C = 0
Equating coefficients of x2 on both sides in (1)
A + B = 0
⇒ – 1 + B = 0
⇒ B = 1
∴ \(\frac{3 x-1}{\left(1-x+x^2\right)(x+2)}\) = \(\frac{-1}{x+2}+\frac{x}{1-x+x^2}\).

Question 3.
\(\frac{x^2-3}{(x+2)\left(x^2+1\right)}\)
Solution:
Let \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}\) = \(\frac{A}{x+2}+\frac{B x+C}{x^2+1}\)
⇒ A(x2 + 1) + (Bx + C) (x + 2) = x2 – 3
Substituting x = – 2 in (1), we get
5A = 1
⇒ A = \(\frac{1}{54}\)
Substituting x = 0 in (1), we get
A + 2C = – 3
⇒ 2C = – 3 – \(\frac{1}{5}\)
C = \(\frac{-8}{5}\)
Equating coefficients of x2 on both sides,
we get A + B = 1
⇒ B = 1 – \(\frac{1}{5}\)
⇒ B = \(\frac{4}{5}\)
∴ \(\frac{x^2-3}{(x+2)\left(x^2+1\right)}\) = \(\frac{1}{5(x+2)}+\frac{4}{5} \frac{(x-2)}{\left(x^2+1\right)}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(b)

Question 4.
\(\frac{x^2+1}{\left(x^2+x+1\right)^2}\)
Solution:
Let \(\frac{x^2+1}{\left(x^2+x+1\right)^2}\) = \(\frac{A x+B}{\left(x^2+x+1\right)}+\frac{C x+D}{\left(x^2+x+1\right)^2}\)
⇒ (Ax + B) (x2 + x + 1) + Cx + D = x2 + 1 …………(1)
Equating coefficient of x3 on both sides in (1) we get, A = 0.
Equating coefficient of x2 on both sides in (1) we get A + B = 1
⇒ B = 1.
Equating coefficient of x on both sides in (1)
we get A + B + C = 0.
⇒ 0 + 1 + C = 0
⇒ C = – 1
Substituting x = 0 in (1),
we get B + D = 1
⇒ D = 0
∴ \(\frac{x^2+1}{\left(x^2+x+1\right)^2}\) = \(\frac{1}{x^2+x+1}-\frac{x}{\left(x^2+x+1\right)^2}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 7 Partial Fractions Ex 7(b)

Question 5.
\(\frac{x^3+x^2+1}{(x-1)\left(x^3-1\right)}\)
Solution:
\(\frac{x^3+x^2+1}{(x-1)\left(x^3-1\right)}\) = \(\frac{x^3+x^2+1}{(x-1)^2\left(x^2+x+1\right)}\)
Let \(\frac{x^3+x^2+1}{(x-1)^2\left(x^2+x+1\right)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C x+D}{x^2+x+1}\)
∴ A (x2 + x + 1) (x – 1) + B(x2 + x + 1) + (Cx + D) (x – 1)2 = x3 + x2 + 1
⇒ A (x3 – 1) + B (x2 + x + 1) + (Cx + D) (x – 1)2 = x3 + x2 + 1 …………..(1)
Substituting x = 1 in (1),
we get 3B = 3
⇒ B = 1
Equating coefficients of x3 on both sides in (1),
We get A + C = 1
Equating coefficients of x2 on both sides in (1),
weget B – 2C + D = 1
⇒ 2C = D ………….(3)
Equating coefficients of x on both sides in (1)
we get B + C – 2D = 0
⇒ B + C – 4C = 0
⇒ B = 3C
⇒ 1 = 3C
⇒ C = \(\frac{1}{3}\)
Substituting in (2) we get
A = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
also D = 2C
⇒ D = \(\frac{2}{3}\)
∴ \(\frac{x^3+x^2+1}{(x-1)\left(x^3-1\right)}\) = \(\frac{2}{3(x-1)}+\frac{1}{(x-1)^2}+\frac{x+2}{3\left(x^2+x+1\right)}\).

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Telangana TSBIE TS Inter 1st Year Chemistry Study Material 1st Lesson Atomic Structure Textbook Questions and Answers.

TS Inter 1st Year Chemistry Study Material 1st Lesson Atomic Structure

Very Short Answer Type Questions

Question 1.
What is the charge, mass and charge to mass ratio of an electron?
Answer:
i) Charge on electron
= – 1.602 × 10-19 coulomb or
– 4.8 × 10-10 e.s.u.

ii) Mass of electron
= 0.0005485 amu or
– 9.1 × 10-31 kg

iii) Charge-to-mass ratio
= -1.76 × 108 coulomb/gm or
= – 5.28 × 1017 e.s.u/gm

Question 2.
Calculate the charge of one mole of electrons.
Answer:
Charge of one mole of electrons
= 1.602 × 10-19 × 6.023 × 1023 coulomb
= 96500 coulomb
= 1 Faraday.

Question 3.
Calculate the mass of one mole of electrons.
Answer:
Mass of one mole of electrons
= 9.1 × 10-31 × 6.023 × 1023 kg
= 5.48 × 10-7 kg

Question 4.
Calculate the mass of one mole of protons.
Answer:
Mass of one mole of protons
= 1.672 × 10-27 × 6.023 × 1023 kg
= 1.007 × 10-3 kg

Question 5.
Calculate the mass of one mole of neutrons.
Answer:
Mass of one mole of neutrons = 1.675 × 10-27 × 6.023 × 1023
= 1.00885 × 10-3 kg

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 6.
How many neutrons and electrons are present in the nuclei of
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 1
Answer:
Number of electrons is equal to atomic number Z.

Number of neutrons = Mass number (A) – Atomic number (Z).
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 2

Question 7.
What is a black body?
Answer:
The ideal body, which emits and absorbs all frequencies of radiations is called a black body’.

Question 8.
Which part of electromagnetic spectrum does Balmer series belong?
Answer:
The series of lines formed when electron jumps from higher orbits i.e., 3, 4, 5, …………… to the second orbit, is called Balmer series. This series belongs to visible region.

Question 9.
What is an atomic orbital?
Answer:
The three-dimensional region of the space around the nucleus in an atom where the probability of finding an electron is maximum is called atomic orbital’.

Question 10.
When an electron is transferred in hydrogen atom from n = 4 orbit to n = 5 orbit to which spectral series does this belong?
Answer:
When electronic transition tatkes place between 4th to n = 5 orbit, the absorbed frequency is in far I.R. region. It is called, Brackett series.

Question 11.
How many p – electrons are present in sulphur atom?
Answer:
Sulphur, S (Z = 16) 1s²2s²2p63s²3p4
Total no. of ‘p’ electrons is 6 + 4 = 10

Question 12.
What are the values of principal quantum number (n) and azimuthal quantum number (l) for a 3d electron?
Answer:
For a 3d electron, n = 3 and l = 2.

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 13.
What is the complete symbol for the atom with the given atomic number (Z) and atomic mass (A) ?
I) Z = 4, A = 9 II) Z = 17, A = 35 III) Z = 92, A = 233.
Answer:
I) 94Be
II) 3517Cl
III) 23392U

Question 14.
Draw the shape of d orbital.
Answer:
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 3

Question 15.
Draw the shape of dx²-y² orbital.
Answer:
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 4

Question 16.
What is the frequency of radiation of wavelength 600 nm?
Answer:
Frequency υ = \(\frac{c}{\lambda}\)
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 5

Question 17.
What is Zeeman effect?
Answer:
The splitting of spectral lines into a number of closely-spaced lines under the influence of strong magnetic field is called ‘Zeeman effect’.

Question 18.
What is Stark effect?
Answer:
The splitting of spectral lines into a number of closely spaced lines under the influence of strong electric field is called Stark effect’.

Question 19.
To which element does the following electronic configuration correspond?
I) 1s²2s²2p63s²3p¹
II) 1s²2s²2p63s²3p6
III) 1s²2s²2p5
IV) 1s²2s²2p².
Answer:
I) Aluminium
II) Argon
III) Fluorine
IV) Carbon.

Question 20.
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 4000 Å. What is the threshold frequency (υ0)?
Answer:
Velocity of electron is ‘Zero’ means KE = 0,
∴ hυ = hυ0 (or) υ = υ0

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 6

Question 21.
Explain Pauli’s exclusion principle. [Mar. ’19(AP)]
Answer:
Pauli’s exclusion principle :
“No two electrons in an atom can have the same set of four quantum numbers.”

This principle implies that “an orbital can accommodate a maximum of two electrons” with opposite spins
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 7

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 22.
What is Aufbau principle? [Mar. ’19(AP)]
Answer:
In the ground state of the atoms,

“The orbitals are filled up with electrons in the increasing order of their energy.”

The relative order of energy of various orbitals is
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f ………..

Question 23.
What is Hund’s rule?
Answer:
Hund’s rule :
“Pairing up of electrons takes place only when all the available degenerate orbitals in a given sub-shell are filled with one electron each.”

Question 24.
Explain Heisenberg’s uncertainty principle.
Answer:
Heisenberg’s uncertainty principle:
“It is not possible to determine simultaneously and accurately both the position and men mentum (or velocity) of a microscopic moving particle like electron, proton etc.”

Mathematically it can be expressed as,
∆x × ∆p ≥ \(\frac{h}{4 \pi}\)
where
∆x = uncertainty involved in the position
∆p = uncertainty in momentum and h = Planck’s constant.

Question 25.
What is the wavelength of an electron moving with a velocity of 2.0 × 107 ms-1?
Answer:
λ = \(\frac{h}{mv}\)
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 8

Question 26.
An atomic orbital has n = 2, what are the possible values of l and ml?
Answer:
The atomic orbital having n value may be 2s and 2p.
For 2s orbital l = 0, ml = 0.
For 2p orbital l = 1, ml = -1, 0, +1

Question 27.
Which of the following orbitals are possible? (2s, 1p, 3f, 2p)
Answer:
2s and 2p are possible.
1p is not possible because first orbit contains only s-orbital but no p – orbital.

3f is not possible because third orbit contains three orbitals only. They are 3s, 3p and 3d.

Question 28.
The static electric charge on the oil drop is -3.2044 × 10-19C. How many electrons are present on it?
Answer:
Charge on each electron = – 1.6022 × 10-19C
Charge on oil drop = – 3.2044 × 10-19C
∴ Number of electrons
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 9

Question 29.
Arrange the following type of radiation in increasing order of frequency:
(a) X – rays (b) visible radiation (c) micro-wave radiation and (d) radiation from radio waves.
Answer:
The order of frequencies is
Radio waves < microwaves < visible radiation < X – rays.

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 30.
How many electrons in an atom may have n = 4 and in ms = +\(\frac{1}{2}\)?
Answer:
If n = 4 the number of electrons in 4th orbit is 2n² = 2(4)² = 32.
Number of electrons with ms = +\(\frac{1}{2}\) in fourth orbit =16.

Question 31.
How many sub-shells are associated with in = 5?
Answer:
n = 5 orbit contains five sub-shells s, p, d, f and g. i.e., when n = 5; l = 0, 1, 2, 3, 4.

Question 32.
Explain the particle nature of electromagnetic radiations.
Answer:
Electromagnetic radiations contain energy packets (particles) called quantum. The energy associated with each particle is given by
E = hυ

where
h = Planck’s constant
υ = frequency of electromagnetic radiation

The particle nature of electromagnetic radiation is proved by photoelectric effect.

Question 33.
Explain the significance of Heisenberg’s uncertainty principle.
Answer:
Significance of uncertainty principle :

  1. This principle applies for motion of microscopic objects and not so for macroscopic objects.
  2. For massive bodies, the uncertainties have no practical importance.
  3. Microscopic bodies like electrons cannot have fixed paths of motion, according to this principle.

Question 34.
What series of lines are observed in hydrogen spectra?
Answer:
1) Lyman 2) Balmer 3) Paschen 4) Bracket and 5) Pfund series of lines are observed in hydrogen spectra.

Short Answer Questions

Question 1.
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 5 to an energy level with n = 3?
Answer:
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 10

Question 2.
An atom of an element contains 29 electrons and 35 neutrons. Deduce (I) the number of protons and (ii) the electronic configuration of the element.
Answer:
Since the atom contains 29 electrons, the atomic number of element is 29.
∴ The number of protons is also 29.
The element is copper.
Its electronic configuration is 1s²2s²2p63s²3p63d104s¹

Question 3.
Explain giving reasons, which of the following sets of quantum numbers are not possible.
a) n = 0, l = 0, ml = 0, m2 = +\(\frac{1}{2}\)
b) n= 1, l = 0, ml = 0, m2 = –\(\frac{1}{2}\)
c) n = 1, l = 1, ml = 0, m2 = +\(\frac{1}{2}\)
d) n = 2, l = 1, ml = 0, m2 = +\(\frac{1}{2}\)
e) n = 3, l = 3, ml = – 3, m2 = +\(\frac{1}{2}\)
f) n = 3, l = 1, ml = 0, m2 = +\(\frac{1}{2}\)
Answer:
Set (a) is not possible, because zero value is not permissible for n.
Set (c) is not possible, because for n = 1, possible value of l is zero only.
Set (e) is not possible, because for n = 3, permissible values for l are 0, 1, 2 only.

Question 4.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de-Broglie wavelength associated with the electron revolving around the orbit.
Answer:
According to Bohr’s model
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 11
So the circumference of the Bohr orbit (2πr) must be equal to an integral (n) multiple of the de-Broglie wavelength (λ).

Question 5.
The longest wavelength doublet absorption transition is observed at 589.0 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.
Answer:
Frequency υ = \(\frac{c}{\lambda}\)
For the radiation having wavelength 589.0 nm the frequency
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 12
For the radiation having wavelength 589.6 nm the frequency
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 13
Energy of radiation having wavelength 589.0 nm is
E = hυ or h\(\frac{c}{\lambda}\)
h is Planck’s constant 6.626 × 10-34 Js
= 6.626 × 10-34 Js × 5.09 × 1014 Hz
= 33.73 × 10-20 J

Energy of the radiation having wavelength 589.6 is
E = hυ = h\(\frac{c}{\lambda}\)
= 6.626 × 10-34 Js × 5.088 × 1014 Hz
= 33.71 × 10-20 J

The energy difference between two excited states
= 33.73 × 10-20 J – 33.71 × 10-20 J
= 2 × 10-22J

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 6.
What are the main features of quantum mechanical model of an atom?
Answer:
Quantum mechanical model of atom explains the motion of very minute particles which possess both particle and wave nature. Schrodiriger’s equation applies to these type of particles and it is given by
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 14

Salient features of the quantum mechanical model of atom:

  1. The energy of an electron in an atom is quantised. It means, the electron can have only certain values of energy.
  2. The energy levels in an atom are quantised, which are the allowed solutions of Schrodinger wave equation, applied for electron wave.
  3. All the information about the orbiting electron in atom is contained in the orbital wave function, Ψ.
  4. The exact path of an electron cannot be determined accurately. We can only find the probability of the electron at different points in an atom.
  5. Ψ² is the probability density, which is always positive. From the value of Ψ² at different points within the atom, it is possible to find the orbital’ of electron.

Orbital is the three-dimensional space around the nucleus of an atom, where there is high probability of finding the electron.

Question 7.
What is a nodal plane? How many nodal planes are possible for 2p- and 3d- orbitals?
Answer:
The probability to find the electron at a point within the nucleus in an atom is negligible. This point is called nodal point’. ‘The plane passing through this point is called ‘nodal plane’ (or angular node). So nodal plane is the plane of zero electron density.

The number of nodal planes for an orbital is equal to its Azimuthal quantum number 1 value.
Thus,
No. of nodal planes for s – orbital (l = 0) is zero.
No. of nodal planes for p – orbital (l = 1) isl.
No. of nodal planes for d – orbital (l = 2) is 2.
No. of nodal planes for f – orbital (l = 3) is 3.

Question 8.
The Lyman series occurs between 91.2 nm and 121.6 nm, the Balmer series occurs between 364.7 nm and 656.5 nm and the Paschen series occurs between 820.6 nm and 1876 nm. Identify the spectral regions to which these wavelengths correspond.
Answer:

Name of series Spectral Region
1) Lyman Ultra Violet
2) Balmer Visible
3) Paschen Near infrared

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 9.
How are the quantum numbers n, 1, m, for hydrogen atom are obtained?
Answer:
Atomic number (Z) of hydrogen atom = 1.
Its electron configuration is 1s¹. From this configuration, it is evident that the electron in ‘H’ atom involves in 1st orbit with n = 1. For an ‘S’ orbital l = 0, m = 0.
∴ The four quantum number values for ‘H’ atom are n = 1, l = 0, m = 0, s = + ½.

Queestion 10.
A line in Lyman series of hydrogen atom has a wavelength of 1.03 × 107 m. What is the initial energy level of the electron?
Answer:
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 15
Hence by trial and error method, we found, n2 = 3

So the electron belongs to 3rd energy level (M = 3) originally.

Question 11.
If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the momentum of the electron.
Answer:
According to Heisenberg’s uncertainty principle
∆x ∆p = \(\frac{h}{4 \pi}\)
∆x is uncertainty in position
p is uncertainty in momentum
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 16

Question 12.
If the velocity of the electron is 1.6 × 106 ms-1, calculate the de-Broglie wavelength associated with this electron.
Answer:
According to de-Broglie equation λ = \(\frac{h}{mv}\)
λ = wavelength of electron
h = Planck’s constant 6.626 × 10-34 Js
m = mass of electron 9.1 × 10-31 kg
v = velocity of electron 1.6 × 106 ms-1
Substituting these values in de-Broglie equation
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 17

Question 13.
Explain the difference between emission and absorption spectra. (AP ’15)
Answer:

Emission spectra Absorption spectra
1) The spectra is formed when electron jumps from higher orbits to lower orbits. 1) The spectra is formed when electron jumps from lower orbit to higher orbit.
2) It is formed due to emission of energy in quanta. 2) It is formed due to absorption of energy in quanta.
3) It contains bright lines on dark background. 3) It contains dark lines on bright background.
4) It can be continuous or discontinuous. 4) It is always dis-continuous.

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 14.
The quantum numbers of electrons are given below. Arrange them in order of increasing energies. Do any of these combinations have same energy?
a) n = 4, l = 2, ml = – 2, ms = + ½
b) n = 3, l = 2, ml = – 1, ms = – ½
c) n = 4, l = 1, ml = 0, ms = + ½
d) n = 3, l = 1, ml = – 1, ms = – ½
Answer:
Rule :

  1. The relative energy of an electron is given by its (n + l) value, m and s contributions are very little.
  2. If two electrons have the same (n + l) value, electron with lower n value has lower energy.
    a) n = 4, l = 2 (n + l) value = 6
    b) n = 3, l = 2 (n + l) value = 5
    c) n = 4, l = 1 (n + 0 value = 5
    d) n = 3, l = 1 (n + 0 value = 4

Increasing order of energies of electrons is : (4) < (2) < (3) < (1)

Question 15.
The work function for Cesium atom is 1.9 eV. Calculate the threshold frequency of the radiation. If the Cesium element is irradiated with a wavelength of 500 nm, calculate the kinetic energy of the ejected photoelectron.
Answer:
The energy (E) of a 300 nm photon is given by.
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 18
Threshold energy = 1.9 eV × 1.6022 × 10-19
= 3.044 × 10-19 J

Kinetic energy of electron = Energy of photon – Threshold energy
= 3.976 × 10-19 J – 3.044 × 10-19 J
= 9.32 × 10-20 J

Question 16.
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Answer:
Radius of the orbit = 1.3225 nm
= 1.3225 × 10sup>-7 cm = 13.225 × 10-8 cm

∵ The radius of the orbit = 0.529 × n² = 13.225
∴ n = \(\sqrt{\frac{13.225}{0.529}}\) = 5
Radius of the second orbit = 211.6 pm
= 2.116 × 10-8 cm
0.529 × n² = 2.116
or n = 2

The electron transition takes place 5 to 2. So, transition belongs to Balmer series.
It is in the visible region.
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 19
∴ The wavelength of the radiation = 4328 Å

Question 17.
Explain the difference between orbit and orbital.
Answer:

Orbit Orbital
1. It is the circular path around the nucleus in which the electron revolves. 1. It is the region of the space around the nucleus where theprobability of finding the electron is maximum.
2. It represents planar motion of electron. 2. It represents three dimensional motion of electron.
3. Orbits are non- directional. Hence they cannot explain shape of molecules. 3. Orbitals (except S – orbital) have directional character. Hence they can account for shapes of molecules.
4. The maximum no. of electrons in an orbit is given by 2n². 4. An orbital can accommodate a maximum of 2 electrons with opposite spins.

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 18.
Explain the photoelectric effect.
Answer:
Photoelectric effect :
J.J. Thomson and P. Lenard showed that “when a beam of light of suitable wavelength is allowed to fall on the surface of a metal, the electrons are emitted from the surface of the metal”. This phenomenon is called, photoelectric effect.

Explanation of photoelectric effect by Einstein:
Einstein (1905) explained photo-electric effect with the help of Planck’s quantum theory of radiation.

According to quantum theory of radiation, a photon of light of frequency, υ is associated with energy equal to hυ. When a photon of light of frequency υ (υ is higher than threshold frequency) falls on a metal, some of the energy associated with the photon is consumed to separate the electron from the surface and the remaining energy is imparted to the ejected electron, to give it certain velocity, say, equal to v. Due to this velocity, the emitted electron gains kinetic energy equal to \(\frac{1}{2}\)mv². The proton of energy which is consumed to separate the electron is equal to the binding energy of the electron.

This energy is called threshold energy’ or work function’ and is equal to the product of h and threshold frequency (υ0) of a photon of radiation falling on the metal surface. Thus this energy is equal to hu0. This is the minimum energy associated with the photon (of frequency υ0) of the incident radiation which is consumed to separate one electron from the metal surface. The above discussion shows that hυ is equal to the sum of \(\frac{1}{2}\)mv² (K.E. of one electron) and hυ0
i.e., hυ = hυ0 + \(\frac{1}{2}\)mv²

Long Answer Questions

Question 1.
Explain Rutherford’s nuclear model of an atom. What are its drawbacks?
Answer:
In 1910, Rutherford proposed a model of the atom to explain the results of his experiment. It is called, Planetary model or Nuclear model of atom. The main features of the model are given below.

  1. Atom is spherical and mostly hollow.
  2. The positive charge and the mass of the atom is concentrated in a small region at the centre of the atom. The region is called, ‘Nucleus’.
  3. The electrons present outside the nucleus (called, extra-nuclear electrons) are equal in number to the protons inside the nucleus.
  4. Just as the planets revolve around the Sun, the electrons revolve around the nucleus.
  5. The revolving electron is under the in-fluence of two forces – i) The centripetal force of attraction towards nucleus and ii) the centrifugal force directed away from the orbiting path. These two forces being equal and opposite, balance each other and the electron continues to move in its orbit.

Drawbacks in Rutherford’s Model:
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 20

  1. According to electrodynamics, any charged particle moving under the influence of opposite charge should lose energy continuously, come closer and closer to the nucleus in a spiral path and ultimately fall in the nucleus as shown in figure. Then the atom shall collapse due to the merging of electrons with the nucleus. But it is not happening.
  2. If the electrons lose energy continuously, the atomic spectrum should have a continuous band. But the atomic spectrum of the elements are found to contain discrete lines.

Question 2.
Explain briefly the Planck’s quantum theory.
Answer:
Planck’s Quantum Theory:

  1. The emission of radiation is due to vibrations of charged particles (electrons) in the body.
  2. The emission of radiation is not continuous, but in discrete packets of energy called ‘quanta’. This emitted radiation propagates in the form of waves.
  3. The energy associated with each quantum (E) is given by E = hυ, where υ is the frequency of radiation and h is Planck’s constant (h = 6.63 × 10-27 ergs)
  4. A body can emit or absorb either one quantum (hυ) of energy or some whole number multiples of it.

Success of Planck’s Quantum Theory :
This theory successfully explains the black body radiations. A black body is a perfect absorber and also a perfect radiator of radiations.

Explanation of Graph:
The radiations emitted by a hot black body, when passed through a prism, produce a spectrum of different wavelengths.

A plot of the ‘intensity of radiation’ against wavelength’ gives a curve. Such curves obtained at different temperatures of a black body are shown in the graph.

The following conclusions can be drawn from a study of the shapes of the curves.
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 21

  1. At a given temperature, the intensity of radiation increases with wavelength, reaches a maximum and then decreases.
  2. As the temperature increases, the peak of the curve (maximum point in curve) shifts to lower wavelengths.

The wave theory of light’ could not explain the above experimental results, satisfactorily.

Question 3.
What are the postulates of Bohr’s model of hydrogen atom? Discuss the importance of this model to explain various series of line spectra in hydrogen atom. [AP 17, 15; TS 16; IPE 14, 13, Mar. ’19 (AP); Mar.18 (AP & TS)]
Answer:
A) Postulates of Bohr’s theory:
1) Electrons revolve around the nucleus in certain, definite circular paths, called orbits. They are also known as energy levels or shells.

2) Each orbit is associated with a definite amount of energy. Hence these orbits are called energy orbits or energy levels or energy shells.

3) As the value of n increases, the size and energy of the orbit increases.

4) As long as the electrons revolve in the circular orbits, they neither emit nor absorb energy. Hence, these orbits are also called stationary orbits.

5) More than one stationary orbit is possible for any electron and the angular momentum of the revolving electron is quantized. That is, the electrons revolve in those circular orbits in which the angular momentum of the revolving electron is an integral multiple of h/2π.

This can be expressed as mvr = n × \(\frac{h}{2 \pi}\)
where m = mass of the electron;
v = velocity of the electron
r = radius of the orbit;
h – Planck’s constant,
n – Principal quantum number.

6) The energy of electron changes when the electron jumps from one orbit to another orbit. When the electron jumps from a lower orbit to a higher orbit, energy is absorbed. Similarly when the electron jumps from a higher orbit to a lower orbit, energy is emitted. The energy emitted or absorbed is given by the equation, ∆E = E2 – E1 = hυ
where
E1 = Energy of lower orbits;
E2 = Energy of higher orbits;
∆E = Difference in energy;
h = Planck’s constant;
υ = Frequency.

B) Explanation of formation of different lines in various series of Hydrogen atomic spectrum :
Hydrogen atom has only one electron and it revolves around the nucleus in the first orbit. When certain amount of hydrogen gas containing a large number of electrons, is heated or exposed to light energy or is subjected to electric discharge, the different electrons absorb different amount of energy and get excited to different higher orbits. But, they do not stay there for longer periods. Hence, they try to come back to lower orbits (de-excitation). This de-excitation need not necessarily be the same for all the atoms. Some excited electrons may come back from any of the higher energy levels to lower energy level n = 1. Then Lyman series is formed in the U.V. region. Some excited electrons may come back from any of the higher energy levels to lower energy level n = 2. Then

Balmer series is formed in the visible region. Similarly, when the excited electron comes back, from any of the higher energy levels to lower levels 3, 4 & 5, we get Paschen, Bracket and Pfund series respectively in the I.R. region.

The de-excitation process from higher energy level to lower energy level also may take place in a single step or in different steps. For example, the de-excitation process of an electron from 4th energy level to 1st energy level can be as shown below.

Direct step : From 4th level to 1st (i.e.,) 4 → 1
Different steps : 4 → 3 → 2 1; 4 → 3 1; 4 → 2 → 1.

For each electronic transition one line is formed in the spectrum. Thus we get a large number of lines in a given series.
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 22

Question 4.
Explain the success of Bohr’s theory for hydrogen atom.
Answer:
The main success of Bohr’s model:
1. Bohr’s model could explain the stability of an atom:
According to Bohr’s model, an electron revolving in a particular orbit cannot lose energy. An electron will absorb energy when it jumps from lower energy level to higher energy level or lose energy when it jumps from higher energy level to lower energy level. When there is no vacant lower energy level the electron keeps revolving in the same orbit without losing energy and thus explains the stability.

2. Bohr theory helped to calculate the energy of an electron in an orbit:
Basing on Bohr’s model a mathematical relation can be developed to calculate the energy of an electron in nth orbit.
En = \(\frac{-\mathrm{k}^2 2 \pi^2 m \mathrm{e}^4}{\mathrm{n}^2 \mathrm{~h}^2}\)
where m = Mass of electron
e = Charge on electron
h = Planck’s constant
k = Coulombs constant 9.0 × 109 Jm/C²

Substituting the values of m, e, h, π and k in the above expression we get
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 23
Using Bohr’s model the radius of an orbit in hydrogen atom can be calculated
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 24

3. Bohr’s model could explain the atomic spectrum of hydrogen :
The different spectral lines in hydrogen spectrum is due to the emission of energy in the form of light when the electron jumps to lower energy level. When hydrogen gas is supplied from some external source, the electrons in different atoms absorb energy and excited to higher energy levels. When the electron returns back to lower energy levels it gives out energy in the form of quantum equal to the difference of energies between the two energy levels.
E2 – E1 = hυ or υ = \(\frac{E_2-E_1}{h}\)

where E1 is the energy of lower energy level, E2 is the energy of high energy level and h is Planck’s constant.

According to Bohr’s model, in a hydro gen atom (or in any other atom) E2 and E1 can have only certain definite values. From this, it follows that u can have only certain fixed values. Thus Bohr’s model explains why there are certain discrete lines in the spectrum of hydrogen.

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 5.
What are the consequences that lead to the development of quantum mechanical model of an atom?
Answer:
In 1923 Louis de-Broglie proposed that like light matter also has dual character. It exhibits wave as well as particle nature. He derived an equation to calculate the wavelength (λ) of the wave associated with a particle of mass m, moving with velocity v and it is given by
λ = \(\frac{h}{mv}\) (or) λ = \(\frac{h}{p}\)

‘h’ is Planck’s constant and P is momentum of the particle. It may be noted that wave character has significance only in the case of sub-microscopic particles such as electrons.

The precise location of a wave cannot be predicted as a wave is not located at a particular point but extends in space. Thus the wave nature of electron puts some restriction in finding the position precisely. W. Heisenberg put forward this limitation in the form of uncertainty principle. According to Heisenberg,

“It is not possible to determine simultaneously and accurately both the position and momentum of a small moving particle, such as electron.”

It can be expressed mathematically as,
∆x × ∆p ≥ \(\frac{h}{4 \pi}\)
∆x × m ∆v ≥ \(\frac{h}{4 \pi}\)

where ∆x = uncertainty in position
∆p = uncertainty in momentum
m = mass of particle
∆v = uncertainty in velocity
h = Planck’s constant.

The uncertainty principle is due to the direct consequence of dual nature of matter.

A microscopic object such as electron has both observable wave-like and particle-like properties. The behaviour of such particles cannot be explained on the basis of classical mechanics which is based on Newton’s laws of motion. In order to explain the behaviour of electrons a new branch of science called quantum mechanical model of atom was developed. Quantum mechanics is a theoretical science that takes into account the dual nature of matter.

Question 6.
Explain the salient features of quantum mechanical model of an atom.
Answer:
Quantum mechanical model of atom explains the motion of very minute particles which possess both particle and wave nature. Schrodiriger’s equation applies to these type of particles and it is given by

Salient features of the quantum mechanical model of atom:

  1. The energy of an electron in an atom is quantised. It means, the electron can have only certain values of energy.
  2. The energy levels in an atom are quantised, which are the allowed solutions of Schrodinger wave equation, applied for electron wave.
  3. All the information about the orbiting electron in atom is contained in the orbital wave function, Ψ.
  4. The exact path of an electron cannot be determined accurately. We can only find the probability of the electron at different points in an atom.
  5. Ψ² is the probability density, which is always positive. From the value of Ψ² at different points within the atom, it is possible to find the orbital’ of electron.

Orbital is the three-dimensional space around the nucleus of an atom, where there is high probability of finding the electron.

Question 7.
What are the limitations of Bohr’s model of an atom? [AP ’15; ’13]
Answer:
Bohr’s model of atom could explain the
i) stability of atom ii) the line spectrum of hydrogen atom and ions with a single electron (He+, Li2+, Be3+ ……………)
But this model could not explain the following observations.

  1. Bohr’s model cannot explain the spectra of atoms or ions having more than one electron.
  2. When the spectrum of hydrogen was taken with a spectroscope of high resolving power, it was observed that each line is actually a group of fine lines’. Bohr’s model cannot explain this fine structure in the atomic spectra.
  3. It cannot explain Zeeman effect (splitting of spectral lines in the presence of magnetic field) and stark effect (splitting of spectral lines in an electric field).
  4. It also cannot explain the ability of atoms to form molecules by chemical bonds.
  5. It failed to explain the dual nature of electrons and also the geometry and shapes of molecules.

Question 8.
What are the evidences in favour of dual behaviour of electron?
Answer:
The particle nature of electron can be explained by the photoelectric effect and Planck’s Quantum Theory.

When light falls on the surface of a metal, electrons are ejected from the surface with some kinetic energy by absorbing energy in the form of quantum. The kinetic energy of photo electrons is directly proportional to frequency. Kinetic energy of photoelectrons is independent of intensity because increase energy does not affect the energy of photons rather it simply increases the number of photons falling on the surface of metal and hence the number of photoelectrons. This proves the particle nature of electrons.

The wave nature of electrons was verified experimentally by Davisson and Germer by carrying out diffraction experiments with a beam of fast moving electrons. This wave nature of electrons is utilised in the construction of electron microscope.

From the evidences of photoelectric effect in favour of particle nature and also from the evidences of experiments conducted by Davisson and Germer in favour of wave nature, we can conclude that electron has dual nature.

Question 9.
How are the quantum numbers n, l and ml arrived? Explain the significance of these quantum numbers. [AP & TS ’15, ’16 TS Mar. ’19, ’17) (IPE ’14, ’13, ’10, ’06, ’04]
Answer:
To describe an electron completely four quantum numbers were predicted.

They are

  1. Principal quantum number, n
  2. Azimuthal quantum number, l
  3. Magnetic quantum number, ml and
  4. Spin quantum number, ms.

a) Principal Quantum Number:

  1. This was proposed by Neils Bohr.
  2. It is denoted by the letter ‘n’.
  3. It represents the circular orbits around the nucleus.
  4. As the value of n increases the size and energy of the orbit increases.
  5. According to number method n has the values 1, 2, 3, ……… According to letter method ‘n’ can be represented by the letters K, L, M, ………
  6. In any orbit, the number of sub-orbits = n,
    number of orbitals = n² ;
    number of electrons =2n².
  7. This Quantum number describes the size and energy of the orbit.

b) Azimuthal Quantum Number:

  1. This was proposed by Sommerfeld.
  2. It is also known as Angular momentum quantum number.
  3. It is denoted by the letter ‘l’.
  4. This quantum number represents the sub-levels present in the main levels.
  5. The sub-levels are s, p, d and f.
  6. The l values of s, p, d and f sub-levels are 0, 1, 2 and 3 respectively.
  7. The first main level contains only one sub-level and it is s. The second main level contains s, p sub-levels. The third main level contains s, p, d sub-levels. The fourth main level contains s, p, d and f sub-levels.
Main level Sub-levels Sub-sub-levels
1 s 1s
2 s 2s
p 2p
3 s 3s
p 3p
d 3d
4 s 4s
p 4p
d 4d
f 4f

8) The relative energy values of Is, 2s, 2p etc., can be calculated by adding up their n and 1 values.
Ex :1. Energy value of Is =1+0 = 1
2. Energy value of 2p = 2 + 1 = 3
3. Energy value of 3d = 3 + 2 = 5
4. Energy value of 4f =4 + 3 = 7
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 25

9) This Quantum number describes the shape of the orbital.

Sub-level Shape
s spherical
p dumb-bell
d double dumb-bell
f four fold dumb-bell

c) Magnetic Quantum Number:

  1. This was proposed by Lande.
  2. It is denoted by the letter ‘m’.
  3. This quantum number describes the sub – sub levels or orbitals present in a given sub – level.
  4. ‘ml‘ has values from -l to +l through ‘0’.
  5. The total number of ‘m’ values for a given value of T is ( 2l + 1).
  6. All the orbitals present in a given sublevel possesses the same energy values, because they possess the same n and l values.
  7. This quantum number describes the orientation of the orbitals in space.

d) Spin Quantum Number:

  1. Uhlenbeck and Goudsmit proposed it.
  2. It is denoted by ms.
  3. This quantum number describes the spin of the revolving electron.
  4. ‘ms‘ value of clockwise electron is +½ and that of anticlockwise electron is -½.
  5. Clockwise revolving electron is represented by +½ and anticlockwise revolving electron is represented by -½ .
  6. This quantum number describes the direction of spin of the revolving electron.

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 10.
Explain the dual behaviour of matter. Discuss its significance to microscopic particles like electrons.
Answer:
de-Broglie’s hypothesis:
The light is found to exhibit wave nature as well as particle nature (dual nature). Based on this idea of dual nature of light’, de Broglie in 1924 proposed that “all micro-particles including the electron moving with high velocity are associated with dual nature (i.e.,) both particle and wave nature.

De-broglie derived an expression for the wavelength of the moving electron.

Expression for de-Broglie wavelengths:
According to ’Planck’s quantum theory’, energy of a photon,
E = hυ; But υ = \(\frac{c}{\lambda}\)
∴ E = h.\(\frac{c}{\lambda}\) ………… (1)

Einstein’s mass-energy equivalence equation is E = mc² ………….. (2)
combining eqns. (1) and (2),
\(\frac{h.c}{\lambda}\) = mc² or λ = \(\frac{h}{mc}=\frac{h}{p}\)
(p = me = momentum)

This equation is applicable to photons as well as to all microparticles, moving with high speed.
∴ We can write in general, λ = \(\frac{h}{p}=\frac{h}{mv}\)

where m = mass of the microparticle and v = its velocity
‘λ’, is called, de-Broglie wavelength or material wavelength.

Significance of de-Broglie’s concept:
According to Bohr’s theory, electron revolves in an orbit in which its angular momentum (mvr) is an integral multiple of \(\frac{h}{2 \pi}\). Bohr assumed electron as a particle. Hence his equation can be taken as mvr = n(\(\frac{h}{2 \pi}\)) where n = a whole number.

According to de Broglie, electron behaves as a standing (or stationary) wave which extends round the nucleus in a circular orbit. If the ends of the electron wave meet to give a regular series of crests and troughs, the electron wave is said to be in phase’. It means, there is constructive interference of electron waves and the electron motion has a character of standing wave or nonenergy radiating motion. Always it is a necessary condition to get an electron wave in phase’ such that the circumference of the Bohr’s orbit (= 2πr) is equal to the whole number of multiples of the wavelength (λ) of the electron wave.
nλ = 2πr
λ = \(\frac{2 \pi r}{n}\)
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 26

This is Bohr’s equation which stipulates that “the angular momentum of an electron moving round the nucleus is an integral multiple of \(\frac{h}{2 \pi r}\)“. This shows that, de-Broglie’s theory and Bohr’s theory are in agreement with each other.

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 11.
What are various ranges of electromagnetic radiation? Explain the characteristics of electromagngtic radiation.
Answer:
According to Maxwell, “electromagnetic radiation”, is made up of electromagnetic waves. An electromagnetic wave propagating through space is a combination of two components, one is varying electric field and the other is the varying magnetic field. These two fields are perpendicular to each other and are also perpendicular to the direction of propagation of wave.

Characteristics of electromagnetic radiations :
1) These are produced by oscillating charged particles in a body.

2) The radiations can pass through vacuum also. So medium for transmission is not required.
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 27

3) The wavelength (λ) is the distance between two neighbouring crests or troughs of the wave. Unit for λ is cm, m, nm, Å.

4) Frequency of the wave (υ) is the number of waves which cross a particular point in one second.
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 27.1
Unit of frequency is cycles per sec (cps) or hertz (Hz)
(1 Hz = 1 cps).
The frequency is inversely proportional to its wavelength.
υ ∝ \(\frac{1}{\lambda}\)
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 28

5) Velocity (c) of a wave is the distance travelled by the wave in one second.
Velocity = wavelength × frequency
⇒ c = λ × υ. Velocity of all electromagnetic radiations in space or in vacuum is the same and is equal to 3 × 1010 cm/s.

6) Wave number (\(\overline{\mathrm{υ}}\)) is the number of wavelengths per centimeter. It is reciprocal of wavelength.
\(\overline{\mathrm{υ}}\) = \(\frac{1}{\lambda}\) Unit of \(\overline{\mathrm{υ}}\) is cm-1.

7) Amplitude (A) is the height of the crest or depth of trough of a wave. It determines the intensity or brightness of the light.

8) Electromagnetic spectrum is the spectrum which shows the wavelengths or frequencies or wave numbers of various regions of electromagnetic radiations.
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 29

Question 12.
Define atomic orbital. Explain the shapes of s, p and d orbitals with the help of diagrams.
Answer:
Atomic orbital :
The three-dimensional space around the nucleus in an atom where the probability of finding an electron is maximum is called an atomic orbital.

Shapes of s, p and d orbitals :
The shapes of the orbitals are the angular distribution curves’, obtained as one of the solutions of Schrodinger wave equation.

1) Shape of s – orbital:
i) s orbital is spherical in shape.
ii) As the value of principal quantum number increases, size of s orbital increases. Thus, sizes are : 1s < 2s < 3s …………
iii) An ‘s’ orbital has no directional property. s-orbital
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 30

2) Shape of p-orbital :
Shape of p-orbital is dumb-bell. There are three p – orbitals. They are px, py and pz orbitals. Each orbital has two lobes. These lobes are oriented along their respective axes. Each p-orbital has one nodal plane.
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 31

3) Shape of d – orbital :
d – orbital has double dumb-bell shape. There are five d – orbitals. They are dxy, dyz, dxz, dx²-y² and d – Each d orbital has 4 lobes. In dxy, orbital, the lobes are placed in – between the x and y axes, similar is the case with other orbitals, dyz and dxz. In d orbital, two lobes lie along z axis and there is a ring of electron-cloud around the centre. In dx²-y² orbital, the lobes lie along x and y-axes. Each d orbital has two nodal planes.
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 32
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 33

Question 13.
Explain the boundary surfaces for three 2p orbitals and five 3d orbitals diagrammatically.
Answer:
Shape of p – orbital:
Shape of p – orbital is dumb-bell. There are three 2p – orbitals. They are 2px, 2py and 2pz orbitals. Each orbital has two lobes. These lobes are oriented along their respective axes. Each p orbital has one nodal plane.
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 34

Shape of d – orbital:
d – orbital has double dumb-bell shape. There are five d – orbitals. They are dxy, dyz, dxz, dx²-y² and . Each d orbital has 4 lobes. In dxy orbital, the lobes are placed in – between the x and y axes, similar is the case with other orbitals, dyz and dxz. In d orbital, two lobes lie along z axis and there is a ring of electron-cloud around the centre. Inx²-y², orbital, the lobes lie along x and y-axes. Each d orbital has two nodal planes.
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 35
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 36

Question 14.
Illustrate the reasons for the stability of completely filled and half-filled subshells.
Answer:
The completely filled and half filled orbitals give stability to the atoms due to following reasons.

1. Completely filled and half filled orbitals give spherical symmetry to the atoms.
2. Also they give more number of Coulom- bic exchange energies.

These can be explained by taking the electronic configurations of Chromium and Copper as examples.

The expected configurations for Cr and Cu are :
Cr (Z = 24) 1s²2s²2p63s²3p64s²3d4
Cu (Z = 29) 1s²2s²2p63s²3p64s²3d9.

The two sub-shells 4s and 3d differ slightly in their energies. Then the electron shifts from a sub-shell of lower energy (4s) to a sub-shell of higher energy (3d), provided such a shift results in all orbitals of the same sub-shell of higher energy (here, 3d) getting either completely filled or half- filled. So the electronic configuration of Cr and Cu are
Cr (Z = 24) 1s²2s²2p63s²3p64s¹3d5
Cu (Z = 29) 1s²2s²2p63s²3p64s¹3d10

These configurations are more stable.

The completely filled and half-filled subshells are more stable for the following reasons:

1) There is symmetrical distribution of electrons in the sub-shells. So they are more stable.
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 37
The 3d orbitals are completely filled.

2) When two or more electrons with the same spin are present in the degenerate orbitals, these electrons exchange their positions. Then some energy is released, called exchange energy’. Then the energy of the atom decreases and it becomes more stable. The more the number of ‘exchange pair’ of electrons, the greater would be the decrease of energy and hence the greater is the stability of the atom. In the above configurations of Cr and Cu, there are more ‘exchange pairs’ than in their previous configurations.
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 38
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 39
Possible exchange for a d5 configuration

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 15.
Explain emission and absorption spectra. Discuss the general description of line spectra in hydrogen atom.
Answer:
1) Emission spectrum:
The spectrum produced by emitted radiation is called emission spectrum. This spectrum contains bright lines on dark background. This spectrum is of two types. They are (a) Continuous spectrum (b) Discontinuous spectrum.

a) Continuous spectrum :
This spectrum contains a continuous band of colours like rainbow. This spectrum is given by molecules and is also known as Molecular spectrum.

b) Discontinuous spectrum :
This spectrum consists of sharp, distinct lines. This spectrum is given by atoms and is also known as atomic spectrum.

Absorption spectrum:
The spectrum produced by absorbed radiation is known as absorption spectrum. This spectrum contains dark lines on bright background.
The position of the lines or bands in both emission and absorption spectra is same.

2) Explanation of formation of different lines in various series of Hydrogen atomic spectrum :
Hydrogen atom has only one electron and it revolves around the nucleus in the first orbit. When certain amount of hydrogen gas containing a large number of electrons, is heated or exposed to light energy or is subjected to electric discharge, the different electrons absorb different amount of energy and get excited to different higher orbits. But, they do not stay there for longer periods. Hence, they try to come back to lower orbits (de-excitation). This de-excitation need not necessarily be the same for all the atoms. Some excited electrons may come back from any of the higher energy levels to lower energy level n = 1. Then Lyman series is formed in the U.V. region. Some excited electrons may come back from any of the higher energy levels to lower energy level n = 2. Then

Balmer series is formed in the visible region. Similarly, when the excited electron comes back, from any of the higher energy levels to lower levels 3, 4 & 5, we get Paschen, Bracket and Pfund series respectively in the I.R. region.

The de-excitation process from higher energy level to lower energy level also may take place in a single step or in different steps. For example, the de-excitation process of an electron from 4th energy level to 1st energy level can be as shown below.

Direct step : From 4th level to 1st (i.e.,) 4 → 1
Different steps : 4 → 3 → 2 1; 4 → 3 1; 4 → 2 → 1.

For each electronic transition one line is formed in the spectrum. Thus we get a large number of lines in a given series.

Additional Questions & Answers

Question 1.
Calculate the number of protons, neutrons and electrons in 8035Br.
Answer:
In this case, 8035Br , Z = 35, A = 80, species is neutral.
Number of protons = number of electrons = Z = 35
Number of neutrons = A – Z = 80 – 35 = 45.

Question 2.
The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species.
Answer:
The atomic number is equal to number of protons = 16. The element is sulphur (S).

Atomic mass number = number of protons + number of neutrons = 16 + 16 = 32

Species is not neutral as the number of protons is not equal to electrons. It is anion (negatively charged) with charge equal to excess electrons = 18 – 16 = 2. Symbol is 3216S2-.

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 3.
The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz (kilo hertz). Calculate the wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to?
Answer:
The wavelength, λ, is equal to c/v, where c is the speed of electromagnetic radiation in vacuum and v is the frequency. Substituting the given values, we have
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 40
This is a characteristic radiowave wavelength.

Question 4.
The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these wavelengths in frequencies (Hz), (1nm = 10-9m)
Answer:
Frequency of violet light
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 41
The range of visible spectrum is from 4.0 × 1014 to 7.5 × 1014 Hz in terms of frequency units.

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 5.
Calculate (a) wave number and (b) frequency of yellow radiation having wavelength 5800 A.
Answer:
a) Calculation of wave number (\(\overline{\mathrm{ν}}\))
λ = 5800Å = 5800 × 10-8cm = 5800 × 1010 m

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 42
b) Calculation of the frequency (v)
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 43

Question 6.
Calculate energy of one mole of photons of radiation whose frequency is 5 × 1010 Hz.
Answer:
Energy (E) of one photon is given by the expression
E = hυ
h = 6.626 × 10-34 J s
V = 5 × 1014 s-1 (given)
E = (6.626 × 10-34 J s) (5 × 1014 s-1)
= 3.313 × 10-19 J

Energy of pne mole of photons
= (3.313 × 10-19 J) × (6.022 × 1023 mol-1)
= 199.51 kJ mol-1

Question 7.
A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb.
Answer:
Power of the bulb = 100 watt = 100 J s-1
Energy of one photon E = hν = hc/λ
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 44

Question 8.
When electromagnetic radiation of wavelength 300 nm Calls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68 × 105 J mol-1. What is the minimum, energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted?
Answer:
The energy (E) of a 300 nm photon is given by
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 45

The energy of one mole of photons
= 6.626 × 10-19 J × 6.022 × 1023 mol-1
= 3.99 × 105 J mol-1

The minimum energy needed to remove a mole of electrons from sodium
= (3.99-1.68)105J mol-1
= 2.31 × 105 J mol-1

The minimum energy for one electron
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 46
(This corresponds to green light)

Question 9.
The threshold frequency ν0 for a metal is 7.0 × 1014s-1. Calculate the kinetic energy of an electron emitted when radiation of frequency ν = 1.0 × 1015 s-1 hits the metal.
Answer:
According to Einstein’s equation Kinetic energy = \(\frac{1}{2}\)mev² = h(ν – ν0)
= (6.626 × 10-34 Js) (1.0 × 1015 s-1 – 7.0 × 1014 s-1)
= (6.626 × 10-34 Js) (10.0 × 1014 s-1 – 7.0 × 1014 s-1)
= (6.626 × 10-34 Js) (3.0 × 1014 s-1)
= 1.988 × 10-19 J

Question 10.
What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom?
Answer:
Since ni = 5 and nf = 2, this transition gives rise to a spectral line in the visible region of the Balmer series.
∆E = 2.18 × 10-18 J (\(\frac{1}{5^2}-\frac{1}{2^2}\))
= -4.58 × 10-19 J

It is an emission energy The frequency of the photon (taking energy in terms of magnitude) is given by
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 47

Question 11.
Calculate the energy associated with the first orbit of He+. What is the radius of this orbit?
Answer:
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 48
The radius of the orbit is given by
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 49

Question 12.
What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m s-1?
Answer:
According to de Brogile equation
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 50

Question 13.
The mass of an electron is 9.1 × 10-31 kg. If its K.E. is 3.0 × 10-25 J, calculate its wavelength.
Answer:
Since K. E. = \(\frac{1}{2}\) mv²

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 51

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 14.
Calculate the mass of a photon with wavelength 3.6 Å.
Answer:
λ = 3.6 Å = 3.6 × 10-10m
Velocity of photon = velocity of light
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 52

Question 15.
A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 Å. What is the uncertainty involved in the measurement of its velocity?
Answer:
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 53

Question 16.
A golf ball has a mass of 40g, and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in the position.
Answer:
The uncertainty in the speed is 2%, i.e.,
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 54

This is nearly ~ 1018 times smaller than the diameter of a typical atomic nucleus. As mentioned earlier for large particles, the uncertainty principle sets no meaningful limit to the precision of measurements.

Question 17.
What is the total number of orbitals associated with the principal quantum number n = 3?
Answer:
For n = 3, the possible values of l are 0, 1 and 2. Thus there is one 3s orbital (n = 3, l = 0 and ml = 0); there are three 3p orbitals (n = 3, l = 1 and m; = -1, 0, +1); there are five 3d orbitals (n = 3, l = 2 and ml = -2, -1, 0, +1, +2).

Therefore, the total number of orbitals is 1 +3 + 5 = 9

The same value can also be obtained by using the relation;
number of orbitals = n², i.e., 3² = 9.

TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure

Question 18.
Using s, p, d, f notations, describe the orbital with the following quantum numbers
(a) n = 2, l = 1,
(b) n = 4, l = 0,
(c) n = 5, 1 = 3,
(d) n = 3, l = 2
Answer:
TS Inter 1st Year Chemistry Study Material Chapter 1 Atomic Structure 55

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Telangana TSBIE TS Inter 1st Year English Study Material 2nd Lesson A Red Red Rose Textbook Questions and Answers.

TS Inter 1st Year English Study Material 2nd Lesson A Red Red Rose

Annotations (Section – A, Q.No. 2, Marks : 4)

Question 1.
O my Luve’s like a red, red rose.
That’s newly sprung in June.
Answer:
Introduction:
This couplet is taken from the poem, A Red Red Rose written by Robert Burns. It is one of the best lyrics of English poetry. It blends the eternity of love with the mortality of life.

Context & Explanation:
The poet begins by using a simile to compare his love to a rose. In other words, his love is like a flower that has just bloomed in June. His love is fresh and is bursting with life. His feelings are very profound.

Critical Comment:
It is an address to the speaker’s lover to whom he swears eternal love and loyalty.

కవి పరిచయం :
రాబర్ట్ బర్న్స్ అంత్యానుప్రాసయుక్తముగా వ్రాసిన ‘ఒక ఎర్ర ఎర్ర గులాబి’ పద్యం నుండి ఈ పంక్తులు తీసుకోబడినవి. ఇతను 18వ శతాబ్దపు స్కాటిష్ కవి. ఈ పద్యంలో ఒక యువకుడి భావోద్వేగ తీవ్రత వర్ణింపబడినది.

సందర్భం :
ఇది పద్యము యొక్క ప్రారంభ ద్విపద. ఒక యువప్రేమికుడు తన ప్రియురాలికి తన హృదయాన్ని వివరిస్తున్నాడు. అతను తన ప్రేమ ఎర్ర గులాబి అంత అందంగాను మరియు కోమలంగాను ఉందని ప్రకటించుకుంటున్నాడు. తన ప్రేమ జూన్లో వికసించిన గులాబీ అంత తాజాగా ఉందని. గులాబి పువ్వు విశ్వవ్యాప్తంగా ఆమోదించబడిన ప్రేమ చిహ్నము. గులాబి యొక్క సౌకుమార్యము మరియు సౌందర్యము ప్రేమ లక్షణాలను ప్రముఖంగా ప్రతిబింబిస్తాయి. ఉపమానములు కవిత యొక్క కళాసంపదను పెంచుతాయి. వివరణ : ఇక్కడ ఒక యువ ప్రేమికుడు తన ప్రియురాలికి తన హృదయాన్ని వివరిస్తున్నాడు.

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Question 2.
O my Luve’s like the melodie
That’s sweetly pla’d in tune.
Answer:
Introduction:
This couplet is taken from the beautiful lyric A Red Red Rose, written by Robert Burns. It is one of the best lyrics of English poetry. It blends the eternity of love with the mortality of life.

Context & Explanation:
The poet compares his love to a melody that is sweetly played in tune. His love is a song that is sung just so right in fact that it’s kind of sweet. His feelings are very profound.

Critical Comment:
Here, the poet compares his beloved to a sweet melody which is nice to hear.

కవి పరిచయం :
రాబర్ట్ బర్న్స్ అంత్యానుప్రాసయుక్తముగా వ్రాసిన ‘ఒక ఎర్ర ఎర్ర గులాబి’ పద్యం నుండి ఈ పంక్తులు తీసుకోబడినవి. ఇతను 18వ శతాబ్దపు స్కాటిష్ కవి. ఈ పద్యంలో ఒక యువకుడి భావోద్వేగ తీవ్రత వర్ణింపబడినది.

సందర్భం :
‘ఒక ఎర్ర ఎర్ర గులాబి’ ఒక మధుర ప్రేమ గేయము. లోతైన ప్రేమలో పడ్డ ఒక యువకుడు తన ప్రియురాలికి తన హృదయాన్ని చిత్రించుతున్నాడు. ఆమె పట్ల తన ప్రేమ ఎంత గాఢమైనదో తెలపాలని కోరుకుంటున్నాడు. ఆ బలమైన కోరిక అతనిని ఉపమానముల వాడుక వైపు నడిపిస్తుంది. తన ప్రేమ ఎర్ర గులాబి లాంటిది అంటాడు అతను. ఆయన ఇంకా అదనంగా (ఇక్కడ) తన ప్రేమ తీయగా పాడబడిన మధుర గీతిక అంటున్నారు. ప్రేమకు, గులాబీలకు మరియు సంగీతానికి మధ్య బంధం బలమైనది. అది ఇక్కడ మరింత బలోపేతం చేయబడింది.

వివరణ :
ఇక్కడ కవి తన ప్రేమ తియ్యగా పాడబడిన మధురగీతికగా పోలుస్తున్నాడు.

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Question 3.
And I will Luve thee still, my dear,
Till a’ the seas gang dry :
Answer:
Introduction:
This couplet is extracted from the beautiful lyric A Red Red Rose, written by Robert Burns. It is one of the best lyrics of English poetry. It blends the eternity of love with the mortality of life.

Context & Explanation :
The speaker says he will love his bonny lass until all the seas dry up. The word ‘a’ is the shortened form of all. It is very common in Scots English. Gang does not refer to a group of people. It is an old word that means ‘go or a walk. The seas will probably never gang dry. So, the speaker seems to be saying that he will love his lass forever.

Critical Comment:
Here, the poet makes several promises to love his beloved forever.

కవి పరిచయం :
రాబర్ట్ బర్న్స్ అంత్యానుప్రాసయుక్తముగా వ్రాసిన ‘ఒక ఎర్ర ఎర్ర గులాబి’ పద్యం నుండి ఈ పంక్తులు తీసుకోబడినవి. ఇతను 18వ శతాబ్దపు స్కాటిష్ కవి. ఈ పద్యంలో ఒక యువకుడి భావోద్వేగ తీవ్రత వర్ణింపబడినది.

సందర్భం :
“ప్రేమ విచిత్రము” అన్నారు విజ్ఞులు. ప్రేమకు తర్కం తెలియదు. ఇక్కడ యువ ప్రేమికుడికి ఒకే ఒక లక్ష్యము ఉంది. తన ప్రియురాలికి తను ఆమెను ఎంత గాఢంగా ప్రేమిస్తున్నాడో నొక్కి చెప్పాలని అతను కోరుకుంటున్నాడు. ఆ విషయాన్ని నిరూపించటానికి అతను ఉపమానాలు మరియు అతిశయోక్తులు వాడుతున్నాడు. ఇచ్చిన ద్విపదలో ఆయన అంటున్నారు ఇలా : సముద్రాలన్నీ ఎండిపోయినా కూడా, నా ప్రేమ కొనసాగుతూనే ఉంది. సముద్రాలన్నీ నిజంగా ఎప్పటికైనా పూర్తిగా ఆవిరి అవుతాయా ? ఆ విషయం పట్ల అతనికి బాధ అసలు లేదు. అతని ఏకైక పని తన మనసును ఆమెకు చూపటం.

వివరణ :
ఇక్కడ కవి తన ప్రియురాలికి తన రకరకాల వాగ్దానాలను ప్రకటిస్తున్నాడు.

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Question 4.
And fare thee weel, my only
Luve and fare thee weel a while!
Answer:
Introduction:
This couplet is taken from the beautiful lyric, ‘A Red Red Rose’ written by Robert Burns. It is one of the best lyrics of English poetry. It blends the eternity of love with the mortality of life.

Context & Explanation :
The speaker says that he will love his beloved forever. Even after the seas get dried up, all the rocks melt, and the sands of life exhaust their love stays alive. It will last forever. For the present, the speaker says good bye only to return soon, though the journey is to a far off place the poem blends the eternity of love with the mortality of life.

Critical Comment:
The poet makes several promises to love his beloved forever.

కవి పరిచయం :
రాబర్ట్ బర్న్ అంత్యానుప్రాసయుక్తముగా వ్రాసిన ‘ఒక ఎర్ర ఎర్ర గులాబి’ పద్యం నుండి ఈ పంక్తులు తీసుకోబడినవి. ఇతను 18వ శతాబ్దపు స్కాటిష్ కవి. ఈ పద్యంలో ఒక యువకుడి భావోద్వేగ తీవ్రత వర్ణింపబడినది.

సందర్భం :
మనసు మార్గాలు విచిత్రము. మరీ ప్రేమికుడి మనసు మార్గాలు మరింత విచిత్రము. ప్రేమలో పడ్డ యువకుడు తన ప్రియురాలి పట్ల తన ప్రేమ ఎంత లోతైనదో పదే, పదే నొక్కి చెబుతాడు. పద్యం యొక్క మొదటి మూడు స్టాంజాల నిండా అతని ప్రేమకు మద్దతుగా అతిశయోక్తులు మరియు ఉపమానాలే. సముద్రాలు ఎండిపోయినా, పర్వతాలు కరిగిపోయినా, జీవితాలు అనే ఇసుక అంతరించినా నా ప్రేమ కొనసాగుతూనే ఉంటుందని చెప్పాడు. ఇక ఇప్పుడు ఆ ప్రేమికుడు తన ప్రియురాలికి వీడ్కోలు పలుకుతున్నాడు, కొద్ది కాలానికి మాత్రమే అయినప్పటికి అతను ఒక వెయ్యి మైళ్ళ దూర ప్రయాణం చేపడుతున్నాడు. అతను ఆమెకు హామీ ఇస్తున్నాడు తను మళ్ళీ తప్పక ఆమె దగ్గరకు వస్తానని. “Fare thee well” అంటే “మీకు వీడ్కోలు” అని.

వివరణ :
ఇక్కడ కవి తన ప్రియురాలికి తన రకరకాల వాగ్దానాలను ప్రకటిస్తున్నాడు.

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Paragraph Questions & Answers (Section – A, Q.No. 4, Marks : 4)

Question 1.
How is the feeling of love expressed in the poem A Red Red Rose ?
Answer:
The poem A Red Red Rose is written by Robert Burns. It is one of the best lyrics of English poetry. It blends the eternity of love with the mortality of life. It is an address to the speaker’s lover to whom he swears eternal love and loyalty.

The speaker shares his romantic love for his beloved in this poem. His feelings are very profound. He compares his beloved with a fresh and beautiful rose sprung in June and to a sweet melody as well. He also makes several promises to live his beloved forever. He makes a promise that he will return to her life after their temporary separation. He promises to be with her, no matter how long the journey takes.
A Rose speaks of Love silently

రాబర్ట్ బర్న్స్ అనే స్కాటిష్ కవి. ‘ఒక ఎర్ర ఎర్ర గులాబి’ అనే పద్యాన్ని రచించాడు. ఆంగ్ల పద్యసాహిత్యంలో ఇది ఒక ఉత్తమమైన పద్యం. ఇందులో కథకుడు ప్రియురాలితో తన ప్రేమను వ్యక్తం చేస్తున్నాడు.

ప్రియుడు తన ప్రేమను ప్రియురాలికి తెలుపుతున్నాడు. అతని అనుభూతులు పంచుకుంటున్నాడు. అతను తన ప్రేమను జూన్లో వికసించే ఒక తాజా ఎర్రని గులాబితో పోలుస్తున్నాడు. అతను తన ప్రేయసికి రకరకాల వాగ్దానాలు చేస్తున్నాడు. వారి ఎడబాటు తాత్కాలికమేనని మళ్ళీ తప్పక ఆమె దగ్గరకు వస్తానని వాగ్దానం చేస్తున్నాడు. ప్రస్తుతం వీడ్కోలు పలుకుతున్నానని తెలిపాడు.

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Question 2.
Why is love compared to a Red Red Rose ?
Answer:
The poem, ‘A Red Red Rose’ is written by Robert Burns. He is one of the leading voices of Scotland in English literature. The present poem pictures his love for his beloved. His love is as beautiful as a fresh rose that has just bloomed in June. It is fresh and bursting with life. Here love is compared to a red rose because red rose has been an ancient symbol of love in almost all cultures. In this case, rose is newly spring in June. So, we can understand that his love is always at the starting point. Robert uses his rose with the meaning that it is very strong and passionate. It shows how strong is the speaker’s feeling.

రాబర్ట్ బర్న్స్ అనే స్కాటిష్ కవి. ‘ఒక ఎర్ర ఎర్ర గులాబి’ అనే పద్యాన్ని రచించాడు. ఆంగ్ల పద్యసాహిత్యంలో ఇది ఒక ఉత్తమమైన పద్యం. ఇందులో ప్రియుడు ప్రియురాలితో తన ప్రేమను వ్యక్తం చేస్తున్నాడు.

ప్రియుడు తన ప్రేమను ప్రియురాలికి తెలుపుతున్నాడు. అతని అనుభూతులు పంచుకుంటున్నాడు. అతను తన ప్రేమను జూన్లో వికసించే ఒక తాజా ఎర్రని గులాబితో పోలుస్తున్నాడు. ఇక్కడ తన ప్రేమను ఒక ఎర్ర ఎర్ర గులాబితో పోలుస్తూ అన్ని సంస్కృతులలో పురాతన కాలం నుంచి ఎర్రగులాబి అనేది ప్రేమకు చిహ్నంగా పేర్కొన్నాడు. ఇక్కడ గులాబి జూన్లో నూతనముగా వికసించినది. రాబర్ట్ గులాబిని బలమైన ప్రేమకు తార్కాణంగా చెప్పాడు. కథకుని భావాలు తన ప్రేమ పట్ల ఎంత బలంగా ఉన్నాయో ఈ పద్యంలో తెలుస్తున్నాయి.

Question 3.
What does the speaker promise in A Red Red Rose ? * (Imp, Model Paper)
Answer:
The poem, A Red Red Rose, is written by Robert Burns. It is one of the best lyrics of English poetry. The speaker shares his romantic lone for his beloved. He promises different things to his beloned. He vones to love his beloved until the seas have dried up, the fire of the sun has melted the ice, and human life is over. He uses these examples to express his feelings. Thus, promises his eternal love to his borny lass and that no matter how far he might go, he will always return to her aside.
TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose 2
రాబర్ట్ బర్న్స్న గొప్ప ప్రేమ కవిగా పరిగణిస్తారు. అత్యంత విస్తృతంగా ఆమోదించబడిన ప్రేమ చిహ్నం గులాబి. బర్న్స్ యొక్క కవిత ‘ఒక ఎర్ర ఎర్ర గులాబి’ పేరు నుండి చివరి పంక్తి వరకు అదొక ప్రేమ పద్యము అని నిరూపించుకుంటుంది. ఒక యువ ప్రేమికుడు ఈ కవితలో కథకుడు. అతను పదే పదే నొక్కి చెబుతాడు తన ప్రేమ ఒక ఎర్ర గులాబి లాంటిది మరియు ఒక మధుర గేయం లాంటిది అని.

అతను ఇంకా ఇంకా తన ప్రేమ కొనసాగుతుందని, సముద్రాలు ఎండిపోయినా, పర్వతాలు కరిగిపోయినా, జీవితం అనే ఇసుక అంతరించిపోయినా కూడా. ఆమెను చాలా గాఢంగా, లోతుగా ప్రేమిస్తున్నానని నొక్కి చెబుతాడు. గులాబీలు, సంగీతం ప్రేమకు ప్రతినిధులుగా చక్కగా ఉంటాయి. అందుకే ఇక్కడ ప్రేమికుడు తన ప్రేమ ఎర్ర గులాబి లాంటిది అంటాడు. కవిత పేరు కూడా ఈ విషయాన్ని నొక్కి చెబుతుంది.

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Question 4.
Describe the speaker’s devotion to his beloved as expressed in the last two lines of A Red Red Rose.
Answer:
Robert Burns poem A Red Red Rose pictures a young speaker’s love for his beloved. The poem blends the eternity of love with the mortality of life. Especially, in the last two lines, the speaker is completely devoted to his beloved. The has promised his sweet heart that he will return to her wherever he goes, no matter what the distance. Even if her relationship is ten thousand miles away, his love will never die. He will continue to love her. All in all these lines represent the immortality of his love for his beloved.

రాబర్ట్ బర్న్స్ అనే స్కాటిష్ కవి. ‘ఒక ఎర్ర ఎర్ర గులాబి’ అనే పద్యాన్ని రచించాడు. ఆంగ్ల పద్యసాహిత్యంలో ఇది ఒక ఉత్తమమైన పద్యం. ఇందులో కథకుడు ప్రియురాలితో తన ప్రేమను వ్యక్తం చేస్తున్నాడు. చివరిగా ఆ ప్రేమికుడు తన ప్రియురాలికి వీడ్కోలు పలుకుతున్నాడు, కొద్దికాలానికి మాత్రమే అయినప్పటికి. అతను ఒక వెయ్యి మైళ్ళ దూర ప్రయాణం చేపడుతున్నాడు. అతను ఆమెకు హామీ ఇస్తున్నాడు తను మళ్ళీ తప్పక ఆమె దగ్గరకు వస్తానని. “Fare thee well” అంటే “మీకు వీడ్కోలు” అని.

A Red Red Rose Summary in English

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose 1

Robert Burns is a great lyrical poet. He became truly the national poet of Scotland. The last years of his life were fruitful in the lyrical songs that gave him not merely a national but a universal reputation. In the present poem “A Red Red Rose” he describes his love for his beloved. This is a simple, but sincere poem in which he pours out his intense love for his beloved. He describes his pious and ardent love in a heart-rending and picturesque manner.

The poet points out that love is newly emerging feeling fully bloomed like a pretty rose in lovely spring. It is filled with the warmth of June, the summer. He says that his love is fully grown, bloomed like a lovely rose, sprung in June. It is like a sweet melody played in a passionate tune.

He endows his emotion with a concrete form he sees vividly in his beloved’s image. He declares the immortality of his love. He says that his love will remain till the seas get dried, till the rocks melt with the sun, and till death snatches him away from his sweetheart.

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

Finally, he overcomes all his grief. It is because he realizes and convinces the beloved that this parting is not an end. And they will be united again after crossing the path of death covering a distance of ten thousand miles. Thus the ravages of time will fail to bring any change upon his pious feelings. He bids his beloved farewell physically. He bids it only temporarily as he is sure to get united with her immortally in the other world beyond the limits of life and death-the physical concepts. Burns very convincingly assures, his beloved that he will reach her through the distance between them were ten thousand miles, symbolically.

A Red Red Rose Summary in Telugu

Lyric అనగా గేయము అని అర్థం. దీనిని లయబద్ధముగా పాడవచ్చును. ఆంగ్ల భాషలో Robert Burns ఖ్యాతి గాంచిన గేయ రచయిత. ఇతను Scotland జాతీయ కవిగా కూడా గుర్తింపు పొందాడు. అతని చివరి కాలంలోని గేయరచనలు జాతీయ స్థాయిలోనే కాక ప్రపంచ స్థాయిలో కూడా పేరు తెచ్చాయి. ప్రస్తుత పద్యంలో కవి తన ప్రేయసి పట్లున్న తన ప్రేమను వివరిస్తున్నాడు. తన ప్రేయసిని గులాబీతో పోల్చి పద్య రచన చేశాడు.

తన ప్రేయసిని వసంత ఋతువులో అప్పుడే వికసించిన గులాబీతో పోల్చాడు. ఆ గులాబీ పరిమళాలు వెదజల్లుతూ ఆకర్షణీయంగా, సుందరంగా, కోమలంగా, లలితముగా ఎలా ఉందో తన ప్రేయసి కూడా అలాగే ఉందని వర్ణించినాడు. తన ప్రేమకు చావులేదని, సూర్యుడు, చంద్రుడు, భూమి, ఆకాశము, సముద్రములు ఉన్నంతకాలము వారి ప్రేమ కూడా చిరకాలమని వివరంగా వర్ణిస్తాడు.

తన ప్రేయసితో విడిపోయినప్పటికీ అది కేవలం తాత్కాలికమని, స్వర్గలోకములో ఆమెను కలుసుకుంటానని లేకుంటే మరియొక జన్మలోనైనా ఆమెను కలుస్తానని అతని ఆశ. మనస్సులు ఒకటైనవేళ, శరీరములు వేరైనప్పటికీ ప్రేమ శాశ్వతమైనదని అభివర్ణిస్తాడు.

A Red Red Rose Summary in Hindi

स्कॉट लैड – कवि रॉबर्ट बर्न्स से 18 वीं शती में लिखित प्रेम गीत है यह ‘एक लाल लाल गुलाब’ – ‘A Red Red Rose’ यह क्षाव्य गीत है, जिसमें झलकता है कि यौवन में प्रेम कितना भावोद्वेग भरित होता है । लयबद्ध रुप से बढ़नेवाली यह कविता पाठकों से अपार प्रेम पाती है । कथक अपनी प्रिया को अपना प्यार सोलह पंक्तियों में बताता है ।

TS Inter 1st Year English Study Material Chapter 2 A Red Red Rose

“मेरा प्रेम जून में विकसित ताजा लाल गुलब जैसा है । जितना तेरा सौंदर्थ और आनंद है, तेरे प्रति मेरा प्रेम उतना गहरा है । तेरे प्रति मेरा प्रेम शाखत है । समुद्र सूख जाने पर भी, पर्वत गल जाने पर भी जीवन की रेत मिट जाने पर भी मेरा पेर प्रेम तो जारी रहता है। फिर भी, कुछ समय के तुझे विदा करता हूँ जरुर वापस आता हूँ। मेरी यात्रा दस हज़ार मील की होने पर भी लौट है । आता तब तक विदाई ।

Meanings and Explanations

Melody (n) / mel / adi/ (మెలడీ) : sweet sounds, మధురగీతం , मधुर गीत
Bonny (adj) / bani / (బోని) : healthy looking, with glow of health, ఆరోగ్యాంగా అందంగా ఉన్న , कमनीय, रमणीय
Lass (n)/aæs/(ల్యాస్) : girl, యువతి , युवती, किशोरी
Art : are, ఉన్నారు, है, हैं, हो
Thou (pron)/ ðau / (దౌ) : you, నీవు, మీరు,, आप, तुम
Gang dry : get dried, ఎండిపోవుట , अदुश्य होना
Fare thee well : Good-bye, may god bless you, వీడ్కోలు , तुझे विदा
Sprung(v-pp) / spraŋ / స్స్పంగ్) : bloomed, వికసించిన, పూసెను,, खिला हुआ
Weel : well, ఆరోగ్యవంతమైన , स्वस्थ
Luve (n)/lav / (లవ్) : love, ప్రేమ, प्यार, प्रेम, मुहबत
Play’d : played, ఆడటం, खेलना

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 10 Properties of Triangles Important Questions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

I.

Question 1.
In ΔABC, If a=3,b=4 and sin A=3/4, find the angle B.
Solution :
By sine rule,\(\frac{a}{\sin A}=\frac{b}{\sin B}\)
⇒ sinB =\(\frac{b \sin A}{a}=\frac{4 \times 3 / 4}{3}\) = 1
⇒ sinB = 1 = B = 90°

Question 2.
If the lengths of the sides of a triangle are 3, 4, 5, find the circumradius of the triangle.
Solution:
Since 32+42= 52 the triangle is right angled and hypotenuse = 5 = circum diameter.
∴ Circum radius = \(\frac{5}{2}\) cm.

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

Question 3.
lf a=6,b=5,c=9, then find the angle A.
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 1

Question 4.
In a Δ ABC, show that ∑(b +c)cosA = 2s
Solution:
LH.S. = ∑ (b + c) cos A
= (b + c) cos A+ (c + a) cos B + (a + b) cos C
= (bcosA+ acos B)+(ccosA + acosC) + (b cos C + c cos B)
= c+b+a
= a+b+c=2s=R.H.S.

Question 5.
In a Δ ABC, if
(a) (a+b+c)(b+c- a) = 3bc, findA.
Solution:
Given (a + b + c)(b + c-a) = 3bc
⇒ (2s) 2(s – a) = 3 bc
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 2

(b) If a = 4, b = 5, c = 7, find cos \(\frac{B}{2}\)
Solution:
2s = a+b+c = 4+5+7=16
⇒ s = 8
∴ s – b=8-5=3 and
\(\cos \frac{B}{2}=\sqrt{\frac{s(s-b)}{a c}}=\sqrt{\frac{8 \times 3}{4 \times 7}}=\sqrt{\frac{6}{7}}\)

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

Question 6.
If ΔABC, find \(b \cos ^2 \frac{C}{2}+c \cos ^2 \frac{B}{2}\)
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 3

Question 7.
If \(\tan \frac{\mathrm{A}}{2}=\frac{5}{6}\) and \(\tan \frac{\mathrm{C}}{2}=\frac{2}{5}\)
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 4
⇒ 3s – 3b = s
⇒ 2s = 3b ⇒ a+b +c = 3b ⇒ a = 2b
⇒ a, b, c are in A.P.

Question 8.
If \(\cot \frac{A}{2}=\frac{b+c}{a}\),find angle B
Solution:
\(\cot \frac{A}{2}=\frac{b+c}{a}\)
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 5

Question 9.
If tan \(\left(\frac{\mathrm{C}-\mathrm{A}}{2}\right)=\mathrm{k} \cot \frac{\mathrm{B}}{2}\),find K
Solution:
Using Napiers rule,
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 6

Question 10
In ΔABC, Show that \(\frac{b^2-c^2}{a^2}=\frac{\sin (B-C)}{\sin (B+C)}\)
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 7

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

Question 11.
Show that \((b-c)^2 \cos ^2 \frac{A}{2}+(b+c)^2 \sin ^2 \frac{A}{2}=a^2\)
Solution:
L.H.S= \((b-c)^2 \cos ^2 \frac{A}{2}+(b+c)^2 \sin ^2 \frac{A}{2}\)
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 8

Question 12.
If ΔABC, Prove that \(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{r}\)
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 10

Question 13.
Show that r r1 r2 r3 = Δ2
Solution:
L.H.S. r. r1 . r2 . r3
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 11

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

Question 14.
In an equilateral triangle, find the value of  \(\frac{\mathbf{r}}{\mathbf{R}}\)
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 12

Question 15.
The perimeter of ΔABC is 12 cm and it’s in radius is 1 cm. Then find the area of the triangle.
Solution:
Given 2s = 12 ⇒ s = 6 cm. ; r = 1 cm
Area of Δ ABC,
Δ = rs = (1) (6) = 6sq.cm

Question 16.
Show that r r1 = (s – b) (s – c).
Solution:
L.H.S. = r. r1
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 13

Question 17.
In ΔABC, is = 6sq. cm and s = 1.5cm find ‘r’
Solution:
\(\mathrm{r}=\frac{\Delta}{\mathrm{s}}=\frac{6}{1.5}=\frac{6}{3 / 2}=4 \mathrm{~cm}\)

Question 18.
Show that \(r r_1 \cot \frac{A}{2}=\Delta\)
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 14

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

II.

Question 1.
In a \(\Delta \mathrm{ABC}\),prove that \(\tan \left(\frac{B-C}{2}\right)=\frac{b-c}{b+c} \cot \frac{A}{2}\)
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 15
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 16

Question 2.
Prove that cot A+cot B+ cot C \(=\frac{a^2+b^2+c^2}{4 \Delta}\)
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 17
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 23

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

Question 3.
Show that r + r3 + r1 – r2 = 4R cos B.
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 18

Question 4.
In a ΔABC, if r1= 8, r2=12, r3 = 24, find a, b, c.
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 19

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

Question 5.
If the sides of a triangle are 13, 14, 15, then find the circum diameter.
Solution:
Let a= 13,b= 14,c= 15
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 20

Question 6.
Show that
a2cotA+b2cotB+c2cotC = \(\frac{\mathbf{a b c}}{\mathbf{R}}\)
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 21

Question 7.
Prove that a(b cos C – c cos B) = b2 – c2.
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 22

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

Question 8.
Show that \(\frac{c-b \cos A}{b-c \cos A}=\frac{\cos B}{\cos C}\)
Solution:
We have c = a cos B + b cos A and
b = a cos C + c cos A
\(\text { L.H.S. }=\frac{c-b \cos A}{b-c \cos A}\)
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 24

Question 9.
Show that b2 sin 2C + c2sin 2B = 2bc sin A.
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 26

Question 10.
Show that
\(a \cos ^2 \frac{A}{2}+b \cos ^2 \frac{B}{2}+c \cos ^2 \frac{C}{2}=s+\frac{\Delta}{R}\)
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 27
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 28

Question 11.
In a ΔABC, if acosA=bcosB, prove that the triangle is either isosceles or right angled.
Solution:
a cos A = b cos B
⇒ 2R sin A cos A = 2R sin B cos B
⇒ sin 2A = sin 2B
⇒ sin 2A = sin (180° – 2B)
⇒ 2A = 2B or 2A = 180° – 2B
⇒ A=B or A=90°- B
⇒ A=B or A+B=90°
⇒  c = 90°
∴ The triangle is isosceles or right angled.

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

Question 12.
If \(\cot \frac{A}{2}: \cot \frac{B}{2}: \cot \frac{C}{2}\) = 3 : 5 : 7, Show that a : b : c =6 : 5 : 4.
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 29
Thens – a = 3ks  – b = 5ks – c= 7k
∴ 3s – (a + b + c) = 15k
= 3s – 2s= 15k=s= 15k
∴ a = s – 3k = a = 12k
b = s – 5k=b= 10k
c = s – 7k c = 8k
∴ a : b : c = 12k : 10k : 8k = 6 : 5 : 4.

Question 13.
Prove that a3cos(B – C)+b3cos(C – A) + cos3 cos (A – B) = 3 abc.
Solution:
∑a3cos(B-C)
= ∑ a2. a cos (B – C)
=∑ a2 2R sin A cos(B – C)
= ∑ a2 2sin(B + C) cos (B – C)
= \(R \Sigma a^2\left(2 \frac{b}{2 R} \cos B+2 \frac{c}{2 R} \cos C\right)\)
= ∑ a2(bcos B +ccosC)
= a2 (bcos B + ccos C) + b2(ccos C + acosA) + c2 (a cos A + b cos B)
= ab (a cos B+ b cos A) + bc (b cos C + c cos B) + ca(ccos A+acos C)
= abc + bca + cab
= 3 abc = R.H.S.

Question 14.
If P1 P2 p3 are the altitudes of a ΔABC to the opposite sides show that
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 30
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 31
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 32

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

Question 15.
The angle of elevation of the top point P of the vertical tower PQ of height h from a point A is 450 and from a point B Is 600, where B is a point at a distance 30 meters from the point A measured along the line AB which makes an angle 30° with AQ. Find the height of the tower.
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 33

Question 16.
Two trees A and B are on the same side of a tiver. From a point C in the river, the distances of the trees A and B are 250 m and 300 m respectively. If the angle C is 450, find the distance between the trees (use \(\sqrt{2}\) = 1.414).
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 34
Given AC = 300 m and BC = 250 m and in the Δ ABC, using cosine rule
AB2 = AC2 + BC2 – 2AC.BC. cos 45°
= (300)2 + (250)2 -2 (300) (250) – \(\frac{1}{\sqrt{2}}\)
= 46450
∴ AB = 215.5 m (approximately)

Question 17.
Express \(\frac{a \cos A+b \cos B+c \cos C}{a+b+c}\) in terms of R and r.
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 35
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 36

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

Question 18.
If a = 13, b = 14, c = 15, find r1.
Solution:
\(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\mathrm{s} \Rightarrow \mathrm{s}=\frac{13+14+15}{2}=\frac{42}{2}=21\)
s – a = 21- 13=8
s – b = 21 – 14 = 7 and
s – c = 21 – 15 = 6
Δ2 = 21 x 8 x 7 6
Δ = 7 x 12 = 84sq. units
r1 = \(\frac{\Delta}{s-a}=\frac{84}{8}\) = 10.5 units

Question 19.
If r r2 = r1 r3, then find B.
Solution:
Given r r2 = r1 r3
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 37

Question 20.
In a Δ ABC, show that the sides a, b, and c are in A.P. If and only If r1, r2, r3 are in H.P.
Solution :
r1, r2, r3 are in H.P.
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 38
⇔ s – a, s – b, s – c are in A.P
⇔ – a, – b, – c are in AP.
⇔ a, b, c are in A.P.

Question 21.
If A = 90°, show that 2 (r + R) = b + c
Solution:
L.H.S. = 2 (r + R)
= 2r + 2R
= 2(s – a) tan\(\frac{\mathrm{A}}{2}\) + 2R . 1
= 2 (s – a) tan 45° + 2R sinA (∵ A = 90°)
= (2s – 2a) + a
= 2s – a = a + b + c – a = b + c = R.H.S.

Question 22.
Show that
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 39
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 40

Question 23.
Prove that Σ (r1 + r) \(\tan \left(\frac{B-C}{2}\right)=0\)
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 41

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

Question 24.
Show that \(\left(r_1+r_2\right) \sec ^2 \frac{C}{2}=\left(r_2+r_3\right) \sec ^2 \frac{A}{2} =\left(r_3+r_1\right) \sec ^2 \frac{B}{2}\)
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 42
Hence the triangle is right angled at A.

Question 25.
Show that \(\frac{a b-r_1 r_2}{r_3}=\frac{b c-r_2 r_3}{r_1}=\frac{c a-r_3 r_1}{r_2}\)
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 44
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 45

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

III.

Question 1.
If A, A1, A2, A3 are the areas of in circle and exercise of a triangle respectively then prove that
\(\frac{1}{\sqrt{A_1}}+\frac{1}{\sqrt{A_2}}+\frac{1}{\sqrt{A_3}}=\frac{1}{\sqrt{A^2}}\)
Solution:
Let r1, r2, r3 and r be the radii of excircies and incircie respectively whose areas are A1, A2, A3 and A.
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 46

Question 2.
In a Δ ABC prove that
\(\frac{r_1}{b c}+\frac{r_2}{c a}+\frac{r_3}{a b}=\frac{1}{r}-\frac{1}{2 R}\)
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 47
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 48
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 49

Question 3.
If (r2 – r1) (r3 – r3) = 2r2r3, show that A=90°
Solution:
Given (r2 – r1) (r3 – r1) = 2r2r3
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 50

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

Question 4.
Prove that \(\frac{r_1\left(r_2+r_3\right)}{\sqrt{r_1 r_2+r_2 r_3+r_3 r_1}}=a\)
Solution:
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 51
TS Inter 1st Year Maths 1A Properties of Triangles Important Questions 52

TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure

Telangana TSBIE TS Inter 1st Year Chemistry Study Material 3rd Lesson Chemical Bonding and Molecular Structure Textbook Questions and Answers.

TS Inter 1st Year Chemistry Study Material 3rd Lesson Chemical Bonding and Molecular Structure

Very Short Answer Type Questions

Question 1.
What is octet rule?
Answer:
Atoms having octet (8 electrons) in their valence shell e.g.inert gases (except He) are stable. Atoms which do not have octet in their valence, shell will try to get octet either by losing op by gaining or by sharing electrons to acquire stability. This is known as octet rule.

Question 2.
Write Lewis dot structures for S and S2-.
Answer:
Lewis dot structures of S and S2-

TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 1

Question 3.
Write the possible resonance structures for SO3.
Answer:
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 2

Question 4.
Predict the change, if any, in hybridization of Al atom in the following reaction AlCl3 + Cl → AlCl4.
Answer:
1) AlCl3 is mostly covalent.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 3
undergoes sp² hybridisation. The molecule is trigonal planar. It has 3 bond pair of electrons and no lone pair of electrons.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 4
3) In AlCl3, the central atom of Al has vacant p – orbitals. So it can accept a lone pair of electrons from Cl ion and form the complex AlCl4 ion.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 5
∴ Al(sp²) Cl³ → Al(sp³) Cl

AlCl4 has tetrahedral structure. Al atom undergoes sp³ hybridisation. There are 4 bond pair of electrons and no lone pair of electrons in AlCl4 ion.

TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure

Question 5.
Which of the two ions Ca2+ or Zn2+ is more stable and why?
Answer:
Ca2+ ion is more stable than Zn2+ ion.

Reason :
An ion with ‘inert gas configuration’ is more stable than an ion with ‘pseudo inert gas configuration’.

Ca2+ (2,8,8) has inert gas configuration i.e., electronoctet’ in the valence shell.

Zn2+ (2, 8, 18) has 18 electrons in the outermost energy level. It has pseudo inert gas configuration.

Question 6.
Cl ion is more stable than Cl atom – Why?
Answer:
An atom or ion with inert gas configuration (ns²np6) is comparatively more stable.
Cl(Z) = 17 1s²2s²2p63s²3p5
Cl 1s² 2s² 2p6 3s²3p6
So Cl is more stable than Cl.

Question 7.
Why argon does not form Ar2 molecule?
Answer:
Electronic configuration of 18Ar is 1s² 2s² 2p6 3s² 3p6.

Since the atoms of argon contain eight electrons in their valence shell they are stable and do not participate in bonding. So Argon do not form Ar., molecules.

Question 8.
What is the best possible arrangement of four bond pairs in the valence shell of an atom to minimise repulsions?
Answer:
In order to have minimum repulsions between the four bond pairs present in the valence shell of an atom the best possible arrangement is tetrahedron with bond angle 109° 28′.

Question 9.
If A and B are two different atoms, when does AB molecule become covalent?
Answer:
In order to form a covalent bond in between the atoms A & B.

  1. The electronegativity difference between them should be less than 1.9.
  2. A & B should share one or more electron pairs mutually.

Question 10.
What is meant by localized orbitals?
Answer:
The orbitals which are concentrated between the two nuclei of bonded atoms are called localised orbitals.

Question 11.
How many Sigma and Pi bonds are present in (a) C2H2 and (b) C2H4? [AP ’15]
Answer:
a) The structure of C2 H2 is
H – C ≡ C – H
C2 H2 contain 3 sigma and 2 pi bonds,

b) The structure of C2H4 is
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 6
C2H4 contain 5 sigma and one pi bonds.

Question 12.
Is there any change in the hybridization of Boron and Nitrogen atoms as a result of the following reaction?
BF3 + NH3 → F3BNH3
Answer:
Formation of Ammonia – Boron trifluoride:
In NH3 molecule, the central ‘N’ atom contains a lone pair of electrons in one of its sp³ hybrid orbitals. Before reaction with NH3, the central boron atom of BF3 will be in its sp² hybridised state while reacting with NH3, in order to accept the lone pair of electrons from N atom of NH3, the central ‘B’ atom of BF3 changes its hybridisation from sp² to sp³. But, there will be no change in the hybridised state of N atom of NH3.

Short Answer Questions

Question 1.
Explain Kossel – Lewis approach to Chemical bonding.
Answer:
Lewis considered the atom in terms of a positively charged ‘Kernel’. Kernel contains inner electrons and the nucleus. Lewis assumed that the outer shell can accommodate a maximum of eight electrons which occupy the eight corners of a cube surrounding the Kernel. He assumed this type of arrangement in noble gases and hence they are stable. The atoms which do not have this type of arrangement achieve the stable octet by combining with other atoms by sharing of electrons forming a covalent bond.

Kossel proposed that the highly electronegative elements like halogens gain electrons and convert into anions. The highly electropositive alkali metals lose electrons and convert into positive ions. During their conversion into ions they get the noble gas configuration, i.e., octet and thus get stability. Now the positive and negative ions unite together by electrostatic attraction between them.

Thus Lewis proposed the covalent bond formation and Kossel proposed the ionic bond formation.

Question 2.
Write the general properties of Ionic Compounds.
Answer:
Properties of Ionic compounds:

  1. In general Ionic compounds are solids.
  2. They possess high melting and boiling points.
  3. Ionic compounds are soluble in polar solvents like water but insoluble in or-ganic solvents like benzene, chloroform, carbon tetrachloride, etc.
  4. Ionic compounds in solid state, are bad conductors of electricity, due to absence of free ions.
  5. In fused state or in their aqueous solu-tions, free ions are present. Then, they are good conductors of electricity.
  6. Reactions between ionic compounds are very fast.
  7. They do not exhibit isomerism because, ionic bond is non-directional.

TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure

Question 3.
State Fajan’s rules, and give suitable examples. [AP Mar. ’19; AP ’16 ’15]
Answer:
Fajan’s rules are applied to predict the nature of bond in a molecule.

Fajan’s rules:
1) As the size of cation increases, the ionic nature of the bond increases.
Ex: The ionic nature among alkali metal ions follows the order : Ll+ < Na+ < K+ < Rb+ < Cs+

2) Ionic bond is favoured with a small anion.
Ex : Out of CaF2 and Cal2, CaF2 is more ionic, as the size of F ion is smaller than that of l ion. size : F < I.

3) Smaller charges on cation or anion or both the ions, favours the formation of ionic bond.
Ex: NaCl is more ionic than AlCl3 as the charge on Na (Na+) is less than that on Al (Al+3).

4) Cations with inert gas configuration form ionic bonds, while cations with pseudo inert gas configuration favour covalent bond formation.
Ex : NaCl is ionic, since Na+ (2, 8, 8) has inert gas configuration.
CuCl is more covalent, since Cu+ (2, 8, 8) has pseudo inert gas configuration.

Question 4.
What is Octet rule? Briefly explain its significance and limitations.
Answer:
Octet Rule :
In order to attain stability, “Atoms show tendency to have 8 electrons in their outermost shell”.

Significance of octet rule :
It provides an explanation to the question, ‘why do atoms combine?’. It also explains why some atoms are chemically inert. It is able to explain, the cause of formation of different types of bonds, ionic and covalent. The main points of the octet theory are :

  1. Atoms with 8 electrons in the outermost shell (2 for He) are chemically stable. So they do not take part in chemical reactions.
  2. An atom with less than 8 electrons in outermost shell is reactive and so combines with other atoms. Atoms with less than 4 electrons in the outermost shell tend to lose them, while those having more than
  3. Atoms enter into chemical reactions either by losing, by gaining or by sharing electrons in the outermost shell tend to gain the electrons, so that they get the nearest inert gas configuration.
  4. In this process of bond formation, every atom tries to acquire 8 electrons in its outermost shell.
  5. The chemical reactivity of an atom is measured by its tendency to lose, gain or share electrons.

Limitations:

  1. Octet rule couldn’t explain the shape of molecule.
  2. Octet rule is not satisfied for molecules having odd number of electrons.
    Ex: No.
  3. It could not explain the stability of molecules in which central atom has less or more than 8 electrons.
    Ex : In BeCl2, Be has 4 electrons in its valence shell.

Question 5.
Write the resonance structures for NO2 and NO3.
Answer:
Resonance structures of NO2.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 7
Resonance structures of NO3.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 8

Question 6.
Use Lewis symbols to show electron transfer between the following pairs of atoms to form cations and anions : (a) K and S (b) Ca and O (c) Al and N.
Answer:
a) Electronic configurations of K and S
19K = 1s² 2s² 2p6 3s² 3p6 4s¹
16S = 1s² 2s² 2p6 3s² 3p4
Lewis symbols for K, S and their ions.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 9

b) Electronic configurations of Ca and O
20Ca = 1s² 2s² 2p6 3s² 3p6 4s²
8O = 1s² 2s² 2p4
Lewis symbols of Ca, O and their ions.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 10

c) Electronic configuration of A1 and N
13Al = 1s² 2s² 2p6 3s² 3p¹
7N = 1s² 2s² 2p³
Lewis symbols of Al, N and their ions.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 11

Question 7.
Explain why H2O has dipolemoment while CO2 does not have.
Answer:
Dipolemoment of a polyatomic molecule is a vector quantity.
H2O molecule has angular structure with two
O – H bonds, oriented at an angle of 104.5°.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 12

So the dipolemoment of the water molecule is the resultant of the dipolemoments of two O – H bonds.

The dipolemoment in the case of CO2 is zero. This is because the two equal bond dipoles point in opposite directions and cancel the effect on each other.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 13

TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure

Question 8.
Define Dipolemoment. Write its application [AP Mar. ’17]
Answer:
Dipolemoment is the product of the magnitude of the charge and the distance between the centres of positive and negative charges. It is designated by µ.

Dipolemoment (µ) = Charge (Q) × distance (r) between the charges.
Units of dipolemoment are Debye units (D)

Applications:
1. Dipolemoment is useful in calculating the percent ionic character of polar covalent bonds.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 14

2. Dipolemoment is useful in predicting the shapes of molecules, e.g.,

  1. In the case of AX2 type molecule if dipolemoment is zero they have linear structure but if they have some dipolemoment they are angular.
  2. AX3 type molecules with dipolemoment zero are planar triangular in shape but if they have some dipolemoment they have pyramidal shape.
  3. AX4 type molecules with dipolemoment zero have symmetric tetrahedral structure but if they have some dipolemoment they have distorted tetrahedral shape.
  4. In geometrical isomers trans isomers have zero dipolemoment. While cis isomers have some dipole-moment.

3. It is useful to predict the polarity of molecule.

  1. Molecular with dipolemoment is greater than zero are polar.
  2. Molecular with dipolemoment is equal to zero are non polar.

Question 9.
Explain why BeF2, molecule has zero dipolemoment although the Be-F bonds are polar.
Answer:
In BeF2 molecule the Be-F bonds have dipolemoment due to the difference in the electronegativities of beryllium and fluorine.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 15

But in polyatomic molecules the dipolemoment of a molecule is the vector sum of the dipolemoments of various bonds.

In the case of BeF2 the dipolemoment becomes zero because the two equal bond dipoles point in opposite direction and cancel the effect each other.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 16

Question 10.
Explain the structure of CH4 molecule.
Answer:
CH4 (Methane):
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 17
In the formation of methane molecule, the central carbon atom undergoes sp³ hybridisation in its excited state 1s²2s¹ 2p¹x, 2p¹y, 2p¹z. As a result of which four sp³ hybrid orbitals will form on it. All of them contain an electron, each. Now, these four hybrid orbitals overlap Head – Head with Is orbitals of four hydrogen atoms forming CH4 molecule. The structure of the molecule is tetrahedral. The bond angle is 109°28′. Since there are no lone pairs, there is no distortion in the shape of the molecule.

Question 11.
Explain Polar Covalent bond with a suitable example.
Answer:
When a covalent bond is formed between two similar atoms, for example in H2, O2 Cl2, N2 or F2 the shared pair of electrons is equally attracted by the two atoms. So the shared electron pair is situated exactly between the two identical nuclei. The bond so formed is non-polar covalent bond.

If a bond is formed between atoms of different elements to form heteronuclear molecule like HF the shared electron pair between the two atoms gets displaced more towards fluorine since the electronegativity of fluorine is more than that of hydrogen. As a result partial negative charge develops on fluorine atom while an equal amount of partial positive charge develops on hydrogen atom.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 18

The resultant covalent bond becomes polar covalent bond. The direction of the displacement of electron is represented with an arrow mark on the bond.

TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure

Question 12.
Explain the shape and bond angle in BC13 molecule in terms of Valence Bond Theory.
Answer:
In BCl3 the central atom is boron. Its atomic number is 5. Therefore its ground state electron configuration is 1s²2s² 2p¹x 2p°y 2p°z.

From this configuration, it is evident that it exhibits monovalency. The first excited state configuration of B is 1s²2s² 2p¹x 2p¹y 2p°z. Since, there are two half-filled orbitals, the valency of Boron is 3.

In order to explain the formation of BCl3 molecule sp² hybridisation is to be assumed to boron atom in its excited state I. As a result of which three sp² hybrid orbitals will form on it. Now, the three sp² hybrid orbitals of B’ atom overlap Head – Head with 3pz orbitals of three Cl atoms forming three sigma bonds. The shape of the molecule is plane triangular and the bond angle is 120°.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 19

Question 13.
What are a and π bonds? Specify the differences between them.
Answer:
Sigma Bond :
A covalent bond formed by end-end overlap of orbitals is called sigma covalent bond.

Pi Bond :
A covalent bond formed by lateral overlap of orbitals is called pi covalent bond.

Sigma bond (σ) pi bond (π)
1) A σ bond is formed by the axial overlap of two half-filled orbitals belonging to the valence shells of the two combining atoms. 1) A π bond is formed by the lateral overlap of orbitals.
2) The σ bonding electron cloud is symmetric about the internuclear axis. 2) The π bonding electron cloud lies above and below the plane of the internuclear axis.
3) It is a strong bond since the extent of overlap is much. 3) It is weaker than σ bond, since the extent of overlap is less.
4) It allows free rotation of atoms or groups about the bond. 4) π bond restricts such free rotation.
5) It can exist independently. 5) It is formed only after a σ bond is formed.
6) It determines the shape of the molecule. 6) The shape of the molecule is independent of such bond.

TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure

Question 14.
Even though nitrogen in ammonia is in sp³ hybridization, the bond angle deviate from 109°28. Explain.
Answer:
In NH3 molecule, three sp³ hybrid orbitals contain bond pairs and one sp³ hybrid orbital contains a lone pair of electrons. There will be repulsions between the lone pair and bond pair electrons. These are stronger than bp – bp repulsions, (l.p – b.p > b.p – b.p). So the bond pairs are pushed closer together. Hence the bond angle decreases from tetrahedral angle. The molecule as-sumes a pyramidal shape.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 20

Question 15.
Show how a double and triple bond are formed between carbon atoms in
(a) C2 H4 and (b) C2 H2 respectively.
Answer:
a) C2H4 (Ethylene) :
In the formation of ethylene molecule, the two carbon atoms will undergo sp² hybridisation in their excited states. As a result of which sp² hybrid orbitals will form on each of them. All the six hybrid orbitals contain an electron, each. At first, one sp² hybrid orbital of each carbon atom overlap Head-Head forming one sigma bond in between them. Now, the remaining four sp² hybrid orbitals overlap Head-Head with Is’ orbitals of four H’ atoms forming four sigma bonds. Now, the unhybridised pure 2pz orbitals of each carbon atom overlap side-wise forming π bond in between them. Thus a double bond is formed in between the two carbon atoms.

In total there are five sigma bonds and one pi bond in ethylene molecule. The molecule as a whole is planar.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 21

b) 2H2 (Acetylene) :
In the formation of acetylene molecule, the two carbon atoms undergo sp hybridisation in their excited states (1s² 2s¹ 2p¹x, 2p¹y, 2p¹z). As a result of which two sp hybrid orbitals form on each of them. All of them contain an electron, each. One sp hybrid orbital of each carbon atom overlaps Head – Head forming a sigma bond in-between them.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 22

Now, the remaining two ‘sp’ hybrid orbitals overlap Head – Head with ‘Is’ orbitals of two H – atoms forming two sigma bonds. Now’, the unhybridised, pure 2py and 2pz orbitals of each carbon atom overlap side – wise forming tw’o pi bonds in-between them.

Thus, a triple bond is formed between the two carbon atoms. In total there are three sigma bonds and two pi bonds in acetylene molecule. The bond angle is 180°. The molecule is linear.

Question 16.
Explain the hybridization involved in PCl5 molecule. [TS ’16, ’15; Mar. ’11; Mar. ’18 (TS)]
Answer:
In the formation of PCl5 molecule the central phosphorus atom undergoes sp³d hybridisation.
Ex: Formation of PCl5 molecule
Atomic number of (P) = 15
∴ Ground state electron configuration
= 1s²2s²2p63s²3p³

Excited state configuration
= 1s²2s²2p63s¹3p¹x 3p¹y 3p¹z 3d¹
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 23

In the formation of PCl5 molecule the central ‘P’ atom undergoes sp³d hybridization in its excited state. As a result of which five sp d hybrid orbitals will form on it. All the five hybrid orbitals contain an electron, each. Now, the five sp³d hybrid orbitals overlap Head-Head with half-filled 3pz orbitals of five chlorine atoms forming five sigma bonds. The shape of the molecule is trigonal bipyramidal with bond angles 120° and 90°.

Question 17.
Explain the hybridization involved in SF6 molecule. [TS, AP Mar. 19; (IPE 14, 10)
Answer:
In the formation of SF6 molecule the central ‘S’ atom undergoes sp³ d² hybridization.
Ex : Formation of SF6 molecule :
Atomic number of Sulphur =16
∴ Ground state configuration of ‘S’
= 1s²2s²2p63s²3p4

Excited state I configuration
= 1s²2s²2p63s² 3p¹x 3p¹x 3p¹x 3d¹

Excited state II configuration
= 1s²2s²2p63s¹3p¹x 3p¹y 3p¹z 3dx²-y² 3d

TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 24

In the formation of SF6 molecule, the central ‘S’ atom undergoes sp³d² hybridization in its excited state II configuration. As a result of which six hybrid orbitals form on it. All of them contain an electron, each. Now all these six hybrid orbits overlap Head-Head with half-filled 2pz orbitals of six ‘F’ atoms forming six o bonds. The shape of the molecule is octahedral with a bond angle of 90° and 180°.

Question 18.
Explain the formation of Coordinate Covalent bond with one example. [AP ’16; Mar. ’13, ’11]
Answer:
Co-ordinate covalent bond is a special type of covalent bond and in which the shared pair of electrons are contributed by only one of the bonded atoms. This bond is represented by an arrow mark (“→”), pointing towards the acceptor atom. The atom which contributes electron pair for sharing is called donor atom and the atom which accepts the electron pair is called acceptor.

In order to form a co-ordinate bond between two atoms, one of the atoms should possess lone pair of electrons and the other atom should possess empty orbital.

Ex : Formation of Ammonium ion (NH+4): Ammonium ion is formed by the union of NH3 molecule with H+ ion. In NH3 molecule the central ‘N’ atom has one lone pair of electrons and H+ ion has empty orbital. Hence, N- atom of NH3 molecule donates its lone pair to the empty orbital of H+ ion, forming a co-ordinate covalent bond between them.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 25

Question 19.
Which hybrid orbitals are used by Carbon atoms in the following molecules?
(a) CH3 – CH3
(b) CH3 – CH = CH2
(c) CH3 – CH2 – OH
(d) CH3 – CHO
Answer:
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 26
In CH3 – CH3 both C1 and C2 atoms are involved in sp³ hybridisation.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 27
In CH3 – CH = CH2 the C1 atom is involved in sp³ hybridisation. C2 and C3 atoms afe involved in sp² hybridisation.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 28
In CH3 – CH2 – OH both C1 and C2 atoms are involved in sp³ hybridisation.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 29
In CH3 – CHO C1 atom is involved in sp³ hybridisation and C2 atom is involved in sp² hybridisation.

Question 20.
What is Hydrogen bond? Explain the different types of Hydrogen bonds with example. [AP ’17, ’16, ’15; TS ’16; Mar. ’18 (AP & TS)]
Answer:
The concept of H – bond is due to Huggins, Latimer and Rodebush.

Def: “The electrostatic force of attraction between a partially charged hydrogen atom of a molecule and a highly electronegative atom of the same molecule or different molecule is known as H – bond.”

Explanation:
1) The molecule having H – bonds should have a highly electronegative atom like F, O or N directly linked to H – atom by a covalent bond.

2) H – bond is stronger than van der Waal’s forces but weaker than ionic and covalent bonds.
van der waal’s forces < H – bond

Energy : 2 – 10 kJ mol-1 2 – 40 kJ mol-1 < Covalent bond < Ionic bond 200 – 400 kJ mol-1

3) The strength of H – bond increases with the EN of the atom attached covalently to H atom.

Types of H – bonds : 2 types. 1) Intramolecular H – bonding and 2) Inter-molecular H – bonding

1) Intramolecular H – bonding :
The H – bonding formed between H atom of a molecule and a highly electronegative atom of the same molecule, is called intramolecular hydrogen bond.
Ex:
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 30

2) Inter-molecular H – bonding :
The H – bonding formed in between H – atom of one molecule and a highly electronegative atom of another molecule of same substance or different substances is called intermolecular hydrogen bond.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 31

Consequences of H – bonding:

  1. Highly viscous nature of H2SO4, HNO3, H3PO4 etc, is due to H – bonding.
  2. The high b.p. of water is due to H – bonds in it.
  3. H2O is a liquid but H2S is a gas, since H – bond is present in H2O only.

TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure

Question 21.
Explain the formation H2 molecule on the basis of Valence Bond theory.
Answer:
This theory explains the shapes of covalent molecules as well as the directions of the bonds in them.

Important postulates:

  1. Covalent bond is formed by the overlap of two atomic orbitals.
  2. The overlapping orbitals contain unpaired electrons with opposite spins.
  3. Each of the bonded atoms retains its own atomic orbitals. But the electron pair in the overlapping orbitals is shared by both the atoms.
  4. The greater the extent of overlap of orbitals, the stronger is the covalent bond formed.
  5. Except ‘s’ orbitals, the remaining all atomic orbitals are directional. So the bonds formed due to their overlap are also directional. This determines the shape of the molecule.
  6. When the atomic orbitals of the two atoms overlap, then electron density increases on their internuclear axis. This binds the two atoms, firmly. Thus the molecule is stabilised.

Ex : Formation of Hydrogen molecule (s – s overlap) :
Hydrogen molecule is formed due to overlap of 1 s – orbitals of two hydrogen atoms. The bond formed between the two H – atoms is called σs-s bond and the type of overlapping is s – s overlap.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 32

Question 22.
Using Molecular Orbital Theory explain why the B2 molecule is paramagnetic?
Answer:
Electronic configuration of boron atom is 1s² 2s² 2p¹ There are ten electrons in B2 molecule. In terms of molecular orbital theory the electronic configuration of B2 molecule is
(σ Is)² (σ* Is)² (σ 2s)² (σ* 2s)² (π2p¹ =π2p¹)

The presence of unpaired electrons in the two π bonding molecular orbitals explains its paramagnetic nature.

Question 23.
Write the important conditions necessary for linear combination of atomic orbitals.
Answer:
The necessary conditions for the effective overlap of atomic orbitals to form molecular orbitals are

  1. The combining atomic orbitals must have the same or almost the same energy.
    Ex: 1s – 1s overlap is permitted but not 1s – 2s or 2s – 2p overlap.
  2. The combining atomic orbitals must be able to overlap to the maximum extent to form m.o’s since if the overlapping is more, there will be a greater electron density between the nuclei of the combining atoms and hence the bond formed between them will be stronger.
  3. The combining atomic orbitals must have the same symmetry about the molecular axis.

It means that, atomic orbitals having the same or nearly the same energy will not overlap, if they do not have the same symmetry. Thus the allowed over laps are : s – s, s – px, px – px, py – py and Pz – Pz.

In the other types of overlaps, s – py s – pz, px – py and px – pz. no m.o is formed since the atomic orbitals do not have the same symmetry. They are nonbonding m.o’s.

Question 24.
What is meant by the term Bond order? Calculate the bond orders in the following. [TS ’15; IPE ’14]
(a) N2 (b) O2 (c) O2 and (d) O2
Answer:
Bond order is the number of bonds between two atoms. It is one half the difference between the number of electrons present in the bonding and the antibonding orbitals.
Bond order = \(\frac{1}{2}\) [Nb – Na]

Where Nb is the number of electrons in bonding molecular orbitals while Na is the number of electrons in antibonding mole- culars orbitals. For calculating bond order electrons present in valence only are considered.

a) MO electronic configuration of N2 molecule is
g 1s)² (σg* 1s)² (σg 2s²) (σg * 2s²) (π2p² = π2P²) (σg 2p)²
Bond order = \(\frac{10-4}{2}=\frac{6}{2}\) = 3

b) MO electronic configuration of 02 molecule is
(σ 1s)² (σ * 1s)² (σ 2s)² (σ * 2s)² (σ 2p2)² (π2p²x π2p²y) (π * 2p) (π * 2p¹)
Bond order = \(\frac{1}{2}\)[Nb – Na] = \(\frac{1}{2}\) (10 – 6) = 2

c) MO electronic configuration of O+2 molecule is
(σ 1s)² (σ * 1s)² (σ 2s)² (σ * 2s)² (σ 2p2)² (π2p² π2p²y) (π * 2p¹)
Bond order = \(\frac{1}{2}\)[Nb – Na] = \(\frac{1}{2}\) (10 – 5) = 2.5

d) MO electronic configuration of O2 molecule is,
(σ 1s)² (σ * 1s)² (a 2s)² (σ * 2s)² (σ 2p2)² (π2p²x π2p²y)(π * 2p2x (π* 2p¹y)
Bond order = \(\frac{1}{2}\)[Nb – Na] = \(\frac{1}{2}\) (10 – 7) = 1.5

Question 25.
Of BF3 and NF3, dipolemoment is observed for NF3 and not for BF3. Why?
Answer:
In polyatomic molecules, the dipolemoment of the molecule is the resultant of vector sum of the dipolemoment of various bonds.

BF3 molecule has symmetric planar triangular structure. The three equal bond dipoles point to the corners of a triangle at an angle of 120° and cancel the effect on each other.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 33

The three bond moments give a net sum of zero as the resultant of any two is equal and opposite to the third.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 34

In NF3 the three N – F bonds have dipolemoments. Also the lone pair present on nitrogen atom show dipolemoment in the opposite direction to the resultant N – F bond dipolemoments.

Though the orbital dipolemoment due to lone pair and the resultant dipole moment of three N – F bonds are in opposite direction there remains some resultant dipolemoment since they do not cancel completely.

Question 26.
Even though both NH3 and NF3 are pyramidal, NH3 has a higher dipolemoment compared to NF3 Why? [AP ’16]
Answer:
Both NH3 and NF3 have pyramidal shape with a lone pair of electrons on nitrogen atom. Although fluorine is more electronegative than nitrogen, the resultant dipole moment in NH3 is greater than in NF3.

In the case of NH3 the orbital dipolemoment due to the lone pair electrons and the resultant dipolemoment of the three N – H bonds are acting in the same direction. This enhances the dipolemoment of molecule.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 35

In NF3 the orbital dipolemoment is in the direction opposite to the resultant dipolemoment of the three N – F bonds. The orbital dipolemoment due to lone pair electrons decreases this effect of the resultant N – F bond moments. This results in the decrease of,dipolemoment in NF3 than in NH3.

Question 27.
How do you predict the shapes of the following molecules making use of VSEPR theory? [AP ’15]
(a) XeF4 (b) BrF5 (c) ClF3 and (d) ICl4
Answer:
a) In XeF4 the xenon atom has 8 electrons in its outer orbit. Of these 8 electrons 4 electrons are participated in 4 bonds with 4 fluorine atoms the remaining 4 electrons will present as 2 lone pairs. Thus in XeF4 around Xe atom there are 4 bp and 2 lp a total of 6 electron pairs. These six electron pairs are arranged octahedrally around Xe atoms. To minimize the repulsions between electron pairs the two lp occupies the opposite corners of octahedron. So the shape of XeF4 is square planar.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 36

b) In BrFs, Br atom has 5 bond pair and one lone pair. These six electron pairs are arranged octahedrally around the Br atom. The lone pair occupies one corner of octahedron. So the shape of BrF5 is square pyramid.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 37

c) In ClF3 around Cl atom there are 3 bond pair and 2 lone pair. These 5 electron pairs are arranged in trigonal bipyramid structure. To minimise the repulsions the lone pair always occupies the equatorial positions in trigonal bipyramid structure. So the shape of ClF3 becomes T – shape.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 38

d) In ICl4 ion there are four bond pair and 2 lone pair a total of six electron pairs. These are arranged octahedrally around I atom. To minimise the repulsions between electron pairs the two lp occupy the opposite corners of the octahedron. So ICl4 ion have square planar structure.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 39

Long Answer Questions

Question 1.
Explain the formation of Ionic Bond with a suitable example.
Answer:
“The electrostatic force of attraction between oppositely charged ions formed by the transfer of electrons” is called Ionic Bond.

Orbital concept of bonding :
The method of formation of ionic bond in sodium chloride can be shown by the following orbital model:

Electronic configurations are:
Na (Z = 11) 1s²2s²2p63s¹; Cl (Z = 17) 1s²2s²2p63s²3p5
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 40
In the formation of sodium chloride, the 3s electron of Na atom is transferred to the half-filled 3p orbital of Cl atom. It gets paired up with it.

TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 41

The Na+ and Cl ions thus formed will have the inert gas configurations of neon and argon and get attracted by coulofnbic electrostatic forces of attraction, forming Na+Cl, ionic compound.

TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure

Question 2.
Explain the factors favourable for the formation of Ionic Compounds.
Answer:
Factors that favour the formation of ionic bond:

a) Factors favourable for the cation formation :
i) Low IP :
Atom with low IP loses electrons readily. So it forms cations readily. K+ (IP 496 KJ mol-1) ion forms readily than Na+ (IP 520 KJ mol-1) ion.

ii) Low charge on the ion :
Formation of an ion with low charge requires lower IP than formation of an ion with high charge. Thus out of Na+, Mg2+ and Al3+, the ease of formation Of these ions is Na+ > Mg+ > Al3+.

iii) Large atomic size :
The force of attraction between the nucleus and the valence electrons is less in case of larger atoms than in case of smaller atoms. So larger atoms can lose electrons more readily than smaller atoms.
Ex : Cs+ is formed more easily than any of the other alkali metal ions.

iv) Formation of cation with inert gas configuration :
Cations with inert gas con-figuration are easily formed.
Ex: Out of Ca2+ and Zn2+, Ca2+ ion (2,8,8) is easily formed than Zn2+ ion (2, 8, 18) with pseudo inert gas configuration.

b) Factors favourable for anion formation:
i) High EN and high EA :
An atom with high EN and high EA has a greater tendency to accept electrons and so it can readily form anion.
Ex : The ease of formation of anions is
F > O2- >N-3

ii) Small atomic size: Small atoms can hold strongly the electrons gained by it.
Ex : The order of ionic nature in halide ions is F > Cl > Br > I
F is the smallest and I is the largest atom, among them.

iii) Low charge on the ion :
Stability of ion decreases with increase in its charge. Anion with low – ve charge is formed more readily than anion with high -ve charge.
Ex: F ion is easily formed than O2- ion.

Question 3.
Draw Lewis Structures for the following molecules.
(a) H2S (b) SiCl4
(c) BeF2 and (d) HCOOH
Answer:
In the Lewis structures, only the valence electrons are indicated by dots (•), crosses (x) or circles (O).
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 42
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 43

Question 4.
Write notes on (a) Bond Angle (b) Bond Enthalpy (c) Bond length and (d) Bond order.
Answer:
a) Bond angle :
The angle between the orbitals containing bonding electron pairs around the central atom in a molecule, complex ion is called bond angle. Bond angle is expressed in degrees.

i) Bond angle depends on the presence of lone pair electrons with increase in the number of lone pairs the bond angle decreases, eg. in the molecules which contain four electron pairs the bond angles are follows.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 44
ii) Bond angle decreases with increase in the electronegativity of the bonded atom and increases with increase in the electronegativity of central atom. eg.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 45
iii) Bond angle depends on the hydridisation.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 46

b) Bond enthalpy :
The amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state is called bond enthalpy.

More is the bond enthalpy, stronger the bond and more is the stability of the molecule.

In the case of poly atomic molecules the bond enthalpy is the average of the similar bond enthalpies is taken, eg. in H2O to break the two O – H bonds is not same. The energy required for breaking first bond is 502 kJ mol-1 and for the second bond is 427 kJ mol-1. So the average value 464.5 kJ mol-1 is considered as the bond enthalpy.

c) Bond length :
It is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.

Bond length is equal to the sum of the covalent radii of two similar atom in the bond. In hetero atomic molecules, the bond length can be calculated using the formula.

dAB = rA + rB + c (XA – XB). Where XA and XB are the electronegativities of atom A and B. The value of C depends on the nature of atoms participating in the bond.

Bond lengths depend on the number of bonds between atoms, type of hybridisation etc.

d) Bond order :
It is defined as the number of bonds between the two atoms in a molecule.
Eg : Bond orders in H2 is 1, O2 = 2 and N2 = 3.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 47

Isoelectronic molecules and ions have identical bond orders, eg. F2 and O2-2 have bond order 1. N2, CO and NO+ have bond order 3.

More the bond order more is the bond enthalpy, lesser the bond length, more the stability.

Question 5.
Give an account of VSEPR Theory, and its applications.
Answer:
VSEPR theory was proposed by Gillespy and Nyholm. It predicts the shapes of molecules Without reference to the hybridisation. The important points in the theory are :

  1. The shape of a molecule depends upon the number of electron pairs (bond pairs) present on the central atom. If there are only ‘bond pairs’ and no ‘lone pairs’ on the central atom, the molecules assumes a regular geometrical shape.
  2. If there are ‘lone pairs’ also on the central atom, then the structure gets distorted and the bond angle changes.
  3. A ‘lone pair of electrons’ occupy more space around the central atom than a bond pair, because the ‘lone pair’ is attracted by only one nucleus, while the ‘bond pair- is attracted by two nuclei.
  4. The electron pairs orient in space so as to have minimum repulsions among them, which determines the ‘bond angles’.
  5. The order of repulsion between various electron pairs is
    l.p – 1.p > l.p – b.p > b.p – b.p
  6. The magnitude of repulsion between bond pairs of electrons depends on the electronegativity differences between the central atom and the other atoms.
  7. The order of repulsion between different bonds is :
    Triple bond > Double bond > Single bond

TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 48

Applications:
1. Structure of NH3 molecule :
In NH3 molecule, the central ‘N’ atom contains 3 bond pairs one lone pair. In total, there are four pair of electrons. Hence, according to VSEPR theory, the expected structure of the molecule in tetrahedral with a bond angle of 109°281. But, because of the presence of lone pair, the shape of the molecule gets distorted. Therefore, the real structure of the molecule is pyramidal with a bond angle of 107°.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 49

2. Structure of H2O molecule :
In H2O molecule, the central ‘O’ atom contains two bond pair and two lone pairs. In total there are four pair of electrons. Hence, according to VSEPR theory, the expected structure of the molecule is tetrahedral with a bond angle of 109°28¹. But, because of the presence of lone pair, the shape of the molecule gets distorted. Therefore, the real structure of the molecule is V-shape with a bond angle of 104.5°.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 50

Question 6.
How do you explain the geometry of the molecules on the basis of Valence bond Theory?
Answer:
The valence bond theory explains the shape, the formation and directional properties of bonds in poly atomic molecules like BeCl2, BCl3, CH4, NH3, H2O, etc. in terms of overlap and hybridisation of atomic orbitals.

Eg., in BeCl2, Beds the central atom. Its atomic number is 4. Therefore its ground state electron configuration is 1s² 2s². In order to explain the formation of BeCl2, SP hybridisation is to be assumed for Be in its excited state. The excited state configuration of Be is 1s²2s¹ 2p¹x 2p°y 2p°z. As a result of which two sp hybrid orbitals will form on it. Now the two sp hybrid orbitals overlap Head-Head with 3pz orbitals of two chlorine atoms forming two sigma bonds.

The molecule is linear and the bond angle is 180°.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 51

In BCl3 the central atom is boron. Its atomic number is 5; Therefore its ground state electron configuration is 1s² 2s²
2p¹x 2p°y 2p°z. From this configuration it is evident that it exhibits mono-valency. The first excited state configuration of B is 1s²2s¹2p¹x 2p°y 2p°z. Since, there are two half-filled orbitals, the valency of Boron is 3.

In order to explain the formation of BCl3 molecule sp hybridisation is to be assumed to boron atom in its excited state I. As a result of which three sp2 hybrid orbitals will form on it. Now, the three sp² hybrid orbitals of ‘B’ atom overlap Head – Head with 3pz orbitals of three Cl atoms forming three sigma bonds. The shape of the molecule is plane triangular and the bond angle is 120°.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 52

CH4 (Methane) :
In CH4, the central atom is carbon. Its atomic number is 6. Therefore the electron configuration is 1s² 2s² 2p² (2px¹, 2py¹, 2pz° ). In order to explain the formation of four bonds excited stated configu-ration (1s² 2s¹ 2px¹ 2py¹ 2pz¹) is to be taken. From this configuration, we can say that carbon can form four bonds. [Ex. CH4]
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 53

In the formation of methane molecule, the central carbon atom undergoes sp³ hybridisation in its excited state (1s² 2s¹ 2p¹x 2p¹y 2p¹z). As a result of which four sp hybrid orbitals form on it. Each of them contains an unpaired electron. Now, these four hybrid orbitals overlap Head- Head with Is orbitals of four hydrogen atoms forming CH4 molecule. The structure of the molecule is tetrahedral. The bond angle is 109°28′. Since there are no lone pairs, there is no distortion in the structure of the molecule.

NH3 (Ammonia) in NH3 molecule, one sp³ orbital contains a lone pair of electrons. There will be repulsions between the lone pair and bond pair electrons. These are stronger than bp – bp repulsions. (l.p – b.p > b.p – b.p). So the bond pairs are pushed together. Hence the bond angle decreases from tetrahedral angle. The molecule assumes a pyramidal shape.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 54

TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure

Question 7.
‘What do you understand by Hybridisation? Explain different types of hybridisation involving s and p orbitals. [Mar. ’18 (AP) (TS Mar. 17, 15; Mar. 13]
Answer:
Hybridisation:
“Intermixing of atomic orbitals of almost equal energies of an atom and their redistribution into an equal number of identical orbitals is called, ‘hybridisation’.

Examples :
Let us see the three types of hybridisations – sp³, sp² and sp.

1) sp³ Hybridisation :
The phenomenon of intermixing of one s’ orbital and three p’ orbitals forming four sp³’ hybrid orbitals is called sp³ hybridisation.

Each of the sp³ hybrid orbitals possess \(\frac{1}{4}\) of s – character and \(\frac{3}{4}\) of p character. The bond angle in-between any two sp³ hybrid orbitals is 109°28′. The shape of the molecule in which the central atom undergoes sp³ hybridisation is tetrahedral.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 55

Example :
Formation of Methane (CH4 molecule :
In the formation of methane molecule the central carbon atom undergoes sp³ hybridisation in its excited state. As a result of which four sp³ hybrid orbitals will form on it. All the four sp³ hybrid orbitals possess bond pair of electrons. Now, the four sp³ hybrid orbitals overlap head-head with ‘1s’ orbitals of four H’ atoms forming CH4 molecule. The shape of the molecule is tetrahedral with a bond angle of 109°28′.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 56

2) sp² Hybridisation:
The phenomenon of intermixing of one s’ orbital and two p’ orbitals forming three sp²’ hybrid orbitals is called sp² hybridisation.

Each of the sp² hybrid orbitals possesses \(\frac{1}{3}\) s – character and \(\frac{2}{3}\) p – character. The bond angle in between any two sp² hybrid orbitals is 120°. The shape of the molecule in which the central atom undergoes sp² hybridisation is plane triangular.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 56

Example:
Formation of Boron trichloride (BCl3) molecule :
In the formation of BCl3 molecule the central boron atom undergoes sp² hybridisation in its excited state (1s² 2s¹ 2px¹, 2py¹, 2pz°). As a result of which three sp² hybrid orbitals will form on it. All the three hybrid orbitals possess an electron each. Now, the three sp2 hybrid orbitals overlap head – head with half-filled ‘3pz‘ orbitals of three chlorine atoms forming BCl3 molecule. The shape of the molecule is plane triangular with a bond angle of 120°.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 58

3) sp Hybridisation:
The phenomenon of intermixing of one s’ orbital and one p’ orbital of an atom forming two ‘sp’ hybrid orbitals is called sp hybridisation.

Each of the sp hybrid orbitals possesses \(\frac{1}{2}\) s – character and \(\frac{1}{2}\) p – character. The bond angle in-between the two hybrid orbitals is 180°. The shape of the molecule in which the central atom undergoes sp hybridisation is linear.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 59

Example : Formation of Beryllium chloride (BeCl2) :
In the formation of BeCl2 molecule the central Be’ atom undergoes sp hybridisation in its excited state (1s² 2s¹ 2px¹, 2py°, 2pz°) As a result of which two sp hybrid orbitals will form on it.

The two hybrid orbitals contain an electron, each. Now, the two sp hybrid orbitals overlap head-head with half-filled ‘3pz’ orbitals of two chlorine atoms forming BeCl2 molecule. The shape of the molecule is linear with a bond angle of 180°.
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 59

Question 8.
Write the salient features of Molecular Orbital Theory.
Answer:
M.O.T. was proposed by Hund and Mulliken.

  1. Formation of MOs : The atomic orbitals of the bonded atoms with similar energies and symmetrical with respect to the inter-nuclear axis lose their identities, merge together and give rise to a set of new orbitals, named, ‘molecular orbitals’.
  2. Now in the molecule, the electrons will be influenced by the nuclei of all atoms, since the m.o. covers all the nuclei.
  3. The m.o.’s are formed by LCAO method.
  4. The number of m.o’s produced is equal to the number of a.o’s that combine. ,
  5. Half of these m.o’s possess lower energy than the a.o’s from which they form. They are called, bonding m.o’s. The remaining half will have higher energy. They are called, anti bonding m.o’s
  6. The shape of m.o’s is determined by the shapes of a.o’s from which they form.
  7. Axial overlap of two a.o’s gives a m.o’s while parallel overlap gives n m.o’s.
  8. Based on spectroscopic studies, the relative order of energy of the m.o’s of ten m.o’s of Li2,
    Be2, B2, C2 and N2 are :
    TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 61
  9. Electrons occupy the m.o’s which reflects Aufbau principle, Hund’s rule of maximum multiplicity and Pauli’s exclusion principle.
  10. Electrons present in bonding m.o’s help to stabilise the bond, while those present in anti bonding m.o.’s more than destabilize the bond.
  11. The inner orbitals which do not take part in bonding are called, ‘non-bonding orbitals’.

TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure

Question 9.
Give the Molecular Orbital Energy diagram of (a) N2 and (b) O2 Calculate the respective bond order. Write the magnetic nature of N2 and O2 molecules. [TS Mar. ’19 (IPE ’14)]
Answer:
a) Molecular Orbital Energy diagram of N2:
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 62
Electronic configuration of molecular or-bital of N2:
σ 1s², σ* 1s², σ 2s², σ* 2s². π2p²y = π2p²z, σ 2p²x

Calculation of bond order of N2 :
Number of bonding electrons (Nb) = 10
Number of anti-bonding electrons (Na) = 41
Bond order = \(\frac{1}{2}\) (Nb – Na)
= \(\frac{1}{2}\)(10 – 4) = 3
Nitrogen is diamagnetic since all the electrons in N2 molecule are paired.

b) Molecular Orbital Energy diagram of O2:
TS Inter 1st Year Chemistry Study Material Chapter 3 Chemical Bonding and Molecular Structure 63
Electronic configuration of molecular orbital of O2 :
σ 1s², σ* 1s², σ 2s², σ* 2s². π2p²x, π2p²y = π2p²z, π* 2p¹y = π* 2p¹z

Calculation of bond order of O2 :
Number of bonding electrons (Nb) = 10
Number of anti-bonding electrons (Na) = 6
Bond order = \(\frac{1}{2}\) (Nb – Na) = \(\frac{1}{2}\)(10 – 6) = 2

Oxygen is paramagnetic as there are two unpaired electrons in two antibonding n π* 2py and π* 2pz molecular orbitals. So oxygen is diamagnetic.

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Telangana TSBIE TS Inter 1st Year Chemistry Study Material 2nd Lesson Classification of Elements and Periodicity in Properties Textbook Questions and Answers.

TS Inter 1st Year Chemistry Study Material 2nd Lesson Classification of Elements and Periodicity in Properties

Very Short Answer Type Questions

Question 1.
What is the difference in the approach between Mendeleev’s periodic law and the modern periodic law?
Answer:
Mendeleev’s periodic law is based on atomic weights while the modern periodic law is based on electronic configuration.

Mendeleev’s periodic law states that the physical and chemical properties of the elements are periodic functions of their atomic weight.

Modern periodic law states that the physical and chemical properties of the elements are periodic functions of their outer electronic configuration.

Question 2.
In terms of period and group, where would you locate the element with Z = 114?
Answer:
The electronic configuration of the element with Z = 114 is
1s²2s²2p 3s²3p63d10
4s²4p64d104f14 5s²5p65d105f14
6s²6p66d18 7s²7p²

Since in the atom of the element with Z = 114 electrons are filled in 7 orbits it should belong to 7th period. It has 4 electrons in the outermost orbit, so it should belong to IV A group.

Question 3.
Write the atomic number of the element, present in the third period and seventeenth group of the periodic table.
Answer:
The elements in 17th group are halogens. The halogen in the third period is chlorine. Its atomic number is 17.

Question 4.
Which element do you think would have been named by
a) Lawrence Berkeley Laboratory
b) Seaborg’s group
Answer:
a) Lawrencium (Z = 103)
b) Seaborgium (Z = 106)

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 5.
Why do elements in the same group have similar physical and chemical properties?
Answer:
All the elements in a group have same outer electronic configuration which is responsible for the similar physical and chemical properties with some gradation.

Question 6.
What are representative elements? Give their valence shell configuration.
Answer:
s and p block of elements, excluding ‘O’ group are called ‘representative elements’. Their valence configuration is ns1-2np0-5.

Question 7.
Justify the position of f-block elements in the periodic table.
Answer:

  1. These are present in two series, 4f and 5f series. In these elements, the differ-entiating electron enters in the (n – 2)f sub-level.
  2. All these elements have similar outer configuration (n-2)f1-14 (n – 1)d0-1 ns². So all the 4f series and 5f series elements show similar properties.

For the above reasons, these elements are grouped together as lanthanides and actinides respectively and placed sepa-rately below the main body of the periodic table. These two series of elements belong to III B group only, in the 6th and 7th periods respectively.

Question 8.
An element ‘X’ has atomic number 34. Give its position in the periodic table.
Answer:
According to Bohr – Bury principle, the electrons are distributed in the various shells as : 2, 8, 18, 6. Total number of shells represents the period and the number of valence electrons represents the group. So it belongs to 4th period and VIth A group (Z = 34)

Question 9.
What factors impart characteristic properties to the transition elements?
Answer:

  1. Small size of atom
  2. High nuclear charge
  3. Variable valency and
  4. Availability of’d’ orbitals for bonding

Due to these reasons, transition elements exhibit characteristic properties.

Question 10.
Give the outer shells configuration of d- block and f-block elements.
Answer:
The outer shell configuration of d-block el-ements is (n – 1)d1-10 ns1 or 2.

The outer shell electronic configuration of f-block elements is
(n – 2)f1-14 (n – 1)d0 or 1 ns².

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 11.
State and give one example for Dobereiner’s law of triads and Newland’s law of octaves.
Answer:
Dobereiner’s law of triads:
When a group of three elements of similar properties are arranged in increasing order of their atomic weight the atomic weight of the middle element is an arithmetic mean of the other two elements e.g.,

Element Atomic weight
Li 7
Na 23
K 39

Newland’s law of octaves: When elements are arranged in the increasing order of their atomic weights, every eighth element has properties similar to the first element analogous to every eighth note that resembles the first in octaves of music.
TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 1

Question 12.
Name the anomalous pairs of elements in the Mendeleev’s periodic table.
Answer:
In Mendeleev’s periodic table elements are arranged in the increasing order of their atomic weights. But few elements are arranged in reverse order. They are called anomalous pairs. They are Argon and Potassium, Tellurium and Iodine, Cobalt and Nickel.

Question 13.
How does atomic radius vary in a period and in a group? How do you explain the variation?
Answer:
As we go down in a group, atomic radius gradually increases.

Reason :
As we go down in a group, new shells are opened and electrons which enter in them are attracted weakly by the nucleus. As we go from left to right in a period, atomic radius gradually decreases.

Reason :
As we go along a period the electrons enter in the same outer shell but at the same time nuclear charge increases. Hence electrons are attracted strongly.

Question 14.
Among N-3, O-2, F, Na+, Mg+2 and Al+3
a) What is common in them?
b) Arrange them in the increasing ionic radii.
Answer:
a) All the ions contain same number of electrons (10 electrons).
b) Al+3 < Mg2+ < Na+ < F < O-2 < N-3

Question 15.
What is the significance of the term isolated gaseous atom while defining the ionization enthalpy.
Hint: Requirement for comparison.
Answer:
To compare the ionization enthalpy of different elements only isolated gaseous atoms must be considered.

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 16.
Energy of an electron in the ground state of the hydrogen atom is – 2.18 × 10-18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol-1.
Answer:
Since the hydrogen atom is in ground state the electron is present in 1st orbit.

Ionization potential = E – E1
= 0 – (- 2.18 × 10-18 J) = 2.18 × 10-18 J/atom
or 2.18 × 10-18 J × 6.023 × 1023
= 13.13 × 105 J / mole

Question 17.
Ionization enthalpy1 (IE1) of O is less than that of N-explain.
Answer:
In N, the 2p orbitals are half-filled (1s²2s² 2p¹x 2p¹y, 2p¹z) and more stable. Hence the IE1 of N is high. In O (1s²2s² 2p²x 2p²y, 2p²z), there are repulsions among the paired up 2p electrons, which lowers the I.E.

Question 18.
Which in each pair of elements has a more negative electron gain enthalpy?
a) O or F
b) F or Cl
Answer:
a) In a period from left to right electron gain enthalpy increases. So F has more electron gain enthalpy than O.

b) Due to small size of fluorine atom the electron repulsions are more. So some energy is required to overcome the repulsion. So electron gain enthalpy of chlorine is more than fluorine.

Question 19.
What are the major differences between metals and non-metals?
Answer:
1) Metals are usually solids at room temperature. Metals have high melting and boiling points. Metals are good conductors of heat and electricity. Metals are malleable and ductile.

2) Non-metals are usually solids or gases at room temperature. Non-metals have low melting and boiling points. Non-metals are poor conductors of heat and electricity. Most non metallic solids are brittle and are neither malleable nor ductile.

Question 20.
Use the periodic table to identify elements.
a) With 5 electrons in the outer subshell
b) Would tend to lose two electrons
c) Would tend to gain two electrons.
Answer:
a) The number of electrons in the outer orbit is equal to the group number of periodic table. Since the outer subshell contains five electrons it should belong to Vth A group.

b) Since the element tends to lose two electrons, its atom contains two electrons in the outer orbit. So the element belongs to second group.

c) The elements tend to gain two electrons to get octet to acquire stability. Since the element tends to gain electrons it should have 6 electrons in its outer orbit. So the element belongs to 6th A group.

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 21.
Give the outer electronic configuration of s, p, d, and f-block elements.
Answer:
s – block elements ns¹ or ns²
p – block elements ns np¹ to ns² np6
d – block elements (n – 1)d1-10 ns1 or 2
f – block elements (n – 2)f1-14 (n – 1)d0 or 1 ns²

Question 22.
Write the increasing order of the metallic character among the elements B, Al, Mg, and K.
Answer:
B < Al < Mg < K

Question 23.
Write the correct increasing order of non-metallic character for B, C, N, F, and Si.
Answer:
More the electronegativity and ionization energies more is the non-metallic character. So the correct increasing order of non- metallic character is
Si < B < C < N < F

Question 24.
Write the correct increasing order of chemical reactivity in terms of oxidizing property for N, O, F and Cl.
Answer:
More the electronegativity and electron gain enthalpies more is the oxidation power.
N < O < F < Cl

Question 25.
What is electronegativity? How is this useful in understanding the nature of elements?
Answer:
The relative tendency of an atom in a covalent molecule to attract the shared pair of electrons towards itself is called electronegativity.

The elements having more electronegativity values are more non-metallic and the elements having less electronegativity values are more metallic.

Question 26.
What is screening effect? How is it related to IE?
Answer:
The electrons present in the inner shells act as screens between the nucleus and the valence shell electrons, in other words, these electrons partially neutralise the force of attraction of the nucleus, over the valence electrons. This is called, shielding or screening effect’.

When the screening effect increases (No. of inner electron-shells increase), the IP decreases.

The magnitude of screening effect ∝ \(\frac{1}{I.P}\).

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 27.
How are electronegativity and metallic & non-metallic characters related?
Answer:
Electronegativity provides a means of predicting the nature of elements. Electronegativity is directly related to the non-metallic character of elements and inversely related to the metallic character of elements.

The electronegativity increases across a period from left to right. So metallic character decreases and non-metallic character increases.

In group electronegativity decreases from top to bottom. So the metallic character increases down and the non-metallic character increases.

Question 28.
What is the valency possible to arsenic with respect to oxygen and hydrogen?
Answer:
With respect to oxygen, the possible valency is 3 and 5 in As2O3 and As2O5 respectively. With respect to hydrogen, the valency of arsenic is 3 in AsH3.

Question 29.
What is an amphoteric oxide? Give the formula of an amphoteric oxide formed by an element of group – 13.
Answer:
The oxide which reacts both with acids and bases forming salts is called an amphoteric oxide. In 3rd group, Al2O3 and Ga2O3 are amphoteric oxides.

Question 30.
Name the most electronegative element. Is it also having the highest electron gain enthalpy? Why or Why not?
Answer:
The most electronegative element is fluorine (4.0). Fluorine do not have the highest EA value. Among halogens, Cl has the highest EA value. The EA values are in the order, Cl > F > Br > I > At.

Reason :
The size of F atom is small, compared to the size of Cl atom. The addition of one electron to F atom produces high electron density around it. Then the electron-electron repulsions increase in F atom. Because of this, F atom shows lesser tendency to attract another electron towards it and form F ion. Hence F has low EA than that of Cl.

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 31.
What is diagonal relation? Give one pair of elements, that have this relation.
Answer:
The first element of a group shows similarities in properties with the second element of the next group. This is called, diagonal relationship’.
TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 2
But this relation continues effectively upto the IV group only.

Question 32.
How does the nature of oxides vary in the third period?
Answer:
The basic nature decreases and the acidic nature increases from Na2O to Cl2O7, as shown below.
TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 3

Question 33.
Radii of iron atom and its ions follow Fe > Fe2+ > Fe3+ – explain.
Answer:
When a neutral atom loses an electron, a positive ion (cation) forms. In this ion, there will be less number of electrons than protons. So the nuclear attraction on the valence electrons would increase. Then the size decreases.

As more and more electrons are removed, the nuclear attraction on the valence shell of electrons increases more and more. Consequently, the ionic size decreases more and more.
So, sizes : Fe > Fe2+ > Fe3+.

Question 34.
IE2 > IE1 for a given element – why?
Answer:
By the removal of an electron from a neutral atom, a uni +ve ion is formed. In this uni + ve ion the number of protons are more than the number of electrons. Hence the nuclear attractions will be more in uni + ve ion. As a result of which more amount of energy is required to remove an electron from the uni + ve ion. Hence, 2nd IP values are always greater than first IP values.

Question 35.
What is lanthanide contraction? Give one of its consequences.
Answer:
The steady decrease of atomic or ionic size from left to right in lanthanides, as the atomic number increases, is called Lanthanide contraction’.

In lanthanides, the differentiating electron enters the (n – 2) f subshell. Due to their peculiar shapes, f orbitals do not provided proper shielding for the valence electrons from the nuclear attraction. Consequently, the atomic or ionic size decreases gradually from left to right in lanthanides.

The decrease in contraction is more regular in Ln+3 ions than in Ln atom.

Consequences:

  1. Because of the lanthanide contraction, the hardness, m.p, b.p of the elements increase from Ce to Lu.
  2. There are more similarities between 4d and 5d series of elements than between 3d and 4d series. The main reason is, ‘lanthanide contraction’.
  3. The effect of lanthanide contraction is felt even in the post lanthanides. As a result, atomic sizes in the pairs of elements Zr/Hf; Nb/Ta; Mo/W are the same. So these pairs of elements possess similar properties. Zr and Hf resemble each other so much that their isolation is a quite difficult process.

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 36.
What is the atomic number of the element, having maximum number of unpaired 2p electrons? To which group does it belong?
Answer:
A p orbital can accommodate maximum of 3 unpaired electrons. So the electronic configuration is 1s² 2s² 2p³. The atomic number of element is 7 and the element is nitrogen. It belongs to V th group.

Question 37.
Sodium is strongly metallic, while chlorine is strongly non-metallic – explain.
Answer:
Sodium is more electropositive due to its small ionisation energy. Hence it is strongly metallic.

Chlorine has highest electron affinity and also more electronegativity. So it is more non-metallic.

Question 38.
Why are zero group elements called noble gases or inert gases?
Answer:
Zero group elements have stable ns²np6 outer electronic configuration. Helium has completely filled 1s² configuration. So these elements are chemically inert and hence they are called inert gases. Recently it was found that these elements are also participating in chemical reactions but they are less reactive like noble metals such as gold and platinum. Hence they are also called noble gases.

Question 39.
Select in each pair, the one having lower ionization energy and explain the reason.
a) I and l
b) Brand K
c)Li and Li+
d) Ba and Sr
e) O and S
f) Be and B
g) N and O
Answer:
a) I.P value of l is less : Reason : The size of I is greater than I
b) I.P value of K is less: Reason : K is electro +ve element whereas Br is electro -ve element
c) I.P value of Li is less : Reason : The size of Li is greater than Li+
d) I.P value of Ba is less : Reason: The size of Ba is greater than Sr
e) I.P value of S is less : Reason : The size of S atom is greater than 0 atom
f) I.P value of B is less : Reason : Be has fully filled atomic orbitals.
g) I.P value of 0 is less : Reason : N has half-filled atomic orbitals.

Question 40.
IE1 of O < IE1 of N but IE2 of O > IE2 of N – Explain.
Answer:
In N the 2p orbitals are half-filled (1s²2s² 2P¹xyz). The half-filled orbitals are more stable. In O (1s² 2s² 2p²x 2p¹x 2p¹y, 2p¹z) there are repulsions among the paired 2p electrons. So IP1 of O is less than N.

After removing one electron O+ has stable half-filled 2p³ electronic configuration but N+ has 2p² electronic configuration. So IE2 of O is greater than IE2 of N.

Question 41.
Na+ has higher value of ionization energy than Ne, though both have same electronic configuration – Explain.
Answer:
Na+ has more number of protons (11) than Ne (10). So the nucleus of Na+ holds the electrons strongly. Hence the ionisation energy of Na+ is greater than Ne.

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 42.
Which in each pair of elements has a more electronegative gain enthalpy? Explain.
a) N or O
b) F or Cl
Answer:
a) Since N has stable halffilled p³ configuration it has less electron gain enthalpy than O.
b) Electron gain enthalpy of fluorine is less than chlorine because there is repulsion between electrons in small fluorine atoms.

Question 43.
Electron affinity of chlorine is more than . that of fluorine – explain.
Answer:
The size of fluorine atom is small. The addition of electron to small fluorine atom results in electron-electron repulsions. To overcome these repulsions some of the energy liberated due to addition of electron is consumed by fluorine. So the electron gain enthalpy decreases. These electron-electron repulsions are absent in bigger chlorine atom. So the electron gain enthalpy of fluorine is less than chlorine.

Question 44.
Which in each has higher electron affinity?
a) F or Cl
b) O or O
c) Na+ or F
d)For F
Answer:
a) F
b) O
c) Na+
d) F

Question 45.
Arrange the following in order of increasing ionic radius:
a) Cl, P-3, S-2, F
b) Al+3, Mg+2, Na+, O-2, F
c) Na+, Mg+2, K+
Answer:
a) F < Cl < S-2 < P-3
b) Al3+ < Mg2+ < Na+ < F < O2-
c) Mg2+ < Na+ < K+

Question 46.
Mg+2 is smaller than O-2 in size, though both have same electronic configuration – explain.
Answer:
Mg2+ and O-2 are isoelectronic but Mg2+ has more number of protons (12) than O-2 (8). Due to more attractive power of nucleus of Mg2+ it is smaller than O-2.

Question 47.
Among the elements B, Al, C, and Si
a) Which has the highest first ionization enthalpy?
b) Which has the most negative electron gain enthalpy?
c) Which has the largest atomic radius?
d) Which has the most metallic character?
Answer:
a) C
b) C
c) Al
d) Al

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 48.
Consider the elements N, P, O and S and arrange them in order of:
a) Increasing first ionization enthalpy
b) Increasing negative electron gain enthalpy
c) Increasing non-metallic character.
a) S < P < O < N
b) N < P < O < S
c) P < S < N < O

Question 49.
Arrange in given order:
a) Increasing EA : O, S and Se
b) Increasing IE1 : Na, K and Rb
c) Increasing radius: I, I+ and I
d) Increasing electronegativity: F, Cl, Br, I
e) Increasing EA : F, Cl, Br, I
f) Increasing radius: Fe, Fe+2, Fe+3
Answer:
a) 0 < Se < S
b) Rb < K < Na
c) I < I < I+
d) I < Br < Cl < F
e) I < Br < F < Cl
f) Fe+3 < Fe+2 < Fe

Question 50.
a) Name the element with highest ionization enthalpy.
b) Name the family with highest value of ionization enthalpy.
c) Which element possesses highest electron affinity?
d) Name unknown elements at the time of Mendeleef.
e) Name any two typical elements.
Answer:
a) Helium
b) Zero group
c) Chlorine
d) Eka Boron – Scandium
Eka Aluminium – Gallium
Eka Silicon – Germanium
Eka Manganese – Technisium
e) Sodium and magnesium.

Question 51.
a) Name any two bridge elements.
b) Name two pairs showing diagonal relationship.
c) Name two transition elements.
d) Name two rare earths.
e) Name two transuranic elements.
Answer:
a) Sodium and magnesium
b) Li, Mg; Be. Al
c) Chromium and Copper
d) Cerium and Lutesium
e) Neptunium and Plutonium

Short Answer Questions

Question 1.
On the basis of quantum numbers, justify that the 6th period of the periodic table should have 32 elements.
Answer:
The atoms of 6 th period elements contain 6 orbits containing the electrons.

In the long form periodic table, every period starts with the filling of a new orbit. So sixth period starts with the filling of principal quantum number 6. The sixth period should end with the filling of p-sub-shell of the same orbit. Thus the outer electronic configuration of last element is 6s²6p6. Before the 6p orbital starts filling the subshells to be filled with electrons are 6s, 4f and 5d.

The electrons that can be filled in 6s are 2, 4f are 14, 5d are 10 and 6p are 6, a total of 32 electrons. Thus the number of elements that can present in the 6th period are 32 .elements.

Question 2.
How did Mosley’s work on atomic numbers show that atomic number is a fundamental property better than atomic weight?
Answer:
Mosley (1913) obtained the X – ray spectrum of different elements and found that the frequency v of the lines in the spectrum is related with the atomic number (Z) of the element, as : √υ = a(Z – b)

a, b are constants for any selected series of lines in the X – ray spectrum (K, L, M, …… etc. series) of the element. The element is used as anticathode in the X – ray tube. Z = at. no. of the element (used as anticathode).

When Mosley plotted a graph between √υ × 108 (on Y – axis) and the atomic number Z (on X – axis) of the elements for which the X- rays were obtained, he got a straight line.

He also plotted a graph between √υ × 108 and atomic weight of the element. He did not get a straight line.
TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 4

These observations indicated that atomic number is a fundamental property of an atom and not its atomic weight. Actually, the frequency (υ) of X – rays depends on the internal structure of the atom which is related to the number of electrons in the atom (atomic number).

Question 3.
State modern periodic law. How many groups and periods are present in the long form of the periodic table?
Answer:
Modern periodic law :
Modern periodic law was predicted by Mosley. It states that “The physical and chemical properties of the elements are periodic functions of their atomic numbers.”

Groups and periods :
In the long form of the periodic table given by Bohr, there are 18 groups (vertical columns) and 7 periods (horizontal rows).

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 4.
Why are f-block elements placed below the main table?
Answer:
To accommodate the d-block elements the s-block and p-block elements are separated. There remains a gap between the s-block and p-block elements in the 2nd and 3rd periods.

If the f-block elements have to be accommodated in the periodic table again the size of the periodic table should be increased by separating the s-block and p-block. Then the table looks awkward. To avoid this the f-block elements are placed below the main table.

Question 5.
Mention the number of elements present in each of the periods in the long form periodic table.
Answer:
First period contains two elements only. Second and third periods contain eight elements.

Fourth and fifth periods contain eighteen elements only.

Sixth period consists of 32 elements. Seventh period is an incomplete period and contains 29 elements.

Question 6.
Give the outer orbit general electronic configuration of
a) Noble gases
b) Representative elements
c) Transition elements
d) Inner transition elements.
Answer:
General outer configurations :
a) Noble gases : ns²np6 (ns² for He)
b) Representative elements : ns1-2 np0-5
c) Transition elements : (n – 1)d1-10ns1-2
d) Inner transition elements : (n – 2)f1-14 (n – 1)d0-1ns².

Question 7.
Give any four characteristic properties of transition elements.
Answer:
Characteristic properties of d – block elements :

  1. They are hard and heavy metals.
  2. They possess high density, melting and boiling points.
  3. They are good electrical and thermal conductors.
  4. They exhibit variable oxidation states.
  5. Most of these are metals and their ions are paramagnetic.
  6. They and their ions exhibit colour.
  7. They and their oxides act as catalysts.
  8. They form alloys.

Question 8.
What are rare earths and transuranic elements?
Answer:
The 14 elements from cerium (Z = 58) to Lutesium are called rare earth elements because their abundance in the earth crust is very less. The properties of all these 14 elements are similar to lanthanum. So they Are called lanthanides or lanthanons or lanthanoids. These are called 4f series elements because in these elements the valence electron enters into 4f orbital.

The elements after uranium (Z = 92) in the periodic table are called transuranic elements. These elements do not occur in the nature. They are man made, synthetic and artificial. They are all radioactive and disintegrate into other elements. These elements are called 5f series elements, because in these elements the valence electron enters into 5f orbital.

Question 9.
What is isoelectronic series? Name a series that will be isoelectronic with each of the following atoms or ions.
a) F b) Ar c) He d) Rb+
Answer:
Ions having the same number of electrons but different number of protons is called isoelectronic series. In isoelectronic series the size of the ion decreases with increases in atomic number.
a) N-3 O-2 F Na+ Mg2+ Al3+
b) P-3 S-2 Cl Ar K+ Ca2+ Sc3+
c) H He Li+ Be2+
d) As3- Se2- Br Rb+ Sr2+

Question 10.
Explain why cation is smaller and anion is larger in radii than their parent atoms.
Answer:
When an electron is removed from a neutral atom, cation is formed. The nuclear charge in both the cation and in its parent atom is the same. But the number of electrons in the cation are less than in its parent atom.

Hence, the nuclear attractions, in a cation will be more than in its parent atom. As a result of which the electron cloud of cat-ion shrinks. Hence the size of a cation is always smaller than its parent atom.
TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 5

When an electron is added to a neutral atom, anion is formed. The nuclear charge in both the anion and in its parent atom is the same. But, the number of electrons in the anion are more than in its parent atom. Hence the nuclear attractions, in the anion will be less than in its parent atom. As a result of which the electron Cloud expands. Hence, the size of anion is always greater than its parent atom.

Question 11.
Arrange the second period elements in the increasing order of their first ionization enthalpies. Explain why Be has higher IE, than B.
Answer:
Li < B < Be < C < O < N < F
In Be the electron to be removed is from 2s orbital whereas in B the electron to be removed is from 2p. The penetration of a 2s electron to the nucleus is more than that of 2p electron, thus 2s electron is strongly attracted by the nucleus than 2p electron. The 2s electron is shielded by only one orbital Is in Be but 2p electron is shielded by two orbitals Is and 2s. So the IE1 of Be is more than IE1 of B.

Question 12.
IE1 of Na is less than that of Mg but IE2 of Na is higher than that of Mg – explain.
Answer:
Sodium atom has only one electron in its outer orbit and by losing that electron it gets stability by acquiring octet. Further in the nucleus of sodium there are less number of protons than in magnesium. So nuclear attraction on the electrons in sodium is less than in magnesium. So IE1 of Na is less than IE1 of Mg.

In Na+ the outer orbit has stable octet. To remove the electron from stable octet more energy is required. But in Mg2+ there is one more electron outside the stable octet i.e., in 3s orbital. To remove the 3s electron the energy required is less. So IE2 of Na is greater than IE2 of Mg.

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 13.
What are the various factors due to which the IE of the main group elements tends to decrease down a group?
Answer:
1) Atomic size :
With increase in the atomic size, the distance from the nucleus to the outer electrons increases. So the attraction of the nucleus on outer electrons decreases. Hence IE decreases.

2) Nuclear charge :
With increase in nuclear charge i.e., effective nuclear charge, attraction of the nucleus on the outer electrons increases. So IE in-creases.

3) Screening effect or Shielding effect :
The inner orbits shield the nuclear attraction on the outer electrons. So with increase in the inner electrons shielding effect increases and thus IE decreases.

4) Extent of penetration of valence shell into inner electron :
The penetrating power of the orbitals towards the nucleus is in the order s > p > d > f. Nuclear attraction on the electrons in these orbitals also will be in the same order. So to remove an electron from different orbitals of the same orbit the energy required is in the order s > p > d

5) Number of charges on the ion :
With increase in the number of positive charges on an ion the nuclear attraction on the electrons increases. So IE increases.

6) Electronic configuration :
Atoms having octet in the outer orbit, or exactly half-filled and completely filled orbitals give stability to the atom. The energy required from the stable electronic configurations will be more.

Question 14.
The first ionization enthalpy values On k.J mol-1) of group 13 elements are :
TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 6
How do you explain this deviation from the general trend?
Answer:
Generally I. P. values decrease from top to bottom in a group. Gallium (Ga) has more I.P. than Indium due to poor shielding effect of 3d electrons. Thallium has more l.P. than Indium due to poor shielding effect of 4f – electrons.

Question 15.
Would you expect the second electron gain enthalpy of oxygen as positive, more negative or less negative than the first? Justify.
Answer:
The second electron gain enthalpy of oxygen is always positive.

The amount of energy released when an electron is added to a neutral isolated gaseous atom is called electron gain enthalpy
X(g) + e → X(g) ; ΔH = -ve

As result of adding an electron to the neutral atom it converts into uninegative ion. It becomes difficult to add one more electron to this uninegative ion as there is repulsion between the negative charge on the ion and negative charge of electron. So it requires some energy to overcome the repulsion and to add an electron to the uninegative ion.
X(g) + e → X2-(g) ; ΔH = +ve

So the second electron gain enthalpy of oxygen is always positive.

Question 16.
What is the basic difference between the electron gain enthalpy and electropositivity?
Answer:
The amount of energy released when an electron is added to a neutral gaseous isolated atom is called electron gain enthalpy.
X(g) + e → X(g) ; ΔH = -ve

Electropositivity is the tendency to lose Electron. It is directly related to the metallic character. More the electropositive character of an element more the tendency to lose the electron and thus the element is more metallic in nature.

If electron gain enthalpy is more, the element has more tendency to gain electron. So that element will have more non-metallic character.

Elements having more electropositive character can act as strong reducing agents while the elements having more electron gain enthalpies will act as oxidising character.

Question 17.
Would you expect IE1 for two isotopes of the same element to be the same or different? Justify.
Answer:
Isotopes of the same element have same I.E values. Though the isotopes of an element have different atomic weights they have same atomic number. Since the nuclei of isotopes of same element contain same number of protons their nuclear charge is same. Also their atomic sizes are same. So the nuclear attraction on the electrons in different isotopes of same element is same. Hence the IE1 of the isotopes of the same element are same.

Question 18.
Increasing order of reactivity among group – 1 elements is Li < Na < K < Rb < Cs, whereas among group -17 elements it is F > Cl > Br > I – explain.
Answer:
Reactivity of group-I elements is proportional to metallic nature and metallic nature increases from Li to Cs. So, the order of reactivity is Li < Na < K < Rb < Cs.

Reactivity of group -17 elements is proportional to non-metallic nature. Non-metallic nature decreases from F to I. So, the order of reactivity is F > Cl > Br > I.

Question 19.
Assign the position of the element having outer electronic configuration.
a) ns²np4 for n = 3
b) (n – 1)d²ns² for n = 4
Answer:
a) n = 3 indicates the atoms of the element have electrons in three orbits. So the element belongs to third period.

Its outer electronic configuration ns²np4 indicates the presence of 4 electrons in its outer orbit. So the element belongs to VI(A) group of the periodic table as the number of electrons in the outer orbit is equal to group number. The element is silicon with outer electron configuration 3s² 3p².

b) n = 4 indicates the element belongs to 4th period as the number of orbits filled with electrons is equal to period number. Since there are two electrons (n – 1)d orbital it belongs to d-block elements. Its outer electron configuration is 3d²4s². The element is Titanium and it is in the IV(B) group of the periodic table.

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 20.
Predict the formulae of the stable binary compounds that would be formed by the combination of the following pairs of elements.
a) Li and O
b) Mg and N
c) Al and I
d) Si and O
e) P and Cl
f) Element with atomic number 30 and Cl
Answer:
a) Valency of Li is 1 and that of O is 2. So the formula of compound is Li2G.
b) Valency of Mg is 2 and that of N is 3. So the formula of compound is Mg3N2.
c) Valency of Aluminium is 3 and that of I is 1. So the formula of the compound is AlI3.
d) The valency of silicon is 4 and that of oxygen is 2. So the formula of the compound is SiO2.
e) The phosphorous exhibits two types of valencies 3 and 5 but the compound with Cl in + 3 oxidation state is stable. So the formula of stable binary compound is PCl3.
f) The element with atomic number 30 is Zinc. Its valency is 2. So the formula of its binary compound with Cl is ZnCl2.

Question 21.
Write a note on the variation of metallic nature in a group and in a period.
Answer:
In any group of the periodic table as we move from top to bottom electropositive character increases. So metallic character increases.

In any period of the periodic table electronegativity increases. So metallic nature decreases.

Question 22.
How does the covalent radius increase in group 7?
Answer:
In every group, the differentiating electron enters in to a new orbit. So, the number of orbits in the atom of an element increases down the group. With the increase in the atomic number, the nuclear charge also increases, resulting in more attraction on the valence electrons. But the increase in the size of the atom exceeds the nuclear attraction. As a result the atomic radius (or) co-valent radius increases down the group.

Question 23.
Which element of 3rd period has the highest I.E1? Explain the variation of I.E1 in this period.
Answer:
In each period the last element i.e., the inert gas element has the highest I.E. In 3rd period, it is Argon.

Variation of I.E. in 3rd period :
The increasing order of IE of elements of 3rd period is Na < Al < Mg < Si < S < P < Cl < Ar. Mg and P have greater IE values than those of A1 and S respectively.

Reason:
In Mg, the 3s electrons are paired up (1s²2s²2p63s²). So the outermost electron i.e., 3s electron is paired up and moreover it is present in the s – orbital, which is more penetrating. So it requires more energy to remove this electron. So Mg has higher I.E., than expected.

In P, the 3p orbitals are half-filled (1s²2s²2p63s²3p¹x3p¹y3p¹z). So the atom is quite stable. So it. has higher I.E. than expected.

Question 24.
What is valency of an element? How does it vary with respect to hydrogen in the third period.
Answer:
Valency:
Valency is the combining capacity. It is the number of atoms of hydrogen or chlorine or any monovalent atom, with which one atom of the element combines.

Generally, all the elements in a group show the same valency. In case of s-block of elements, valency = group number.

In case of p – block of elements,
Valency = group number, or (8 – group no.)

Variation of valency w.r.t hydrogen, in a period :
In representative elements, the valency increases from 1 to 4 and then decreases to 1 from left to right.
Ex: Valency of 3rd period of elements :
TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 7

Question 25.
What is diagonal relationship? Give a pair of elements having diagonal relationship. Why do they show this relation?
Answer:
The first element of a group shows similarities in properties with the second element of the next group. This is called, ‘diagonal relationship’.

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 8
But this relation continues effectively upto the IV group only.

Question 26.
What is Lanthanide Contraction? What are its consequences?
Answer:
The steady decrease of atomic or ionic size from left to right in lanthanides, as the atomic number increases, is called ‘Lanthanide Contraction’.

In lanthanides, the differentiating electron enters the (n – 2) f subshell. Due to their peculiar shapes, f orbitals do not provide proper shielding for the valence electrons from the nuclear attraction. Consequently, the atomic or ionic size decreases gradually from left to right in lanthanides.

The decrease in contraction is more regular in Ln+3 ions than in Ln atom.

Consequences:

  1. Because of the lanthanide contraction, the hardness, m.p, b.p of the elements increase from Ce to Lu.
  2. There are more similarities between 4d and 5d series of elements, than between 3d and 4d series. The main reason is, lanthanide contraction’.
  3. The effect of lanthanide contraction is felt even in the post lanthanides. As a result, atomic sizes in the pairs of elements Zr/Hf; Nb / Ta; Mo/W are the same. So these pairs of elements possess similar properties. Zr and Hf resemble each other so much that their isolation is a quite difficult process.

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 27.
The first IP of lithium is 5.41 eV and electron affinity of Cl is – 3.61 eV. Calculate ∆H in kJ mol-1 for the reaction : Li(g) + Cl(g) → Li+(g) + Cl(g)
Answer:
leV = 1.602 × 10-22 kJ/atom.
or leV / atom = 96.48 kJ/mol.
TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 9

Question 28.
How many Cl atoms can you ionize in the process Cl → Cl+ + e by the energy liberated for the process Cl + e → Cl for one Avogadro number of atoms. Given IP = 13.0 eV, and EA = 3.60 eV. Avogadro number = 6 × 1023.
Answer:
The energy liberated during the addition of electron to neutral isolated gaseous atom is electron affinity. For chlorine

Cl(g) + e → Cl(g) EA = 3.60 eV/atom
For one mole i.e., Avogadro number of chlorine atoms
3.60 × 6 × 1023 = 21.6 × 1023 eV.

The ionisation energy of chlorine atom
Cl → Cl+ + e+ is 13.0eV
The number of Cl atoms that ionise with 21.6 × 1023 eV is
\(\frac{21.6\times10^{23}}{13.0}\) = 1.662 × 1023

Question 29.
The electron affinity of chlorine is 3.7 eV. How much energy in kcal is released when 2g of chlorine atoms is completely converted to Cl ions in the gaseous state? (1 eV = 23.06 kcal).
Answer:
1 eV / atom = 23.06 k cal / mol.
In the conversion of one mol. of Cl atoms into Cl ions the energy liberated is 3.7 × 23.06 k.cal/mol.
The energy liberated by 35.5 g of Cl = 3.7 × 23.06 k.cal.
∴ The energy liberated by 2g of Cl = ?
\(\frac{2\times3.7\times23.06}{35.5}\) = 4.8069 k.cal.

Long Answer Questions

Question 1.
Discuss the classification of elements by Mendeleev.
Answer:
The periodic classification of elements bused on ‘atomic weights’ was done by Lothar Meyeer (Germany) and Mendeleev (Russia), independently.

Mendeleev’s periodic law:
“The physical and chemical properties of elements and their compounds are a periodic function of their atomic weights”.

Mendeleev arranged the then known 65 elements in a periodic table. He did not blindly follow the atomic weight but gave more importance to their chemical properties in arranging them in the table.

Explanation of the periodic law:
When the elements are arranged in the increasing order of their atomic weights, elements with similar properties appear again and again, at regular intervals, just like the days, weeks, months, seasons, etc. repeat at regular intervals of time. This is called periodicity of properties.

Mendeleev’s Table:
Mendeleev introduced a periodic table containing the then known 65 elements. In this table, while arranging the elements, he gave importance not only to their atomic weights, but also to their physical and chemical properties. This table was defective in some respects. Then he introduced another table, after rectifying the defects of that table. It is called, ‘Short form of periodic table’. He named the horizontal rows as ‘periods’ and the vertical columns, as ‘groups’. It has in all 9 groups, I to VIII and a ‘O’ group. The first 7 groups were divided into A and B subgroups. There are 7 periods in the table.

The VIII group contains three triads, namely, (Fe, Co, Ni); (Ru, Rh, Pd) and (Os, lr, Pt).

Merits of Mendeleev’s table :

  1. Actually, it formed the basis for the development of other modern periodic tables.
  2. Mendeleev left some vacant spaces in his periodic table, for the then unknown elements. But he predicted the properties of those elements. Later on, when these elements were discovered, they exactly fitted into those vacant places having properties, predicted by Mendeleev. Ex : Eka-boron (scandium), Eka-silicon (germanium), Eka-aluminium (gallium) etc.
  3. ‘O’ group elements were not known at the time of Mendeleev. Later when they were discovered, they found a proper place in that table under ‘O’ group of elements. Similarly, the radioactive elements.
  4. In case of these pairs of elements Tellurium – Iodine, Argon – Potassium and Cobalt – Nickel, there is a reversal of the trend. The first element has higher atomic weight than the second one. These are called, anomalous pairs.

However, based on their atomic numbers, and chemical properties, this arrangement proved quite justified.

Drawbacks of Mendeleev’s periodic table:

  1. Dissimilar elements were placed in the same group.
    Ex : The coinage metals Cu, Ag and Au are placed along with the alkali metals K, Rb, Cs etc. in the I group. The only common property among them is that they are all univalent (valency = 1).
  2. The 14 rare earths having different atomic weights are kept in the same place.
  3. Hydrogen could not be given a proper place, as it resembles both alkali met-als and halogens in its properties.

Question 2.
From a study of properties of neighbouring elements, the properties of an unknown element can be predicted – Justify with an example.
Answer:
Mendeleev left some gaps in the periodic table. He predicted the properties of these unknown elements after studying the properties of the neighbouring elements. When he proposed the periodic table, the elements scandium (Z = 21), gallium (Z = 31) and germanium (Z = 32) were unknown. Not only he predicted their properties but also named them eka-boron, eka-alumi- nium and eka-silicon respectively. The three elements were soon discovered within 20 years of his pronouncement of the ‘periodic law’ (during his life time). It is astonishing to see that when the properties of these elements were studied, these showed a remarkable agreement with those predicted by Mendeleev.

In addition to the above three elements, Mendeleev predicted the discovery of some more elements.

Examples:

Properties of Eka-silicon predicted by Mendeleev (1871) Properties of Germanium discovered by Winkler (1886)
1. Atomic weight 72 Atomic weight 72.6
2. Specific gravity 5.5 Sp. gravity 5.46
3. Colour – dirty grey Colour – greyish white
4. Specific heat 0.073 Specific heat 0.076
5. Oxide – ESO2 less basic than Sn O2, but more basic than Si O2 Sp. gr. 4.70, refractory Oxide – Ge O2, slightly basic Sp. gr. 4.70, refractory
6. Chloride – ESCl4 a liquid, B.P. below 100°C Chloride – GeCl4, liquid B.P. 86.5°C

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 3.
Discuss the construction of long form periodic table.
Answer:
The important characteristic property of an element is found to be its ‘atomic number’ and not atomic weight. Accordingly, the periodic law was modified – “The physical and chemical properties of the elements are periodic functions of their atomic numbers”. Later on it was found that while deciding the properties of an element, its electronic configuration plays a very important role. So Bohr constructed the long form of the periodic table, based on the electronic configurations of the elements. The periodic law can be stated as “The physical and chemical properties of the elements are periodic functions of their electronic configurations”.

Salient features:

  1. This table is prepared based on a fun-damental property “atomic number”.
  2. This table can be easily studied, remem-bered and reproduced.
  3. Similarities, differences and trends in properties are more clearly reflected in this table.
  4. Vertical columns are known as groups and horizontal rows are called periods.
  5. There are seven periods in this table. The first period consists of two elements only. Second and third periods contain 8 elements each. Fourth and fifth periods contain 18 elements each. Sixth period consists of 32 elements. Seventh period is an incomplete period and consists of 19 elements.
  6. There are eighteen groups in this table. They are designated IA, IIA, IIIB, IVB, VB, VIB, VIIB, VIII, IB, IIB, 1IIA, IVA, VA, VIA, VILA, 0 (American Convention of naming).
  7. The elements in IA, IIA, IIIA, IVA, VA, VIA and VILA are known as representative elements or normal elements.
  8. The elements in IB, IIB, IIIB, IVB, VB, VIB, VIIB and VIII are called Transition elements.
  9. Zero group elements are placed at the extreme right of the table. These are called inert gases or noble gases. They possess stable ns np configuration.
  10. Short periods are broken and long periods are extended to accommodate transition elements.
  11. Lanthanides and actinides are placed separately at the bottom of the periodic table.
  12. Based on the entrance of differentiating electron, the table is divided into four blocks. They are s – block, p – block, d – block and f – block. In the elements of s – block, differentiating electron enters into s – orbital. Similarly in the elements of p – block, d – block and f – block, the differentiating electron enters into p, d and f – orbitals respectively.
  13. Based on complete and incomplete electron shells and chemical properties, the elements are classified into four types. They are 1) Type I (Inert gas elements) 2) Type II (Representative elements) 3) Type III (Transition elements) 4) Type IV (Inner transition elements).
  14. All the elements in a group possess similar properties, because they possess the same valence electron configuration.

Question 4.
Discuss the relation between the number of electrons filled into the sub energy levels of an orbit and the maximum number of elements present in a period.
Answer:
Construction of periods :
a) 1st period (short period) consists of two elements H1 and He2. The K shell (n = 1) can have a maximum of two electrons only. So there are only two elements in this period.

b) 2nd period (short period) contains 8 elements namely Li3 to Ne10.

Li atom has completed K – shell and a new shell, L starts with one electron in the remaining elements of the period i.e., from Be to Ne, the L shell is gradually filled up till Ne is reached. In Ne, the K as well as L shells are both completed.

Thus, the period ends. In these elements, the second energy level (L) gets gradually filled with a maximum of 8 electrons. So this period contains.8 elements.

c) 3rd period (short period) also contains 8 elements, i.e., Na11 to Ar18.

The M shell starts filling up with sodium. The shell builds up steadily, until argon is reached. After argon, the differentiating electron does not enter M shell (i.e., 3rd shell) but enters a new shell i.e., N shell (4th shell). Hence the 3rd period has only 8 elements.

d) 4th period (long period) has 18 elements which are K19 and Kr36.

The N shell starts filling in K (potassium), K has configuration 2, 8, 8, 1. In Ca, another electron enters the N shell. It has the configuration 2, 8, 8, 2.

Starting with the next element Sc (Z = 21), the penultimate M shell is expanded till it is complete with 18 electrons. In Cr and Cu, only one electron is present in the outermost shell i.e., Nth shell, while all the other elements contain 2 electrons. With Zn, the M shell is complete.

Then the extra electron enters the outermost (Nth) shell in the successive elements Ga to Kr. In this 4th period, 4s, 3d, 4p levels are successively filled with electrons. Hence this period has 18 elements.

e) 5th period follows closely the sequence in 4th period. It starts filling with 5s level. It ends when the 5p level is complete. The sublevels 5s, 4d and 5p get filled up. Then the total number of electrons filled into these levels is 18 and hence there are 18 elements in this period. The period starts with Rb and ends with Xe.

f) 6th period is long. In it, 6s, 4f, 5d and 6p energy levels get filled, which includes the 14 lanthanides. The total number of electrons filled in these levels is 32. Hence the total number of elements in the period is 32.

g) 7th period is incomplete, which includes the 14 actinides. There are nearly 20 elements in this period.

Question 5.
Write an essay on s, p, d and f block elements. [AP Mar. ’19; (AP, ’17, ’15; TS ’15]
Answer:
Depending upon the entering of differentiating electron into the atomic orbitals, the elements of long form of periodic table are divided into four blocks. They are s, p, d and f – blocks.

s – Block:

  1. The elements, in which the differentiating electron enters the s – sub-level are called s-block elements.
  2. In this block, there are two groups. They are IA and IIA.
  3. The valence electron configuration of elements of IA group is ns¹ and that of IIA is ns².
  4. The elements of LA are Alkali metals and IIA are Alkaline earth metals.
  5. These are metals and are highly reactive.

p – Block:

  1. The elements in which the differentiating electron enters the p – sub-level are called p – block elements.
  2. In this block, there are six groups. They are IIIA, IVA, VA, VIA. VIIA, and ‘O’.
  3. The valence – electron configuration of the elements of these groups varies from ns² np¹ to ns² np6.
  4. It includes metals, non-metals, metalloids, noble gases.

d – Block:

  1. The elements in which the differentiating electron enters into d – sub-level are called d – block elements.
  2. In this block, there are ten groups. They are IIIB, IVB, VB, VIB, VIIB, VIII, IB, IIB.
  3. The general outer configuration of d – block elements is (n – 1)d1-10 ns1-2.
  4. These elements are also known as Transition elements.
  5. These are all metals having high H.P and B.P.

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 10

f – Block:

  1. The elements in which the differentiating electron enters into f – orbital are called f – block elements.
  2. In this block, there are 14 groups. They have no designation.
  3. Their general outer configuration is, (n – 2)1-14 (n – 1)d0-1ns².
  4. The f – block elements are arranged separately at the bottom in two rows.
  5. The elements of first row are called lanthanides and in these elements the valence electron enters into 4f orbital.
  6. The elements of second row are called actinides and in these elements the valence electron enters into 5f orbital.
  7. All the f – block elements are also known as Inner Transition elements.
  8. There are all metals.

Question 6.
Relate the electronic configuration of elements and their properties in the classification of elements.
Answer:
All the elements are divided into four types on the basis of their properties and electronic configurations. They are : (i) Noble gas elements (ii) Representative elements (iii) Transition elements and (iv) Inner transition elements.

i) Noble gas elements :
Elements in which the outermosts and p sub-shells are completely filled are called inert gas elements. “0” group elements belong to this type. Due to completely filled shells they show chemical inertness and possess more stability. The general outer configuration of these elements is ns²np6 (except helium.) The configuration of helium is 1s². They are called noble gases.

ii) Representative elements :
Elements in which the outermost s and p sub shells are incompletely filled are called representative elements. These elements are so named because they represent most of the chemical reactions known.

Elements of s – block and p – block (except ‘O’ group) belong to this type. The general outer electronic configuration of these elements is ns¹, ns², ns²np¹ to ns² np5. These elements enter into chemical reactions either by losing or gaining or sharing of electrons. Many of the non-metals, metalloids and some metals belong to this type.

iii) Transition elements:
Elements in which the outermost and penultimate shells are partially filled are called transition elements.

They belong to d-block. Their general configuration is (n – 1)d1-9 ns1-2. They are so named because there is a gradation from electro + ve nature to electro – ve nature. They possess the following characteristic properties.

  1. They exhibit variable oxidation states.
  2. They form coloured compounds.
  3. They are all paramagnetic.
  4. They and their oxides acts as catalysts.
  5. They form alloys and interstitial compounds.
  6. They form complex compounds.

iv) Inner transition elements:
Elements in which the outermost, penultimate and antipenultimate shells are partially filled are called Inner transition elements.

They belong to f – block and are placed separately at the bottom of the table. Their general outer configuration is (n – 2)f1-14 (n – 1)d0,1 ns1-2. They are so named as they represent a transition of physical and chemical properties among them. There are two series of inner transition elements corresponding to 4f and 5f series. 4f series are called lanthanides and elements of 5f series are called actinides. Lanthanides are also called rare earths. Majority of actinides are synthetic.

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 7.
What is a periodic property? How the following properties vary in a group and in a period? Explain a) Atomic radius b) Electron gain enthalpy. [TS ’16, ’15; AP ’15; IPE ’14, ’11, ’10; Mar. ’18 (AP & TS)]
Answer:
The repetition of similar chemical properties of elements at regular intervals with increasing atomic number is called periodic property or periodicity.

a) Atomic radius :
In a period, as we go from left to right, the atomic radius gradually decreases.

Reason :
As we go from left to right in a period, the differentiating electron enters into the same orbit but at the same time, the nuclear charge increases. As a result of which the effective nuclear charge over the outermost electrons increases, leading to a decrease in the atomic radius.

As we go from top to bottom in a group, the atomic radius gradually increases.

Reason :
As we go from top to bottom in a group, the valence electrons enter into new shells, even though the nuclear charge increases. As a result of which, the effective nuclear charge over the outermost electrons decreases, leading to n increase in the atomic radius.

b) Electron gain enthalpy or Electron affinity :
In a period, as we go from left to right, Electron affinity gradually increases.

Reason:
As we go from left to right in a period, the atomic radius gradually decreases. As a result of which, the attraction between the added electron and the nucleus increases. Hence, electron affinity increases in a period.

In a group as we go from top to bottom, the electron affinity values gradually decrease.

Reason :
As we go down a group, the atomic radius gradually increases. As a result of which, the attraction between theadded electron and the nucleus decreases. Hence, electron affinity decreases in a group.

Question 8.
What is a periodic property? How the following properties vary in a group and in a period? Explain a) IP b) EN. [AP, TS ’16, ’15 ; IPE ’14, ’11, ’10; Mar. ’18 (AP & TS)]
Answer:
The repetition of similar physical and chemical properties of elements at regular intervals with increasing atomic number is called periodic property or periodicity.

a) Trends of I.P.:
As we go from left to right in a period, atomic radius gradually decreases. As a result of which the nuclear attractions on the valence electrons gradually increases. So, the energy required to remove the outer electrons gradually increases. In other words, as we go from left to right in a period, the I.P. values increase.

As we go from top to bottom in a group, atomic radius gradually increases. As a result of which the nuclear attraction on the valence electrons gradually decreases. So, the energy required to remove the outer electrons gradually decreases. In other words, in a group the I.P. values gradually decrease from top to bottom.

b) Electronegativity :
As we go from left to right in a period, the electronegativity gradually increases.

Reason:
In a period, as we go from left to right, atomic radius gradually decreases. As a result of which the tendency of attraction of nucleus on the bonded pair increases. Hence the values of electronegativity increases in a period.

As we go from top to bottom in a group, the electronegativity values gradually decrease.

Reason :
As we go from top to bottom in a group, the atomic radius gradually increases. As a result of which the tendency of attraction of nucleus on the bonded pair decreases. Hence, the values of electrone-gativity decreases in a group.

Question 9.
Write a note on
a) Atomic radius
b) Metallic radius
c) Covalent radius.
Answer:
Atomic radius refer to both metallic radius and covalent radius.

a) Atomic radius :
As the atomic radius increases, the distance between the nucleus and the outermost electrons increases. Hence, the effective nuclear charge on the outermost electrons decreases. As a result of which the energy required to remove the electrons decreases. From this it is evident that as atomic radius increases I.P decreases. Similarly as atomic radius decreases I.P increases.

b) Crystal radius (metallic radius) :
It is defined as one-half of the inter-nuclear distance between two adjacent atoms of a metal.
Ex : Crystal radius of sodium = \(\frac{3.72}{2}\) = 1.86 Å
Crystal radius of potassium = \(\frac{4.62}{2}\) = 2.31 Å

c) Covalent radius :
It is defined as one – half of the equilibrium distance between the nuclei of two atoms bonded by a covalent bond.
Ex: Covalent radius of hydrogen = 0.37 Å
Covalent radius of chlorine = 0.99 Å
TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 11
AB = covalent radius = 0.99 Å
CD = van der Waal’s radius = 1.8 Å
Covalent, van der Waal’s radii of chlorine

van der Waal’s radius:
It is defined as one-half of the inter-nuclear distance between atoms of two adjacent molecules of an element bonded by van der Waal’s forces facing each other, in solid state.
Ex : van der Waal’s radius of hydrogen = 1.2 Å
van der Waal’s radius of chlorine = 1.80 Å
Note:
van der Waal’s radius of an atom is 40% larger than its covalent radius.

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 10.
Define IE1 and IE2. Why is IE2 > IE1 for a given atom? Discuss the factors that effect IE of an element. [TS Mar. ’19; AP, TS 16; Mar. ’13]
Answer:
The minimum amount of energy required to remove an electron present in the outermost orbit of a neutral, isolated gaseous atom is called ionisation potential. It is denoted by I1 and is measured in kilocalories or electron volts.
M(g) + I1 → M(g)+ + e

The amount of energy required to remove another electron from a uni +ve ion is called second ionisation potential. It is denoted as I2.
M(g)+ + I2 → M(g)+ + e

Ionisation energy is measured in ev/atom (or) k cal/mole (or) KJ/mole.

The uni +ve ion formed, by the removal of an electron from a neutral atom will have more nuclear attractions over the electron cloud because the number of protons will be more than the number of electrons. As a result of which more energy is required to remove an electron from this uni +ve ion than a neutral atom. Hence second I.P values are always greater than first I.P values.

Factors affecting Ionisation potential:
1. Atomic radius :
As the atomic radius increases, the distance between the nucleus and the outermost electrons increases. Hence, the effective nuclear charge on the outermost electrons decreases. As a result of which the energy required to remove the electrons decreases. From this it is evident that as atomic radius increases I.P decreases. Similarly as atomic radius decreases I.P increases.

2. Nuclear Charge :
As nuclear charge increases, the nuclear attractions over the valence electrons increases. So, more amount of energy is required to remove these electrons. From this it is evident that as nuclear charges increases I.P. increases. Similarly as nuclear charge decreases I.P. decreases.

3. Screening effect (or) Shielding effect:
In multielectron atoms, the electrons present in the inner shells screen the electron present in the outermost shells from the nuclear attractions. This effect is known as screening effect. Screening effect depends upon the number of screens (no. of inner orbits). As the screening effect increases, ionisation potential decreases. Similarly as the screening effect decreases, I.P. increases.

4. Extent of penetration of orbitals of valence electrons :
As the penetration power of orbitals increases, IP also increases. Order of penetration of orbitals is S > P > d > f. Hence, order of IP is also S > P > d > f.

5. Electron configuration :
Atoms of elements with half-filled or completely filled electron configuration are more stable. Such atoms have more IP values.

Question 11.
How do the following properties change in group – 1 and in the third period? Explain with example.
a) Atomic radius
b) IE
c) EA
d) Nature of oxides.
Answer:
a) Atomic radius :
As we move from top to bottom of group – 1 elements the atomic radius gradually increases. As we move from top to bottom in group -1 elements the number of orbits filling with electrons increases. So the atomic radius increases from top to bottom in group -1 elements.

In a period from left to right the atomic radius gradually decreases.

In the 3rd period, the differentiating electron enters into the same 3rd orbit. The electrons entering into the same orbit have less shielding power but the nuclear charge is increasing. Consequently the effective nuclear charge increases. So atomic radius decreases.

b) Ionisation Energy :
As we move from top to bottom of group -1 elements the atomic radius increases gradually. So the distance from the nucleus to the outer electrons increases. This results in the decrease in nuclear attraction on outer electrons. Hence ionisation energy decreases from Li to Cs as we move down the group.

While moving from left to right along 3rd period atomic radius decreases and effective nuclear charge increases gradually. So ionisation energy increases in the 3rd period from Na to Cl.

c) Electron Affinity :
As we move down the group -1 elements atomic radius increases gradually. Due to this reason the attraction between the added electron and the nucleus decreases. Hence electron affinity decreases from Li to Cs.

In the 3rd period while we are moving from left to right the atomic radius decreases and effective nuclear charge increases gradually. So the attraction between added electron and the nucleus increases gradually. Hence electron affinity increases from Na to Cl with certain exceptions.

d) Nature of oxides :
All the elements of 1A group are called alkali metals. Their oxides are basic in nature. They dissolve in water to give basic solutions.
Ex : Na2O, K2O etc.
Na2O + H2O → 2NaOH

The basic nature of these oxides increases from top to bottom in LA group.
In 3rd period the basic nature of oxides decreases and acidic nature increases from left to right.
TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 12

Question 12.
Define electron gain enthalpy. How it varies in a group and in a period? Why is the electron gain enthalpy of O or F is less negative than that of the succeeding element in the group?
Answer:
The amount of energy released when an electron is added to a neutral isolated gaseous atom is called Electron affinity.
X(g) + e → X(g) + energy

As a result of adding up an electron to neutral atom, it gets converted into uni – ve ion. It is difficult to add up another electron to this uni – ve ion, because of the repulsions between the electrons already present and the electron to be added. Hence, in order to add up another electron, energy is to be given to overcome the repulsive forces. That is why second electron affinity values are always +ve.
X(g) + e → X(g) + energy

In a period as we go from left to right, the electron affinity values gradually increases. In a group as we go from top to bottom, electron affinity values gradually decreases.

Electron affinity values of inert gases is ‘O’. The element with highest electron affinity value is chlorine.

Oxygen and fluorine atoms are smaller than their succeeding elements sulphur and chlorine. In oxygen and fluorine atoms the outermost orbit is second orbit when an electron is added to these small second orbit there will be electron-electron repulsions. To overcome these repulsions some energy is consumed from the energy liberated due to attraction on the added electron and nucleus. This results in the liberation of less energy or less negative electron gain enthalpies in 0 and F than that of the succeeding elements in the group.

Question 13.
a) What is electronegativity?
b) How does it vary in a group and in a period?
Answer:
a) The relative tendency of an atom in a covalent molecule to attract the shared pair of electrons towards itself is called its electronegativity’.

Pauling’s scale :
In this method, E.N. is calculated from the bond energies. The following expression is used.
XA – XB = 0.208 √∆

where XA = E.N. of atom A; XB = E.N. of atom B; ∆ = Bond polarity, where ∆ is in kCal/mole.

Bond polarity = Experimental bond energy – Theoretical bond energy.
∆ = EA-B – \(\frac{1}{2}\)(EA-A + EB-B)
where
EA-B = Bond energy of A – B molecule (experimental)
EA-A = Bond energy of A – A molecule
EB-B = Bond energy of B – B molecule

Metals are electropositive. So they possess low E.N. values. Non-metals are electronegative. So they possess high E.N. values.

b) In a group from top to bottom electronegativity decreases.

In a period from left to right electronegativity increases.

TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties

Question 14.
Explain the following
a) Valency b) Diagonal relationship c) Variation of nature of oxides in the Group -1.
Answer:
a) Valency :
Valency is the combining capacity. It is the number of atoms of hydrogen or chlorine or any monovalent atom, with which one atom of the element combines.

Generally all the elements in a group show the same valency. In case of s-block elements valency = group number.
In case of p – block of elements,
Valency = group number, or (8 – group no.)

Variation of valency w.r.t hydrogen, in a period :
In representative elements, the valency increases from 1 to 4 and then decreases to 1 from left to right.
Ex : Valency of 3rd period of elements :
TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 13

b) Diagonal relationship :
The first element of a group shows similarities in properties with the second element of the next group. This is called, diagonal relationship’.
TS Inter 1st Year Chemistry Study Material Chapter 2 Classification of Elements and Periodicity in Properties 14
But this relation continues effectively upto the IV group only,

c) Variation of nature of oxides in the Group -1 :
All the elements of IA group are called alkali metals. These oxides are basic in nature. They dissolve in water giving alkaline solutions.
Na2O + H2O → 2NaOH

The basic nature of these oxides increases from top to bottom in group -1 elements.

TS Inter First Year Maths 1B Transformation of Axes Important Questions

Students must practice these Maths 1B Important Questions TS Inter First Year Maths 1B Transformation of Axes Important Questions to help strengthen their preparations for exams.

TS Inter First Year Maths 1B Transformation of Axes Important Questions

Question 1.
When the origin is shifted to (3, 4) by the translation of axes, find the transformed equation of 2x2 + 4xy + 5y2 = 0. [Mar. ’00]
Solution:
Given equation is 2x2 + 4xy + 5y2 = 0 ……..(1)
Let (X, Y) be the new coordinates of the point (x, y).
Given a point, (h, k) = (3, 4)
By the translation of axes,
x = X + h, y = Y + k
x = X + 3, y = Y + 4
The transformed equation of (1) is,
2(X + 3)2 + 4(X + 3)(Y + 4) + 5(Y + 4)2 = 0
⇒ 2(X2 + 6X + 9) + 4(XY + 4X + 3Y + 12) + 5(Y2 + 8Y + 16) = 0
⇒ 2X2 + 12X + 18 + 4XY + 16X + 12Y + 48 + 5Y2 + 40Y + 80 = 0
⇒ 2X2 + 5Y2 + 28X + 52Y – 4XY + 146 = 0
⇒ 2X2 + 4XY + 5Y2 + 28X + 52Y + 146 = 0

Question 2.
When the origin is shifted to the point (2, 3), the transformed equation of a curve is x2 + 3xy – 2y2 + 17x – 7y – 11 = 0. Find the original equation of the curve. [Mar. ’16 (TS), ’13, ’11, ’10, ’09; May ’15 (AP), ’09]
Solution:
The given equation is,
X2 + 3XY – 2Y2 + 17X – 7Y – 11 = 0 ………(1)
Let (x, y) be the original coordinates of the point (X, Y).
Given a point, (h, k) = (2, 3).
By the translation of axes,
x = X + h, y = Y + k
⇒ x = X + 2, y = Y + 3
⇒ X = x – 2, Y = y – 3
The original equation of (1) is
(x – 2)2 + 3(x – 2)(y – 3) – 2(y – 3)2 + 17(x – 2) – 7(y – 3) – 11 = 0
⇒ x2 + 4 – 4x + 3(xy – 3x – 2y + 6) – 2(y2 + 9 – 6y) + 17x – 34 – 7y + 21 – 11 = 0
⇒ x2 + 4 – 4x + 3xy – 9x – 6y + 18 – 2y2 – 18 + 12y + 17x – 34 – 7y + 21 – 11 = 0
⇒ x2 + 3xy – 2y2 + 4x – y – 20 = 0
∴ The original equation is x2 + 3xy – 2y2 + 4x – y – 20 = 0

TS Inter First Year Maths 1B Transformation of Axes Important Questions

Question 3.
Find the point to which the origin is to be shifted so as to remove the first-degree terms from the equation 4x2 + 9y2 – 8x + 36y + 4 = 0. [May ’03]
Solution:
Given equation is 4x2 + 9y2 – 8x + 36y + 4 = 0
Comparing the given equation with ax2 + 2hxy + by2 + 2gx + 2fy + c = 0
We get a = 4, h = 0, b = 9, g = -4, f = 18, c = 4.
TS Inter First Year Maths 1B Transformation of Axes Important Questions Q3

Question 4.
Find the point to which the origin is to be shifted by the translation of axes 50 as to remove the first-degree terms from the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 where h2 ≠ ab. [May ’97]
Solution:
Given equation is ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 …….(1)
Let (α, β) be the required point.
Let (X, Y) be the new coordinates of the point (x, y).
∴ x = X + α, y = Y + β
Now substitute the values of x, y in equation (1)
∴ The transformed equation is a(X + α)2 + 2h(X + α)(Y + β) + b(Y + β)2 + 2g(X + α) + 2f(Y + β) + c = 0
⇒ a(X2 + α2 + 2Xα) + 2h(XY + βX + αY + αβ) + b(Y2 + β2 + 2βY) + 2gX + 2gα + 2fY + 2fβ + c = 0
⇒ aX2 + aα2 + 2aXα + 2hXY + 2hβX + 2hαY + 2hαβ + bY2 + bβ2 + 2bβY + 2gX + 2gα + 2fy + 2fβ + c = 0
⇒ aX2 + 2hXY + bY2 + 2X(aα + hβ + g) + 2Y(hα + bβ + f) + aα2 + 2hαβ + bβ2 + 2gα + 2fβ + c = 0
Since, x, y terms are eliminated
aα + hβ + g = 0 ……..(2)
hα + bβ + f = 0 ……….(3)
Solving (2) & (3)
TS Inter First Year Maths 1B Transformation of Axes Important Questions Q4

Question 5.
Prove that the angle of rotation of the axes to eliminate xy term from the equation ax2 + 2hxy + by2 = 0 is \(\frac{1}{2} \tan ^{-1}\left(\frac{2 h}{a-b}\right)\) where a ≠ b and \(\frac{\pi}{4}\) if a = b. [Mar. ’13 (Old); ’13, ’06, ’02]
Solution:
x = X cos θ – Y sin θ
y = X sin θ + Y cos θ
Given equation is ax2 + 2hxy + by2 = 0 ……..(1)
∴ The transformed equation of (1) is
a(X cos θ – Y sin θ)2 + 2h(X cos θ – Y sin θ) (X sin θ + Y cos θ) + b(X sin θ + Y cos θ)2 = 0
⇒ a[X2 cos2θ + Y2 sin2θ – 2X cos θ . Y sin θ] + 2h(X2 cos θ . sin θ + XY cos2θ – XY sin2θ + Y2 sin θ cos θ + b[X2 sin2θ + Y2 cos2θ + 2XY sin θ cos θ] = 0
⇒ [aX2 cos2θ + aY2 sin2θ – 2aX cos θ . Y sin θ + 2hX2 cos θ . sin θ + 2hXY cos2θ – 2hXY sin2θ + 2hY2 sin θ cos θ + bX2 sin2θ + bY2 cos2θ + 2bXY sin θ cos θ] = 0
⇒ X2[a cos2θ + 2h cos θ . sin θ + b sin2θ) + XY (-2a sin θ cos θ + 2h cos2θ – 2h sin2θ + 2b sin θ cos θ) + Y2 (a sin2θ – 2h sin θ cos θ + b cos2θ] = 0
Since xy term is eliminated, then,
⇒ -2a sin θ . cos θ + 2h cos2θ – 2h sin2θ + 2b sin θ cos θ = 0
⇒ -a sin 2θ + 2h(cos2θ – sin2θ) + b sin 2θ = 0
⇒ -a sin 2θ + 2h cos 2θ + b sin 2θ = 0
⇒ 2h cos 2θ = a sin 2θ – b sin 2θ
⇒ 2h cos 2θ = (a – b) (sin 2θ)
TS Inter First Year Maths 1B Transformation of Axes Important Questions Q5

Question 6.
Find the angle through which the axes are to be rotated so as to remove the xy term in the equation x2 + 4xy + y2 – 2x + 2y – 6 – 0. [May ’04, ’96]
Solution:
Given equation is x2 + 4xy + y2 – 2x + 2y – 6 = 0
By comparing the given equation with ax2 + 2hxy + by2 + 2gx + 2fy + c = 0,
we get a = 1, h = 2, b = 1
The required angle of rotation
TS Inter First Year Maths 1B Transformation of Axes Important Questions Q6

TS Inter First Year Maths 1B Transformation of Axes Important Questions

Question 7.
When the axes are rotated through an angle of 60°, the new coordinates of the point are (3, 4). Find their original coordinates. [May ’03]
Solution:
Let (x, y) be the original coordinates of (X, Y).
Given that, the angle of rotation θ = 60°
(X, Y) = (3, 4) then
x = X cos θ – Y sin θ
= 3(cos 60°) – 4(sin 60°)
TS Inter First Year Maths 1B Transformation of Axes Important Questions Q7

Question 8.
When the axes are rotated through an angle of 45°, the transformed equation of a curve is 17x2 – 16xy + 17y2 = 225. Find the original equation of the curve. [May ’15 (TS), ’12, ’10, ’01; Mar. ’15 (TS), ’08]
Solution:
Given equation is 17x2 – 16xy + 17y2 = 225 ………(1)
Angle of rotation θ = 45°
Let (x, y) be the original coordinates of (X, Y) then
X = x cos θ + y sin θ
= x cos 45° + y sin 45°
TS Inter First Year Maths 1B Transformation of Axes Important Questions Q8
50x2 + 18y2 = 450
25x2 + 9y2 = 225

Question 9.
When the axes are rotated through an angle α, find the transformed equation of x cos α + y sin α = P. [Mar. ’19 (TS); Mar. ’14, ’00; May ’07]
Solution:
Given equation is x cos α + y sin α = P ………..(1)
Let (X, Y) be the new coordinates of (x, y)
Given that, the angle of rotation θ = α, then
x = X cos θ + Y sin θ = X cos α + Y sin α
y = X sin θ + Y cos θ = X sin α + Y cos α
The transformed equation of (1) is
(X cos α – Y sin α) cos α + (X sin α + Y cos α) sin α = P
⇒ X cos2α – Y sin α cos α + X sin2α + Y cos α sin α = P
⇒ X cos2α + X sin2α = P
⇒ X(cos2α + sin2α) = P
⇒ X(1) = P
⇒ X = P

Question 10.
When the axes are rotated through an angle \(\frac{\pi}{6}\), find the transformed equation of x2 + 2√xy – y2 = 2a2. [Mar. ’15 (AP); May ’13, ’06, ’03; Mar. ’12, ’07, ’03; B.P.; Mar. ’18 (TS)]
Solution:
Given equation is x2 + 2√3 xy – y2 = 2a2 ………(1)
Let (X, Y) be the new coordinate of (x, y).
Given that, angle of rotation θ = \(\frac{\pi}{6}\) = 30°, then
x = X cos θ – Y sin θ
= X cos 30° – Y sin 30°
TS Inter First Year Maths 1B Transformation of Axes Important Questions Q10
TS Inter First Year Maths 1B Transformation of Axes Important Questions Q10.1

TS Inter First Year Maths 1B Transformation of Axes Important Questions

Question 11.
When the axes are rotated through an angle \(\frac{\pi}{4}\), find the transformed equation 3x2 + 10xy + 3y2 = 9. [Mar. (AP & TS) ’17; May ’14, ’11, ’05; Mar. ’08]
Solution:
Given equation is, 3x2 + 10xy + 3y2 = 9 ……(1)
Let (X, Y) be the new coordinates of (x, y).
Given that, angle of rotation θ = \(\frac{\pi}{4}\) = 45°
x = X cos θ – Y sin θ
= X cos 45° – Y sin 45°
TS Inter First Year Maths 1B Transformation of Axes Important Questions Q11

Some More Maths 1B Transformation of Axes Important Questions

Question 12.
When the origin is shifted to (-1, 2) by the translation of axes find the transformed equation of x2 + y2 + 2x – 4y + 1 = 0.
Solution:
Given equation is x2 + y2 + 2x – 4y + 1 = 0 ……(1)
Let (X, Y) be the new coordinates of the point (x, y).
Given that, (h, k) = (-1, 2)
By the translation of axes,
x = X + h, y = Y + k
⇒ x = X- 1, y = Y + 2
The transformed equation of (1) is
(X – 1)2 + (Y + 2)2 + 2(X – 2) – 4(Y + 2) + 1 = 0
⇒ X2 – 2X + 1 + Y2 + 4 + 4Y + 2X – 2 – 4Y – 8 + 1 = 0
⇒ X2 + Y2 – 4 = 0
∴ The transformed equation is X2 + Y2 – 4 = 0

Question 13.
When the origin is shifted to (-1, 2) by the translation of axes, find the transformed equation of 2x2 + y2 – 4x + 4y = 0.
Solution:
Given equation is 2x2 + y2 – 4x + 4y = 0 …..(1)
Let (X, Y) be the new coordinates of the point (x, y).
Given that, (h, k) = (-1, 2)
By the translation of axes,
x = X + h, y = Y + k
⇒ x = X – 1, y = Y + 2
The transformed equation of (1) is,
2(X – 1)2 + (Y + 2)2 – 4(X – 1) + 4(Y + 2) = 0
⇒ 2(X2 – 2X + 1) + (Y2 + 4 + 4Y) – 4X + 4 + 4Y + 8 = 0
⇒ 2X2 – 4X + 2 + Y2 + 4 + 4Y – 4X + 4 + 4Y + 8 = 0
⇒ 2X2 + Y2 – 8X + 8Y + 18 = 0
∴ The transformed equation is 2X2 + Y2 – 8X + 8Y + 18 = 0

TS Inter First Year Maths 1B Transformation of Axes Important Questions

Question 14.
When the origin is shifted to the point (3, -4), the transformed equation of a curve is x2 + y2 = 4. Find the original equation of the curve.
Solution:
Given equation is X2 + Y2 = 4 ………(1)
Let (x, y) be the original coordinates of the point (X, Y).
Given point, (h, k) = (3, -4)
By the translation of axes,
x = X + h, y = Y + k
⇒ x = X + 3, y = Y – 4
⇒ X = x – 3, Y = y + 4
The original equation of (1) is (x – 3)2 + (y + 4)2 = 4
⇒ x2 + 9 – 6x + y2 + 16 + 8y = 4
⇒ x2 + y2 – 6x + 8y + 25 = 4
⇒ x2 + y2 – 6x + 8y + 25 – 4 = 0
⇒ x2 + y2 – 6x + 8y + 21 = 0
∴ The original equation is x2 + y2 – 6x + 8y + 21 = 0.

Question 15.
When the origin is shifted to the point (-1, 2), the transformed equation of a curve is x2 + 2y2 + 16 = 0. Find the original equation of the curve.
Solution:
Given equation is X2 + 2Y2 + 16 = 0 …….(1)
Let (x, y) be the original coordinates of the point (X, Y).
Given a point, (h, k) = (-1, 2)
By the translation of axes,
x = X + h, y = Y + k
⇒ x = X – 1, y = Y + 2
⇒ X = x + 1, Y = y – 2
The original equation of (1) is (x + 1)2 + 2(y – 2)2 + 16 = 0
⇒ x2 + 2x + 1 + 2(y2 + 4 – 4y) + 16 = 0
⇒ x2 + 2x + 1 + 2y2 + 8 – 8y + 16 = 0
⇒ x2 + 2y2 + 2x – 8y + 25 = 0
∴ The original equation is x2 + 2y2 – 2x – 8y + 25 = 0.

TS Inter First Year Maths 1B Transformation of Axes Important Questions

Question 16.
When the origin is shifted to (-2, -3) and the axes are rotated through an angle of 45°, find the transformed equation of 2x2 + 4xy – 5y2 + 20x – 22y – 14 = 0.
Solution:
Given equation is 2x2 + 4xy – 5y2 + 20x – 22y – 14 = 0 ………(1)
Let (X, Y) be the new coordinates of (x, y).
Given that, (h, k) = (-2, -3)
angle of rotation θ = 45°
x = X cos θ – Y sin θ + h
= X cos 45 – Y sin 45 + (-2)
TS Inter First Year Maths 1B Transformation of Axes Important Questions Some more Q5
TS Inter First Year Maths 1B Transformation of Axes Important Questions Some more Q5.1
TS Inter First Year Maths 1B Transformation of Axes Important Questions Some more Q5.2

TS Inter First Year Maths 1B Locus Important Questions

Students must practice these Maths 1B Important Questions TS Inter First Year Maths 1B Locus Important Questions to help strengthen their preparations for exams.

TS Inter First Year Maths 1B Locus Important Questions

Question 1.
Find the equation of the locus of a point that is at a distance 5 from (-2, 3), in the XOY plane. [Mar. ’02; May ’95]
Solution:
Let A (-2, 3) be the given point.
Let P(x, y) be any point on the locus.
The given geometric condition is PA = 5
TS Inter First Year Maths 1B Locus Important Questions SAQ Q1
Squaring on both sides
⇒ 13 + x2 + y2 + 4x – 6y = 25
⇒ x2 + y2 + 4x – 6y = 25 – 13
⇒ x2 + y2 + 4x – 6y = 12
∴ The equation of locus of P(x, y) is x2 + y2 + 4x – 6y = 12.

Question 2.
Find the equation of the locus of a point P such that the distance of P from the origin is twice the distance of P from A(1, 2). [Mar. ’12; May ’05, ’97]
Solution:
O(0, 0), A(1, 2) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is OP = 2PA
TS Inter First Year Maths 1B Locus Important Questions SAQ Q2
⇒ x2 + y2 = 4[(x – 1)2 + (y – 2)2]
⇒ x2 + y2 = 4(x2 – 2x + 1 + y2 – 4y + 4)
⇒ x2 + y2 = 4x2 – 8x + 4 + 4y2 – 16y + 16
⇒ 3x2 + 3y2 – 8x – 16y + 20 = 0
∴ The equation of the locus of P(x, y) is 3x2 + 3y2 – 8x – 16y + 20 = 0.

TS Inter First Year Maths 1B Locus Important Questions

Question 3.
Find the equation of the locus of a point equidistant from A(2, 0) and the Y-axis. (May ’03)
Solution:
A(2, 0) is the given point.
Let P(x, y) be any point on the locus.
TS Inter First Year Maths 1B Locus Important Questions SAQ Q3
The distance of P from Y-axis = PN = |x|
The given geometric condition is PA = |PN|
\(\sqrt{(x-2)^2+(y-0)^2}=|x|\)
Squaring on both sides
⇒ \(\left(\sqrt{(x-2)^2+(y-0)^2}\right)^2=(|x|)^2\)
⇒ (x – 2)2 + (y)2 = x2
⇒ x2 – 4x + 4 + y2 = x2
⇒ y2 – 4x + 4 = 0
∴ The equation of the locus of P(x, y) is y2 – 4x + 4 = 0.

Question 4.
Find the equation of the locus of a point P, the square whose distance from the origin is 4 times its y-coordinate. [Mar. ’00]
Solution:
O(0, 0) is the origin.
Let P(x, y) be any point on the locus.
The distance of P from the X-axis = y.
TS Inter First Year Maths 1B Locus Important Questions SAQ Q4
The given geometric condition is OP2 = 4 PM
⇒ \(\left(\sqrt{(x-0)^2+(y-0)^2}\right)^2=4 y\)
⇒ x2 + y2 = 4y
⇒ x2 + y2 – 4y = 0
∴ The equation of the locus of P(x, y) is x2 + y2 – 4y = 0.

Question 5.
Find the equation of locus of a point P such that PA2 + PB2 = 2c2, where A = (a, 0), B = (-a, 0), and 0 < |a| < |c|. [Mar. ’00]
Solution:
A = (a, 0), B = (-a, 0) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is PA2 + PB2 = 2c2
⇒ \(\left[\sqrt{(x-a)^2+(y-0)^2}\right]^2\) + \(\left[\sqrt{(x+a)^2+(y-0)^2}\right]^2\) = 2c2
⇒ (x – a)2 + y2 + (x + a)2 + y2 = 2c2
⇒ x2 – 2ax + a2 + y2 + x2 + 2ax + a2 + y2 = 2c2
⇒ 2x2 + 2y2 + 2a2 = 2c2
⇒ x2 + y2 + a2 = c2
⇒ x2 + y2 = c2 – a2
∴ The equation of the locus of P(x, y) is x2 + y2 = c2 – a2

Question 6.
Find the equation of locus of P, if the line segment joining (2, 3) and (-1, 5) subtends a right angle at P. [Mar. ’13, ’05, ’03; May ’12, ’04, ’03, ’02]
Solution:
Let A(2, 3), B(-1, 5) are the given points.
Let P(x, y) be any point on the locus
The given geometric condition is ∠APB = 90°
TS Inter First Year Maths 1B Locus Important Questions SAQ Q6
Then, PA2 + PB2 = AB2
⇒ \(\left[\sqrt{(x-2)^2+(y-3)^2}\right]^2+\left[\sqrt{(x+1)^2+(y-5)^2}\right]^2\) = \(\left[\sqrt{(2+1)^2+(3-5)^2}\right]^2\)
⇒ (x – 2)2 + (y – 3)2 + (x + 1)2 + (y – 5)2 = (2 + 1)2 + (3 – 5)2
⇒ x2 + 4 – 4x + y2 + 9 – 6y + x2 + 1 + 2x + y2 + 25 – 10y = 9 + 4
⇒ 2x2 + 2y2 – 2x – 16y + 26 = 0
⇒ x2 + y2 – x – 8y + 13 = 0
∴ The equation of the locus of P(x, y) is x2 + y2 – x – 8y + 13 = 0.

TS Inter First Year Maths 1B Locus Important Questions

Question 7.
Find the equation of the locus of P, if A = (4, 0), B = (-4, 0) and |PA – PB| = 4. [May ’13]
Solution:
A = (4, 0), B = (-4, 0) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is |PA – PB| = 4
⇒ PA – PB = ±4
⇒ PA = ±4 + PB
Squaring on both sides
⇒ (PA)2 = (±4 + PB)2
⇒ PA2 = 16 + PB2 ± 8PB
⇒ (x – 4)2 + (y – 0)2 = 16 + (x + 4)2 + (y – 0)2 ± 8PB
⇒ x2 – 8x + 16 + y2 = 16 + x2 + 8x + 16y2 ± 8PB
⇒ -8x = 8x + 16 ± 8PB
⇒ -8x – 8x – 16 = ±8PB
⇒ -16 – 16x = ±8PB
⇒ -2 – 2x = ±PB
Squaring on both sides
⇒ (-2 – 2x)2 = (±PB)2
⇒ 4 + 8x + 4x2 = PB2
⇒ 4 + 4x2 + 8x = (x + 4)2 + (y – 0)2
⇒ 4 + 4x2 + 8x = x2 + 8x + 16 + y2
⇒ 3x2 – y2 – 12 = 0
∴ The equation of the locus of P(x, y) is 3x2 – y2 – 12 = 0.

Question 8.
Find the equation of the locus of a point, the sum of whose distances from (0, 2) and (0, -2) is 6. [Mar. ’16 (TS)]
Solution:
Let A = (0, 2), B = (0, -2) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is PA + PB = 6
⇒ PA = 6 – PB
Squaring on both sides
⇒ PA2 = (6 – PB)2
⇒ PA2 = 36 – 12PB + PB2
⇒ (x – 0)2 + (y – 2)2 = 36 + (x – 0)2 + (y + 2)2 – 12PB
⇒ x2 + y2 + 4 – 4y = 36 + x2 + y2 + 4 + 4y – 12PB
⇒ -4y = 36 + 4y – 12PB
⇒ 12PB = 36 + 4y + 4y
⇒ 12PB = 36 + 8y
⇒ 3PB = 9 + 2y
Squaring on both sides
⇒ (3PB)2 = (9 + 2y)2
⇒ 9PB2 = (9 + 2y)2
⇒ 9[(x – 0)2 + (y + 2)2] = 81 + 36y + 4y2
⇒ 9[x2 + y2 + 4y + 4] = 81 + 36y + 4y2
⇒ 9x2 + 9y2 + 36y + 36 = 81 + 36y + 4y2
⇒ 9x2 + 9y2 + 36 – 81 – 4y2 = 0
⇒ 9x2 + 5y2 – 45 = 0
∴ The equation of the locus of P(x, y) is 9x2 + 5y2 – 45 = 0.

Question 9.
A(2, 3) and B(-3, 4) are two given points. Find the equation of the locus of P, so that the area of the triangle PAB is 8.5. [Mar. ’11]
Solution:
A(2, 3), B(-3, 4) are the points given.
Let P(x, y) be any point on the locus.
The given geometric condition is an area of the triangle, PAB = 8.5
TS Inter First Year Maths 1B Locus Important Questions SAQ Q9
∴ The equation of the locus of P(x, y) is (x + 5y) (x + 5y – 34) = 0
⇒ x2 + 5xy – 34x + 5xy + 25y2 – 170y = 0
⇒ x2 + 10xy + 25y2 – 34x – 170y = 0

TS Inter First Year Maths 1B Locus Important Questions

Question 10.
Find the equation of the locus of p, if the ratio of the distances from p to A(5, -4) and B(7, 6) is 2 : 3. [Mar. ’14, ’98; May ’08, ’01]
Solution:
Let A(5, -4), B(7, 6) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is PA : PB = 2 : 3
⇒ \(\frac{\mathrm{PA}}{\mathrm{PB}}=\frac{2}{3}\)
⇒ 3PA = 2PB
TS Inter First Year Maths 1B Locus Important Questions SAQ Q10
⇒ 9[x2 – 10x + 25 + y2 + 16 + 8y] = 4[x2 – 14x + 49 + y2 – 12y + 36]
⇒ 9x2 – 90x + 225 + 9y2 + 144 + 72y = 4x2 – 56x + 196 – 4y2 – 48y + 144
⇒ 9x2 – 90x + 225 + 9y2 + 144 + 72y – 4x2 + 56x – 196 – 4y2 + 48y – 144 = 0
⇒ 5x2 + 5y2 – 34x + 120y + 29 = 0
∴ The equation of the locus of P(x, y) is 5x2 + 5y2 – 34x + 120y + 29 = 0

Question 11.
A(1, 2), B(2, -3) and C(-2, 3) are three points. A point p moves such that PA2 + PB2 = 2PC2. Show that the equation to the locus of P is 7x – 7y + 4 = 0. [Mar. ’19 (T.S); Mar. ’17 (AP); May ’15 (A.P); ’07]
Solution:
A(1, 2), B(2, -3), C(-2, 3) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is PA2 + PB2 = 2PC2
⇒ [(x – 1)2 + (y – 2)2] + [(x – 2)2 + (y + 3)2] = 2[(x + 2)2 + (y – 3)2]
⇒ x2 – 2x + 1 + y2 + 4 – 4y + x2 + 4 – 4x + y2 + 9 + 6y = 2[x2 + 4 + 4x + y2 + 9 – 6y]
⇒ 2x2 + 2y2 – 6x + 2y + 18 = 2x2 + 8 + 8x + 2y2 + 18 – 12y
⇒ -6x + 2y = 8 + 8x – 12y
⇒ 8 + 8x – 12y + 6x – 2y = 0
⇒ 14x – 14y + 8 = 0
⇒ 7x – 7y + 4 = 0
∴ The equation of the locus of P(x, y) is 7x – 7y + 4 = 0.

Some More Maths 1B Locus Important Questions

Question 12.
Find the equation of the locus of a point that is at a distance 5 from A(4, -3).
Solution:
A(4, -3) is the given point.
Let P(x, y) be any point on the locus.
TS Inter First Year Maths 1B Locus Important Questions Some More Q1
The given geometric condition is PA = 5
⇒ \(\sqrt{(x-4)^2+(y+3)^2}\) = 5
Squaring on both sides
⇒ \(\left(\sqrt{(x-4)^2+(y+3)^2}\right)^2\) = (5)2
⇒ (x – 4)2 + (y + 3)2 = 25
⇒ x2 – 8x + 16 + y2 + 9 + 6y = 25
⇒ x2 + y2 – 8x + 6y = 0
∴ The equation of locus of P(x, y) is x2 + y2 – 8x + 6y = 0

Question 13.
Find the equation of the locus of a point P, if the distance of P from A(3, 0) is twice the distance of P from B(-3, 0).
Solution:
A(3, 0), B(-3, 0) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is PA = 2PB
TS Inter First Year Maths 1B Locus Important Questions Some More Q2
⇒ (x – 3)2 + (y – 0)2 = 4[(x + 3)2 + (y – 0)2]
⇒ x2 – 6x + 9 + y2 = 4[x2 + 6x + 9 + y2]
⇒ x2 – 6x + 9 + y2 = 4×2 + 24x + 36 + 4y2
⇒ 4x2 + 24x + 36 + 4y2 – x2 + 6x – 9 – y2 = 0
⇒ 3x2 + 3y2 + 30x + 27 = 0
⇒ x2 + y2 + 10x + 9 = 0
∴ The equation of the locus of P(x, y) is x2 + y2 + 10x + 9 = 0.

TS Inter First Year Maths 1B Locus Important Questions

Question 14.
The ends of the hypotenuse of a right-angled triangle are (0, 6) and (6, 0). Find the equation of the locus of its third vertex. [June ’10; Mar. ’08; Sept. ’00]
Solution:
Let A(0, 6), B(6, 0) are the given points.
Let P(x, y) be the third vertex
TS Inter First Year Maths 1B Locus Important Questions Some More Q3
The given geometric condition is ∠APB = 90°
⇒ PA2 + PB2 = AB2
⇒ \(\left[\sqrt{(x-0)^2+(y-6)^2}\right]^2+\left[\sqrt{(x-6)^2+(y-0)^2}\right]^2\) = \(\left[\sqrt{(0-6)^2+(6-0)^2}\right]^2\)
⇒ x2 + (y – 6)2 + (x – 6)2 + y2 = (-6)2 + (6)2
⇒ x2 + y2 – 12y + 36 + x2 – 12x + 36 + y2 = 36 + 36
⇒ 2x2 + 2y2 – 12x – 12y + 72 = 72
⇒ 2x2 + 2y2 – 12x – 12y = 0
⇒ x2 + y2 – 6x – 6y = 0
∴ The equation of the locus of P(x, y) is x2 + y2 – 6x – 6y = 0.

Question 15.
Find the locus of the third vertex of a right-angled triangle, the ends of whose hypotenuse are (4, 0) and (0, 4). [Mar. ’13(old); Mar. ’18 (TS)]
Solution:
Let A(4, 0), B(0, 4) are the given points.
Let P(x, y) be the third vertex.
The given geometric condition is ∠APB = 90°
⇒ PA2 + PB2 = AB2
TS Inter First Year Maths 1B Locus Important Questions Some More Q4
⇒ (x – 4)2 + y2 + x2 + (y – 4)2 = (4)2 + (-4)2
⇒ x2 – 8x + 16 + y2 + x2 + y2 – 8y + 16 = 16 + 16
⇒ 2x2 + 2y2 – 8x – 8y = 0
⇒ x2 + y2 – 4y – 4x = 0
⇒ x2 + y2 – 4x – 4y = 0
∴ The equation of the locus of P(x, y) is x2 + y2 – 4x – 4y = 0.

Question 16.
Find the equation of the locus of a point, the difference of whose distances from (-5, 0) and (5, 0) is 8.
Solution:
Let A(-5, 0), B(5, 0) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is |PA – PB| = 8
⇒ PA – PB = ±8
⇒ PA = ±8 + PB
Squaring on both sides
⇒ PA2 = (±8 + PB)2
⇒ PA2 = (±8)2 + PB2 + 2(±8) PB = 64 + PB2 ± 16PB
⇒ (x + 5)2 + (y – 0)2 = 64 + (x – 5)2 + (y – 0)2 ± 16PB
⇒ x2 + 25 + 10x + y2 = 64 + x2 – 10x + 25 + y2 ± 16PB
⇒ 10x = 64 – 10x ± 16PB
⇒ 10x – 64 + 10x = ±16PB
⇒ 20x – 64 = ±16PB
⇒ 5x – 16 = ±4PB
Squaring on both sides
⇒ (5x – 16)2 = (±4PB)2
⇒ 25x2 + 256 – 160x = 16PB2
⇒ 25x2 + 256 – 160x = 16[(x – 5)2 + (y – 0)2]
⇒ 25x2 + 256 – 160x = 16[x2 + 25 – 10x + y2]
⇒ 25x2 + 256 – 160x = 16x2 + 400 – 160x + 16y2
⇒ 25x2 + 256 – 16x2 – 400 – 16y2 = 0
⇒ 9x2 – 16y2 – 144 = 0
⇒ 9x2 – 16y2 = 144
∴ The equation of the locus of P(x, y) is 9×2 – 16y2 = 114.

TS Inter First Year Maths 1B Locus Important Questions

Question 17.
Find the equation of the locus of P, if A = (2, 3), B = (2, -3), and PA + PB = 8. [Mar. ’08, ’03]
Solution:
Let A = (2, 3) and B = (2, -3) are the given points..
Let P(x, y) be any point on the locus.
The given geometric condition is PA + PB = 8
⇒ PA = 8 – PB
Squaring on both sides
⇒ PA2 = (8 – PB)2
⇒ PA2 = 64 – 16PB + PB2
⇒ (x – 2)2 + (y – 3)2 = (x – 2)2 + (y + 3)2 + 64 – 16PB
⇒ x2 – 4x + 4 + y2 – 6y + 9 = x2 – 4x + 4 + y2 + 6y + 9 + 64 – 16PB
⇒ -6y = 6y + 64 – 16PB
⇒ 16PB = 6y + 6y + 64
⇒ 16PB = 12y + 64
⇒ 4PB = 3y + 16
Squaring on both sides
⇒ 16PB2 = (3y + 16)2
⇒ 16PB2 = 9y2 + 256 + 96y
⇒ 16[(x – 2)2 + (y + 3)2] = 9y2 + 256 + 96y
⇒ 16[x2 – 4x + 4 + y2 + 9 + 6y] = 9y2 + 256 + 96y
⇒ 16x2 – 64x + 64 + 16y2 + 144 + 96y – 9y2 – 256 – 96y = 0
⇒ 16x2 + 7y2 – 64x – 48 = 0
∴ The equation of the locus of point P(x, y) is 16x2 + 7y2 – 64x – 48 = 0.

Question 18
A(5, 3) and B(3, -2) are two fixed points. Find the equation of the locus of P, so that the area of triangle PAB is 9. [Mar. ’09, ’06]
Solution:
Let A(5, 3), B(3, -2) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is Area of the triangle PAB = 9
⇒ \(\frac{1}{2}\left|\begin{array}{ll}
x_1-x_2 & y_1-y_2 \\
x_1-x_3 & y_1-y_3
\end{array}\right|\) = 9
⇒ \(\frac{1}{2}\left|\begin{array}{ll}
x-5 & y-3 \\
x-3 & y+2
\end{array}\right|\) = 9
⇒ \(\left|\begin{array}{ll}
x-5 & y-3 \\
x-3 & y+2
\end{array}\right|\) = 18
⇒ |(x – 5)(y + 2) – (x – 3) (y – 3)| = 18
⇒ |xy – 5y + 2x – 10 – xy + 3x + 3y – 9| = 18
⇒ |5x – 2y – 19| = 18
⇒ (5x – 2y – 19) = ±18
(-) 5x – 2y – 19 = -18
5x – 2y – 19 + 18 = 0
5x – 2y – 1 = 0
(+) 5x – 2y – 19 = 18
5x – 2y – 19 – 18 = 0
5x – 2y – 37 = 0
∴ The equation of the locus of point P(x, y) is (5x – 2y – 1) (5x – 2y – 37) = 0
⇒ 25x2 – 10xy – 185x – 10xy + 4y2 + 74y – 5x + 2y + 37 = 0
⇒ 25x2 – 20xy + 4y2 – 190x + 76y + 37 = 0

Question 19.
Find the equation of the locus of a point, which forms a triangle of area 2 with the points A(1, 1) and B(-2, 3). [May ’13(old)]
Solution:
Let A(1, 1), B(-2, 3) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is Area of the triangle PAB = 2
⇒ \(\frac{1}{2}\left|\begin{array}{ll}
x_1-x_2 & y_1-y_2 \\
x_1-x_3 & y_1-y_3
\end{array}\right|\) = 2
⇒ \(\frac{1}{2}\left|\begin{array}{ll}
x-1 & y-1 \\
x+2 & y-3
\end{array}\right|\) = 2
⇒ |(x – 1)(y – 3) – (y – 1) (x + 2)| = 4
⇒ |xy – 3x – y + 3 – xy – 2y + x + 2| = 4
⇒ |-2x – 3y + 5 | = 4
⇒ -2x – 3y + 5 = ±4
(+) -2x – 3y + 5 = 4
2x + 3y – 5 + 4 = 0
2x + 3y – 1 = 0
(-) -2x – 3y + 5 = -4
2x + 3y – 5 – 4 = 0
2x + 3y – 9 = 0
∴ The equation of the locus of P(x, y) is (2x + 3y – 1) (2x + 3y – 9) = 0.
⇒ 4x2 + 6xy – 18x + 6xy + 9y2 – 27y – 2x – 3y + 9 = 0
⇒ 4x2 + 12xy + 9y2 – 20x – 30y + 9 = 0

TS Inter First Year Maths 1B Locus Important Questions

Question 20.
If the distance from p to the points (2, 3) and (2, -3) are in the ratio 2 : 3, then find the equation of the locus of P. [May ’15 (TS)]
Solution:
Let A(2, 3), B(2, -3) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is PA : PB = 2 : 3
TS Inter First Year Maths 1B Locus Important Questions Some More Q9
⇒ 9[x2 + 4 – 4x + y2 + 9 – 6y] = 4[x2 + 4 – 4x + y2 + 9 + 6y]
⇒ 9x2 + 36 – 36x + 9y2 + 81 – 54y – 4x2 + 16 – 16x + 4y2 + 36 + 24y
⇒ 9x2 + 36 – 36x + 9y2 + 81 – 54y – 4x2 – 16 + 16x – 4y2 – 36 – 24y = 0
⇒ 5x2 + 5y2 – 20x – 78y + 65 = 0
∴ The equation of the locus of P(x, y) is 5x2 + 5y2 – 20x – 78y + 65 = 0.

Question 21.
Find the equation of the locus of a point that is equidistant from the coordinate axes.
Solution:
P(x, y) is any point on the locus.
Let M, and N are the projections drawn from P to the X-axis and Y-axis respectively.
The distance of P from the X-axis is |PM| = |y|
The distance of P from the Y-axis is |PN| = |x|
The given geometric condition is |PM| = |PN|
|y| = |x|
TS Inter First Year Maths 1B Locus Important Questions Some More Q10
Squaring on both sides |y| = |x|
⇒ y2 = x2
⇒ x2 – y2 = 0
∴ The equation of the locus of P(x, y) is x2 – y2 = 0.

TS Inter First Year Maths 1B Locus Important Questions

Question 22.
Find the equation of the locus of a point that is equidistant from points A(-3, 2) and B(0, 4).
Solution:
A(-3, 2) and B(0, 4) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is PA = PB
TS Inter First Year Maths 1B Locus Important Questions Some More Q11
⇒ (x + 3)2 + (y – 2)2 = (x – 0)2 + (y – 4)2
⇒ x2 + 6x + 9 + y2 – 4y + 4 = x2 + y2 + 16 – 8y
⇒ 13 + 6x – 4y = 16 – 8y
⇒ 13 + 6x – 4y – 16 + 8y = 0
⇒ 6x + 4y – 3 = 0
∴ The equation of the locus of P(x, y) is 6x + 4y – 3 = 0.

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Students must practice these Maths 1B Important Questions TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 1.
If Q(h, k) is foot of the perpendicular from P(x1, y1) on the straight line ax + by + c = 0 then show that \(\frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-\left(a x_1+b y_1+c\right)}{a^2+b^2}\). [May ’14, ’07; Mar. ’03]
Solution:
Let A(x1, y1), P(h, k)
‘P’ lies in ax + by + c = 0 then
ah + bk + c = 0
ah + bk = -c
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q1.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q1.2

Question 2.
Find the foot of the perpendicular from (-1, 3) on the straight line 5x – y – 18 = 0. [May ’13 (old), ’07: Mar. ’03]
Solution:
Given the equation of the straight line is 5x – y – 18 – 0
Comparing with ax + by + c = 0, we get
a = 5, b = -1, c = -18
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q2
Let the given point A(x1, y1) = (-1, 3)
Let (h, k) is the foot perpendicular from the point A(-1, 3) on line 5x – y – 18 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q2.1
∴ Foot of the perpendicular P(h, k) = (4, 2)

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 3.
If Q(h, k) is the image of the point p(x1, y1) with respect to the straight line ax + by + c = 0 then, show that \(\frac{h-\mathbf{x}_1}{\mathbf{a}}=\frac{\mathbf{k}-\mathbf{y}_1}{\mathbf{b}}=\frac{-2\left(a \mathbf{x}_1+\mathbf{b y}_1+\mathbf{c}\right)}{a^2+b^2}\). [Mar. ’19 (AP); Mar. ’13, ’04, ’93; May ’06, ’01]
Solution:
Let A(x1, y1), B(h, k)
‘C’ is the midpoint of \(\overline{\mathrm{AB}}\) then
C = \(\left(\frac{\mathrm{x}_1+\mathrm{h}}{2}, \frac{\mathrm{y}_1+\mathrm{k}}{2}\right)\)
‘B’ is the image of A, then midpoint ‘C’ lies on ax + by + c = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q3.1

Question 4.
Find the image of (1, -2) with respect to the straight line 2x – 3y + 5 = 0. [Mar. ’13]
Solution:
Given the equation of the straight line is 2x – 3y + 5 = 0
Comparing with ax + by + c = 0, we get
a = 2, b = -3, c = 5
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q4
Let the given point A(x1, y1) = (1, -2)
Now, B(h, k) be the image of A(1, -2) with respect to the straight line 2x – 3y + 5 = 0
\(\frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-2\left(a x_1+b y_1+c\right)}{a^2+b^2}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q4.1
∴ Image of B(h, k) = (-3, 4)

Question 5.
Find the equation of the straight lines passing through the point (-3, 2) and making an angle of 45° with the straight line 3x – y + 4 = 0. [Mar. ’09]
Solution:
Given, equation of the straight line is 3x – y + 4 = 0 ……..(1)
slope of the line (1) is m = \(\frac{-3}{-1}\) = 3
Let the given point A = (-3, 2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q5
Let the slope of the required line is m2 = m
∴ The equation of the required line is passed through (-3, 2), and having slope m is y – y1 = m(x – x1)
y – 2 = m(x + 3) …….(2)
y – 2 = mx + 3m
mx – y + (3m + 2) = 0
Given that the angle between the lines is 45°,
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q5.1
Squaring on both sides
⇒ 10(m2 + 1) = 2(9m2 + 6m + 1)
⇒ 10m2 + 10 = 18m2 + 12m + 2
⇒ 8m2 + 12m – 8 = 0
⇒ 2m2 + 3m – 2 = 0
⇒ 2m2 + 4m – m – 2 = 0
⇒ (m + 2) (2m – 1) = 0
⇒ m + 2 = 0 (or) 2m – 1 = 0
⇒ m = -2 (or) \(\frac{1}{2}\)
If m = -2,
(2) ⇒ y – 2 = m(x + 3)
y – 2 = -2x – 6
2x + y + 4 = 0
If m = \(\frac{1}{2}\),
(2) ⇒ y – 2 = \(\frac{1}{2}\)(x + 3)
2y – 4 = x + 3
x – 2y + 7 = 0
∴ The equation of the straight line is 2x + y + 4 = 0
∴ The equation of straight line is x – 2y + 7 = 0
∴ The required equations of the straight line are, 2x + y + 4 = 0, x – 2y + 7 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 6.
Find the equations of the straight lines passing through the point of intersection of the lines 3x + 2y + 4 = 0, 2x + 5y = 1 and whose distance from (2, -1) is 2. [May ’09; Mar. ’09]
Solution:
Given equations of the straight lines are
3x + 2y + 4 = 0 ……….(1)
2x + 5y – 1 = 0 ………(2)
Let the given point A = (2, -1)
∴ The equation of the straight line passing through the point of intersection of lines is (1) & (2) is L1 + λL2 = 0
(3x + 2y + 4) + λ(2x + 5y – 1) = 0 ……(3)
3x + 2y + 4 + 2λx + 5λy – λ = 0
(3 + 2λ)x + (2 + 5λ)y + (4 – λ) = 0
Given that the perpendicular distance from point A(2, -1) to the line (3) is 2.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q6
squaring on both sides
(-λ + 4)2 = 29λ2 + 32λ + 13
λ2 + 16 – 8λ = 29λ2 + 32λ + 13
28λ2 + 40λ – 3 = 0
28λ2 + 42λ – 2λ – 3 = 0
14λ(2λ + 3) – 1(2λ + 3) = 0
(2λ + 3) (14λ – 1) = 0
(2λ + 3) = 0, 14λ – 1 = 0
λ = \(\frac{-3}{2}\); λ = \(\frac{1}{14}\)
Case I: If λ = \(\frac{-3}{2}\), then
The equation of the straight line is from (3)
(3x + 2y + 4) + (\(\frac{-3}{2}\)) (2x + 5y – 1) = 0
\(\frac{6 x+4 y+8-6 x-15 y+3}{2}=0\)
-11y + 11 = 0
y – 1 = 0
If λ = \(\frac{1}{14}\), then
The equation of the straight line is from (3)
(3x + 2y + 4) + (\(\frac{1}{14}\)) (2x + 5y – 1) = 0
\(\frac{42 x+28 y+56+2 x+5 y-1}{14}=0\)
44x + 33y + 55 = 0
4x + 3y + 5 = 0
∴ The required equations of the straight lines are y – 1 = 0 & 4x + 3y + 5 = 0

Question 7.
Find the orthocentre of the triangle whose vertices are (-5, -7), (13, 2), and (-5, 6). [Mar. ’16 (AP) ’12, ’03; May ’98]
Solution:
Let A(-5, -7), B(13, 2), C(-5, 6) are the given points.
The equation of the altitude \(\overline{\mathbf{A D}}\).
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q7
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q7.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q7.2

Question 8.
Find the circumcentre of the triangle whose vertices are (-2, 3), (2, -1) and (4, 0). [Mar. ’17 (AP), ’13 (Old), ’11; May ’15 (TS), ’02, ’92]
Solution:
Let A(-2, 3), B(2, -1), C(4, 0) are the given points
Let S(α, β) be the circumcentre of the triangle ABC
then, SA = SB = SC
Now, SA = SB
⇒ SA2 = SB2
⇒ (α + 2)2 + (β – 3)2 = (α – 2)2 + (β + 1)2
⇒ α2 + 4 + 4α + β2 + 9 – 6β = α2 + 4 – 4α + β2 + 1 + 2β
⇒ 4α + 9 – 6β + 4α – 1 – 2β = 0
⇒ 8α – 8β + 8 = 0
⇒ α – β + 1 = 0 …….(1)
Also SB = SC
⇒ SB2 = SC2
⇒ (α – 2)2 + (β + 1)2 = (α – 4)2 + (β – 0)2
⇒ α2 + 4 – 4α + β2 + 2β + 1 = α2 + 16 – 8α + β2
⇒ 4 – 4α + 2β + 1 – 16 + 8α = 0
⇒ 4α + 2β – 11 = 0 ……(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q8

Question 9.
Find the orthocentre of the triangle whose sides are given by x + y + 10 = 0, x – y – 2 = 0, and 2x + y – 7 = 0. [May ’13]
Solution:
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q9
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q9.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q9.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q9.3

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 10.
Find the circumcentre of the triangle whose sides are 3x – y – 5 = 0, x + 2y – 4 = 0 and 5x + 3y + 1 = 0. [May ’11, ’05; Mar. ’06]
Solution:
Given, the equations of the straight lines are
3x – y – 5 = 0 ……..(1)
x + 2y – 4 = 0 ……..(2)
5x + 3y + 1 = 0 ……..(3)
Vertex A: Solving (1) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q10
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q10.1
Let S(α, β) be the circumcentre then SA = SB = SC
Now, SA = SB
⇒ SA2 = SB2
⇒ (α – 1)2 + (β + 2)2 = (α – 2)2 + (β – 1)2
⇒ α2 – 2α + 1 + β2 + 4 + 4β = α2 + 4 – 4α + β2 + 1 – 2β
⇒ -2α + 4β + 4α + 2β = 0
⇒ 2α + 6β = 0
⇒ α + 3β = 0 ……..(4)
Also SB = SC
⇒ SB2 = SC2
⇒ (α – 2)2 + (β – 1)2 = (α + 2)2 + (β – 3)2
⇒ α2 – 2α + 1 + β2 – 2β + 1 = α2 + 4α + 4 + β2 – 6β + 9
⇒ 4α + 9 – 6β + 4α – 1 + 2β = 0
⇒ 8α – 4β – 8 = 0
⇒ 2α – β + 2 = 0 ……..(5)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q10.2

Question 11.
Find the equations of the straight lines passing through the point (1, 2) and make an angle of 60° with the line √3x + y + 2 = 0. [May ’03; B.P.]
Solution:
Given, equation of the straight line is √3x + y + 2 = 0 ……….(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q11
Let the given point A(x1, y1) = (1, 2)
Given that θ = 60°
Let the slope of the required straight line is = m
∴ The equation of the straight line is passed through A(1, 2) and having slope ‘m’ is y – y1 = m(x – x1)
y – 2 = m(x – 1) …….(2)
y – 2 = mx – m
mx – y – m + 2 = 0
Let ‘θ’ be the angle between the lines (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q11.1
Squaring on both sides
⇒ m2 + 1 = (√3m – 1)2
⇒ m2 + 1 = 3m2 + 1 – 2√3m
⇒ 2m2 – 2√3m = 0
⇒ m2 – √3m = 0
⇒ m(m – √3) = 0
⇒ m = 0 (or) m – √3 = 0
⇒ m = 0 (or) m = √3
Case 1: If m = 0 then, the equation of a required straight line is,
from (2), y – 2 = 0(x – 1)
y – 2 = 0
y = 2
Case 2: If m = √3 then, the equation of a required straight line is
from (2), y – 2 = √3(x – 1)
y – 2 = √3x – √3
√3x – y + 2 – √3 = 0
∴ The equations of the required straight lines are y = 2 and √3x – y + 2 – √3 = 0

Question 12.
If p and q are the lengths of the perpendiculars from the origin to the straight lines x sec α + y cosec α = a and x cos α – y sin α = a cos 2α prove that 4p2 + q2 = a2. [Mar. ’13 (Old), ’08; May ’13]
Solution:
Given, the equations of the straight lines are
x sec α + y cosec α – a = 0 …….(1)
x cos α – y sin α = a cos 2α ……….(2)
Let the point p(x1, y1) = (0, 0)
Now, p = the perpendicular distance from the point p(0, 0) to the straight line (1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q12
q = the perpendicular distance from the point p(0, 0) to the straight line (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q12.1
L.H.S = 4p2 + q2
= 4(a sin α cos α)2 + (a cos 2α)2
= 4a2 sin2α cos2α + a2 cos2
= a2(2 sin α cos α)2 + a2 cos2
= a2 sin2 2α + a2 cos2
= a2(sin2 2α + cos2 2α)
= a2
= R.H.S
∴ 4p2 + q2 = a2

Some More Maths 1B Straight Lines Important Questions

Question 13.
Find the equation of the straight line which makes an angle 135° with the positive X-axis measured counterclockwise and passing through the point (-2, 3).
Solution:
Given that, the inclination of a line is θ = 135°
The slope of a straight line is, m = tan θ
= tan 135°
= tan (180° – 45°)
= -tan 45°
= -1
Let the given point A(x1, y1) = (-2, 3)
∴ The equation of the straight line passing through the point (-2, 3) and having slope ‘-1’ is,
y – y1 = m(x – x1)
y – 3 = -1(x + 2)
y – 3 = -x – 2
x + y – 1 = 0

Question 14.
Find the equation of the straight line passing through the point (2, 3) and making non-zero intercepts on the axes of coordinates whose sum is zero. [Mar. ’12; May ’09]
Solution:
The equation of a straight line in the intercept form is
\(\frac{x}{a}+\frac{y}{b}\) = 1 ………(1)
Given that, the sum of the intercepts = 0
a + b = 0
b = -a
From (1), \(\frac{x}{a}+\frac{y}{-a}=1\)
x – y = a ……(2)
Since equation (2) passes through the point (2, 3) then,
2 – 3= a
∴ a = -1
Substitute the value of ‘a’ in equation (2)
x – y = -1
x – y + 1 = 0
∴ The equation is x – y + 1 = 0.

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 15.
Find the equation of the straight line passing through points (4, -3) and perpendicular to the line passing through points (1, 1) and (2, 3).
Solution:
Let A(1, 1), B(2, 3), C(4, -3) are the given points
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q3
Slope of the perpendicular line = \(\frac{-1}{m}=\frac{-1}{2}\)
∴ The equation of the straight line passing through C(4, -3) and having slope \(\frac{-1}{2}\) is y – y1 = \(\frac{1}{m}\) (x – x1)
y + 3 = \(\frac{-1}{2}\) (x – 4)
2y + 6 = -x + 4
x + 2y+ 6 – 4 = 0
x + 2y + 2 = 0

Question 16.
Transform the equation 3x + 4y + 12 = 0 into (i) slope-intercept form (ii) intercept form (iii) normal form.
Solution:
(i) Slope-intercept form:
Given, the equation of the straight line is 3x + 4y + 12= 0
4y = -3x – 12
y = \(\frac{-3 x}{4}-\frac{12}{4}=\left(\frac{-3}{4}\right) x+(-3)\)
which is in the form of y = mx + c
Slope, m = \(\frac{-3}{4}\), y-intercept, c = -3
(ii) Intercept form:
Given, the equation of the straight line is 3x + 4y + 12 =0
3x + 4y = -12 × 1
\(\frac{3 x}{-12}+\frac{4 y}{-12}=1\)
\(\frac{x}{-4}+\frac{y}{-3}=1\)
which is in the form of \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept, a = -4, y-intercept, b = -3
(iii) Normal form:
Given the equation of the straight line is, 3x + 4y + 12 = 0
3x + 4y= -12
-3x – 4y = 12
On dividing both sides with \(\sqrt{a^2+b^2}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q4
which is in the form x cos α + y sin α = p
∴ cos α = \(\frac{-3}{5}\), sin α = \(\frac{-4}{5}\), p = \(\frac{12}{5}\)

Question 17.
Find the distance between the parallel straight lines 5x – 3y – 4 = 0, 10x – 6y – 9 = 0. [Mar. ’13(old), ’09, ’02; May ’12]
Solution:
Given, the equations of the straight lines are
5x – 3y – 4 = 0
10x – 6y – 9 = 0 …….(2)
2(5x – 3y – 4) = 0
10x – 6y – 8 = 0 ………(1)
Comparing (1) with ax + by + c1 = 0, we get
a = 10, b = -6, c1 = -8
Comparing (2) with ax + by + c2 = 0, we get
a = 10, b = -6, c2 = -9
Distance between the parallel lines =
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q5

Question 18.
Find the value of p, if the straight lines 3x + 7y – 1 = 0 and 7x – py + 3 = 0 are mutually perpendicular.
Solution:
Given, the equations of the straight lines are
3x + 7y – 1 = 0 ……..(1)
7x – py + 3 = 0 ……..(2)
Slope of the line (1) is m1 = \(\frac{-3}{7}\)
Slope of the line (2) is m2 = \(\frac{-7}{-p}=\frac{7}{p}\)
Since the given lines are perpendicular then,
m1 × m2 = -1
\(\frac{-3}{7} \times \frac{7}{p}\) = -1
p = 3

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 19.
Find the value of p, if the lines 3x + 4y = 5, 2x + 3y = 4, px + 4y = 6 are concurrent. [May ’06]
Solution:
Given, the equations of the straight lines are,
3x + 4y – 5 = 0 ……..(1)
2x + 3y – 4 = 0 ……..(2)
px + 4y – 6 = 0 …….(3)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q7
∴ The point of intersection of the straight lines (1) & (2) is (-1, 2)
Now, given lines are concurrent then, the point of intersection lies on the line
p(-1) + 4(2) – 6 = 0
-p + 8 – 6 = 0
-p + 2 = 0
p = 2

Question 20.
Find the value of p, if the lines 4x – 3y – 7 = 0, 2x + py + 2 = 0, 6x + 5y – 1 = 0 are concurrent.
Solution:
Given, the equations of the straight lines are,
4x – 3y – 7 = 0 …….(1)
2x + py + 2 = 0 ……..(2)
6x + 5y – 1 = 0 ……….(3)
Solving (1) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q8
∴ The point of intersection of the straight lines (1) & (3) is, (1, -1).
Since the given lines are concurrent, then the point of intersection (1, -1) lies on line (2).
2(1) + p(-1) + 2 = 0
2 – p + 2 = 0
4 – p = 0
p = 4

Question 21.
Show that the lines 2x + y – 3 = 0, 3x + 2y – 2 = 0 and 2x – 3y – 23 = 0 are concurrent and find the point of concurrent.
Solution:
Given, the equations of the straight lines are,
2x + y – 3 = 0 …….(1)
3x + 2y – 2 = 0 ……….(2)
2x – 3y – 23 = 0 ………(3)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q9
∴ The point of intersection of the straight lines (1) & (2) is (4, -5).
Now substitute the point (4, -5) in equation (3)
2(4) – 3(-5) – 23 = 0
8 + 15 – 23 = 0
23 – 23 = 0
0 = 0
∴ Given lines are concurrent.
∴ Point of concurrence = (4, -5).

Question 22.
A straight line meets the coordinate axes in A and B. Find the equation of the straight line, when \(\overline{\mathbf{A B}}\) is divided in the ratio 2 : 3 at (-5, 2).
Solution:
The equation of the straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) ……..(1)
The straight line (1) meets the X-axis at A(a, 0), Y-axis at B(0, b).
Now, P(-5, 2) divides \(\overline{\mathbf{A B}}\) in the ratio 2 : 3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q10
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q10.1

Question 23.
A straight line meets the coordinate axes in A and B. Find the equation of the straight line when \(\overline{\mathbf{A B}}\) is divided in the ratio 1 : 2 at (-5, 4).
Solution:
The equation of the straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) ……..(1)
The straight line (1) meets the X-axis at A(a, 0), Y-axis at B(0, b).
Now, P(-5, 4) divides \(\overline{\mathbf{A B}}\) in the ratio 1 : 2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q11
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q11.1

Question 24.
If non-zero numbers a, b, c are in harmonic progression, then show that the equation \(\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0\) represents a family of concurrent lines and find the point of concurrency.
Solution:
Given that, a, b, c are in harmonic progression.
then, \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\)
\(\frac{2}{b}-\frac{1}{a}=\frac{1}{c}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q12
bx + ay + 2a – b = 0
b(x – 1) + a(y + 2) = 0
(x – 1) + \(\frac{a}{b}\) (y + 2) = 0
This is of the form L1 + λL2 = 0
∴ \(\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0\) represents a set of lines passing through the point of intersection of
L1 = x – 1 = 0 ……..(1)
L2 = y + 2 = 0 ………(2)
Solve (1) & (2)
from (1), x – 1 = 0 ⇒ x = 1
from (2) y + 2 = 0 ⇒ y = -2
∴ The point of concurrence = (1, -2).
∴ \(\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0\) represents a set of concurrent lines.

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 25.
If a, b, c are in arithmetic progression, then show that the equation ax + by + c = 0 represents a family of concurrent lines and find the point of concurrency.
Solution:
Given that, a, b, c are in arithmetic progression, then
2b = a + c
c = 2b – a
Now, ax + by + c = 0
ax + by + (2b – a) = 0
ax + by + 2b – a = 0
a(x – 1) + b(y + 2) = 0
x – 1 + \(\frac{b}{a}\) (y + 2) = 0
This is of the form L1 + λL2 = 0
∴ ax + by + c = 0 represents a set of lines passing through the point of intersection of
L1 = x – 1 = 0 …….(1)
L2 = y + 2 = 0 ……..(2)
Solve (1) & (2)
from (1), x – 1 = 0 ⇒ x = 1
from (2), y + 2 = 0 ⇒ y = -2
∴ The point of concurrence = (1, -2).
∴ ax + by + c = 0 represents a set of concurrent lines.

Question 26.
A straight line parallels the line y = √3x passes through Q(2, 3), and cuts the line 2x + 4y – 27 = 0 at p then, finds the length of PQ.
Solution:
Given the equation of the straight line is y = √3x
This is of the form y = mx
⇒ m = √3
⇒ tan θ = √3
⇒ θ = 60°
Since \(\overline{\mathrm{PQ}}\) is parallel to the straight line y = √3x
∴ Slope of \(\overline{\mathrm{PQ}}\) = √3
Inclination of a straight line \(\overline{\mathrm{PQ}}\), 0 = 60°
Given point Q(x1, y1) = (2, 3)
∴ The coordinates of P = (x1 + r cos θ, y1 + r sin θ)
= (2 + r cos 60°, 3 + r sin 60°)
= (2 + r(\(\frac{1}{2}\)), 3 + \(\frac{\sqrt{3} r}{2}\)) where |r| = PQ
Since the point, ‘p’ lies on the straight line 2x + 4y – 27 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q14

Question 27.
A straight line with slope 1 passes through Q(-3, 5) and meets the line x + y – 6 = 0 at P. Find the distance PQ.
Solution:
Given, equation of the straight line is x + y – 6 = 0 ……….(1)
The slope of the straight line m = 1
Given point Q(x1, y1) = (-3, 5)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q15
∴ The equation of the straight line \(\overline{\mathrm{PQ}}\) having slope ‘1’ and passes through the point Q(-3, 5) is (y – y1) = m(x – x1)
y – 5 = 1(x + 3)
y – 5 = x + 3
x – y + 8 = 0 …….(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q15.1

Question 28.
A straight line through Q(2, 3) makes an angle \(\frac{3 \pi}{4}\) with the negative direction of the X-axis. If the straight line intersects the line x + y – 7 = 0 at P, find the distance PQ.
Solution:
Given, equation of the straight line is x + y – 7 = 0 ……..(1)
Given point Q(x1, y1) = (2, 3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q16
Given that the straight line makes an angle \(\frac{3 \pi}{4}\) with the ‘-ve’ direction of the X-axis, then the straight line makes an angle π – \(\frac{3 \pi}{4}\) = \(\frac{\pi}{4}\) = 45° with the positive direction of X-axis.
∴ The slope of the straight line is m = tan θ
= tan 45°
= 1
The equation of the straight line \(\overline{\mathrm{PQ}}\) having slope ‘1’ and passing through the point Q(2, 3) is,
y – y1 = m(x – x1)
y – 3 = 1(x – 2)
y – 3 = x – 2
x – y + 1 = 0 …….(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q16.1

Question 29.
Find the foot of the perpendicular drawn from (4, 1) upon the straight line 3x – 4y + 12 = 0. [May ’08; June ’02]
Solution:
Given the equation of the straight line is 3x – 4y + 12 = 0.
Comparing with ax + by + c = 0, we get
a = 3, b = -4, c = 12
Let the given point A(x1, y1) = (4, 1)
Let (h, k) is the foot perpendicular from the point A(4, 1) on the line 3x – 4y + 12 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q17
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q17.1

Question 30.
Find the foot of the perpendicular drawn from (3, 0) upon the straight line 5x + 12y – 41 = 0.
Solution:
Given the equation of the straight line is 5x + 12y – 41 = 0
Comparing with ax + by + c = 0, we get
a = 5, b = 12, c = -41
Let the given point A(x1, y1) = (3, 0)
Let (h, k) is the foot of the perpendicular from the point A(3, 0) on the line 5x + 12y – 41 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q18
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q18.1

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 31.
Find the image of the point (1, 2) in the straight line 3x + 4y – 1 = 0. [Mar. ’07, ’02, ’97; May ’04, ’97]
Solution:
Given the equation of the straight line is 3x + 4y – 1 = 0
Comparing with ax + by + c = 0, we get
a = 3, b = 4, c = -1
Let the given point A(x1, y1) = (1, 2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q19
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q19.1

Question 32.
x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining points A, B. If A = (-1, -3) find the coordinates of B. [Mar. ’13 (Old)]
Solution:
Given, equation of the straight line is x – 3y – 5 = 0
Comparing with ax + by + c = 0, we get
a = 1, b = -3, c = -5
Let the given point A(x1, y1) = (-1, -3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q20
x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining the points A, B then, B(h, k) is the image of A(-1, -3) with respect to the straight line x – 3y – 5 = 0.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q20.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q20.2

Question 33.
If 2x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining (3, -4) and (α, β), find α + β.
Solution:
Given, the equation of the straight line is 2x – 3y – 5 = 0
Comparing with ax + by + c = 0, we get
a = 2, b = -3, c = -5
Let the given point A(x1, y1) = (3, -4)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q21
2x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining the points A, B then, B(α, β) is the image of A(3, -4) with respect to the straight line 2x – 3y – 5 = 0.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q21.1
∴ Image of A(3, 4) is (α, β) = (-1, 2)
∴ α + β = -1 + 2 = 1

Question 34.
Find the orthocentre of the triangle whose vertices are (-2, -1), (6, -1), and (2, 5). [May ’12; Mar. ’07, ’04]
Solution:
Let the given vertices are A(-2, -1), B(6, -1) & C(2, 5)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q22
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q22.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q22.2

Question 35.
Find the orthocentre of the triangle whose vertices are (5, -2), (-1, 2), and (1, 4).
Solution:
Let the given points are A(5, -2), B(-1, 2) & C(1, 4)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q23
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q23.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q23.2

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 36.
Find the circumcentre of the triangle whose vertices are (1, 3), (0, -2), and (-3, 1). [Mar. ’18 (AP); May ’06]
Solution:
Let A(1, 3), B(0, -2), C(-3, 1) are the given points.
Let S(α, β) be the circumcentre of the triangle ABC then, SA = SB = SC
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q24
Now, SA = SB
⇒ SA2 = SB2
⇒ (α – 1)2 + (β – 3)2 = (α – 0)2 + (β + 2)2
⇒ α2 – 2α + 1 + β2 + 9 – 6β = α2 + β2 + 4 + 4β
⇒ 4 + 4β + 2α – 9 + 6β – 1 = 0
⇒ 2α + 10β – 6 = 0
⇒ α + 5β – 3 = 0 ……(1)
Also SB = SC
⇒ SB2 = SC2
⇒ (α – 0)2 + (β + 2)2 = (α + 3)2 + (β – 1)2
⇒ α2 + β2 + 4 + 4β = α2 + 9 + 6α + β2 + 1 – 2β
⇒ 9 + 6α – 2β + 1 – 4 – 4β = 0
⇒ 6α – 6β + 6 = 0
⇒ α – β + 1 = 0 ………(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q24.1

Question 37.
Find the circumcentre of the triangle whose vertices are (1, 3), (-3, 5), and (5, -1). [Mar. ’18 (TS)]
Solution:
Let A(1, 3), B(-3, 5), C(5, -1) are the given points.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q25
Let S(α, β) be the circumcentre of the triangle ABC then, SA = SB = SC
Now, SA = SB
⇒ SA2 = SB2
⇒ (α – 1)2 + (β – 3)2 = (α + 3)2 + (β – 5)2
⇒ α2 – 2α + 1 + β2 + 9 – 6β = α2 + 9 + 6α + β2 + 25 – 10β
⇒ 6α + 25 – 10β – 1 + 2α + 6β = 0
⇒ 8α – 4β + 24 = 0
⇒ 2α – β + 6 = 0 ……(1)
Also SB = SC
⇒ SB2 = SC2
⇒ (α + 3)2 + (β – 5)2 = (α – 5)2 + (β + 1)2
⇒ α2 + 9 + 6α + 25 + β2 – 10β = α2 + 25 – 10α + β2 + 1 + 2β
⇒ 9 + 6α – 10β + 10α – 1 – 2β = 0
⇒ 16α – 12β + 8 = 0
⇒ 4α – 3β + 2 = 0 ………(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q25.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q25.2
Circumcentre S = (-8, -10)

Question 38.
Find the circumcentre of the triangle whose vertices are (1, 0), (-1, 2), and (3, 2). [May ’13 (Old)]
Solution:
Let A(1, 0), B(-1, 2), C(3, 2) are the given points.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q26
Let S(α, β) be the circumcentre of the triangle ABC then, SA = SB = SC
Now, SA = SB
⇒ SA2 = SB2
⇒ (α – 1)2 + (β – 0)2 = (α + 1)2 + (β – 2)2
⇒ α2 – 2α + 1 + β2 = α2 + 2α + 1 + β2 + 4 – 4β
⇒ 2α – 4β + 4 + 1 – 1 + 2α = 0
⇒ 4α – 4β + 4 = 0
⇒ α – β + 1 = 0 ……….(1)
Also SB = SC
⇒ SB2 = SC2
⇒ (α + 1)2 + (β – 2)2 = (α – 3)2 + (β – 2)2
⇒ α2 + 2α + 1 + β2 + 4 – 4β = α2 + 9 – 6α + β2 + 4 – 4β
⇒ 1 + 2α – 9 + 6α = 0
⇒ 8α – 8 = 0
⇒ α – 1 = 0
⇒ α = 1
Substitute α = 1 in (1), we get
1 – β + 1 = 0
⇒ -β = -2
⇒ β = 2
∴ Circumcentre S = (1, 2)

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 39.
If the equations of the sides of a triangle are 7x + y – 10 = 0, x – 2y + 5 = 0, and x + y + 2 = 0, find the orthocenter of the triangle. [May ’09, ’00, ’97]
Solution:
Given, the equations of the straight lines are
7x + y – 10 = 0 …….(1)
x – 2y + 5 = 0 …….(2)
x + y + 2 = 0 ……….(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q27
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q27.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q27.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q27.3

Question 40.
Find the orthocentre of the triangle whose sides are given by 4x – 7y + 10 = 0, x + y = 5, and 7x + 4y = 15.
Solution:
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28.3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28.4
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28.5

Question 41.
Find the orthocentre of the triangle formed by the lines x + 2y = 0, 4x + 3y – 5 = 0 and 3x + y = 0. [Mar. ’10]
Solution:
Given, the equations of the straight lines are,
x + 2y = 0 ……..(1)
4x + 3y – 5z = 0 ……….(2)
3x + y = 0 ………(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29.3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29.4
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29.5

Question 42.
Find the circumcentre of the triangle formed by the straight lines x + y = 0, 2x + y + 5 = 0, and x – y = 2.
Solution:
Given, the equations of the straight lines are
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q30
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q30.1
∴ Vertex C = (-1, -3)
Let S(α, β) be the circumcentre of the triangle then, SA = SB = SC
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q30.2
Now, SA = SB
⇒ SA2 = SB2
⇒ (α – 1)2 + (β + 1)2 = (α + 5)2 + (β – 5)2
⇒ α2 – 2α + 1 + β2 + 1 + 2β = α2 + 25 + 10α + β2 + 25 – 10β
⇒ 50 + 10α – 10β – 1 + 2α – 2β – 1 = 0
⇒ 12α – 12β + 48 = 0
⇒ α – β + 4 = 0 ……..(4)
Also SB = SC
⇒ SB2 = SC2
⇒ (α + 5)2 + (β – 5)2 = (α + 1)2 + (β + 3)2
⇒ α2 + 25 + 10α + β2 + 25 – 10β = α2 + 2α + 1 + β2 + 6β + 9
⇒ 25 + 25 + 10α – 10β – 1 = -2α – 9 – 6β = 0
⇒ 8α – 16β + 40 = 0
⇒ α – 2β + 5 = 0 ……..(5)
Solving (4) & (5)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q30.3
α = -3; β = 1
∴ Circumcentre S = (-3, 1)

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 43.
Find the circumcentre of the triangle formed by the straight lines x + y + 2 = 0, 5x – y – 2 = 0, and x – 2y + 5 = 0. [May ’08]
Solution:
Given, the equations of the straight lines are
x + y + 2 = 0 ………(1)
5x – y – 2 = 0 ………(2)
x – 2y + 5 = 0 ………(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q31
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q31.1
Let S(α, β) be the circumcentre of the triangle then, SA = SB = SC
Now, SA = SB
⇒ SA2 = SB2
⇒ (α + 3)2 + (β – 1)2 = (α – 0)2 + (β + 2)2
⇒ α2 + 9 + 6α + β2 + 1 – 2β = α2 + β2 + 4 + 4β
⇒ 6α – 2β + 9 + 1 – 4 – 4β = 0
⇒ 6α – 6β + 6 = 0
⇒ α – β + 1 = 0 ……..(4)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q31.2
Also SB = SC
⇒ SB2 = SC2
⇒ (α + 0)2 + (β + 2)2 = (α – 1)2 + (β – 3)2
⇒ α2 + β2 + 4 + 4β = α2 – 2a + 1 + β2 – 6β + 9
⇒ 4 + 4β – 1 + 2α – 9 + 6β = 0
⇒ 2α + 10β – 6 = 0
⇒ α + 5β – 3 = 0 ……..(5)
Solving (4) & (5)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q31.3

Question 44.
Find the equations of the straight lines passing through the point (-10, 4) and make an angle θ with the line x – 2y = 10 such that tan θ = 2.
Solution:
Given, equation of the straight line is x – 2y – 10 = 0 ……..(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q32
Let the given point P(x1, y1) = (-10, 4)
Given that, tan θ = 2
cos θ = \(\frac{1}{\sqrt{5}}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q32.1
Let the slope of the required straight line = m.
∴ The equation of the straight line is passed through the point P(-10, 4) and having slope ‘m’ is y – y1 = m(x – x1)
y – 4 = m(x + 10)
y – 4 = mx + 10m
mx – y + 10m + 4 = 0
If ‘θ’ be the angle between the lines (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q32.2
Squaring on both sides
5(m2 + 1) = 5(m + 2)2
5m2 + 5 = 5m2 + 20 + 20m (or) m = \(\frac{1}{0}\)
20m + 15 = 0
20m = -15
m = \(\frac{-3}{4}\) or m = \(\frac{1}{0}\)
Case 1: If m = \(\frac{-3}{4}\) then, the equation of the straight line is,
from (2), y – 4 = \(\frac{-3}{4}\) (x + 10)
4y – 16 = -3x – 30
3x + 4y – 16 + 30 = 0
3x + 4y + 14 = 0
Case 2: If m = \(\frac{1}{0}\) then, the equation of straight line is
from (2), y – 4 = \(\frac{1}{0}\) (x + 10)
0 = x + 10
x + 10 = 0
∴ Required equations of the straight lines are 3x + 4y + 10 = 0, x + 10 = 0

Question 45.
The base of an equilateral triangle is x + y – 2 = 0 and the opposite vertex is (2, -1). Find the equations of the remaining sides. [Mar. ’02]
Solution:
Given that base of an equilateral triangle is, x + y – 2 = 0 ……(1)
Let the opposite vertex is, A = (2, -1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q33
Since, Triangle ABC is an equilateral triangle B then,
A = B = C = 60°
Let the slope of the side \(\overline{\mathrm{AB}}\) = m
∴ The equation of the side \(\overline{\mathrm{AB}}\) passing through A(2, -1) and having slope ‘m’ is y – y1 = m(x – x1)
y + 1 = m(x – 2) ………(2)
y + 1 = mx – 2m
mx – y – 2m – 1 = 0
Let ‘θ’ be the angle between the lines (1) & (2) then,
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q33.1
∴ The equations of the required straight lines are
from (2), y + 1 = 2 ± √3(x – 2).

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 46.
Prove that the points (1, 11), (2, 15), and (-3, -5) are collinear and find the equation of the straight line containing them.
Solution:
Let A(1, 11), B(2, 15), C(-3, -5) are the given points.
The equation of the straight line \(\overline{\mathrm{AB}}\) is,
(y – y1) (x2 – x1) = (x – x1)(y2 – y1)
⇒ (y – 11) (2 – 1) = (x – 1) (15 – 11)
⇒ (y – 11) (1) = (x – 1) (4)
⇒ y – 11 = 4x – 4
⇒ 4x – y + 7 = 0 ……….(1)
Now, substituting the point C(-3, -5) in equation (1)
4(-3) – (-5) + 7 = 0
⇒ -12 + 5 + 7 = 0
⇒ -7 + 7 = 0
⇒ 0 = 0
∴ The point C(-3, -5) lie on the straight line 4x – y + 7 = 0.
∴ Given points are collinear.
∴ The equation of the straight line is 4x – y + 7 = 0.

Question 47.
Find the angle which the straight line y = √3x – 4 makes with the Y-axis. [Mar. ’19 (AP & TS)]
Solution:
Given the equation of the straight line is y = √3x – 4
⇒ √3x – y – 4 = 0
Comparing the given equation with ax + by + c = 0, then
a = √3, b = -1, c = -4
The slope of a straight line √3x – y – 4 = 0 is
m = \(\frac{-a}{b}=\frac{-\sqrt{3}}{-1}\) = √3
tan θ = √3
∴ θ = 60° = \(\frac{\pi}{3}\)
The angle which the straight line y = √3x – 4 makes with X-axis = \(\frac{\pi}{3}\)
The angle which the straight line y = √3x – 4 makes with the Y-axis is = 180° – (90° + 60°)
= 180° – 150°
= 30°
= \(\frac{\pi}{6}\)

Question 48.
Write the equations of the straight lines parallel to the X-axis and (i) at a distance of 3 units above the X-axis and (ii) at a distance of 4 units below the X-axis.
Solution:
(i) The equation of the straight lines parallel to the X-axis, at a distance of 3 units above the X-axis is y = 3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q3
(ii) The equation of the straight line parallel to the X-axis at a distance of 4 units below the X-axis is y = -4
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q3.1

Question 49.
Write the equations of the straight lines parallel to the Y-axis and (i) at a distance of 2 units from the Y-axis to the right of it (ii) at a distance of 5 units from the Y-axis to the left of it.
Solution:
(i) The equation of the straight line parallel to the Y-axis and at a distance of 2 units from the Y-axis to the right of it is x = 2.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q4
(ii) The equation of the straight line parallel to the Y-axis and at a distance of 5 units from the Y-axis to the left of it is x = -5.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q4.1

Question 50.
Find the equation of the straight line which makes an angle \(\frac{\pi}{4}\) with the positive X-axis in the positive direction and which passes through the point (0, 0).
Solution:
Given that, inclination of a straight line is θ = \(\frac{\pi}{4}\)
The slope of a line is m = tan θ
= tan \(\frac{\pi}{4}\)
= tan 45°
= 1
Let the given point A(x1, y1) = (0, 0)
∴ The equation of the straight line passing through A(0, 0) and having slope ‘1’ is y – y1 = m(x – x1)
⇒ y – o = 1(x – 0)
⇒ y = x – o
⇒ y = x
⇒ x – y = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 51.
Find the equation of the straight line which makes an angle of 135° with the positive X – axis in the positive direction and which pass through the point (3, -2).
Solution:
Given that, the inclination of a straight line is θ = 135°
The slope of a line is m = tan θ
= tan 135°
= tan (180° – 45°)
= tan 45°
= -1
Let the given point A(x1, y1) = (3, -2)
∴ The equation of the straight line passing through A(3, -2) and having slope ‘-1’ is y – y1 = m(x – x1)
⇒ y + 2 = -1(x – 3)
⇒ y + 2 = -x + 3
⇒ x + y + 2 – 3 = 0
⇒ x + y – 1 = 0

Question 52.
Find the equation of the straight line which makes an angle of 150° with the positive X-axis in the positive direction and the Y-intercept is 2.
Solution:
Given the inclination of a straight line is θ = 150°
The slope of a line is m = tan θ
= tan 150°
= -cot 60°
= \(\frac{-1}{\sqrt{3}}\)
y-intercept, c = 2.
The equation of the straight line having slope \(\frac{-1}{\sqrt{3}}\) and y-intercept ‘2’ is y = mx + c
⇒ y = \(\frac{-1}{\sqrt{3}}\)x + 2
⇒ y = \(\frac{-x+2 \sqrt{3}}{\sqrt{3}}\)
⇒ x + √3y – 2√3 = 0

Question 53.
Find the equation of the straight line which makes an angle \(\tan ^{-1}\left(\frac{2}{3}\right)\) with the positive X-axis in the positive direction and the y-intercept is 3.
Solution:
Given, inclination of a straight line is θ = \(\tan ^{-1}\left(\frac{2}{3}\right)\)
tan θ = \(\frac{2}{3}\)
Slope of a line is m = tan θ = \(\frac{2}{3}\)
y-intercept, c = 3
∴ The equation of a straight line having slope \(\frac{2}{3}\) and y-intercept ‘3’ is y = mx + c
⇒ y = \(\frac{2}{3}\) (x) + 3
⇒ y = \(\frac{2 x+9}{3}\)
⇒ 3y = 2x + 9
⇒ 2x – 3y + 9 = 0

Question 54.
Prove that the points (a, b + c), (b, c + a), (c, a + b) are collinear and find the equation of the straight line containing them.
Solution:
Let A(a, b + c), B(b , c + a), C(c, a + b) are the given points
The equation of the straight line \(\overline{\mathrm{AB}}\) is (y – y1) (x2 – x1) = (x – x1)(y2 – y1)
⇒ (y – b – c) (b – a) = (x – a) (c + a – b – c)
⇒ (y – b – c) (b – a) = (x – a) (a – b)
⇒ (y – b – c) (b – a) = -(x – a) (b – a)
⇒ y – b – c = -x + a
⇒ x + y – a – b – c = 0 ……..(1)
Now, Substituting the point C(c, a + b) in equation (1)
x + y – a – b – c = 0
⇒ c + a + b – a – b – c = o
⇒ 0 = 0
∴ Point C(c, a + b) lies on the straight line x + y – a – b – c = 0.
∴ Given points are collinear.
∴ The straight line equation is x + y – a – b – c = 0.

Question 55.
A(10, 4), B(-4, 9) and C(-2, -1) are the vertices of a triangle. Find the equations of (i) \(\overline{\mathrm{AB}}\) (ii) the median through A (iii) the altitude through B (iv) the perpendicular bisector of the side \(\overline{\mathrm{AB}}\).
Solution:
(i) Equation of the straight line \(\overline{\mathrm{AB}}\) is (y – y1) (x2 – x1)= (x – x1)(y2 – y1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q10
⇒ (y – 4)(-4 – 10) = (x – 10) (9 – 4)
⇒ (y – 4) (-14) = (x – 10) (5)
⇒ -14y + 56 = 5x – 50
⇒ 5x + 14y – 106 = 0
(ii) The equation of the median through ‘A’:
Since ‘D’ is the midpoint of BC, then
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q10.1
The equation of the median through A is the equation of the straight line \(\overline{\mathrm{AD}}\),
(y – y1)(x2 – x1) = (x – x1)(y2 – y1)
⇒ (y – 4) (-3 – 10) = (x – 10)(4 – 4)
⇒ (y – 4) (-13) = (x – 10)(0)
⇒ (y – 4)(-13) = 0
⇒ y – 4 = 0
(iii) The equation of the altitude through ‘B’:
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q10.2
= \(\frac{5}{12}\)
Since \(\overline{\mathrm{BE}} \perp \overline{\mathrm{AC}}\), then
slope of \(\overline{\mathrm{BE}}\) = \(\frac{-1}{m}=\frac{-1}{\frac{5}{12}}=\frac{-12}{5}\)
The equation of the altitude through B is, the equation of the straight line passing through B(-4, 9) and having slope \(\frac{-12}{5}\) is y – y1 = m(x – x1)
⇒ y – 9 = \(\frac{-12}{5}\) (x + 4)
⇒ 5y – 45 = -12x – 48
⇒ 12x + 5y + 3 = 0
(iv) The equation of the perpendicular bisector of side \(\overline{\mathrm{AB}}\):
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q10.3
∴ The equation of the perpendicular bisector of \(\overline{\mathrm{AB}}\) is, the equation of the straight line passing through F(3, \(\frac{13}{2}\)) and having slope \(\frac{14}{5}\) is y – y1 = m(x – x1)
⇒ y – \(\frac{13}{2}\) = \(\frac{14}{5}\)(x – 3)
⇒ \(\frac{2 y-13}{2}=\frac{14}{5}(x-3)\)
⇒ 10y – 65 = 28x – 84
⇒ 28x – 10y – 19 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 56.
A straight line passing through A(1, -2) makes an angle \(\tan ^{-1}\left(\frac{4}{3}\right)\) with the positive direction of the X-axis in the anti-clockwise sense. Find the points on the straight line whose distance from A is 5.
Solution:
Given point (x1, y1) = A(1, -2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q11
The inclination of the straight line is θ = \(\tan ^{-1}\left(\frac{4}{3}\right)\)
tan θ = \(\frac{4}{3}\)
sin θ = \(\frac{4}{5}\)
cos θ = \(\frac{3}{5}\)
Distance, |r| = 5 units
Required point = (x1 + |r| cos θ, y1 + |r| sin θ)
= (x1 ± r cos θ, y1 ± r sin θ)
= (1 ± 5 . \(\frac{3}{5}\), -2 ± 5 . \(\frac{4}{5}\))
= (1 ± 3, -2 ± 4)
= (1 + 3, (-2) + 4), (1 – 3, -2 – 4)
= (4, 2), (-2, -6)

Question 57.
Find the sum of the squares of the intercepts of the line 4x – 3y = 12 on the axes of coordinates. [Mar. ’18 (AP)]
Solution:
Given, the equation of the straight line is 4x – 3y = 12
\(\frac{x}{3}+\frac{y}{-4}=1\) which is of the form \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept, a = 3, y-intercept, b = -4
Now, the Sum of the squares of the intercepts = a2 + b2
= 32 + (-4)2
= 9 + 16
= 25

Question 58.
The intercepts of a straight line on the axes of coordinates are a and b. If p is the length of the perpendicular drawn from the origin to this line, write the value of p in terms of a and b.
Solution:
The intercepts of a straight line on the coordinate axes are a, b.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q13
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q13.1
This is of the form x cos α + y sin α = p
∴ p = the length of the perpendicular drawn from the origin to the line = \(\frac{|a b|}{\sqrt{a^2+b^2}}\)

Question 59.
If the product of the intercepts made by the straight line x tan α + y sec α = 1 (0 ≤ α ≤ \(\frac{\pi}{2}\)) on the coordinate axes is equal to sin α, find α.
Solution:
Given, the equation of the straight line is x tan α + y sec α = 1
\(\frac{x}{\frac{1}{\tan \alpha}}+\frac{y}{\frac{1}{\sec \alpha}}=1\)
\(\frac{x}{\cot \alpha}+\frac{y}{\cos \alpha}=1\)
This is of the form \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept (a) = cot α
y-intercept (b) = cos α
Given that, a product of the intercepts is equal to sin α.
cot α × cos α = sin α
⇒ \(\frac{\cos \alpha}{\sin \alpha}\) × cos α = sin α
⇒ \(\frac{\cos ^2 \alpha}{\sin ^2 \alpha}\) = 1
⇒ cot2α = 1
⇒ tan2α = 1
⇒ tan α = 1
⇒ α = \(\frac{\pi}{2}\) (∵ 0 ≤ α ≤ \(\frac{\pi}{2}\))

Question 60.
A straight line passing through A(-2, 1) makes an angle of 30° with \(\overline{\mathrm{OX}}\) in the positive direction. Find the points on the straight line whose distance from A is 4 units.
Solution:
Given point A(x1, y1) = (-2, 1)
Inclination of straight line = 30°
distance |r| = 4
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q15
∴ Required points = (x1 + |r| cos θ, y1 + |r| sin θ)
= (x1 ± r cos θ, y1 ± r sin θ)
= (-2 ± 4 cos 30°, 1 ± 4 sin 30°)
= (-2 ± 4\(\left(\frac{\sqrt{3}}{2}\right)\), 1 ± 4(\(\frac{1}{2}\)))
= (-2 ± 2√3 ,1 ± 2)
= (-2 + 2√3, 1 + 2), (-2 – 2√3, +1 – 2)
= (-2 + 2√3, 3), (-2 + 2√3, -1)

Question 61.
A straight line whose inclination with the positive direction of the X-axis measured in the anti-clockwise sense is \(\frac{\pi}{3}\) makes a positive intercept on the Y-axis. If the straight line is at a distance of 4 from the origin; find its equation.
Solution:
Given that, inclination of a straight line is θ = \(\frac{\pi}{3}\)
Let ‘l’ is the required straight line.
Now, ‘N’ is the foot of the perpendicular from the origin to the straight line ‘L’
∴ ∠XON = 150°
∴ a = 150°
Given that, p = 4
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q16
The equation of the straight line ‘L’ in the normal form is x cos α + y sin α = p
⇒ x cos (150°) + y sin (150°) = 4
⇒ x cos(180 – 30) + y sin (180 – 30) = 4
⇒ -x cos 30° + y sin 30° = 4
⇒ \(-x\left(\frac{\sqrt{3}}{2}\right)+y\left(\frac{1}{2}\right)=4\)
⇒ -√3x + y = 8
⇒ √3x – y + 8 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 62.
Find the ratio in which the straight line 2x + 3y – 10 = 0 divides the join of the points (2, 3) and (2, 10).
Solution:
Given equation of the straight line is L = 2x + 3y – 10 = 0
Comparing the equation with ax + by + c = 0, we get
a = 2, b = 3, c = -10
Let the given points are A(x1, y1) = (2, 3) and B(x2, y2) = (2, 10)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q17

Question 63.
State whether (3, 2) and (-4, -3) are on the same side or opposite sides of the straight line 2x – 3y + 4 = 0.
Solution:
Given equation of the straight line is L = 2x – 3y + 4 = 0
Let the given points are A(x1, y1) = (3, 2) and B(x2, y2) = (-4, -3)
Now, l11 = L(3, 2)
= 2(3) – 3(2) + 4
= 6 – 6 + 4
= 4 > 0
l22 = L(-4, -3)
= 2(-4) – 3(-3) + 4
= -8 + 9 + 4
= 5 > 0
Since l11 0 and l22 > 0, the given points are on the same side of the straight line, L = 0.

Question 64.
Find the ratio’s in which (i) the X-axis and (ii) the Y-axis divide the line segment \(\overline{\mathrm{AB}}\) joining A(2, -3) and B(3, -6).
Solution:
Given points are A(x1, y1) = (2, -3) and B(x2, y2) = (3, -6)
(i) The X-axis divides the line segment AB in the ratio is,
-y1 : y2 = -(-3) : -6
= 3 : -6
= 1 : -2
= -1 : 2
(ii) Y-axis divides \(\overline{\mathrm{AB}}\) in the ratio is,
-x1 : x2 = -2 : 3

Question 65.
Find the equation of the straight line passing through the point of intersection of the lines x + y + 1 = 0 and 2x – y + 5 = 0 and containing the point (5, -2).
Solution:
Given, the equations of the straight lines are
x + y + 1 = 0 ……….(1)
2x – y + 5 = 0 ………..(2)
Let the given point be (5, -2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q20
∴ The point of intersection of lines (1) & (2) is A = (-2, 1)
Now, the equation of the straight line \(\overline{\mathrm{AB}}\) is (y – y1)(x2 – x1) = (x – x1)(y2 – y1)
⇒ (y – 1) (5 + 2) = (x + 2) (-2 – 1)
⇒ (y – 1)(7) = (x + 2)(-3)
⇒ 7y – 7 = -3x – 6
⇒ 3x + 7y – 1 = 0

Question 66.
Find the ratio in which the straight line 3x + 4y = 6 divides the join of the points (2, -1) and (1, 1).
Solution:
Given equation of the straight line is L = 3x + 4y – 6 = 0
Comparing the equation with ax + by + c = 0, we get
a = 3, b = 4, c = -6
Let the given points are A(x1, y1) = (2, -1) and B(x2, y2) = (1, 1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q21

Question 67.
Transform the equation (2 + 5k)x – 3(1 + 2k)y + (2 – k) = 0 into the form L1 + λL2 = 0 and find the point of concurrency of the family of straight lines represented by the equation.
Solution:
Given, equation is (2 + 5k)x – 3(1 + 2k)y + (2 – k) = 0
⇒ 2x + 5kx – 3y – 6ky + 2 – k = 0
⇒ (2x – 3y + 2) + k(5x – 6y – 1) = 0
This is of the form l1 + λl2 = 0
where l1 = 2x – 3y + 2 = 0 ……..(1)
l2 = 5x – 6y – 1 = 0 ……..(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q22
∴ Point of concurrence = (5, 4)

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 68.
Transform the equation (k + 1)x + (k + 2)y + 5 = 0, into the form L1 + λL2 = 0 and find the point of concurrency of the family of straight lines represented by the equation.
Solution:
Given, equation is (k + 1)x + (k + 2)y + 5 = 0
⇒ kx + x + ky + 2y + 5 = 0
⇒ (x + 2y + 5) + k(x + y) = 0
This is of the form l1 + λl2 = 0
where l1 = x + 2y + 5 = 0 ………(1)
l2 = x + y = 0 ……..(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q23
∴ Point of concurrence = (5, -5)

Question 69.
Find the area of the triangle formed by the straight line x – 4y + 2 = 0 and the coordinate axes.
Solution:
Given, equation of the straight line is x – 4y + 2 = 0 ………(1)
Comparing with ax + by + c = 0, we get
a = 1, b = -4, c = 2
∴ The area of the triangle formed by the straight line (1) & the co-ordinate axis = \(\frac{c^2}{2|a b|}=\frac{2^2}{2|1(-4)|}\) = \(\frac{1}{2}\)

Question 70.
Find the area of the triangle formed by the straight line 3x – 4y + 12 = 0 and the coordinate axes.
Solution:
Given, equation of the straight line is 3x – 4y + 12 = 0 ………(1)
Comparing with ax + by + c = 0, we get
a = 3, b = -4, c = 12
∴ The area of the triangle formed by the straight line and the co-ordinate axis = \(\frac{c^2}{2|a b|}=\frac{12^2}{2|3(-4)|}=\frac{12^2}{2|-12|}=\frac{144}{2(12)}\)
= 6 sq. units.

Question 71.
Find the equation of the straight line passing through the points (-1, 2) and (5, -1) and the area of the triangle formed by it with the axes of coordinates. [May ’15 (TS)]
Solution:
Let the given points are A(x1, y1) = (-1, 2) & B(x2, y2) = (5, -1)
The equation of the straight line passing through the points A(-1, 2) & B(5, -1) is (y – y1)(x2 – x1) = (x – x1)(y2 – y1)
⇒ (y – 2) (5 + 1) = (x + 1) (-1 – 2)
⇒ (y – 2)(6) = (x + 1)(-3)
⇒ 2y – 4 = -x – 1
⇒ x + 2y – 3 = 0 ……..(1)
Comparing equation (1) with ax + by + c = 0, we get
a = 1, b = 2, c = -3
∴ The area of the triangle formed by the straight line (1) and the co-ordinate axis = \(\frac{c^2}{2|a b|}=\frac{9}{2|1(2)|}=\frac{9}{2(2)}=\frac{9}{4}\) sq. units

Question 72.
Find the set of values of ‘a’ if the points (1, 2) and (3, 4) lie to the same side of the straight line 3x – 5y + a = 0.
Solution:
Given, equation of the straight line is 3x – 5y + a = 0 ……..(1)
Comparing (1) with ax + by + c = 0, we get
a = 3, b = -5, c = a
Let the given points are A(x1, y1) = (1, 2) & B(x2, y2) = (3, 4)
Given that, the points (1, 2) & (3, 4) lie on the same direction 3x – 5y + a = 0, then -l11 : l22 < 0
⇒ \(\frac{-\mathrm{L}_{11}}{\mathrm{~L}_{22}}\) < 0
⇒ \(\frac{-\left(a x_1+b y_1+c\right)}{\left(a x_2+b y_2+c\right)}\) < 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q27

Question 73.
Find the angle between the lines 2x + y + 4 = 0 and y – 3x = 7.
Solution:
Given, the equation of the straight lines is
2x + y + 4 = 0 ……..(1)
y – 3x – 7 = 0
⇒ 3x – y + 7 = 0 …….(2)
Comparing (1) with a1x + b1y + c1 = 0, we get
a1 = 2, b1 = 1, c1 = 4
Comparing (2) with a2x + b2y + c2 = 0, we get
a2 = 3, b2 = -1, c2 = 7
If ‘θ’ is the angle between the lines (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q28

Question 74.
Find the angle between the lines √3x + y + 1 = 0 and x + 1 = 0.
Solution:
Given, the equation of the straight lines are
√3x + y + 1 = 0 …….(1)
x + 1 = 0 ………(2)
Comparing (1) with a1x + b1y + c1 = 0, we get
a1 = √3, b1 = 1, c1 = 1
Comparing (2) with a2x + b2y + c2 = 0, we get
a2 = 1, b2 = 0, c2 = 1
If ‘θ’ is the angle between the lines (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q29

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 75.
Find the angle between the lines ax + by = a + b, a(x – y) + b(x + y) = 2b.
Solution:
Given, the equation of the straight lines are
ax + by = a + b
ax + by – a – b = 0 …….(1)
a(x – y) + b(x + y) = 2b
ax – ay + bx + by = 2b
(a + b)x + (-a + b)y – 2b = 0 ………(2)
Comparing (1) with a1x + b1y + c1 = 0, we get
a1 = a, b1 = b, c1 = -a – b
Comparing (2) with a2x + b2y + c2 = 0, we get
a2 = a + b, b2 = b – a, c2 = -2b
If ‘θ’ is the angle between the lines (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q30

Question 76.
Find the equation of a straight line perpendicular to the line 5x – 3y + 1 = 0 and pass through the point (4, -3). [Mar. ’15 (TS)]
Solution:
Given, the equation of the straight line is 5x – 3y + 1 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q31
Slope of the given line is m = \(\frac{-5}{-3}=\frac{5}{3}\)
Slope of the perpendicular line = \(\frac{-1}{m}=\frac{-1}{\frac{5}{3}}=\frac{-3}{5}\)
Let the given point A(x1, y1) = (4, -3)
Equation of the straight line passing through (4, -3) and having slope \(\frac{-3}{5}\) is,
y – y1 = \(\frac{-1}{m}\)(x – x1)
⇒ y + 3 = \(\frac{-3}{5}\)(x – 4)
⇒ 5y + 15 = -3x + 12
⇒ 3x + 5y + 3 = 0

Question 77.
Find the value of k, if the straight line 6x – 10y + 3 = 0 and kx – 5y + 8 = 0 are parallel.
Solution:
Given, the equation of the straight line is
6x – 10y + 3 = 0 …….(1)
kx – 5y + 8 = 0 ……(2)
Slope of the line (1) is m1 = \(\frac{-a}{b}=\frac{-6}{-10}=\frac{3}{5}\)
Slope of the line (2) is m2 = \(\frac{-k}{-5}=\frac{k}{5}\)
Since the given lines are parallel then m1 = m2
⇒ \(\frac{3}{5}\) = \(\frac{k}{5}\)
⇒ k = 3

Question 78.
(-4, 5) is a vertex of a square and one of its diagonals is 7x – y + 8 = 0. Find the equation of the other diagonal.
Solution:
Let ABCD be a square.
Let the given point be A(-4, 5).
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q33
Equation of the diagonal \(\overline{\mathrm{AC}}\) is, 7x – y + 8 = 0 is given
Slope of the diagonal, \(\overline{\mathrm{AC}}\) = 7
Since in a square diagonals \(\overline{\mathrm{AC}}\) & \(\overline{\mathrm{BD}}\) are perpendicular
then slope of the diagonal \(\overline{\mathrm{BD}}\) = \(\frac{-1}{m}=\frac{-1}{7}\)
∴ The equation of the diagonal \(\overline{\mathrm{BD}}\) is y – y1 = \(\frac{-1}{m}\)(x – x1)
⇒ y – 5 = \(\frac{-1}{7}\)(x + 4)
⇒ 7y – 35 = -x – 4
⇒ x + 7y – 31 = 0

Question 79.
A(-1, 1), B(5, 3) are opposite vertices of a square in the XY plane. Find the equation of the other diagonal (not passing through A, B) of the square.
Solution:
Let ABCD be a square.
Given that, opposite vertices of a square are A(-1, 1) and B(5, 3).
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q34
The equation of straight line passing through E & slope -3 is y – y1 = \(\frac{-1}{m}\)(x – x1)
⇒ y – 2 = -3(x – 2)
⇒ y – 2 = -3x + 6
⇒ 3x + y – 8 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 80.
Show that lines x – 7y – 22 = 0, 3x + 4y + 9 = 0 and 7x + y – 54 = 0 form a right angled isosceles triangle.
Solution:
Given, the equations of the straight lines are
x – 7y – 22 = 0 …..(1)
3x + 4y + 9 = 0 …….(2)
7x + y – 54 = 0 ………(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q35
Let ‘A’ be the angle between the lines (1) & (3) then,
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q35.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q35.2
∴ A = 90°, B = 45°, C = 45°
∴ Given lines form a right angle isosceles triangle.

Question 81.
If ab > 0, find the area of the rhombus enclosed by the four straight lines ax ± by ± c = 0.
Solution:
Given, the equations of the straight lines are
ax + by + c = 0 ……..(1)
ax + by – c = 0 ………(2)
ax – by + c = 0 ……..(3)
ax – by – c = 0 ………(4)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q36
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q36.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q36.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q36.3

Question 82.
Find the area of the parallelogram whose sides are 3x + 4y + 5 = 0, 3x + 4y – 2 = 0, 2x + 3y + 1 = 0 and 2x + 3y – 7 = 0.
Solution:
Given, the equations of the straight lines are
3x + 4y + 5 = 0 …….(1)
3x + 4y – 2 = 0 ………(2)
2x + 3y + 1 = 0 ……….(3)
2x + 3y – 7 = 0 ………..(4)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q37
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q37.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q37.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q37.3

Question 83.
Find the incentre of the triangle whose vertices are (1, √3), (2, 0), and (0, 0).
Solution:
Given, A (x1, y1) = (1, √3), B(x2, y2) = (2, 0), C(x3, y3) = (0, 0)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q38
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q38.1

Question 84.
Find the incentre of the triangle whose sides are x = 1, y = 1, and x + y = 1.
Solution:
Given, the equation of the straight lines are
x = 1 …….(1)
y = 1 ………(2)
x + y = 1 ……..(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q39
Vertex A:
Solving (1) & (3)
from (1), x = 1
from (3), x + y = 1
⇒ 1 + y = 1
⇒ y = 0
∴ Vertex A = (1, 0)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q39.1
Vertex B:
Solving (1) & (2)
from (1), x = 1
from (2), y = 1
∴ Vertex B = (1, 1)
Vertex C:
Solving (2) & (3)
from (2), y = 1
from (3), x + y = 1
⇒ x + 1 = 1
⇒ x = 0
∴ Vertex C = (0, 1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q39.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q39.3

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 85.
Find the values of k, if the angle between the straight lines kx + y + 9 = 0 and 3x – y + 4 = 0 is \(\frac{\pi}{4}\).
Solution:
Given equations are
kx + y + 9 = 0 ……..(1)
3x – y + 4 = 0 ………(2)
Comparing (1) with a1x + b1y + c1 = 0, we get
a1 = k, b1 = 1, c1 = 9
Comparing (2) with a2x + b2y + c2 = 0, we get
a2 = 3, b2 = -1, c2 = 4
Given that, θ = \(\frac{\pi}{4}\)
Let ‘θ’ is the angle between the lines (1) & (2) then
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q40
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q40.1
Squaring on both sides
⇒ (k2 + 1)(10) = 2(3k – 1)2
⇒ 10k2 + 10 = 18k2 + 2 – 12k
⇒ 8k2 – 12k – 8 = 0
⇒ 2k2 – 3k – 2 = 0
⇒ 2k2 – 4k + k – 2 = 0
⇒ (k – 2)(2k + 1) = 0
⇒ k = 2 or k = \(\frac{-1}{2}\)

Question 86.
Find the equation of the straight line passing through the origin and also through the point of intersection of the lines 2x – y + 5 = 0 and x + y + 1 = 0.
Solution:
Given equations of the straight lines are
2x – y + 5 = 0 ………(1)
x + y + 1 = 0 ………(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q41
The point of intersection of lines (1) & (2) is P(-2, 1).
The equation of the straight line passing through the point O(0, 0) & P(-2, 1) is (y – y1)(x2 – x1) = (x – x1)(y2 – y1)
⇒ (y – 0) (-2 – 0) = (x – 0) (1 – 0)
⇒ -2y = x
⇒ x + 2y = 0

Question 87.
Find the equation of the straight line perpendicular to the line 2x + 3y = 0 and pass through the point of intersection of the lines x + 3y – 1 = 0 and x – 2y + 4 = 0.
Solution:
Given equations of the straight lines are
x + 3y – 1 = 0 ……..(1)
x – 2y + 4 = 0 …….(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q42
∴ The point of intersection of lines (1) & (2) is P(-2, 1).
Given equation of the straight line 2x + 3y = 0 ……..(3)
Slope of line is m = \(\frac{-2}{3}\)
Since the required line is perpendicular to the line (3),
Slope of required line is \(\frac{-1}{m}=\frac{-1}{\frac{-2}{3}}=\frac{3}{2}\)
The equation of the straight line passing through P(-2, 1) & slope \(\frac{3}{2}\) is y – y1 = \(\frac{-1}{m}\) (x – x1)
⇒ y – 1 = \(\frac{3}{2}\) (x + 2)
⇒ 2y – 2 = 3x + 6
⇒ 3x – 2y + 8 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 88.
Find the value of ‘a’ if the distances of the points (2, 3) and (-4, a) from the straight line 3x + 4y – 8 = 0 are equal.
Solution:
Given the equation of the straight line is 3x + 4y – 8 = 0
Let the given points be P = (2, 3), Q (-4, a)
Now, the perpendicular distance from the point P(2, 3) to the straight line 3x + 4y – 8 = 0 is
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q43
The perpendicular distance from the point Q(-4, a) to the straight line 3x + 4y – 8 = 0 is,
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q43.1
Given that two perpendicular distances are equal then,
2 = \(\frac{|4 a-20|}{5}\)
⇒ |4a – 20| = 10
⇒ 2|2a – 10| = 10
⇒ 2a – 10 = ±5
⇒ 2a – 10 = 5; 2a – 10 = -5
⇒ 2a = 15; 2a = 5
⇒ a = \(\frac{15}{2}\); a = \(\frac{5}{2}\)
∴ a = \(\frac{15}{2}\) (or) \(\frac{5}{2}\)

Question 89.
Two adjacent sides of a parallelogram are given by 4x + 5y = 0 and 7x + 2y = 0 and one diagonal is 11x + 7y = 9. Find the equations of the remaining sides and the other diagonal.
Solution:
Given that, two adjacent sides of a parallelogram are given by
4x + 5y = 0 …….(1)
7x + 2y = 0 ………(2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q44
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q44.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q44.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q44.3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q44.4
⇒ -6y – 8 = 21x – 35
⇒ 21x + 6y – 27 = 0
⇒ 7x + 2y – 9 = 0
The equation of the diagonal \(\overline{\mathrm{BD}}\) is,
(y – y1)(x2 – x1) = (x – x1)(y2 – y1)
⇒ (y – 0)(1 – 0) = (x – 0)(1 – 0)
⇒ y(1) = x(1)
⇒ y = x
⇒ x – y = 0
∴ Two adjacent sides of a parallelogram are 4x + 5y – 9 = 0, 7x + 2y – 9 = 0.
The equation of one of the diagonals is x – y = 0.

Question 90.
Find the incentre of the triangle formed by the straight lines x + 1 = 0, 3x – 4y = 5, and 5x + 12y = 27.
Solution:
Given, the equation of the straight lines are
x + 1 = 0 ………(1)
3x – 4y = 5 ………..(2)
5x + 12y = 27 ………..(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q45
Vertex A:
Solving (1) & (3)
from (1), x + 1 = 0
⇒ x = -1
from (3), 5(-1) + 12y = 27
⇒ -5 + 12y = 27
⇒ 12y = 32
⇒ y = \(\frac{32}{12}\) = \(\frac{8}{3}\)
∴ Vertex A = (-1, \(\frac{8}{3}\))
Vertex B:
Solving (1) & (2)
from (1), x + 1 = 0
⇒ x = -1
from (2), 3(-1) – 4y = 5
⇒ -3 – 4y = 5
⇒ -4y = 8
⇒ y = -2
∴ Vertex B = (-1, -2)
Vertex C:
Solving (2) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q45.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q45.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q45.3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q45.4

Question 91.
Find the equation of the straight line parallel to the line 3x + 4y = 7 and passing through the point of intersecting of the lines x – 2y – 3 = 0, x + 3y – 6 = 0. [Mar. ’16 (TS)]
Solution:
Given, the equation of the straight lines are
x – 2y – 3 = 0 …….(1)
x – 3y – 6 = 0 …….(2)
Solving (1) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q46
∴ The point of intersection of lines (1) & (2) is \(\mathrm{P}\left(\frac{21}{5}, \frac{3}{5}\right)\)
Given equation of the straight line is 3x + 4y = 7 …….(3)
Slope of line is m = \(\frac{-3}{4}\)
Since the required line is parallel to line 3.
Slope of required line = m = \(\frac{-3}{4}\)
∴ The equation of the straight line passing through \(\mathrm{P}\left(\frac{21}{5}, \frac{3}{5}\right)\) & Slope = \(\frac{-3}{4}\) is y – y1 = m(x – x1)
⇒ \(y-\frac{-3}{5}=\frac{-3}{4}\left(x-\frac{21}{5}\right)\)
⇒ \(\frac{5 y-3}{5}=\frac{-3}{4}\left(\frac{5 x-21}{5}\right)\)
⇒ 5y – 3 = \(\frac{-3}{4}\)(5x – 21)
⇒ 20y – 12 = -15x + 63
⇒ 15x + 20y – 75 = 0
⇒ 3x + 4y – 15 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 92.
Find the value of p, if the straight lines 3x + py – 1 = 0, 7x – 3y + 3 = 0, are mutually perpendicular. [Mar. ’19 (TS)]
Solution:
Given lines are
3x + py – 1 = 0 ……..(1)
7x – 3y + 3 = 0 ………(2)
Slope of (1) is m1 = \(\frac{-3}{p}\)
Slope of (2) is m2 = \(\frac{-7}{-3}\) = \(\frac{7}{3}\)
Since (1) & (2) are perpendicular then
m1 × m2 = -1
⇒ \(\frac{-3}{p} \times \frac{7}{3}=-1\)
⇒ \(\frac{-7}{p}\) = -1
⇒ p = 7

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Students must practice these Maths 1B Important Questions TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 1.
Find the condition for the points (a, 0), (h, k), and (0, b), where ab ≠ 0, to be collinear. [Mar. ’10]
Solution:
Let A(a, 0), B(h, k), C (0, b) be the given points.
Slope of \(\overline{\mathrm{AB}}=\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{k}-0}{\mathrm{~h}-\mathrm{a}}=\frac{\mathrm{k}}{\mathrm{h}-\mathrm{a}}\)
Slope of \(\overline{\mathrm{BC}}=\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{b}-\mathrm{k}}{0-\mathrm{h}}=\frac{\mathrm{b}-\mathrm{k}}{-\mathrm{h}}\)
Since points A, B, C are collinear, then
Slope of \(\overline{\mathrm{AB}}\) = Slope of \(\overline{\mathrm{BC}}\)
\(\frac{k}{h-a}=\frac{b-k}{-h}\)
⇒ -hk = (h – a) (b – k)
⇒ -hk = bh – hk – ab + ak
⇒ bh + ak = ab
⇒ \(\frac{b h}{a b}+\frac{a k}{a b}=1\)
\(\frac{h}{a}+\frac{k}{b}=1\), which is the required condition.

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 2.
Find the value of x, if the slope of the line passing through (2, 5) and (x, 3) is 2. [Mar. ’18 (AP & TS); May ’12; B.P.]
Solution:
Let A = (2, 5), B = (x, 3) are the given points.
Given, Slope of \(\overline{\mathrm{AB}}\) = 2
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q2

Question 3.
Find the value of y, if the line joining the points (3, y) and (2, 7) is parallel to the line joining the points (-1, 4) and (0, 6). [Mar. ’17 (TS), ’14, ’08]
Solution:
Let A = (3, y), B = (2, 7), C = (-1, 4), and D = (0, 6) are the given points.
Given,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q3
Since, the lines \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) are parallel then the
Slope of \(\overline{\mathrm{AB}}\) = Slope of \(\overline{\mathrm{CD}}\)
⇒ y – 7 = 2
⇒ y = 9

Question 4.
Find the equation of the straight line which make 150° with the X-axis in the positive direction and which pass through the point (-2, -1). [May ’04]
Solution:
Given that inclination of a straight line is θ = 150°
The slope of a line is, m = tan θ
= tan (150°)
= tan (90° + 60°)
= -cot 60°
= \(\frac{-1}{\sqrt{3}}\)
Let the given point A(x1, y1) is (-2, -1).
∴ The equation of the straight line passing through A(-2, -1) and having slope \(\frac{-1}{\sqrt{3}}\) is y – y1 = m(x – x1)
⇒ y + 1 = latex]\frac{-1}{\sqrt{3}}[/latex] (x + 2)
⇒ √3(y + 1) = -1(x + 2)
⇒ √3y + √3 = -x – 2
⇒ x + √3y + √3 = -2
⇒ x + √3y + √3 + 2 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 5.
Find the equations of the straight lines passing through the origin and making equal angles with the coordinate axes. [May ’05]
Solution:
Let l1, l2 are the equations of the straight lines passing through the origin and making equal angles with the co-ordinate axes
Case I: Inclination of a straight line l1 is θ = 45°
Slope of a line l1 is, m = tan θ = tan 45° = 1
Let the given point O = (0, 0)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q5
∴ The equation of a straight line l1 passing through O(0, 0) and having slope ‘1’ is y – y1 = m(x – x1)
y – 0 = 1(x – 0)
⇒ y = x
⇒ x – y = 0
Case II: Inclination of a straight line l2 is θ = 135°
The slope of a line l2 is, m = tan θ
= tan 135°
= tan (90° + 45°)
= -tan 45°
= -1
Let the given point O = (0, 0)
∴ The equation of a straight line l2 passing through O(0, 0) and having slope ‘-1’ is y – y1 = m(x – x1)
⇒ y – 0 = -1(x – 0)
⇒ y = -x
⇒ x + y = 0
∴ Required equations of the straight lines are x – y = 0; x + y = 0

Question 6.
Find the equation of the straight line passing through (-4, 5) and cutting off equal and non-zero intercepts on the coordinate axes. [Mar. ’15 (TS) ’13 (Old), ’07, ’00; May ’10, ’08]
Solution:
The equation of a straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) …….(1)
Given that, the straight line making equal intercepts on the co-ordinate axis, then a = b
From (1),
\(\frac{x}{a}+\frac{y}{a}=1\)
x + y = a ……….(2)
Since equation (2) passes through the point (-4, 5) then,
-4 + 5 = a
∴ a = 1
Substitute the value of ’a’ in equation (2)
∴ x + y = 1

Question 7.
Find the equation of the straight line passing through (-2, 4) and making non-zero intercepts whose sum is zero. [Mar. ’15 (AP), ’13; May ’15 (TS), ’02]
Solution:
The equation of a straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) ……..(1)
Given that, the straight line-making intercepts whose sum is ‘0’
i.e., a + b = 0
b = -a
From (1)
\(\frac{x}{a}+\frac{y}{-a}=1\)
x – y = -a …….(2)
Since equation (2) passes through the point (-2, 4) then,
-2 – 4 = a
∴ a = -6
Substitute the value of ‘a’ in equation (2)
x – y = -6
x – y + 6 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 8.
Find the equation of the straight line passing through the point (3, -4) and making X and Y-intercepts which are in the ratio 2 : 3. [Mar. ’08]
Solution:
The equation of a straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) ……..(1)
Given that, the ratio of intercepts = 2 : 3
X-intercept = 2a
Y-intercept = 3a
From (1),
\(\frac{x}{2 a}+\frac{y}{3 a}=1\)
\(\frac{3 x+2 y}{6 a}=1\)
3x + 2y = 6a ……..(2)
Since equation (2) passes through point (3, -4) then,
3(3) + 2(-4)= 6a
⇒ 9 – 8 = 6a
⇒ 6a = 1
⇒ a = \(\frac{1}{6}\)
Substitute the value of ‘a’ in equation (2)
3x + 2y = 6(\(\frac{1}{6}\))
∴ 3x + 2y = 1

Question 9.
Find the equation of the straight line passing through the points \(\left(a t_1^2, 2 at_1\right)\) and \(\left(a t_2^2, 2 at_2\right)\). [Mar. ’14. ’04; May ’15 (AP), ’00]
Solution:
Let A\(\left(a t_1^2, 2 at_1\right)\) and B\(\left(a t_2^2, 2 at_2\right)\) are the given points
The equation of the straight line passing through the points A\(\left(a t_1^2, 2 at_1\right)\) and B\(\left(a t_2^2, 2 at_2\right)\) is
(y – y1) (x2 – x1) = (x – x1) (y2 – y1)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q9

Question 10.
Find the equation of the straight line passing through A(-1, 3) and (i) parallel (ii) perpendicular to the straight line passing through B(2, -5) and C(4, 6). [May ’12; Mar. ’11]
Solution:
A(-1, 3), B(2, -5), C(4, 6) are the given points.
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q10
(i) Slope of the parallel line = m = \(\frac{11}{2}\)
∴ The equation of the straight line passing through A(-1, 3) and having slope \(\frac{11}{2}\) is y – y1 = m(x – x1)
y – 3 = \(\frac{11}{2}\)(x + 1)
⇒ 2y – 6 = 11x + 11
⇒ 11x – 2y + 17 = 0
(ii) Slope of the perpendicular line = \(\frac{-1}{m}=\frac{-1}{11 / 2}=\frac{-2}{11}\)
∴ The equation of the straight line passing through A(-1, 3) and having slope \(\frac{-2}{11}\) is y – y1 = \(\frac{-1}{m}\) (x – x1)
y – 3 = \(\frac{-2}{11}\) (x + 1)
⇒ 11y – 33 = -2x – 2
⇒ 2x + 11y – 31 = 0

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 11.
A(10, 4), B(-4, 9) and C(-2, -1) are the vertices of a triangle. Find the equation of the altitude through B. [May ’13 (Old)]
Solution:
Slope of \(\overline{\mathrm{AC}}\) is,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q11
The equation of the altitude through B is, the equation of the straight line passing through B(-4, 9) and having slope \(\frac{-12}{5}\) is
y – y1 = m(x – x1)
y – 9 = \(\frac{-12}{5}\) (x + 4)
5y – 45 = -12x – 48
12x + 5y + 3 = 0

Question 12.
If the portion of a straight line intercepted between the axes of coordinates is bisected at (2p, 2q), write the equation of the straight line. [May ’90]
Solution:
Let a, b be the intercepts of a line.
∴ The line cuts the X-axis at A(a, 0), Y-axis at B(0, b)
Midpoint of \(\overline{\mathrm{AB}}\) is,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q12
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q12.1

Question 13.
Find the angle made by the straight line y = -√3x + 3 with the positive direction of the X-axis measured in the counterclockwise direction. [May ’94]
Solution:
Given, the equation of the straight line is y = -√3x + 3
Comparing this equation with y = mx + c, We get
m = -√3 (∵ m = tan θ)
⇒ tan θ = -√3
⇒ tan θ = tan \(\frac{2 \pi}{3}\)
⇒ θ = \(\frac{2 \pi}{3}\)
∴ The angle made by the straight line is θ = \(\frac{2 \pi}{3}\)

Question 14.
Transform the equation √3x + y = 4 into (i) slope-intercept form (ii) intercept form (iii) normal form. [May ’16 (TS)]
Solution:
Given, the equation of the straight line is √3x + y = 4
(a) Slope-intercept form:
√3x + y = 4
y = -√3x + 4 which is of the form y = mx + c
where, slope (m) = -√3, y-intercept (c) = 4
(b) Intercept form:
Given, the equation of the straight line is,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q14
which is in the form \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept (a) = \(\frac{4}{\sqrt{3}}\), y-intercept (b) = 4
(c) Normal form:
Given the equation of the straight line is √3x + y = 4
On dividing both sides with
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q14.1
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q14.2
which is in the form x cos α + y sin α = p
∴ α = \(\frac{\pi}{6}\), p = 2

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 15.
Transform the equation x + y + 1 = 0 into normal form. [Mar. ’17 (AP), ’08; May ’10; B.P.; Mar. ’18 (TS)]
Solution:
Given, equation of the straight line is x + y + 1 = 0
x + y = -1
– x – y = 1
On dividing both sides with
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q15

Question 16.
Transform the equation 4x – 3y + 12 = 0 into (i) slope-intercept form (ii) intercept form (iii) normal form.
Solution:
(a) Slope-intercept form:
Given equation of the line is 4x – 3y + 12 = 0
3y = 4x + 12
y = \(\frac{4 x+12}{3}=\left(\frac{4}{3}\right) x+4\)
which is in the form of y = mx + c
∴ Slope = \(\frac{4}{3}\), y-intercept = 4
(b) Intercept form:
Given equation is 4x – 3y + 12 = 0
\(\frac{x}{-3}+\frac{y}{4}=1\)
which is in the form of \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept = -3, y-intercept = 4
(c) Normal form:
Given the equation of the straight line is 4x – 3y = -12
-4x + 3y = 12
On dividing both sides by \(\sqrt{a^2+b^2}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q16

Question 17.
Transform the equation x + y – 2 = 0 into (i) slope-intercept form (ii) intercept form (iii) normal form. [Mar. ’12]
Solution:
(a) Slope-intercept form:
Given equation of the straight line is, x + y – 2 = 0
y = -x + 2
which is in the form of y = mx + c
∴ Slope, m = -1, y-intercept, c = 2
(b) Intercept form:
Given equation of the straight line is, x + y – 2 = 0
x + y = 2
\(\frac{x+y}{2}\) = 1
\(\frac{x}{2}+\frac{y}{2}\) = 1
which is of the form \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept (a) = 2, y-intercept (b) = 2
(c) Normal form:
Given equation of the straight line is, x + y – 2 = 0
x + y = 2
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q17
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q17.1

Question 18.
Transform the equation √3x + y + 10 = 0 into (i) slope-intercept form (ii) intercept form (iii) normal form. [May ’04]
Solution:
(a) Slope-intercept form:
Given the equation of the straight line is, √3x + y + 10 = 0
y = -√3x – 10 = -√3x + (-10)
which is in the form of y = mx + c
∴ Slope, m = -√3, y-intercept, c = -10
(b) Intercept form:
Given the equation of the straight line is √3x + y + 10 = 0
√3x + y = -10 × 1
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q18
(c) Normal form:
Given equation of the straight line is, √3x + y + 10 = 0
√3x + y = -10
-√3x – y = 10
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q18.1
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q18.2

Question 19.
If the area of the triangle is formed by the straight lines, x = 0, y = 0, and 3x + 4y = a [a > 0] is ‘6’. Find the value of ‘a’. [May ’11, Mar. ’09, ’07]
Solution:
Given equations of the straight lines are (a > 0) 3x + 4y = a, x = 0 and y = 0
Comparing with ax + by + c = 0, we get
a = 3, b = 4, c = -a
The area of the triangle formed by this line and the co-ordinate axis is equal to \(\frac{c^2}{2|a b|}\)
Given that area of the triangle = 6
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q19

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 20.
Find the value of p, if the straight lines x + p = 0, y + 2 = 0 and 3x + 2y + 5 = 0 are concurrent. [Mar. ’17 (TS), ’13; May ’15 (TS)]
Solution:
Given, the equation of the straight lines
x + p = 0 ……..(1)
y + 2 = 0 ………(2)
3x + 2y + 5 = 0 ………(3)
Solving (2) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q20
∴ Point of intersection of the lines (2) & (3) is (\(\frac{-1}{3}\), -2)
since given lines are concurrent, then, the point of intersection (\(\frac{-1}{3}\), -2) lies on (1)
x + p = 0
⇒ \(\frac{-1}{3}\) + p = 0
⇒ p = \(\frac{1}{3}\)

Question 21.
Find the ratio in which the straight line 2x + 3y = 5 divides the line joining the points (0, 0) and (-2, 1). [Mar. ’14]
Solution:
Given the equation of the straight line is L = 2x + 3y – 5 = 0
Comparing the equation with ax + by + c = 0, we get
a = 2, b = 3, c = 5
Let the given points are A(x1, y1) = (0, 0) and B(x2, y2) = (-2, 1)
Required ratio = \(\frac{-\left(a x_1+b {y}_1+c\right)}{a x_2+b y_2+c}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q21

Question 22.
Find the distance between the parallel straight lines 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0. [Mar. ’19 (AP); May ’13]
Solution:
Given, equations of the straight lines are 3x + 4y – 3 = 0, 6x + 8y – 1 = 0
6x + 8y – 6 = 0 …..(1)
6x + 8y – 1 = 0 …..(2)
Comparing (1) with ax + by + c1 = 0, we get
a = 6, b = 8, c1 = -6
Comparing (2) with ax + by + c2 = 0, we get
a = 6, b = 8, c2 = -1
Distance between the parallel lines =
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q22

Question 23.
Find the equation of ‘k’, if the angle between the straight lines 4x – y + 7 = 0 and kx – 5y – 9 = 0 is 45°. [Mar. ’12, ’08, ’82; May ’11, ’02]
Solution:
Given, the equations of the straight lines are
4x – y + 7 = 0 ……..(1)
kx – 5y – 9 = 0 ……..(2)
Comparing (1) with a1x + b1y + c1 = 0, we get
a1 = 4, b1 = -1, c1 = 7
Comparing (2) with a2x + b2y + c2 = 0, we get
a2 = k, b2 = -5, c2 = -9
Given that, θ = 45°
If ‘θ’ is the angle between the given lines then,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q23
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q23.1
Squaring on both sides
⇒ 17(k2 + 25) = 2(4k + 5)2
⇒ 17k2 + 425 = 2(16k2 + 25 + 40k)
⇒ 17k2 + 425 = 32k2 + 50 + 80k
⇒ 15k2 + 80k – 375 = 0
⇒ 3k2 + 16k – 375 = 0
⇒ 3k2 + 25k – 9k – 75 = 0
⇒ k(3k + 25) – 3(3k + 25) = 0
⇒ (3k + 25) (k – 3) = 0
⇒ 3k + 25 = 0; k – 3 = 0
⇒ 3k = -25; k = 3
⇒ k = 3 or \(\frac{-25}{3}\)

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 24.
Find the equation of the straight line parallel to the line 2x + 3y + 7 = 0 and pass through the point (5, 4). [Mar. ’13, ’03]
Solution:
Given, the equation of the straight line is 2x + 3y + 7 = 0
Given points (5, 4)
The equation of the straight line parallel to 2x + 3y + 7 = 0 is 2x + 3y + k = 0
Since equation (1) passes through the point (5, 4) then,
2(5) + 3(4) + k = 0
⇒ 10 + 12 + k = 0
⇒ 22 + k = 0
⇒ k = -22
∴ The required equation of the straight line is 2x + 3y – 22 = 0

Question 25.
Find the value of k, if the straight lines y – 3kx + 4 = 0 and (2k – 1)x – (8k – 1)y – 6 = 0 are perpendicular. [Mar. ’10]
Solution:
Given, the equations of the straight lines are
y – 3kx + 4 = 0 ………(1)
(2k – 1)x – (8k – 1)y – 6 = 0 ……….(2)
Slope of the line (1) is m1 = \(\frac{-(-3 k)}{1}\) = 3k
Slope of the line (2) is m2 = \(\frac{-(2 k-1)}{-(8 k-1)}=\frac{(2 k-1)}{(8 k-1)}\)
Since the given lines are perpendicular then m1 × m2 = -1
\(3 \mathrm{k}\left(\frac{2 \mathrm{k}-1}{8 \mathrm{k}-1}\right)\) = -1
⇒ 3k(2k – 1) = -1(8k – 1)
⇒ 6k2 – 3k = -8k + 1
⇒ 6k2 + 5k – 1 = 0
⇒ 6k2 + 6k – k – 1 = 0
⇒ 6k(k + 1) – 1(k + 1) = 0
⇒ (k + 1)(6k – 1) = 0
⇒ k + 1 = 0 (or) 6k – 1 = 0
⇒ k = -1 (or) k = \(\frac{1}{6}\)

Question 26.
Find the perpendicular distance from the point (3, 4) to the straight line 3x – 4y + 10 = 0. [Mar. ’16 (AP); May. ’15 (AP)]
Solution:
Given the equation of the straight line is 3x – 4y + 10 = 0
Comparing with ax + by + c = 0, we get
a = 3; b = -4, c = 40
Let the given point p(x1, y1) = (3, 4)
The perpendicular distance from the point (3, 4) to the line 3x – 4y + 10 = 0 is
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q26

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 27.
Find the slopes of the lines x + y = 0 and x – y = 0. [Mar. ’17 (AP)]
Solution:
Slope of x + y = 0 is \(\frac{-a}{b}=\frac{-1}{1}\) = -1
Slope of x – y = 0 is \(\frac{-a}{b}=\frac{-1}{-1}\) = 1

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 9 Hyperbolic Functions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions

I.

Question 1.
Show that \(t \tanh ^{-1}\left(\frac{1}{2}\right)=\frac{1}{2} \log _e^3\)
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 1

Question 2.
If cosh \(x=\frac{5}{2}\) find the values of
(i) cosh (2x)
(ii) sinh (2x)
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 2

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions width=

Question 3.
Prove that for any x ∈ R sinh (3x) =3 sinh (3x) + 3sinh x +4sin3h3 x.
Solution:
LHS : sinh(3x) = sinh (2x + x)
= sinh (2x) cosh x + cosh (2x) sinh x
= (2sinh x cosh x) cosh x + (1 + 2sinh2 x) slnh x
= 2 sinh x cosh2 x + sinh x + 2sinh3 x
= 2sinh x (1 + sinh2 x) + sinh x + 2sinh3 x
= 3sinh x + 4sinh3 x = R.H.S.

Question 4.
Prove that for any x∈R
\(\tanh 3 x=\frac{3 \tanh x+\tanh ^3 x}{1+3 \tanh ^2 x}\)
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 3

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions width=

Question 5.
If cosh x = secθ then prove that \(\tanh ^2\left(\frac{x}{2}\right)=\tan ^2\left(\frac{\theta}{2}\right)\)
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 4
Question 6.
sinhx = 5 then show that \(x=\log _e(5+\sqrt{26})\)
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 5

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions width=

II.

Question 1.
\(\theta \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\) and \(x=\log _{\mathrm{e}}\left(\cot \left(\frac{\pi}{4}+\theta\right)\right)\) that prove that
(i) cosh x = sec2θ
(ii) sin hx= – tan2θ.
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 6
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 7

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions width=

Question 2.
Draw the graph of y = sinh x.
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 8

Question 3.
Draw the graph of y = cosh x.
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 9