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These TS 10th Class Maths Chapter Wise Important Questions Chapter 8 Similar Triangles given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 8 Similar Triangles

Previous Exams Questions

Question 1.
Construct a triangle of sides 4.2 cm, 5.1 cm and 6 cm. Then construct a triangle similar to it, whose sides are \(\frac{2}{3}\) of corresponding sides of the first triangle. (A.P. Mar. ’15)
Solution:

1. Draw a triangle ABC, with sides AB = 4.2 cm, BC = 5.1 cm, CA = 6 cm.
2. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
3. Locate 3 points B₁, B₂, B₃ on BX so that BB₁ = B₁B₂ = B₂B₃
4. Join B₃, C and draw a line through B₂ parallel to B₃C intersecting BC and C¹.
5. Draw a line through C¹ parallel to CA intersect AB at A¹.
6. ∆A¹BC¹ is required triangle.

Question 2.
Is a square similar to a rectangle ? Justify your answer. (A.P. Mar.’15)
Solution:
In a square and rectangle, the corresponding angles are equal. But the corresponding sides are not proportional.

∴ A square and a rectangle are not similar.

Question 3.
In a ∆DEF, A, B and C are mid points of EF, FD and DE respectively. If the area of ∆DEF is 14.4 cm² then find the area of ∆ABC. (T.S. Mar. ’15)
Solution:
Area of ∆ABC = 1/4 of area of ∆DEF
= 1/4 (14.4) = 3.6 cm²

Question 4.
Observe the below diagram and find the values of x and y. (T.S. Mar. ’15)

Solution:
∆ABC ~ AEFC
\(\frac{\mathrm{x}}{6}\) = \(\frac{9}{15}\)
x = \(\frac{9 \times 6}{15}\) = \(\frac{18}{5}\) = 3.6 cm.
∆BDC ~ ABEF
\(\frac{\mathrm{x}}{9.6}\) = \(\frac{\mathrm{y}}{15}\)
⇒ y = \(\frac{3.6 \times 15}{9.6}\) = \(\frac{45}{8}\) = 5.625 cm

Question 5.
Observe the figure given below in ∆FQR if XY // PQ, \(\frac{\mathrm{PX}}{\mathrm{XR}}\) = \(\frac{5}{3}\) and QR = 7.2. Then find the length of RY. (T.S. Mar. ’15)

Solution:
XY || PQ
⇒ \(\frac{\mathrm{PX}}{\mathrm{XR}}\) = \(\frac{\mathrm{QY}}{\mathrm{YR}}\)
⇒ \(\frac{5}{3}\) + 1 = \(\frac{\mathrm{QY}}{\mathrm{RY}}\) + 1
⇒ \(\frac{8}{3}\) = \(\frac{\mathrm{QR}}{\mathrm{RY}}\)
⇒ RY = 7.2 × \(\frac{3}{8}\) = 2.7 cm.

Question 6.
Find the value of ‘x’ in the given figure where ∆ ABC ~ ∆ ADE. (T.S. Mar. ’15)

Solution:
∆ABC ~ ∆ADE
∴ \(\frac{\mathrm{AB}}{\mathrm{AD}}\) = \(\frac{\mathrm{BC}}{\mathrm{DE}}\) = \(\frac{\mathrm{AC}}{\mathrm{AE}}\) ;
BC = x, DE = 5, AE = 3, AC = 9
By substituting \(\frac{\mathrm{BC}}{\mathrm{DE}}\) = \(\frac{\mathrm{AC}}{\mathrm{AE}}\)
∴ \(\frac{\mathrm{x}}{5}\) = \(\frac{9}{3}\) ⇒ x = \(\frac{9 \times 5}{3}\) = 15
∴ x = 15 cm

Question 7.
Construct a triangle of sides 5 cm, 6 cm, and 7 cm then construct a triangle similar to it, whose sides are \(\frac{2}{3}\) of the corresponding sides of the triangle. (T.S. Mar. ’15)

Solution:
Construction steps :

1. Draw a triangle ∆ABC with sides AB = 5 cm, BC = 6 cm and CA = 7 cm.
2. Draw a ray \(\overrightarrow{\mathrm{BX}}\) making an acute angle with BC on the side opposite to vertex A.
3. Locate 3 points B₁, B₂, B₃ on BX. So that BB₁ = B₁B₂ = B₂B₃.
4. Join B₃, C and draw a line through B₂ parallel to B₃ C intersecting BC at C’.
5. Draw a line through C’ parallel to CA intersect AB at A’.
6. ∆A’B’C’ is required triangle.

Additional Questions

Question 1.
In a ∆ ABC, DE || BC and AD = \(\frac{1}{3}\) BD. If BC = 5.6 cm, find DE.
Solution:
Given DE || BC, AD = \(\frac{1}{3}\) BD, BC = 5.6 cm

AD = \(\frac{1}{3}\) BD ⇒ BD = 3 AD …………… (1)
From similar triangles, ABC and ADE
We have \(\frac{\mathrm{AD}}{\mathrm{AB}}\) = \(\frac{\mathrm{DE}}{\mathrm{BC}}\)
⇒ \(\frac{\mathrm{AD}}{\mathrm{AD}+\mathrm{DB}}\) = \(\frac{\mathrm{DE}}{5.6}\)
⇒ \(\frac{\mathrm{AD}}{\mathrm{AD}+3\mathrm{DB}}=\) = \(\frac{\mathrm{DE}}{5.6}\)
[∵ from (1) DB = BD = 3AD]
⇒ \(\frac{\mathrm{AD}}{4 \mathrm{AD}}\) = \(\frac{\mathrm{DE}}{5.6}\)
⇒ \(\frac{1}{4}\) = \(\frac{\mathrm{DE}}{5.6}\)
⇒ DE = \(\frac{5.6}}{4\)
∴ DE = 1.4 cm

Question 2.
In the adjacent figure ∆ABC ~ ∆AHK. If AK = 8 cm, BC = 4.5 cm and HK = 9 cm find AC.
Solution:

Given AK = 8 cm, BC 4.5 cm, HK from similar ∆les, ABC and AHK
We have \(\frac{\mathrm{AC}}{\mathrm{AK}}\) = \(\frac{\mathrm{BC}}{\mathrm{HK}}\)
AC = \(\frac{\mathrm{AK} \times \mathrm{BC}}{\mathrm{HK}}\)
= \(\frac{8 \times 4.5}{9}\)
= 8 × 0.5 = 4
∴ AC = 4 cm

Question 3.
In the below given figure P and Q are points on the sides AB and AC respectively of ∆ABC such that AQ = 3 cm, QC = 5 cm and PQ || BC. Find the ratio of areas of ∆APQ and ∆ABC.

Solution:
Given AQ = 3 cm, QC = 5 cm and PQ || BC.
We know that by theorem,
Ratio of the areas of two similar triangles = Ratio of the squares of their corresponding sides
Now AC = AQ + QC = 3 + 5 = 8 cm
By Theorem \(\frac{\text { area of } \triangle \mathrm{APQ}}{\text { area of } \triangle \mathrm{ABC}}\) = \(\frac{(\mathrm{AQ})^2}{(\mathrm{AC})^2}\)
= \(\frac{3^2}{8^2}\)
= \(\frac{9}{64}\)
∴ Ratio of areas of ∆APQ and ∆ABC = 9 : 64

Question 4.
What value of (5) of x will make DE || AB, in the given figure ?
AD = 5x + 5, CD = x + 2 BE = 3x + 3, CE = x

Solution:
Given in ∆ ABC, DE || AB
AD = 5x + 5, CD = x + 2
BE = 3x + 3, CE = x
y Basic proportional theorem,
If DE || AB, then we have
\(\frac{\mathrm{CD}}{\mathrm{DA}}\) = \(\frac{\mathrm{CE}}{\mathrm{EB}}\)
⇒ \(\frac{x+2}{5 x+5}\) = \(\frac{x}{3 x+3}\)
⇒ (5x + 5) x = (x +2) (3x + 3)
⇒ 5x² + 5x = 3x² + 3x + 6x + 6
⇒ 5x² – 3x² + 5x – 3x – 6x – 6 = 0
⇒ 2x² – 4x – 6 = 0
⇒ 2x² – 6x + 2x – 6 = 0
⇒ 2x (x – 3) + 2(x – 3) = 0
⇒ (x – 3) + (2x + 2) = 0
⇒ x-3 = 0 or 2x + 2 = 0
⇒ x = 3 or 2x = -2
⇒ x = –\(\frac{2}{2}\) = -1
∴ The value of x = 3 will make DE || AB.

Question 5.
∆ABC ~ ∆PQR and their areas are respectively 81 cm² and 144 cm². If QR = 16cm then find BC.
Solution:
\(\frac{\text { area of } \triangle \mathrm{ABC}}{\text { area of } \triangle \mathrm{DEF}}\) = \(\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)^2\)
⇒ \(\frac{81}{144}\) = \(\left(\frac{\mathrm{BC}}{16}\right)^2\)
⇒ \(\frac{9}{12}\) = \(\frac{\mathrm{BC}}{16}\)
⇒ BC = \(\frac{9}{12}\) × 16 = \(\frac{144}{12}\) = 12
∴ BC = 12 cm.

Question 6.
∆ABC ~ ∆DEF, BC = 5 cm, EF = 6 cm and area of ∆DEF = 72 cm². Determine the area of ∆ABC.
Solution:
Given ∆ABC ~ ∆DEF
and BC = 5cm, EF = 6cm
Area of ∆DEF = 72 cm²
We know that areas of two similar triangles are in the ratios of the squares of the corresponding sides.

Question 7.
A ladder 13m long reaches a window of building 12cm above the ground. Determine the distance of the foot of the ladder from the building.
Solution:
In ∆ ABC, B = ∠90°
⇒ AC² = AB² + BC² (By Pythagoras theorem)
Let AC = 13m, AB = 12m, BC = ?

⇒ 13² = 12² + BC²
⇒ 169 = 144 + BC²
⇒ BC² = 169 – 144 = 25
⇒ BC = \(\sqrt{25}\) = 5m
Hence, the foot of the ladder from the building is at a distance of 5m.

Question 8.
The hypotenuse of a right angled triangle is 3 m more than twice of the shortest side. If the third side is 1 m less than the hypotenuse find the sides of the triangle.
Solution:

Let the shortest side = x m = BC
Then hypotenuse = 2x + 3 m = AC
The third side = (2x + 3) – 1
= (2x + 2) m = AB
By phythagoras theorem, we have
AC² = AB² + BC²
⇒ (2x + 3)² = (2x + 2)² + x²
⇒ 4x² + 9 + 12x = 4x² + 4 + 8x + x²
⇒ x² + 8x – 12x + 4 – 9 = 0
⇒ x² – 4x – 5 = 0
⇒ x² – 5x + x – 5 = 0
⇒ x (x – 5) + 1 (x – 5) = 0
⇒ (x – 5) (x + 1) = 0
⇒ x – 5 = 0 or x + 1 = 0
⇒ x = 5 or x = -1
But x can’t be negative x = 5
Hence, the sides of the triangle are
x, 2x + 3, 2x + 2
⇒ x = 5, 2 × 5 + 3, 2 × 5 + 2
= 5, 10 + 3, 10 + 2
i.e. 5, 13, 12

Question 9.
A ladder 25 m long reaches a window which is 15 m above the ground an one side of the road. Keeping its foot at the same point, the ladder is turned to other side of the road to reach a window 20 m high. Find the width of the road.
Solution:

Let A and D be the windows on the either sides of the road
From phythagoras theorem,
AC² = AB² + BC²
⇒ 25² = 15² + BC²
⇒ BC² = 25² – 15² = 625 – 225 = 400
BC = \(\sqrt{400}\) = 20m ……………. (1)
Also CD² = CE² + DE²
25² = CE² + 20²
CE² = 25² – 20² = 625 – 400 = 225
CE = \(\sqrt{225}\) = 15m
Width of the road = BE = BC + CE
= 20 + 15
= 35 m

Question 10.
In the given figure below, If AD ⊥ BC, prove that AB² – BD² = AC² – CD².

Solution:
Given in ∆ ABC, AD ⊥ BC
R.T.P : AB² – BD² = AC² – CD²
Proof : ∆ ABD is a right angled triangle.
We have AB² = AD² + BD² (By pythagaros theorem)
⇒ AB² – BD² = AD² ……………… (1)
∆ ACD is a right angled triangle.
We have AC² = AD² + CD²
⇒ AC² – CD² = AD² ………………. (2)
From (1) and (2), we have
AB² – BD² = AC² – CD²

Question 11.
In an equilateral triangle ABC, if AD is the altitude prove that 3AB² = 4 AD².
Solution:

Given ABC is an equilateral triangle.
Here, AD ⊥ BC and AB = BC = CA
In triangles ADB and ADC
∠ADB = ∠ADC = 90°
Hypotenuse AB = Hypotenuse AC
AD is common
∴ ∆ ADB ≅ ∆ ADC
∴ BD = DC
In ∆ ADB, ∠ADB = 90°(By Phythagoras theorem)
AB² = AD² + BD²
= AD² + \(\left(\frac{\mathrm{BC}}{2}\right)^2\) (∵ BD = CD)
= AD² + \(\frac{\mathrm{BC}^2}{4}\)
AB² = \(\frac{4 A D^2+B C^2}{4}\)
⇒ 4AB² = 4AD² + BC²
⇒ 4AB² – BC² = 4AD²
⇒ 4AB² – AB² = 4AD² (∵ BC = AB)
∴ 3AB² = 4AD²

Question 12.
A wire attached to a vertical pole of height 15 m is 25 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?
Solution:

Let AB = Vertical pole
AC = Wire
B = Base
C = Stake
By Pythagoras theorem
AC² = AB² + BC²
25² = 15² + BC²
BC² = 25² – 15²
BC² = 62² – 225 = 400
BC = \(\sqrt{400}\) = 20 m
∴ The stake should be driven 20 m for the base of the pole so that the wire will be taut.

Question 13.
The larger of two complimentary angles is double the smaller. Find the angles.
Solution:
Let first complementary angle = x
Second complementary angle = 2x
Sum of the two complementary angles = 90°
∴ x + 2x = 90
3x = 90
x = \(\frac{90}{3}\)
x = 30
First angle = 30°, second angle = 60°.

Question 14.
In ∆ ABC, DE || BC and \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{3}{5}\). If AE = 2.1 cm, then find AC.
Solution:
Here, Given DE || BC and \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{3}{5}\)
AE = 2.1
∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{CE}}\)
\(\frac{3}{5}\) = \(\frac{2.1}{\mathrm{CE}}\)

CE = \(\frac{5 \times 2.1}{3}\)
CE = \(\frac{10.5}{3}\) = 3.5
CE = 3.5
Then AC = AE + CE
AC = 2.1 + 3.5
AC = 5.6

Question 15.
What can you say about the ratio of areas of two similar triangles ?
Solution:
The ratio of areas of two similar triangles is( equal to the ratio of the squares of their corresponding sides.
Here

Question 16.
Construct an isosceles triangle whose base is 8 cm and altitude is 4 cm. Then, draw another similar triangle whose side are 1 \(\frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Solution:

1. Draw a lines segment BC = 8 cm
2. Draw the perpendicular bisector PQ of BC intersecting BC at ‘O’
3. Mark a point ‘A’ on PQ such that \(\overline{\mathrm{OA}}\) = 4 cm
4. Join AB and AC to the isosceles triangle ABC
5. Extend BC on either side O’C’ = 1 \(\frac{1}{2}\) times; BC = 12 times.
6. Similarly extend OA So that OA’ = 1 \(\frac{1}{2}\) times A = \(\frac{12}{2}\) = 6
7. Join A’ B’ and A’ C’
8. Now A’ B’ C’ – The required triangle

Question 17.
Give two different examples of pair of 3 similar figures and non similar figures.
Solution:
Similar figures :

1. Any two circles
2. Any two squares

Non-similar figures :

1. A square and a rhombus
2. A square and a rectangle

Question 18.
∆ ABC ~ ∆ DEF and their areas are respectively 64 cm² and 121 cm². If EF = 15.4 cm then find BC.
Solution:
Given area of ∆ABC = 64 cm²
area of ∆DEF = 121 cm²
EF = 15.4 cm, BC = ?
We know that \(\frac{\text { area of } \triangle \mathrm{ABC}}{\text { area of } \triangle \mathrm{DEF}}\)=\(\frac{\mathrm{BC}^2}{\mathrm{EF}^2}\)
⇒ \(\frac{64}{121}\) = \(\frac{\mathrm{BC}^2}{(15.4)^2}\)
⇒ BC² = \(\frac{64}{121}\) × (15.4)²
= \(\frac{64}{121}\) × 15.4 × 15.4
BC² = 125.44
∴ BC = \(\sqrt{125.44}\) = 11.2 cm.

Question 19.
In D ABC, DE || BC and \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{3}{5}\) ; AC = 5.6 find AE.
Solution:

But \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{3}{5}\) , So \(\frac{\mathrm{AE}}{\mathrm{EC}}\) = \(\frac{3}{5}\)
given AC = 5.6
from (1) \(\frac{3}{5}\) = \(\frac{\mathrm{AE}}{\mathrm{AC – AE}}\)
\(\frac{3}{5}\) = \(\frac{\mathrm{AE}}{5.6-\mathrm{AE}}\) (By cross-multiplication)
5 AE = 3 (5.6 – AE)
5 AE = 3 × 5.6 – 3AE
5AE + 3 AE = 3 × 5.6
8AE = 16.8
AE = \(\frac{16.8}{8}\) = 2.1 cm

Question 20.
State and prove basic proportionality theorem.
Solution:
Preposition : If a line is drawn parallel to one side of the triangle to intersect the other sides in distinct points, then the other two sides are divided in the same ratio.

Given : A triangle ABC in which DE || BC and DE intersects AB in D and AC in E.
To Prove : \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\)

Construction : Join BE, CD and draw EF ⊥ AB. DG ⊥ AC.
Proof: In ∆ EAD and ∆ EDB, Here as EF ⊥ AB
therefore EF is the height for both of triangles EAD and EDB.
Now, Area of ∆ EAD = \(\frac{1}{2}\) (base × height)
= \(\frac{1}{2}\) × AD × EF
Area of EDB = \(\frac{1}{2}\) (base × height)
= \(\frac{1}{2}\) × DB × EF

∴ ∆ DBF, ECD are on the same base DE and between the same parallels DE || BC,
We have area of ∆DBE = area of ∆ECD
Hence (1) = (2)
i.e, \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\) (Q.E.D)

Question 21.
Construct a triangle of sides 4 cm, 5 cm and 6 cm then construct a triangle. Similar to it. Whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.
Solution:

Steps of construction :

1. Draw ∆ ABC with AB = 4 cm, BC = 5 cm and CA = 6 cm.
2. Draw a Ray \(\overrightarrow{\mathrm{BX}}\) making an acute angle with BC on the side opposite to vertex A.
3. Mark three points B₁, B₂ and B₃ on \(\overrightarrow{\mathrm{BX}}\) Such that BB₁ = B₁B₂ = B₂B₃.
4. Join B₃, C.
5. Draw a line parallel B₃, C through B₂ meeting BA at A’.
6. ∆ A’ BC’ is required triangle.

Telangana SCERT TS 10th Class Physical Science Study Material Pdf 7th Lesson Classification of Elements- The Periodic Table Textbook Questions and Answers.

TS 10th Class Physical Science 7th Lesson Questions and Answers Classification of Elements- The Periodic Table

Improve Your Learning
I. Reflections on concepts

Question 1.
What are the limitations of Mendeleeff’s periodic table? How could the modern periodic table overcome the limitations of Mendeleeff’s table?
Answer:
Limitations of Mendeleeff’s periodic table: –

1. Position of hydrogen: The position of hydrogen in table is not certain because it can be placed in group IA as well as in group VII A as it resembles both with alkali metals of IA group and halogens of VITA group.
2. Anomalous pair of elements: Certain elements of higher atomic mass precede those with lower atomic mass. Eg: Tellurium proceeds Iodine, potassium placed after Argon. (Atomic mass 40) .
3. DissimIlar elements placed together: Elements with dissimilar properties were placed in same groups as sub-group A and sub-group B.
   Eg: U, Na, and K of the IA group have little resemblance with Cu, Ag, Au of IB group.
4. Some similar elements are separated: Some similar elements like copper and mercury ‘silicon and thallium’ etc., are placed ¡n different groups of the periodic table.

Question 2.
Define the modern periodic law. Discuss the construction of the long form of the periodic table.
Answer:
Modern Periodic Law: The physical and chemical properties of elements are the periodic function of the electronic configurations of their atoms.

1. The modern periodic table has eighteen vertical columns known as groups and seven horizontal rows known as periods.
2. The horizontal rows of elements In a periodic table are called periods. The elements in a period have consecutive atomic numbers. The vertical columns of elements In the table are called groups.
3. Based on which subshell the differentiating electron enters In the atom of the given element the elements are classified as s, p. d, and f block elements.

s-Block elements :- The elements with valence shell configuration ns¹ and ns² are s- Block elements.
p-Block elements: The elements with electronic configuration from ns²np¹ to ns² np⁶ are p-block elements.
s and s block elements together are known as representative elements. d-Block elements or Transition elements:
The elements with valency electronic configuration ns²np⁶ nd¹ to ns²np⁶ (n-1) d¹⁰ are called d-Block elements. These move from left to right in periodic table and we observe a transition of elements from metals to non-metals. Hence these are called transition elements.

f-Block elements or Inner transition elements: The elements In which f-orbitals are being filled in their atoms are called f-Block elements. These elements are called inner transition elements.

Inert Gases: Helium, Neon, Argon, Krypton, Xenon, Radon do not involve In any reaction and they are called Inert gases or noble gases. These are placed In zero group and every period ends with noble gas.

The long periodic table Is divided into 7 periods and 18 groups. First period contains 2 elements second period contains 8 elements. Third period contains 8 elements 4th and 5th periods contain 18 elements each and sixth period contains 32 elements. Seventh period is incomplete.

The 4f elements are called Lanthanolds. Elements from ₅₈Ce to ₇₁Lu. The 5f elements are called Actinoids from ₉₀Th⁵⁸ to ₁₀₃ Lr ⁷¹ These are shown separately at the bottom of the periodic table.

Question 3.
Explain how the elements are classIfied Into s, p, d, and f-block elements In the periodic table and give the advantage of this kind of classification.
Answer:
Depending on the valency shell electronic configuration elements are classified into s, p, d, and f block elements.
s-Block elements: The elements with valence shell electronic configuration ns¹ and ns² are called s-block elements.

p-Block elements: The elements with valence shell electronic configuration ns² np⁶ are called p-block elements. s and p block elements together known as Representative elements.

d-Biock elements : The elements with valence electronic configuration from ns²np⁶ (n -1)d¹ to ns²np⁶(n -1)d¹⁰ are called d-block elements. These are also called as Transition elements.

f-Block elements: The elements In which f-orbitaIs are being filled in their atoms are called f – block elements. These are also known as Inner Transition Elements.

Advantage: This division of elements into blocks Is useful to divide the elements into groups. Every group has the elements with same valence electronic configuration. So they have similar chemical properties.
s-block elements – I A and II A groups
p-block elements – III A to VITI A or zero group.
d-block elements – I B to VIII B groups
f-block elements – Lanthanoids and Actinoids.

Question 4.
Write down the characteristics of the element having atomic number 17.
Electronic configuration …………………….
Period number …………………….
Group number …………………….
Element family …………………….
No. of valence electrons …………………….
Valency …………………….
Metal or Non-metal …………………….
Answer:
Electronic configuration: 1s²2s²2p⁶3s²3p⁵
Period number : 3
Group number: 17
Element family: Halogen (VII A) or 17th group
No. of valency electrons : 2+5 = 7 ( seven )
Valency: 1
Metal or Non-metal : Non-metal

Question 5.
Complete the following table using the periodic table.

Answer:

Question 6.
Complete the following table using the periodic table.

Answer:

Question 7.
Newlands proposed the law of octaves. Mendeleeff suggested eight groups for elements in his table. How do you explain these observations in terms of modern periodic classification?
Answer:

• Newlands proposed the law of octaves.
• According to Newlands law of octaves, every eight element starting from a given one be a repetition of the first with regard to its properties.
• Mendeleeff suggested eight groups for elements in his table.
• The elements in the same group have same properties that means every eight element starting from a given element have same property with that.
• In terms of modern concepts, after completion of one shell the properties of elements are repeated.
• After completion of ns2 np6 configuration the properties of elements are repeated.
• So Newlands law of octaves and Mendeleeff’s suggestion of eight groups for elements are also reliable according to modem concepts.

Question 8.
Why was the basis of classification of elements changed from the atomic mass to the atomic number?
Answer:

1. The first attempt to classify elements was made by Dobereiner.
2. Dobereiner’s attempt gave a clue that atomic masses could be correlated with properties of elements.
3. Newlands’ law of octaves also followed the same basis for classification but this law Is not valid for the elements that had atomic masses higher than calcium.
4. Mendeleeff’s classification also was based on the atomic masses of elements, but it lead to some limitations like Anomalous pair of elements and Dissimilar elements placed together.
5. Mosley by analyzing the X-ray patterns of different elements was able to calculate the number of positive charges in the atoms of respective elements.
6. With this analysis, Mosley realized that the atomic number is more fundamental characteristic of an element than its atomic weight.
7. So, he arranged the elements in the periodic table according to the increasing order of their atomic number.
8. This arrangement eliminated the problem of anomalous senes and dissimilar elements placed together In same group as done by Mendeleeff’s classification.

Question 9.
(a) What is a periodic property? How do the following properties change In a group and period? Explain.
(a) Atomic radius
(b) Ionization energy
(c) Electron affinity
(d) Electronegativity.
Answer:
Periodic property: Periodicity means that when elements are arranged according to their electronic configuration, the periodic properties of elements in the periodic table reoccur at regular intervals – such properties which reoccur are called as periodic properties.

Change of properties In a group and a period.

Explanation:
(a) Atomic radius: Atomic radius of an element Is not possible to measure in its isolated state, because it is not possible to determine the location of the electron that surrounds the nucleus. Hence the distance between nucleus and outermost orbit of an element is taken as atomic radius measured In Pm (Pico Meters).

Variation In group: In a group as we go down, the atomic number of the element Increases, as a new shell Is added. As a result, the distance between nucleus and the outer shell Increases. This Is the reason why the atomic radius increases as we go down in a group.

Variation In period: AtomIc radii of elements decrease across a period from left to right. As we go from left to right, electrons enter into the same main shell or even Inner shell. Therefore there should be no change In the distance between nucleus and outer shell, but nuclear charge Increases. Hence the nuclear attraction on the outer shell increases. As a result the size of the atom decreases.

(b) Ionization enegry: The energy required to remove an electron from the outermost orbit or shell of a neutral gaseous atom is called ionization energy.

Variation in group: As we go down in a group, atomic radius increases. Hence the ionization energy decreases from top to bottom.

Variation In period: As we go from left to right in a period, the atomic radius decreases. Hence the ionization energy increases from left to right In a period but does not follow gradient order as the electron is attracted more towards the nucleus.

(c) Electron affinity: The electron affinity of an element is defined as the energy liberated when an electron Is added to its neutral gaseous atom. Electron affinity of arm atom is also called electron gain enthalpy of that element.

Variation in group: Electron gain enthalpy values decrease as we go down In a group, due to Increase In atomic radius.

Variation In period: Electron gain enthalpy values Increase as we go from left to right in a period, due to decrease in atomic radius.

(d) Electronegativity: The electronegativity of an element Is defined as the relative tendency of its atom to attract electrons towards itself when it is bounded to the atom of another element.

Variation in groups and periods: Electronegativity values of elements decrease as we go down in a group and increase along a period from left to right.

Question 10.
(b) Explain the ionization energy order In the following sets of elements.
(a) Na, Al, Cl
(b) Li, Be, B
(c) C, N, O
(d) F, Ne, Na
(e) Be, Mg, Ca.
Answer:
(a) Na, Al, Cl: All these three elements belong to same period. The order of their atomic size Is Na > Al> Cl. As we move from left to right in a period Ionization energy increases
∴ The order of ionization energy of these elements is Cl> Al> Na.

(b) Li, Be, B: U, Be, B belong to same period. The electronic configuration of Li-1S²2S¹, Be-1S²2S², B-1S²2S²2p¹. The penetration power of 2P is less when compared to 2s. So it is easy to remove electrons from ‘2p’.
∴ The order of ionization energy of these elements is: Be > Li> B.

(c) C, N, O : C, N, O belong to same period. The electronic configuration of C – 1s²2s²2p², N-1s²2s²2p³, O-1s²2s²2p⁴. Nitrogen has half filled degenerate orbital configuration. So, it Is more stable compared to Cl and ‘O’so Its Ionisation energy Is high.
∴ Order of ionization energy of these elements is: N > C> O.

(d) F, Ne, Na: Electronic configuration of F -1S²2s²2p⁵; Ne – 1S²2S²2p⁶; Na – 1s²2s²2p⁶3s¹. Ne is an Inert gas, so It has highest Ionization energy.
‘Na’ has smaller size compared to F. So ¡t ha high ionization energy.
∴ The order of ionization energy is: Ne>F>Na.

(e) Be, Mg, CaP: These elements belong to the same group the atomic size of these elements Is In the order of Ca > Mg > Be. As atomic size increases ionization energy decreases. The order of Ionization energy Is Be > Mg > Ca.

Application Of Concepts

Question 1.
Given below Is the electronic configuration of elements A, B, C, and D.
(A) 1s² 2s² 1. WhIch are the elements coming within the same period?
(B) 1s²2s²2p⁶3s² 2. Which are the ones coming within the same group?
(C) 1s²2s²2p⁶3s² 3p³ 3. Which are the noble gas elements?
(D) 1s²2s²2p⁶ 4. To which group and period does the clement C ‘belong?
Answer:
(1) A and D belong to same penned B and C belong to the same period.
(2) A, B coming in the same group.
(3) D is the noble gas.
(4) C’ belongs to 15th group and third period.

Question 2.
s – block, and p – block elements (except 18th group elements) are sometimes called as ‘Representative elements’ based on their abundant availability In the nature. Is It Justified? Why?
Answer:

• s-block and p-block elements (except 18th group elements) are called ‘Representative elements’.
• All these elements have incomplete outermost shells.
• So they are chemically reactive to obtain stable electronic configuration of noble gases ns²np⁶.
• So, they are abundant In nature n the form of compounds.

Question 3.
The electronic configuration of the elements X, Y, and Z are given below.
(A) X = 2 (B) Y = 2, 6 (C) Z =2, 8, 2
(i) Which element belongs to second period?
Answer:
‘Y’ belongs to the second period. Because differentiating electron enters into second shell.
(ii) Which element belongs to second group?
Answer:
Element ‘Z’ belongs to second group. Because its valence is ‘2’.
(iii) Which element belongs to 18 group?
Answer:
Element X belongs to 18th” group because its 1st shell is completely filled with electrons.

Question 4.
(a) State the number of valence electrons, the group number and the period number of each element given in the following table.

Answer:

(b) State whether the following elements belong to a Group (G), Period (P) or Neither Group nor Period Indicating the letters G or P or N

Answer:

Question 5.
Identify the element that has the larger atomic radius In each pair of the following and mark it with a symbol (✓).
(i) Mg, Ca
(ii) Li, Cs
(iii) N, P
(iv) B, AI
Answer:

1. Mg, Ca: Calcium has large atomic radius. Since Mg and Ca are in same group Ca falling below Mg. Atomic radius increases from top to bottom In a group.
2. Li, Cs: Cs has large atomic radius as Li and Cs are also in the same group and Cs lying below Li.
3. N, P: Phosphorous has large atomic radius. As N and P are also In the same group and P lying below N.
4. B, AI: Al has large atomic radius. B, Al are in same group III A and Al is below B.

Question 6.
Identify the element that has the lower Ionization energy in each pair of the following and mark it with a symbol (✓).
(i) Mg, Na
(ii) Li, O
(iii) Br, F
(iv) K, Br
Answer:

1. Mg, Na: Na has low ionization energy. Since Mg, Na are in same period ionization energy increases from left to right. Na lies left to Mg. So Na has lower ionization energy.
2. Li, O: Li has lower ionization energy. U lies left to ‘O’ in 2nd period.
3. Br, F: Bromine has lower ionization energy; Br & F are in same group. IE decreases from top to bottom. Bromine is below F. So It has lower lE.
4. K, Br: ‘K’ has lower ionization energy. Since it is left to Br in the same period.

Question 7.
How does metallic character change when we move
(i) Down in a group?
(ii) Across a period?
Answer:
(i) Down in a group?
Metallic character increases when we move down the group.
(ii) Across a period?
Answer:
Metallic character decreases when we move along a period.

Question 8.
On the bails of atomic numbers predict to which block the elements with atomic number 9, 37, 46 and 64 belong to?
Answer:

1. Electronic configuration of element with atomic number 9 is 2, 7 – belongs to p-block.
2. Electronic configuration of element with atomic number 37 is 2, 8, 8, 18, 1 – belongs to s- block.
3. Electronic configuration of element with atomic number 46 is 2, 8, 8, 18, 10 – belongs to d – block.
4. Electronic configuration of element with atomic number 64 Is 2, 8, 8, 18, 18, 10 -belongs tO f – block.

Question 9.
Using the periodic table, predict the formula of compound formed between an element X of group 13 and another element Y of group 16.
Answer:
Element X of group 13 Element Y of group 16
Valency of element X = 3 Valency of element Y
18-16=2

Compound: X₂Y₃

Question 10.
An element has atomic number 19. Where would you expect this element in the periodic table and why?
Answer:
The element with atomic number 19 is in 4th’ period and first group of periodic table.
Reason:

1. Electronic configuration : Is² 2s² 2p⁶ 3s² 3p⁶ 4s¹’.
2. The differentiating electron enters Into 4t shell. Hence it belongs to 4th period.
3. The differentiating electron is In ‘s’ orbital. So it belongs to ‘s’ block.
4. The outermost orbital has only one electron. Hence it belongs to first group.

Question 11.
Elements In a group generally possess similar properties, but elements along a period have different properties. How do you explain this statement?
Answer:

1. Physical and chemical properties of elements are related to their electronic configurations, particularly the outer shell configurations.
2. The atoms of the elements in a group possess similar electronic configurations. Therefore, we expect all the elements in a group should have similar chemical properties and there should be a regular gradation in their physical properties from top to bottom.
3. But in a period from left to right, elements get an increase in the atomic number by one unit between any two successive elements.
4. Therefore, the electronic configuration of valence shell of any two elements in a given period is not same. Due to this reason elements along a period possess different chemical properties with regular gradation in their physical properties from left to right.

Question 12.
Name two elements that you would expect to have chemical properties similar to Mg. What is the basis for your choice?
Answer:

• Calcium (Ca) and Stranúum (Sr) are the two elements, which are similar to Mg in chemical properties.
• Because they belong to the same group lIA. Ail the elements which are in the same group have same electronic configuration and same chemical properties.

Question 13.
An element X belongs to 3rd perIod and group 2 of the periodic table. State (a) The no. of valence electrons in it (b) The valency (C) Whether It is metal or a non-metal.
Answer:
The element ‘X’ belongs to 3rd peñad and group 2 is Mg.
(a) The no.of valence electrons 2.
(b) The valency = 2.
(c) It is a metal.

Question 14.
How do you appreciate the role of electronic configuration of the atoms of elements In perIodic classification?
Answer:

1. According to Modern Periodic Iaw the properties of elements are the functions of their atomic number or electronic configuration”.
2. Elements having same number of valence electrons are placed in the same group based on their electronic configuration.
3. So, we can easily locate the position of any element in the periodic table with Its electronic configuration.
4. Hence electronic configuration plays a major role In the preparation of Modern period table. So its role is thoroughly appreciated.

Question 15.
Without knowing the electronic configurations of the atoms of elements Mendeleeff still could arrange the elements nearly close to the arrangements in the modern periodic table. How can you appreciate this?
Answer:

1. Mendeleeff’s vision must be appreciable because he left some gaps predicting the existing of some elements, which are not available at that time.
2. Atomic weights of some elements can be corrected by placing them In correct places.
3. If we compare the long-form periodic table with Mendeleeff’s table, we find many elements with their places unchanged.
4. One can see from these discussions, that the Mendeleeffs accurate prediction and their eventual success made him and his classification of elements famous, Even today his table of elements Is followed with minimum modifications.

Question 16.
Comment on the position of hydrogen in periodic table.
Answer:

1. The configuration of Hydrogen (1s¹) Is responsible for its dual nature.
2. Either the electron can be lost behaving as electropositive elements (H⁺) like alkali metals or the electron can be gained as to complete, is subshell behaving as electronegative element (H^–) like halogens.
3. The position of Hydrogen is not fixed in the periodic table sometimes it is placed with alkali metals and sometimes with halogens.
4. However, Hydrogen exhibits other properties which differ from both alkali metals and halogens.

Higher Order Thinking Questions

Question 1.
How does the positions of elements In the periodic table help you to predict Its chemical properties? Explain with an example.
Answer:

1. The physical and chemical properties of atoms of the elements depend on their electronic configuration, particularly the outer shell configurations.
2. Elements are placed in the periodic table according to the increasing order of their electronic configuration.
3. The elements In a group possess similar electronic configurations. Therefore all the elements in a group should have similar chemical properties.

Eg: Consider K

• It is the element In 4th’ period 1st’ group.
• It is on left side of the periodic table. Hence It Is a metal.
• It is ready to lose an electron to get octet configuration. Hence its reactivity is more.
• It is Alkali metal.
• All alkali metals react with both acids and bases and release H₂ gas.

Question 2.
In period 2, element X is to the right of element Y. Then find which of the elements have
(i) Low nuclear charge
(ii) Low atomic size
(iii) High Ionisation energy
(iv) High electronegativity
(v) More metallic character.
Answer:
X s to the right of Y. So they will be YX
(i) Low nuclear charge: Nuclear charge increases from left to right in a period. So Y has low nuclear charge.
(ii) Low atomic size: Atomic size decreases from left to right in a period. So X has low atomic size.
(iii) High ionization energy: Ionization energy increases across a period from left to right. So X has high ionization energy.
(iv) More metallic character: Metallic nature decreases from left to right. So Y has more metallic character.

Multiple choice questions

Question 1.
Number of elements present in period -2 of the long fom of periodic table ………………. [ ]
(a) 2
(b) 8
(c) 18
(d) 32
Answer:
(b) 8

Question 2.
Nitrogen (Z = 7) is the element of group VAofthe periodic table. Which of the following is the atomic number of the next element in the same group’? [ ]
(a) 9
(b) 14
(c) 15
(d) 17
Answer:
(c) 15

Question 3.
Electron configuration of an atom is 2,8,7. To which of the following elements would it be chemically similar? [ ]
(a) nitrogen(Z=7)
(b) fluorine(Z = 9)
(c) phosphorous(Z = 15)
(d) argon(Z=1 8)
Answer:
(b) fluorine(Z = 9)

Question 4.
Which of the following is the most active metal? [ ]
(a) lithium
(b) sodium
(c) potassium
(d) rubidium
Answer:
(d) rubidium

Suggested Experiments

Question 1.
Aluminium does not react with water at room temperature but reacts with both dil. HCl and NaOH solutions. Verify these statements experimentally. Write your observations with chemical equations. From these observations, can we conclude that Al is a metalloid?
Answer:
1. Aluminum reacts with dii. HCl and releases hydrogen gas with formation of Aluminium chloride.

2. Aluminium reacts with NaOH solution and releases hydrogen gas.

3. Aluminium does not react with water at room temperature. This concludes that the properties of Aluminium in between a metal and non-metal. So it behaves like a metalloid.

Suggested Projects

Question 1.
Collect the Information about reactivity of VIII A group elements (noble gases) from your school library and prepare a report on their special character when compared to other elements of periodic table.
Answer:

1. The elements of Group VIII A (noble gases) are chemically inactive, because all of them have stable “Octet” configurations, except He.
2. He has two electrons in the outermost orbital, i.e., is orbitalis is the maximum, the is orbital can accommodate. Hence “He” is also inactive.
3. The losing or gaining an electron or sharing of electrons is difficult due to high I.E. and zero E.A value of inert gases.
4. Under suitable conditions, the heavier elements of the group form compounds with F₂ and O₂ or XeF₂ or XeO₂F₂.
5. Ar forms coordination compounds with BF₃ whose composition is Ar.n.BF₃
6. Xenon Hexa fluoro platinate (IV)(Xe[PtF₆]) was prepared by N.Barltel. This was the first compounds of inert gases. After this compounds of Xenon with F₂ and O₂ were prepared.
7. The compounds of inert gases have found varied applications in recent times.

Question 2.
Collect information regarding metallic character of elements of IA group and prepare report to support the idea of metallic character increases In a group as we move from top to bottom.
Answer:
Group IA elements are Li, Na, K, Rb, Cs, and Fr
1. Metallic character is the name given to the set of chemical properties associated with elements that are metals.

2. Chemical characteristics of metals include the following

• form cations in ionic compounds with non-metals.
• have ionic halides.
• have ionic hydrides containing the H’ ion.
• have basic oxides.

3. The elements with less electronegative character are metals.
4. All IA group elements have less electronegative characters because they are ready to lose one electron to get octet configuration.
5. If we observe this group, we can find that

Reaction with non-metals:
4M + O₂ → 2M₂O. In this cation is M⁺
Hydrides:

M stands for alkali metal.
Order of ionic nature of hydrides Is LiH < NaH < KH < RbH < CSH
Halides:
2M+X₂ → 2MX
where M stands for alkali metal and X stands for halogen.
All the metal halides are ionic.

Oxides:
4M + O₂ → 2M₂O
But, 4 Li + O₂ →‘ 2 Li₂O₂
All metals form peroxides.
6. From the above reactions we conclude that Group IA elements are metals and their metallic character increases as we go down in the group.

TS 10th Class Physical Science Classification of Elements- The Periodic Table Intext Questions

Page 124

Question 1.
Can you establish the same relationship with the set of elements given in the remaining rows?
Answer:
Yes, we can establish the same relationship with the set of elements given in the remaining rows.

Question 2.
Find average atomic weights of first and third elements ¡n each row and compare It with the atomic weight of the middle element.
Answer:

Question 3.
What do you observe?
Answer:
Dobereiner’s attempts gave a clue that atomic masses could be correlated with properties of elements.

Page 137

Question 4.
Find out the valencies of first 20 elements.
Answer:

(b) How does the valency vary in a period on going from left to right?
Answer:
The valency increases from 1 to 4 and then decreases to zero as we move from left to right in a period with respect to Hydrogen. The valency increases from 1 to 7 with respect to oxygen, in a period from left to right.

(c) How does the valency vary on going down a group?
Answer:
The valency remains same in a group.

Page 139

Question 5.
Do the atom of an element and Its Ion have same size?
Answer:
No, the atom of an element and its ion do not have the same size. The size of cation is less than its neutral atom and size of anion Is greater than its neutral atom.

Question 6.
Which one between Na and Na would have more size? Why?
Answer:

1. The atomic number of sodium is ti and It has 11 protons and ii electrons with outer electron as3s’ whereas Na4 ¡on has il protons but only 10 electrons.
2. The 3s¹ shell of Na has no electron In It.
3. So the outer shell configuration is 2S² 2p⁶.
4. As proton number is moie than electrons the nudeus of Na⁺ ion attracts outer shell electrons with strong nudear force.
5. As a result the Na lori shrinks In size.
6. Therefore, the size of Na ion is less than Na atom.

Question 7.
Which one between Cl and Cl^– would have more size? Why?
Answer:
The electronic configuration of Cl is 1s²2s²2p⁶3s²3p⁵ whereas Cl^– is 1s²2s²2p⁶3s² 3p⁶. In Cl^– , same no.of protons are attracting more number of electrons compared to Cl. Due to decrease in effective nuclear attraction atomic size of anion is more than its neutral atom.

Question 8.
Which one in each of the following pairs Is larger In size?
a) Na, AI
b) Na, Mg⁺²
c) S₂, Cl^–
d) Fe⁺², Fe⁺³
e) C^-4, F
Answer:
a) Na
b) Na
c) S^-2
d) Fe⁺²
e) C^-4

Think And Discuss

Question 1.
What relation about elements did Doberelner wanted to establish?
Answer:
The relative atomic mass of the middle element En each triad is nearer to the average of the relative atomic masses of other two elements. This is what Doberelner wanted to establish.

Question 2.
The densities of Calcium (Ca) and Barium (Ba) are 1.55 and 3.51gm/ cm³ respectively. Based on Doberolner’s law of triads can you give the approximate density of strontium (Sr)?
Answer:
The density of strontium = (1.55 + 3.51 )/2 = 5.06 / 2 = 2.53gm/cm³

Question 3.
Do you know why Newland’s proposed the law of octaves? Explain your answer In terms of the structure of atom.
Answer:
Newlands proposed the law of octaves. According to this law, every 8” element starting from a given elements resembles in its properties to that of starting element with the elements with similar chemical properties are to be present along horizontal.

In modem periodic table, every period starts with a new shell and ends when the shell is filled. We know about octet configuration. Newlands law of octaves Is similar to octet. configuration.

Question 4.
Do you think that Newlands law of octaves is correct? JustIfy.
Answer:
Newlands law of octaves is corretc. Mendeleeff also suggested only 8 groups in his table. Later it was discovered that eighth element has the same properties as the first one.

Question 5.
Why Mendeleeff had to leave certain blank spaces In his periodic table. What is your explanation for this?
Answer:
Based on the arrangement of the elements in the table, he predicted that some elements were missing and left blank spaces at the appropriate places In the table. Mendeleeff believed that some new elements would be discovered definitely. He predicted the properties of these new additional elements in advance purely depending on his table. His predicted properties were almost the same as the observed properties of those elements after their discovery.

Question 6.
What is your understanding about Ea₂O₃ and EsO₂?
Answer:
Ea₂O₃ is the predicted formula of oxide,of the predicted element In eka – Aluminium group, This is similar to Ga₂O₃, the observed property of oxide of Gallum, which was discovered in 1875. EsO₂ is the predicted formula of oxide of predicted element In eka-silicon group. This is similar to GeO₂, the observed property of oxide of Germanium, which was discovered in.1886.

Question 7.
All alkali metals are solids but hydrogen is a gas with diatomic molecules. Do you justify the inclusion of hydrogen in first group with alkali metals?
Answer:
The similarity of hydrogen with alkali metals is ns’ configuration and ability to exhibit +1 oxidation state. Hydrogen, because of its small size and high electronegativity, it is a non-metal with high I.P. It shares one electron with another hydrogen and forms diatomic molecules. Due to weak van der Waals forces of attraction It is a gas whereas in alkali metals, metallic bond is present hence they are solids. Hence, hydrogen cant be included in first group along with alkali metals.

Question 8.
Why lanthanoids and actinoids placed separately at the bottom of the periodic table?
Answer:
Lanthanolds and actinoids are placed separately in the periodic table to maintain its structure and to preserve the principle of classification with similar properties in a single column.

Question 9.
If they are inserted within the table imagine how the table would be?
Answer:
If f-block elements are inserted within the table, the table would be too long.

Question 10.
Second Ionization energy of an element is higher than Its first ionization energy. Why?
Answer:
The required to remove first electron from the outermost orbit or shell of a neutral gaseous atom is called first ionization energy (IE₁).
The energy required to remove an electron from uni-positive ion is called the 2nd ionization energy (IE₂).
The second ionization energy (IE₂) is greater than the first ionization energy (IE₁). On removing an electron from an atom, the uni-positive ion formed will have more effective nudear charge than the neutral atom. This decreases the repulsion between the electrons and at the same time increases the nuclear attraction on the electron in the outer shells. As a result, more energy is required to remove an electron from the uni-positive ion. Hence the second ionization
energy is greater than the first ionization energy.

Question 11.
The calculated electron gain enthalpy values for alkaline earth metals and noble gases are positive. How can you explain this?
Answer:
Due to completely filled s-subshell in alkaline earth metals and completely filled valence shell (i.e. octet) in noble gases, they are highly stable. Hence energy Is required to add an electron I.e their electron gain enthalpy values are positive.

Question 12.
The second-period element, for example, F has less electron gaIn enthalpy than the third-period element of the same group for example ‘Cl’. Why?
Answer:
The atomic size of second period elements decreases and electron number increases as we move from left to right. Because of this elements of N, O, F possess vert small size and high electron density. Hence incoming electron feels repulsion due to existing electrons. Because of this, electron gain enthalpy
is less for second-period element compared to third period of the same group (In 5th’ , 6th and 7th groups).

TS 10th Class Physical Science Classification of Elements- The Periodic Table Activities

Activity 1

Question 1.
Observe the following table and till It.
Answer:

Observations:
a) Can you establish the same relationship with the set of elements given In the remaining rows?
Answer:
Except in the first row, we cannot establish the relationship as said by law of triads.
b) Find average atomic weights of first and third elements in each row and compare It with the atomic weights of the middle element.
Answer:
From table we observe that. except Group A, in the remaining Groups, the average of atomic weights of first and third elements is roughly equal to the middle elements.

c) What do you observe?
Answer:
Dobereiner’s attempt gave a clue that atomic mass could be correlated with properties of elements.

Activity 2

Question 2.
Write the different chemical families and write involved, valency of the family.
Answer:

Question 3.
(a) Find out the valencies of first 20 elements.
(b) How does the valency vary in a period on going from left to right?
(c) How does the valency vary on going down a group?
Answer:

(b) How does the valency vary in a period on going from left to right?
Answer:
The valency increases from 1 to 4 and then decreases to zero as we move from left to right in a period with respect to Hydrogen. The valency increases from 1 to 7 with respect to oxygen, in a period from left to right.

(c) How does the valency vary on going down a group?
Answer:
The valency remains same in a group.

These TS 10th Class Maths Chapter Wise Important Questions Chapter 2 Sets given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 2 Sets

Previous Exams Questions

Question 1.
Write roster and builder form of “The set of all natural numbers which divide 42”. (T.S. Mar. ’15)
Solution:
Factors of 42 = 1, 2, 3, 6, 7, 14, 21, 42
So roster form = {1, 2, 3, 6, 7, 14, 21, 42}
The builder form = (x/x ∈ N, x is a factor of 42 }

Question 2.
Write A = {1, 2, 3, 4} in set builder form. (A.P. June ’15)
Solution:
The given set A = {1, 2, 3, 4}
The set builder form is A = {x/x ∈ N, x < 5 }

Question 3.
Let A = { x/x is an even number }
B = { x/x is an odd number}
C = { x/x is a prime number}
D = { x/x is a multiple of 5 }
Find (A.P. June’15)
i) A ∪ B
ii) A ∩ B
iii) C – D
iv) A ∩ C
Solution:
A = { x : x is an even number }
= {2, 4, 6, 8, 10}
B = { x : x is an odd number}
= {1, 3, 5, 7, 9 }
C = { x : x is a prime number}
= { 2, 3, 5, 7, 11}
D = { x : x is a multiple of 5}
= { 5, 10, 15, 20, 25}

i) A ∪ B = { 2, 4, 6, 8, 10, ……..} ∪ {1, 3, 5, 7, 9, …….}
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, …………}

ii) A ∩ B = { 2, 4, 6, 8, 10, ……….} ∩ {1, 3, 5, 7, 9, ………… }
= { } = Φ

iii) C – D = {2, 3, 5, 7, 11, ……….} – { 5, 10, 15, 20, 25, ………}
= {2, 3, 7, 11, …….}

iv) A ∩ C = { 2, 4, 6, 8, 10, ………….} ∩ {2, 3, 5, 7, 11, ……….}
= {2}

Question 4.
Write all the subsets of B = {p, q} (A.P. Mar ’16)
Solution:
{p}, {q}, {p, q} and { } are the subsets of the given set B = (p, q}
As then n(B) = 2 then number of all subsets = 2ⁿ = 2² = 4.

Question 5.
Write the roster form of the set A. (A.P. Mar. ’16)
A = {x : x = 2n + 1 ∀ n ∈ N}
Solution:
If n = 1 then 2n + 1 = 2(1) + 1
= 2 + 1 = 3
If n = 2 then 2n + 1 = 2(2) + 1
=4 + 1 = 5
If n = 3 then 2n + 1 = 2(3) + 1
= 6 + 1 = 7
So {3, 5, 7, 9, ………..} is the roster form of given set.

Question 6.
If A = {1, 2, 3, 4} and B = (1, 2, 3, 5, 6} then find
(i) A ∩ B
(ii) B ∩ A
(iii) A – B and
(iv) B – C then comment on the above. (A.P. Mar. ’16)
Solution:
Given A = {1, 2, 3, 4} and B = { 1, 2, 3, 5, 6} then
i) A ∩ B = {1, 2, 3, 4} ∩ { 1, 2, 3, 5, 6}
= {1, 2, 3}

ii) B ∩ A = {1, 2, 3, 5, 6} ∩ { 1, 2, 3, 4}
= {1, 2, 3}
So A ∩ B = B ∩ A

iii) A – B = {1, 2, 3, 4} – { 1, 2, 3, 5, 6}
= {4}

iv) B – A = { 1, 2, 3, 5, 6} – {1, 2, 3, 4}
= {5, 6}
So A – B ≠ B – A

Question 7.
Write the Set builder form of A – B where A = { x : x ∈ N and x < 20 } and B = { x : x ∈ N and x ≤ 5 } (T.S. Mar. ’16)
Solution:
Given A = { x : x ∈ N and x < 20 } is
A = {1, 2, 3, …….. 17, 18, 19} and
for B = {x : x ∈ N and x ≤ 5}
B = { 1, 2, 3, 4, 5} then
A – B = {1, 2, 3, ……… 17, 18, 19} – {1, 2, 3, 4, 5}
= {6, 7, 8, …… 18, 19} (T.S. Mar ’15)
The builder form of above A – B is
A – B = {x/x ∈ N and 6 ≤ x ≤ 19}
Or
A – B = {x : x ∈ N and 5 < x < 20}

Question 8.
If A = {x / x ∈ N, x < 6 } and B = { x : x ∈ N, 3 < x < 8} then show that A – B ≠ B – A with the help of venn – diagram.
Solution:
A = {x : x ∈ N, x < 6}
A = {1, 2, 3, 4, 5}
B = {x : x ∈ N, 3 < x < 8}
B = {4, 5, 6, 7}
∴ A – B = {1, 2, 3, 4, 5} – {4, 5, 6, 7}
= {1, 2, 3}

and
B – A = {4, 5, 6, 7} – {1, 2, 3, 4, 5}
= {6, 7}

∴ A – B ≠ B – A

Question 9.
Write the set builder form of A = {1, \(\frac{1}{4}\), \(\frac{1}{9}\), \(\frac{1}{16}\), \(\frac{1}{25}\)} (T.S. Mar. ’16)
Solution:
{\(\frac{1}{1}\), \(\frac{1}{4}\), \(\frac{1}{9}\), \(\frac{1}{16}\), \(\frac{1}{25}\)} are in the form of \(\frac{1}{\mathrm{p}^2}\) whereas p < 6
So, A = { x : x = \(\frac{1}{\mathrm{p}^2}\), p ∈ N, and p < 6} is the set builder form.

Question 10.
If x is set of all factors of 24 and y is set all factors of 36 then find X ∪ Y and X ∩ Y using venn – diagrams. Comment. (T.S. Mar. ’16)
Solution:
X = Set of all factors of 24
= {1, 2, 3, 4, 6, 8, 12, 24}
Y = Set of all factors of 36
= {1, 2, 3, 4, 6, 9, 12, 18, 36}
Venn diagram of X ∪ Y
i)

∴ X ∪ Y = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}

ii)

∴ X ∩ Y = {1, 2, 3, 4, 6, 12}
∴ It is clear X ∪ Y ≠ X ∩ Y

Additional Questions

Question 1.
Which of the following are sets ? Justify your answer.
i) The collection of all months of a year beginning with letter ‘M’
ii) A team of eleven best cricket players of the world.
iii) The collection of all girls in your class.
iv) The collection of all odd integers.
Solution:
i) The months of a year which begin with the letter ‘M’ are March and May.
The required set is {March, May}.
∴ It is a well defined collection of objects.
So, it is a set.
ii) It is not a set, because we cannot determine the eleven best cricket players of the world.
iii) It is a set, because we can divide whether he / she belongs to the given set or not.
iv) It is a set, because we can divide whether the number belongs to the given set or not.

Question 2.
If A = (1, 2, 3, 4} B = (5, 6, 7} and C = (p, q, r, s}, then fill the appropriate symbol, ∈ or ∉ in the blanks.
i) 2 ………… A
ii) 6 …………. C
iii) 4 ………… B
iv) q …………. C
v) 5 …………. B
vi) 8 …………. A
Solution:
i) 2 ∈ A
ii) 6 ∉ C
iii) 4 ∉ B
iv) q ∈ C
v) 5 ∈ B
vi) 8 ∉ A

Question 3.
Express the following statements using symbols.
i) The element p does not belong to ‘A’
ii) ‘q’ is an element of the set ‘B’
iii) 4 belongs to the set of Natural numbers N
iv) 5 belongs to the set prime numbers P
Solution:
i) p ∉ A
ii) q ∈ B
iii) 4 ∈ N
iv) 5 ∈ p

Question 4.
Write the following sets in roaster form.
i) A = (x : x is a natural number less than 8}
ii) B = (x : x is a two digital natural number such that the sum of its digits is 5}
iii) C = (x : x is a prime number which is a divisor of 30}
iv) D = { the set of all letters in the word CRICKET}
Solution:
i) A = (1, 2, 3, 4, 5, 6, 7}
ii) B = (14, 41, 23, 32}
iii) C = (2, 3, 5}
iv) D = { C, E, I, K, R, T}

Question 5.
Write the following sets in the set-builder form.
i) A = {5, 10, 15, 20}
ii) B = {3, 9, 27, 81}
iii) C = {4, 16, 64, 256}
iv) D = {1, 8, 27, 64, 125, ……….. 1000}
Solution:
i) A = {x : x is a multiple of 5 and less than 21}
ii) B = {x : x is a power of 3 and x is less than 5}
iii) C = {x : x is a power of 4, and x is less than 5}
iv) D = {x : x is a cube of natural numbers and not greater than 10}

Question 6.
Match the roaster form with set builder
i) {1, 2, 4, 8} (a) (x : x is an even natural number less than 11}
ii) {3, 5} (b) (x : x is a natural number and divisor of 8}
iii) {L, E, T, R} (c) (x : x is a prime number and a divisor of 15}
iv) (2, 4, 6, 8, 10}(d) (x : x is a letter of the word ‘LETTER’}
Solution:
i) b
ii) c
iii) d
iv) a

Question 7.
If A = {1, 2, 3, 4, 5}; B = {3, 4, 5, 6}, then find A ∪ B and A ∩ B.
Solution:
Given A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6}
A ∪ B = {1, 2, 3, 4, 5, 6}
A ∩ B = {3, 4, 5}.

Question 8.
If A = {4, 5, 6, 7, 8}; B = {7, 8, 9, 10, 11} then find A – B and B – A.
Solution:
Given A = {4, 5, 6, 7, 8},
B = {7, 8, 9, 10, 11}
A – B = {4, 5, 6}
B – A = {9, 10, 11}

Question 9.
Which of the following pairs of sets are disjoint ?
i) A = {3, 4, 5, 6}; B = {4, 6}
ii) A = {a, e, i, o, u}; B = {i, j, k}
iii) A = {5, 10, 15, 20}; B = {8, 16, 24}
iv) A = {1, 3, 5, 7};B = {2, 4, 6, 8}
Solution:
Disjoint sets: Two sets are said to be disjoint sets when they have no elements in common.
i) A and B are not disjoint sets (∵ 4, 6 are common)
ii) A and B are not disjoint sets (∵ i is common)
iii) A and B are disjoint sets (∵ No elements in common)
iv) A and B are disjoint sets (∵ No elements in common)

Question 10.
If A = {3, 4, 5, 6, 7}, B = {2, 4, 6, 8} then find n (A ∪ B).
Solution:
n(A) = 5 (∵ A contains 5 elements)
n(B) = 4 (∵ B contains 4 elements)
But A ∪ B = {2, 3, 4, 5, 6, 7, 8}
Now n(A ∪ B) = 7 (∵ A ∪ B contains 7 elements)

Question 11.
If A and B are two sets and n(A) – 15, n(B) = 25 and n(A ∩ B) = 10, find n (A ∪ B)
Solution:
We Know that n (A ∪ B) = n (A) + n (B) – n(A ∩ B)
∴ n(A ∪ B) = 15 + 25 – 10
n(A ∪ B) = 40 – 10 = 30

Question 12.
Which of the following sets are equal ?
i) A = {3, 4, 5, 6}, B = {7, 8, 9, 10}
ii) C = {a, b, c, d}, D = {d, c, a, b} :
iii) E = {2, 4, 6, 8}, F = {x : x is a positive even number < 10}
iv) G = {5, 10, 15, 20, . . . }, H = {x : x is a multiple of 5}
Solution:
i) A and B are not equal sets
ii) C and D are equal sets
iii) E and F are equal sets
iv) G and H are equal sets

Question 13.
State the reason for the following.
i) {4, 5,6, ……….. 12} ≠ {x : x ∈ N and 4 < x < 12}
ii) {2, 4, 6, 8, 10} ≠ {x : x = 2n + 1 and x ∈ N}
iii) {3, 6, 9, 12} ≠ {x : x is a multiple of 6}
iv) {1, 3, 5, 7, 9} ≠ {x : x is an even number}
Solution:
The first set is {4, 5, 6, … ,12}
Writing the second set in roaster form is {5, 6, 7, …,11}
The 1st and 2nd set have not exactly the same elements.
i) ∴ {4, 5, 6, . . . , 12} ≠ {x : x ∈ N and 4 < x < 12}

ii) The first set is {2, 4, 6, 8, 10}
Writing the 2nd set in roaster form is {3, 5, 7, 9}
∴ {2, 4, 6, 8, 10} ≠ {3, 5, 7, 9}

iii) The first set is {3, 6, 9, 12}
Writing the 2nd set is roaster form is {6, 12, 18, 24, … }
∴ {3, 6, 9, 12} ≠ {6, 12, 18, 24,. . .}

iv) The first set is {1, 3, 5, 7, 9}
Writing the 2nd set in roaster form is {2, 4, 6, 8}
∴ {1, 3, 5, 7, 9} ≠ {2, 4, 6, 8,…}

Question 14.
List all the subsets of the following :
i) A = {a, b}
ii) B = {p, q, r}
iii) {1, 2, 3}
Solution:
i) {a}, {b}, {a, b}, {Φ}
ii) {p}, {q}, {r}, {p, q}, {p, r}, {q, r}, {p, q, r}, {Φ}
iii) {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1,2, 3}, {Φ}

Question 15.
State which of the following sets are empty.
i) Set of integers which lie between 5 and 6.
ii) Set of even natural numbers divisible by 2.
iii) The set of lines which are parallel to the Y-axis.
iv) {x : x is a natural number, x < 6 and x > 8}
Solution:
i) It is empty set, because there is no integer lying between 5 and 6
ii) It is not empty set, because there are even natural numbers which are divisible by 2.
iii) It is not empty set, because there are number of lines parallel to the Y-axis is infinite.
iv) It is empty set, because there is no natural number satisfying this condition.

Question 16.
Which of the following set is finite or infinite ?
i) A = {x : x ∈ N and x < 50}
ii) B = {x : x ∈ N and x ≤ 10}
iii) C = {1³, 2³, 3³, ……}
iv) D = {x : x ∈ N and x is even}
Solution:
i) A = {1,2, 3, 4,…. 49}
This set is finite, because there are 49 numbers possible to count

ii) B = {1, 2, 3, 4, . . . , 10}
This set is finite, because there are 10 numbers possible to count

iii) C = {1³, 2³, 3³, . . . }
This set is infinite, because there are infinite numbers.

iv) D = {2, 4, 6, 8, . . . }
This set is infinite, because there are infinite even numbers.

Question 17.
If A = {1, 2, 3, 4, 5}; B = {4, 5, 6, 7} then find B – A.
Solution:
Given A = {1, 2, 3, 4, 5}; B = {4, 5, 6, 7}
B – A = {4, 5, 6, 7} – {1, 2, 3, 4, 5}
B – A = {6, 7}

Question 18.
If A = {0, 1, 2} and B = {2, 4} then find n (A ∪ B)
Solution:
Given A = {0, 1, 2}, B = {2, 4}
A ∪ B = {0, 1, 2, 4}
∴ n(A ∪ B) = 4

Question 19.
If A’ is the set of all primes below ‘5’ and ‘B’ is the set of all prime factors of ’30’, then is A – B = B – A?
Solution:
Given A = Set of all primes below 5 = {2, 3}
B = Set of all prime factors of 30 = {2, 3, 5}
A – B = { } and B – A = { 5 }
∴ A – B ≠ B – A

Question 20.
Represent the following through Venn- diagram.
i) A – B
ii) B – A
iii) A ∪ B
iv) A ∩ B
Solution:
i) Venn-diagram of A – B is

ii) Venn-diagram of B – A is

iii) Venn-diagram of A ∪ B is

iv) Venn-diagram of A ∩ B is

Question 21.
If A = (x : Y is a Natural number below 10}
B = {x : Y is an even number below 10}
C = {x : Y is an odd number below 10}then
find (i) A – B
(ii) A – C
(iii) B ∪ C
(iv) Also mention the Mutually disjoint sets among (i), (ii) and (iii).
Solution:
Given A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
B = {2, 4, 6, 8}
C = {1, 3, 5, 7, 9}
i) A – B = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}

ii) A – C = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 3, 5, 7, 9}
= {2, 4, 6, 8}

iii) B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9}

iv) (i) and (ii) are disjoint sets because there is no element in common.
(ii) and (iii) are not disjoint sets because there are 2, 4, 6, 8 common elements.
(i) and (iii) are not disjoint sets because there are 1, 3, 5, 7, 9 common elements.

Question 22.
If A = {1, 4, 9, 16, 25, . . . } then write it in set builder form.
Solution:
A = {1, 4, 9, 16, 25, ……..}
= {1², 2², 3², 4², 5², . . . . }
= {x²/ x ∈ N}

Question 23.
If A = {x / x is a prime number and x < 20} then B = {x / 2x + 1, x ∈ w and x < 9} then Find
(i) A ∩ B
(ii) B ∩ A
(iii) A – B
(iv) B – A. What do you observe ?
Solution:
A = {x / x is a prime number and x < 20}
B = {2x + 1, x ∈ w and x < 9}
∴ A = {2, 3, 5, 7, 11, 13, 17, 19}
B = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
1) A ∩ B = {2, 3, 5, 7, 11, 13, 17, 19} ∩ (1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
= {3, 5, 7, 11, 13, 17, 19}

2) B ∩ A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} ∩ {2, 3, 5, 7, 11, 13, 17, 19}
= {3, 5, 7, 11, 13, 17, 19}

3) A – B = {2, 3, 5, 7, 11, 13, 17, 19} – {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
= {2}

4) B – A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} – {2, 3, 5, 7, 11, 13, 17, 19}
= {1, 9, 15}
Note : i) A ∩ B = B ∩ A
ii) A – B ≠ B – A

These TS 10th Class Maths Chapter Wise Important Questions Chapter 10 Mensuration given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 10 Mensuration

Previous Exams Questions

Question 1.
Find the volume and total surface area of a hemisphere whose radius is 35 cm ?
Solution:
Radius of the hemisphere (r) = 35 cm.
Volume of the hemisphere = \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 35 × 35 × 35
= \(\frac{269500}{3}\) cm³ = 89833 \(\frac{1}{3}\) cm³
Total surface area = 3πr²
= 3 × \(\frac{22}{7}\) × 35 × 35 = 11550 cm²

Question 2.
The radius of a conical tent is 5m and its height is 12m. Calculate the length of the canvas used in making the tent if width of canvas is 2 cm.
Solution:
Radius of the conical tent (r) = 5 m.
Height of the tent (h) = 12 m.
Slant height of the cone
(l) = \(\sqrt{r^2+h^2}\)
= \(\sqrt{5^2+12^2}\)
= \(\sqrt{25+144}\) = \(169\) = 13 m
Now, surface area of the tent = πrl
= \(\frac{22}{7}\) × 5 × 13 = \(\frac{1430}{7}\) m
Area of the canvas used = \(\frac{1430}{7}\) m
It is given that the width of the canvas = 2 m
Length of the canvas used = \(\frac{\text { Area }}{\text { Width }}\)
= \(\frac{1430}{7}\) × \(\frac{1}{2}\) = 102.14 m

Question 3.
If a cylinder and cone are of the same radius and height, then how many cones full of milk can fill the cylinder ? Answer with reasons. (Mar. ’15 (T.S)
Solution:
Volume of the cylinder is three times to the volume of the cone if they have same radius and height.

Question 4.
A medicine capsule is in the shape of a cylinder with two hemispheres stock to each of its ends. If the length of cylinder) part is 14mm and the diameter of hemi-sphere is 6 mm, then find the volume of medicine capsule. (Mar. ’15 (T.S)
Solution:
Volume of the capsule.

= volume of the cylinder + 2x volume of the hemisphere
= πr²h + 2 × (\(\frac{2}{3}\) πr³)
= \(\frac{22}{7}\) × 3² × 14 + \(\frac{4}{3}\) × \(\frac{22}{7}\) × 3²
= \(\frac{22}{7}\) × 16
= 16π cubic millimetre.

Question 5.
State the relation between r and l (slant height) of a cone.
Solution:
Slant height (l) = \(\sqrt{r^2+h^2}\)

Question 6.
The radius of a spherical balloon in creases from 7 cm to 14 cm as air is pumped into it. Find the ratio of volumes of balloon before and after pumping thef air. (Mar. ’15 (T.S.))
Solution:
Radius of radius = r₁ : r₂ = 7 : 14 = 1 : 2
Ratio of volumes = r₁³ : r₂³ = 1³ : 2³ = 1 : 8

Question 7.
A conical solid block is exactly fitted inside the cubical box of side ‘a’ then the volume of conical solid block is \(\frac{4}{3}\)πa³. Is this statement true. Justify. (Mar. ’16 (T.S))
Solution:
The cone is exactly fitted inside the cubical box.

So height of cone = a = side of cube a
Radius of cone = \(\frac{\mathrm{a}}{2}\)
Volume of cone = \(\frac{1}{3}\)πr² h
= \(\frac{1}{3}\) . π . \(\left(\frac{\mathrm{a}}{2}\right)^2\) . a = \(\frac{1}{3}\) . π . \(\frac{\mathrm{a}^2}{4}\) . a
= \(\frac{1}{3} \frac{\pi a^3}{4}\) = \(\frac{1}{12}\) πa³
But to say \(\frac{4}{3}\) πa³ is not correct.

Question 8.
If the surface area of a hemisphere is ‘S’ then express ‘r’ interms of ‘s’. (Mar. ’16 (T.S))
Solution:
The surface area of a hemisphere = 2πr² = S
⇒ r² = \(\frac{S}{2 \pi}\)
∴ r = \(\sqrt{\frac{5}{2 \pi}}\)

Question 9.
Find the volume and surface area of a sphere of radius 42 cm.
Solution:
Radius of the sphere (r) = 42 cm
Curved surface area (A) = 4πr²
= 4 × \(\frac{22}{7}\) × 42 × 42 = 22,176 cm²
Volume of sphere V = \(\frac{4}{3}\) πr²
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 42 × 42 × 42
= 3,10,464 cm²

Question 10.
A solid metallic ball of volume 64 cm3 is melted and made into a solid cube. Find the side of solid cube. (Mar. ’16 (T.S))
Solution:
Volume of solid metal = volume of cube
= 64 cm³
∴ Volume of cube = s³ = 64 cm
S = \(\sqrt[3]{64}\) = 4 cm
So side of the cube = 4 cm

Question 11.
DWACRA is supplied cuboidal shaped wax block with measurements 88 cm x 42 cm x 35 cm. From this how many number of cylinderical candles of 2.8 cm diametre and 8 cm of height can be prepared ? (Mar. ’16 (T.S))
Solution:
Shape of wax block = cuboid
Its length (l) = 88 cm
breath (b) =42 cm
height (h) = 35 cm
Then the volume of wax present in block
= lbh = 88 × 42 × 35 cm³ ………….. (1)
Shape of candle = cylinder
Diameter of candle = (d) = 2.8 cm
⇒ radius = r = \(\frac{2.8}{2}\) = 1.4 cm
height (h) = 8 cm
Volume of wax required to make one candle
V = πr²h
= \(\frac{22}{7}\) × 1.4 × 1.4 × 8 cm³
∴ Total number of candles that can be
Total volume of block made = \(\frac{\text { Total volume of block }}{\text { Volume of each candle }}\)
= \(\frac{88 \times 42 \times 35}{\frac{22}{7} \times 1.4 \times 1.4 \times 8}\) = 2625
So, 2625 candles can be made with given measurements.

Additional Questions

Question 1.
A company wants to manufacture 500 hemi-spherical basins from a copper sheet. If the radius of basin is 14 cm, find the area of copper sheet to manufacture the above hemi-spherical basins.
Solution:
Radius of the hemi-spherical basin = r = 14 cm
Surface area of hemi-spherical basin = 2πr²
= 2 × \(\frac{22}{7}\) × 14 × 14
= 1232 cm²
Hence, area of the copper sheet required for one basin = 1232 cm²
∴ Total area of copper sheet required for 500 basins
= 1232 × 500
= 616000 cm²
= 61.6 m²

Question 2.
A toy is in the form of right circular cone whose base radius 5 cm and height is 12cm. Find the area of sheet required to make 10 such toys.
Solution:
Radius of the toy = r = 5 cm
Height of the toy = h = 12 cm
Slant height of the toy = l = \(\sqrt{r^2+h^2}\)
= \(\sqrt{5^2+12^2}\)
= \(\sqrt{25+144}\)
= \(\sqrt{169}\)
= 13 cm
Lateral surface area of the toy = πrl
L.S.A = \(\frac{22}{7}\) × 5 × 13 = 204.28 cm²
∴ Area of the sheet required for 10 caps
= 10 204.28 = 2041.8 cm²

Question 3.
The shape of solid iron rod is a cylindrical. Its height is 20cm and base diameter is 14 cm. Then find the total volume of 60 such rods.
Solution:
Diameter of the cylinder = d = 14 cm
∴ Radius of the base = r = \(\frac{\mathrm{d}}{2}\) = \(\frac{14}{2}\) = 7 cm
Height of the cylinder = h = 20 cm.
Volume of the cylinder = v = πr²h
= \(\frac{22}{7}\) × 7 × 7 × 20
= 3.080 cm³
∴ Total volume of 60 rods = 60 × 3080
= 1,84,800 cm³

Question 4.
A heap of rice is in the form of a cone of diameter 10 m and height 12m. Find its volume. How much Canvas cloth is required to cover the heap ?
Solution:
Diameter of the heap = d = 10 m
∴ Radius = r = \(\frac{\mathrm{d}}{2}\) = \(\frac{10}{2}\) = 5 m
Height of the cone = h = 12 cm
Slant height = l = \(\sqrt{r^2+h^2}\)
= \(\sqrt{5^2+12^2}\)
= \(\sqrt{25+144}\)
= \(\sqrt{169}\)
= 13 cm
Volume of the heap of rice = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 5 × 5 × 12
= 314.29 m³
Canvas cloth required to cover the cap = surface area of the conical heap = πrl
= \(\frac{22}{7}\) × 5 × 13
= 204.29 m²

Question 5.
The curved surface area of a cone is 5170 cm² and its diameter is 84 cm. What is its slant height ?
Solution:
Diameter of the base of the cone = d = 84 cm
∴ Radius of the base = r = \(\frac{\mathrm{d}}{2}\) = \(\frac{84}{2}\) =42 cm
Curved surface area of the cone = πrl
By problem, πrl = 5170
\(\frac{22}{7}\) × 42 × l = 5170
⇒ l = \(\frac{5170 \times 7}{22 \times 42}\) = 39.16cm.
Hence, the slant height of the cone = 39.16 cm.

Question 6.
A cone of height 32 cm and radius 8 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.
Solution:
Given Height of the cone = h = 32 cm.
radius of the cone = r = 8 cm.
Let R = Radius of the sphere.
Since cone resphaes in the form of a sphere Volume of the cone = volume of the sphere
⇒ \(\frac{1}{3}\) πr²h = \(\frac{4}{3}\) πR³
⇒ \(\frac{1}{3}\) × \(\frac{22}{7}\) × 8 × 8 × 32 = \(\frac{4}{3}\) × \(\frac{22}{7}\) × R³
⇒ 8 × 8 × 8 = R³
⇒ R³ = 8³
⇒ R = 8
Hence the radius of the sphere = R = 8 cm.

Question 7.
A metallic sphere of radius 4.9 cm is melted and recast in to the shape of a cylinder of radius 7 cm. Find the height of the cylinder.
Solution:
Let height of cylinder = h
Given radius of sphere = R = 4.9 cm
and Radius of cylinder = r = 7 cm
since sphere is melted in to the shape of a cylinder.
volume of the sphere = volume of the cylinder
⇒ \(\frac{4}{3}\) πR³ = πr²h
⇒ \(\frac{4}{3}\) × \(\frac{22}{7}\) × (4.9)³ = \(\frac{22}{7}\) × 7² × h
⇒ h = \(\frac{4}{3}\) x \(\frac{4.9 \times 4.9 \times 4.9}{7 \times 7}\)
h = 3.2 cm.
Hence the height of the cylinder = h = 3.20 cm

Question 8.
The rain water from a roof of 33m × 30m drains into a cylindrical vessel having diameter of base 4m and height 4.2m. If the vessel is just full, find the rain fall in cm.
Solution:
Dimensions of the roof = 33m × 30m
Diameter of cylindrical vessel = d = 4m
∴ Radius = r = \(\frac{\mathrm{d}}{2}\) = \(\frac{4}{2}\) = 2m
and height of vessel = h = 4.2 m. :
Let H = height of the rain fall
Now, volume of the rain fall = volume of the drained water in the cylindrical vessel
33 × 30 × H = πr²h
⇒ 33 × 30 × H = \(\frac{22}{7}\) × 2 × 2 × 4.2
⇒ H = \(\frac{22}{7}\) × \(\frac{2 \times 2 \times 4.2}{33 \times 30}\)
= 0.053 cm
H = 5.3 cm
∴ Rain fall = 5.3 cm

Question 9.
The radii of the internal and external surfaces of a hollow spherical shell are 2 cm and 4 cm respectively. If it is melted and: recast in to a solid cylinder of radius 2\(\sqrt{2}\) cm. find the height of the cylinder.
Solution:
Given radii of the hollow spherical shell = 2 cm and 4 cm
Let r₁ = 4 cm, r₂ = 2 cm and h = height solid! cylinder and Radius of solid cylinder = r =1
2\(\sqrt{2}\) cm. since hollow spherical shell melt and recast in to solid cylinder.
∴ volume of the follow spherical shell = volume of the solid cylinder
⇒ \(\frac{4 \pi}{3}\) (r₁³ – r₂³) = π – r²h
⇒ \(\frac{4}{3}\) (r₁³ – r₂³) = r²h
⇒ \(\frac{4}{3}\) (4³ – 2³) = (2\(\sqrt{2}\))²h
⇒ \(\frac{4}{3}\) (64 – 8) = 4 × 2 × h
⇒ \(\frac{4}{3}\) × 56 = 8 × h
⇒ h = \(\frac{4}{3}\) × \(\frac{56}{8}\) = \(\frac{28}{3}\) cm
∴ Height of the cylinder = h = \(\frac{28}{3}\) cm

Question 10.
A cylindrical bucket, 36 cm high and with radius of base 16 cm is filled with rice. This bucket is emptied out on the ground and a conical heap of rice is formed. If the height of the comical heap is 27 cm. find the radius d slant height of the heap.
Solution:
Given height of cylindrical bucket = 36 cm = h
and radius = 16 cm = r
Let height of conical heap = 27cm = H
and Radius of conical heap = R
slant height of the heap = I
Since the rice is carried and emptied by cylindrical bucket to form a conical heap,
Volume of cylindrical bucket = volume of conical heap.

Hence, Radius of heap = R = 32 cm
and slant height of the heap = l = 41.86 m

Question 11.
Find the total surface area of a hemi-sphere of radius 2.8 cm.
Solution:
Given radius of hemi-sphere = 2.8 cm
= 3 × \(\frac{22}{7}\) × (2.8)²
= 3 × \(\frac{22}{7}\) × 2.8 × 2.8
= 73.92 cm²

Question 12.
A cone of height 16 cm and radius of base 4 cm is made up modeling clay. A child u9 reshapes it into a sphere find the radius of the sphere.
Solution:
Given height of the cone = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) π × (4)² × 16
= \(\frac{1}{3}\) π × 4 × 4 × 16
Since, the volume of the clay is in the form of the cone and the sphere remains the same, we have
volume of sphere = volume of the cone
\(\frac{4}{3}\) πr³ = \(\frac{1}{3}\) π × 4 × 4 × 16
r³ = \(\frac{1}{3}\) × 4 × 4 × 16 × \(\frac{3}{4}\)
r³ = 64 = 4³
∴ r = 4
∴ The radius of the sphere = 4

Question 13.
A storage tank consists of a circular cylinder with a hemi-sphere stuck on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m find the cost of painting it on the out side at rate of ₹ 20 per m².
Solution:
Total surface area of the tank
= 2 × C.S.A of the hemisphere +C.S.A of cylinder

Hemisphere :
Radius (r) = \(\frac{\text { diamater }}{2}\)
= \(\frac{1.4}{2}\) = 0.7 m
C.S.A of hemi-sphere = 2 πr²
= 2 × \(\frac{22}{7}\) × 0.7 × 0.7
= 3.08 m²
2 × C.S.A = 2 × 3.08 m² = 6.16 m²
Cylinder : Radius (r) = \(\frac{\text { diamater }}{2}\) = \(\frac{1.4}{2}\) = 0.7 m
Height (h) = 8 m
C.S.A. of cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × 0.7 × 8 = 35.2 m²
∴ Total surface area of the cylinder storage tank = 35.2 + 6.16 = 41.36 m²
cost of painting its surface area @ 20 per sq.m is = 41.36 × 20 = ₹ 827.2

Question 14.
How many spherical balls can be made out of a solid cube of lead whose edge measures 44 cm and each ball being 4 cm in diameter.
Solution:
Side of lead cube = 44 cm
Radius of spherical balls = \(\frac{4}{2}\) cm = 2 cm

Now, volume of spherical ball = \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 2³ cm³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 8 cm³
Volume of x spherical balls = \(\frac{4}{3}\) × \(\frac{22}{7}\) × 8 × x cm³
It is clear that volume of x spherical balls = volume of lead cube.
⇒ \(\frac{4}{3}\) × \(\frac{22}{7}\) × 8 × x = 44 × 44 × 44
⇒ x = \(\frac{44 \times 44 \times 44 \times 3 \times 7}{4 \times 22 \times 8}\)
x = 2541
Hence, total number of spherical balls = 2541

Question 15.
Explain the terms in the formula
V = l × b × h
Solution:
Given V = l × b × h
when v = volume
l = length
b = breath
h = height

Question 16.
Self help group wants to manufacture Joker’s caps (conical caps) of 6 cm radius and 8 cm hight. If the available colour paper sheet is 1000 cm² then how many caps can be manufactured from that paper.
Solution:
Radius of Joker’s cap = r = 6 cm
height of Joker’s cap = h = 8 cm
Now l = \(\sqrt{r^2+h^2}\) = \(\sqrt{6^2+8^2}\)
= \(\sqrt{36+64}\) = \(\sqrt{100}\) = 10 cm
Now, surface area of one cap = πrl
= \(\frac{22}{7}\) × 6 × 10 = 188.57 cm²
But area of available colour paper sheet = 1000 cm²
Number of caps Manufactured from that colour paper = \(\frac{\text { Area of colour paper sheet }}{\text { Surface area of are cap }}\)
= \(\frac{1000}{188.57}\) = 5.303 = 5

Question 17.
A cone of height 24 cm and radius of base 6 cm is made up modelling clay. A child reshapes it into a sphere. Find the radius of the sphere.
Solution:
Volume of cone = \(\frac{1}{3}\) π × 6 × 6 × 24 cm³
If r is the radius of the sphere, then its volume is \(\frac{4}{3}\) πr³
Since the volume of clay in the form of the cone and the sphere remains the same, we have
\(\frac{4}{3}\) πr³ = \(\frac{1}{3}\) π × 6 × 6 × 24
r³ = 3 × 3 × 24 = 3 × 3 × 3 × 8
r³ = 3³ × 2³
r = 3 × 2 = 6
Therefore the radius of the sphere is 6 cm.

Question 18.
A rectangular park is to be designed. Its breadth is 3m less than its length. Its area is to be 4 square meters more than the area of park that has already been made in the shape of an isosceles triangle with base as the breadth of the rectangular park and altitude 12m. Find the length and breadth.
Solution:
Let the breadth of the rectangular park be x m.
So, its length = (x + 3) m.
Therefore, the area of the rectangular park
= x(x + 3)m² = (x² + 3x)m².
Now, base of the isosceles triangle = x m.

Therefore, its area = \(\frac{1}{2}\) × x × 12
= 6x m²
According to our requirements,
x² – 3x = 6x + 4
i.e., x² – 3x – 4 = 0
Using the quadratic formula, we get
x = \(\frac{3 \pm \sqrt{25}}{2}\) = \(\frac{3+5}{2}\) = 4 or -1.
But x ≠ -1. Therefore, x = 4.
So, the breadth of the park = 4m and its length will be x + 3 = 4 + 3 = 7m.
Verification :
Area of rectangular park = 28 m²
Area of triangular park = 24 m² = (28 – 4) m².

Question 19.
To calculate the quantity of milk inside a bottle, we need to find out …………… [ ]
a) Area
b) Volume
c) Density
d) Total surface area

Question 20.
The height of right angle triangle is 7 cm less than the base, the length of the diagonal is 17cm, then the length of remaining two sides are …………. [ ]
a) 15cm, 8 cm
b) 12 cm, 5 cm
c) 24 cm, 17 cm
d) All above

These TS 10th Class Maths Chapter Wise Important Questions Chapter 9 Tangents and Secants to a Circle given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 9 Tangents and Secants to a Circle

Previous Exams Questions

Question 1.
What do we call the part a and b in the below circle ? (A.P. June ’15)

Solution:
a is minor segment and ‘b’ is major segment.

Question 2.
How many tangents can be drawn to a circle from a point on the same circle, Why? (T.S. Mar. ’15)
Solution:
The number of tangents that can be drawn to a circle from a point on it is one. This is as per concept of tangents.

Question 3.
Draw a circle with radius 3 cm and construct a pair of tangents from a point 8 cam away from the centre. (T.S. Mar. ’15)
Solution:

AP, BP are joining rays

Question 4.
Construct and measure the length of a pair ? of tangents that are drawn from a point at a distance of 8 cm whose radius is 5 cm. (T.S. Mar. ’16)
Solution:
Steps of construction :

1. Construct a circle with a radius of 5 cm.
2. Take the point ‘p’ in the exterior of the circle which is at a distance of ‘8’ cm from its centre.
3. Construct a perpendicular bisector to OP which meets at M.
4. The draw a circle with a radius of MP or MO from the point M. This circle cuts the previous circle drawn from the centre ‘O’ at the points A and B.
5. Now join the points PA and then PB.
6. PA, PB are the required tangents which are measured 6.2 cm long.
   

OA = 5 cm
OP = 8 cm
AP = PB = 6.2 cm.

Additional Questions

Question 1.
Calculate the length of Tangent from a point 13 cm away from the centre of a circle of radius 5 cm.
Solution:
In ∆OTR ∠OTP = 90°
OT = 5 cm, OP = 13 cm
By Pythagoras theorem

OP² = OT² + PT²
13² = 5² + PT²
⇒ PT² = 13² – 5²
= 169 – 25 = 144
PT = \(\sqrt{144}\) = 12 cm
∴ Length of the required tangent = PT = 12 cm

Question 2.
Two concentric circles of radii 25 cm and 24 cm are drawn. Find the length of the chord of the larger circle which touches the smaller circle.
Solution:
Given two circles of radii 25 cm and 24 cm with common centre.
Let AB be the tangent to the inner circle and chord to the larger circle. Let ‘P’ be the point of contact.

In the figure ∠OPB = 90°
OP = 24 cm, OB = 25 cm
Now OB² = OP² + PB²
25² = 24² + PB²
⇒ 625 = 576 + PB²
⇒ PB2 = 625 – 576 = 49
PB = \(\sqrt{49}\) = 7 cm
Now AB = 2 × PB = 2 × 7 = 14 cm
∴ Length of the chord to the larger circle = AB = 14 cm

Question 3.
Find the area of a quadrant of a circle whose circumference is 88 cm.
Solution:
Given circumference of a circle = 88 cm
Let r = Radius of the circle
Now 2πr = 88
⇒ r = \(\frac{88}{2 \pi}\) = \(\frac{88}{2 \times \frac{22}{7}}\)
= \(\frac{88 \times 7}{2 \times 22}\) = 14 cm
Area of the circle = πr² = \(\frac{22}{7}\) × (14)²
= \(\frac{22}{7}\) × 14 × 14
Area of the quadrant of a circle = \(\frac{1}{4}\) (πr²)
= \(\frac{1}{4}\) × \(\frac{22}{7}\) × 14 × 14
= 154 cm²

Question 4.
The length of the minute hand of a clock is 21 cm. Find the area swept by the volume by the minute hand in 20 minutes.
Solution:
Angle made by minute hand in 1 min = \(\frac{360^{\circ}}{60^{\circ}}\)
= 6°
Angle made by minute hand in 20 min
= 20 × 6° = 120°
The area swept by minute hand is in the shape of a sector with radius.
r = 21 cm and angle x = 120°
Area (A) = \(\frac{\mathbf{x}^{\circ}}{360^{\circ}}\) × πr²
= \(\frac{120^{\circ}}{360^{\circ}}\) × \(\frac{22}{7}\) × 21 × 21
= 462 cm²
∴ Area swept by minute hand in 20 minutes = 462 cm².

Question 5.
Two circle touch internally. The sum of their area is 125π cm² and distance between their centre is 5 cm. Find the radii of the circles.
Solution:
Let r₁ and r₂ be the radii of the given circles.
Sum of the areas of two circles
⇒ πr₁² + πr₂² = 125 π
⇒ π(r₁² + r₂²) = 125 π
⇒ r₁² + r₂² = 125 …………………… (1)
Distance between the centres = r₁ – r₂ (r₁ > r₂)
= 5 cm
Now (r₁ – r₂)² = (r₁² + r₂²) – 2r₁r₂
(5)² = 125 – 2r₁r₂ from (1)
⇒ 2r₁r₂ = 125 – 25
2r₁r₂ = 100 …………….. (2)
Then (r₁ + r₂)² = r₁² + r₂² + 2r₁r₂
= 125 + 100 (∵ from (1) and (2))
(r₁ + r₂)² = 225
⇒ r₁ + r₂ = \(\sqrt{225}\) = 15

and r₁ – r₂ = 5
10 – r₂ = 5
⇒ r₂ = 10 – 5 = 5 cm
⇒ r₁ = \(\frac{10}{2}\) = 10 cm
∴ r₁ = 10 cm and r₂ = 5 cm

Question 6.
Find the area of the shaded region in the figure, given in which two circles with centres A and B touch each other at the point P. If AP = 10 cm and AB = 5 cm.
Solution:
AP denotes the radius of the bigger circle, i.e.,
AP = 10 cm
= r₁ (say)

AB denotes the distance between the centres of two circles i.e
A – B = 5 cm
BP denotes the radius of the smaller Circle i.e.,
BP = AP – AB
= 10 – 5 = 5 cm = r₂ (say)
The area of the shaded region = Area of the bigger circle – Area of the smaller area
= πr₁² – πr₂²
= π(r₁² – r₂²)
= π(10² – 5²2)
= π(100 – 25)
= π(75)
= \(\frac{22}{7}\) × 75
= 235.71 cm²

Question 7.
O is any point inside a rectangle ABCD prove that OB² + OD² = OA² + OC².
Solution:
Here PQ || BC so that
‘P’ is on AB; Q is on CD
If PQ || BC then
PQ ⊥ AB and PQ ⊥ CD
∴ ∠B = ∠C = 90°
∠BPQ = 90°
∠CQP = 90°

So that BPQC and APQD are the rectangles.
From ∆OPB, OB² = BP² + OP² ……………. (1)
and ∆OQD, OD² = OQ² + DQ² ………………. (2)
From ∆OQC, OC² = OQ² + CQ² ……………. (3)
From ∆OAR, OA² = AP² + OP² …………….. (4)
(1) + (2) OB² = BP² + OP²
OD² = OQ² + DQ²
OB² + OD² = BP² + OQ² + OP² + DQ²
From figure BP = CQ and DQ = AP
∴ OB² + OD² = CQ² + OQ² + OP² + AP²
From (3) & (4)
OQ² + CQ² = OC²; AP² + OP² = OA²
∴ OB² + OD² = OC² + OA²

Telangana SCERT TS 10th Class Physical Science Study Material Pdf 6th Lesson Structure of Atom Textbook Questions and Answers.

TS 10th Class Physical Science 6th Lesson Questions and Answers Structure of Atom

Improve Your Learning
I. Reflections on concepts

Question 1.
What information does the electronic configuration of an atom provide?
Answer:

1. The distribution of electrons in shells, sub-shells arid orbitals in an atom is ‘known as electronic configuration.
2. The distribution of electrons In various atomic orbitals provides an understanding of the electronic behaviour of the atom and in turn its reactivity.
3. The short-hand notation is as shown below.

Question 2.
Rainbow is an example for continuous spectrum – explain.
Answer:

1. A rainbow is a natural spectrum appearing in the sky while it is drizzling and the sun is in the east or west.
2. It is caused by dispersion of sunlight by tiny water droplets present in atmosphere. It consists of 7 colours.
3. In a rainbow, there are no sharp boundaries in between colours.
4. Such a spectrum in which there are no sharp boundaries in between colours is known as continuous spectrum.
5. So, rainbow is also a continuous spectrum.

Question 3.
What is an orbital? How is it different from Bohr’s orbit?
Answer:
The region or space around the nucleus where the probability of- finding the electron is maximum is called an orbital.

1. Bohr’s orbit has a definite boundary and fixed energy at different distances from the nucleus. They are circular In shape.
2. OrbitaIs have no definite boundary. It Is a region where we find maximum possibility of electrons. The shape of each orbital is different.
3. Bohr’s orbit can accommodate maximum of (2n²) electrons ¡n it, but each orbital can accommodate only 2 electrons.

Question 4.
Explain the significance of three Quantum numbers in predicting the positions of an electron in an atom.
Answer:
Each electron In an atom Is described by a set of three quantum numbers n, l, and ml.
(1) PrincIpal Quantum number (n): The principal quantum number is used to describe the size and energy of the main shell. It Is denoted by ‘n’. ‘n’ has positive It is use integer values of 1,2,3.
It is used to know the number of orbitals (n²) and electrons in an orbit. (2n²).

As ‘n’ increases the shells become larger and the electrons in those shells are farther from the nucleus.

2. The angular-momentum quantum number (l) :
from ‘0’ to n – 1, for each value of ‘n’. Each ‘l’ value represented one sub-shell. It is used to describe the shape of an orbit.

3. The magnetic quantum number (m_l): The magnetic quantum number (m_l) has integer values between – l and + l including zero.
If l= 0, the possible m_l, value is l
l=1, the possible m_l, value is- 1,0 and 1.

Thus for a certain value of 1, there are (2l + 1) integer values of ml.
These values describe the orientation of the orbital in space relative to the other orbitals in the atom,
Eg: When l = 1, (2l + 1) = 3, that means ni has 3 values namely – 1, 0, 1 or three p orbitals, with different orientations along X, Y, Z axes, labelled as p_x, p_y and p_z orbitals. Predicting the position of an electron in an atom: If the values of n, l and m₁ are 2, 1, -1 respectively the electron is present in 2p_x orbital in L shell.

Question 5.
What is nl^x method? How It is useful?
Answer:
The shorthand notation consists of the principal energy level (n value) the letter representing sub-level (l value) and the number of electrons In the sub-shell is written as superscript nl^x It is useful in writing electron configurations of elements. For example in hydrogen (H), the set of quantum numbers Is n = 1,l = 0, m_l= 0, m_s= 1/2 or -1/2
The electronic configuration is

Question 6.
Which electronic shell Is a higher energy level K or L?
Answer:

1. L shell has higher energy because according to Bohrs theory the shell which Is closer to nucleus has lower energy and the shell which is away from the nucleus has higher energy.
2. K is closer to nucleus. So it has lower energy than L-shell.

Question 7.
What Is absorption spectrum?
Answer:
Absorption spectrum: The spectrum formed by the absorption of energy when electron jumps from lower energy level to higher energy level is called absorption Spectrum. It contains dark lines on bright background.

Question 8.
What is emission spectrum?
Answer:
Emission spectrum is the spectrum of frequencies of electromagnetic radiation due to an atom’s electron making a transition from a high-energy state to low energy state.

Application Of Concepts

Question 1.
Answer the following questions.
a. How many maximum number of electrons that can be accommodated In a principal energy level?
Answer:
The maximum number of electrons that can be accommodated in a principal energy shell is given by the rule 2n², where ‘n’ is the principal quantum number.

b. How many maximum numbers of electrons that can be accommodated In a sub-shell?
Answer:

1. each subshell holds a maximum of twice as many electrons as the number of orbitals in the sub-shell.
2. The maximum number of electrons that can occupy various sub-shells is

c. How many maximum numbers of electrons can be accommodated In an orbital?
Answer:
The maximum number of electrons that can be accommodated In an orbital is 2.

d. How many sub-shells are present In a principal energy shell?
Answer:
The number of sub-shells present in a principal energy shell is equal to principal quantum number ‘n’. Eg:

e. How many spin orientations are possible for an electron in an orbital?
Answer:
Two spin orientations are possible for an electron in an orbital, one clockwise and the other anticlockwise spin. There are represented by + 1/2 and – 1/2.

Question 2.
In an atom the number of electrons In M-shell is equal to the number of electrons in the K and L shells. Answer the following questions.
(a) Which is the outermost shell?
(b)How many electrons are there In its outermost shell?
(C) What is the atomic number of element?
(d) Write the electronic configuration of the element.
Answer:
No. of electrons in M’ shell Is equal to number of electrons In the K and L shells.
∴ No. of electrons n M shell = 2 + 8 = 10
Total electrons in the given atom = 2(K) + 8(L) + 10 (M) = 20
The electronic configuration Is = Is²2s²2p⁶3s²3p⁶4s²3d²

(a) M shell is the outermost shell.
(b) There are ‘2’ electrons in the outermost shell.
(c) The atomic number of the element is 20.
(d) Electronic configuration of the element: Is²2s²2p⁶3s²3p⁴4s²

Question 3.
How many elliptical orbits are there in third Bohr’s orbit?
Answer:

1. To explain the splitting of line spectra, Sommerfeld modified Bohr’s atomic model by adding elliptical orbits.
2. RetaIning the first of Bohr’s circular orbit as such, Sommerfeld added two elliptical orbits to Bohr’s third orbit.

Question 4.
Following orbital diagram shows the electron configuration of nitrogen atom. Which rule does not support this? N (Z = 7)

Answer:

This electronic configuration of nitrogen is not correct.
It does not support Hund’s rule which states, the orbitals of equal energy are occupied by one electron each before pairing of electrons starts.
So the correct electron configuration is:
N (Z=7)

Question 5.
(i) An electron in an atom has the following set of four quantum numbers. To which orbital does it belong?

(ii) Write the four quantum numbers for is’ 1S¹ electron.
Answer:
(i) Given n = 2 and l= 0 represent’s’orbital so the orbital is 2s and s= + ½ so by n1^x method it is 2s¹.
(ii) The four quantum numbers for Is¹’ electron is

Question 6.
The wavelength of a radio wave is 1.0 m. Find its frequency.
Answer:
Given:
Wavelength λ = 1.0 m
We know that velocity of light c = vλ
Where c = 3 x 10⁸ m/s
∴ 3 × 10⁸ = v.1
v= 3 x 10⁸
So, frequency of wave = 3 x 10⁸ Hertz

Question 7.
Which rule Is violated in the electronic configuration is° 25² 2p⁴?
Answer:
In the above electronic configuration Aufbau principle is violated.
(n+ 1) valueof isis 1+0 = 1
(n + 1)valueof2sis2 +0 = 2
Here electrons are filled in 2s Orbital, without completing is orbital which has lower (n + 1) value than 2s. Hence Aufbau principle is violated.

Question 8.
Write the four quantum numbers for the differentiating electron of sodiun (Na atom).
Answer:
(i) Electron configuration of sodium (Z = 11)

(ii) The differentiating electron is in 3s¹. The four quantum numbers of this electron are

Question 8.
Collect the information regarding wavelengths and corresponding frequencies of three primary colours red, blue and green.
Answer:

Red colour has more wavelength among the seven colours of VIBGYOR. So it can be invisible from larger distances also. So it is used In signal lights.

Multiple choice questions

Question 1.
An emission spectrum consists of bright spectral lines on a dark background. Which one of the following does not correspond to the bright spectral lines? [ ]
(A) Frequency of emitted radiation
(B) Wavelength of emitted radiation
(C) Energy of emitted radiations
(D) Velocity of light
Answer:
(D) Velocity of light

Question 2.
The maximum number of electrons that can be accommodated in the L – shell of an atom is …………………. .[ ]
(A) 2
(B) 4
(C) 8
(D) 16
Answer:
(C) 8

Question 3.
If l=l for an atom then the number of orbitals in its sub-shell is ……………………… .[ ]
(A) 1
(B) 2
(C) 3
(D) 0
Answer:
(C) 3

Question 4.
The quantum number which explains about size and energy of the orbit or shell is: …………………….. . [ ]
(A) n
(B) l
(C) m_l
(D) m_s
Answer:
(A) n

Suggested Projects

Question 1.
Collect the information of historical development of atomic theory.
Answer:
The first atomic theory was proposed by. John Dalton based on law of conservation of mass and law of constant proportions.

The important postulates of his theory are:

1. Atoms were indivisible.
2. Atoms of an element are all identical to each other and different from the atoms of other elements.
   Later on various experiments conducted by Thomson, Goldstein etc. proved that atom is divisible and consists of sub-atomic particles like electrons, protons and neutrons. Based on this J.J.

Thomson proposed a model of atom in 1898. According to Thomson,

1. An atom is considered to be a sphere of uniform positive charge and electrons are embedded in It.
2. The total mass of the atom is considered to be uniformly distributed throughout the atom.
3. The negative and the positive charges are supposed to be balanced out and the atom as a whole is electrically neutral.

This model is also called as plum pudding model or watermelon model. Thomson’s student Ernest Rutherford conducted alpha particle scattering experiment got the results which were not in favour of Thomson’s model.

Based on his experiment, Rutherford proposed a model of atom. According to him,

1. All the positively charged material In an atom formed a small dense Centre called the nucleus of the atom. The electrons were not a part of nucleus.
2. Negatively charged electrons revolve around the nucleus is well-defined orbits like planets revolve around the sun.
3. The size of nucleus is very small as compared to the size of the atom.

This model could not account for stability of atom, as revolving electron must lose energy and eventually crash into the nucleus, as a result matter would not exist in the it he form that we see it now. In 1913 Nells Bohr proposed another model to overcome Rutherford’s defect According to Bohr.

• The electrons revolve around the nuleus In a discrete orbits called stationary, orbits.
• While revolving in these discrete orbits the electrons do not radiate enery and this helps that the electrons do not crash into the nucleus.
• Theseorbitsorshellsare represented by K, L, M, N, ……………………………. or the numbers 1,2,3, …………………….
  This model could not predict the spectra of atoms later Bohr along with Sommerfeld proposed the Bohr-Sommerfeld model of atom to account for the spectra of atoms.

Question 2.
Collect the information about the scientists who developed the atomic theories.
Answer:
(a) John Dalton

1. Chemist John Dalton was born on September 6, 1766, In Eagles field, England.
2. During his early career, he identified the hereditary nature of red-green colour blindness.
3. In 1803, he revealed the concept of Daltons’ law of partial pressures.
4. Also in the 1800s, he was the first scientist to explain the behaviour of atoms in terms of the measurement weight.
5. During the early 1800s, Dalton also postulated a law of thermal expansion that illustrated the heating and cooling reaction of gases to expansion and compression.
6. After suffering a second stroke, Dalton died quietly on the evening of July 26, 1844, at his home in Manchester England.

(b) J.J. Thomson:

1. J.J. Thomson was born on December 18, 1856, in Cheetham Hill, England and went on to attend Trinity College at Cambridge, where he would come to head the Cavendish Laboratory.
2. His research In Cathode rays led to the discovery of electrons and he pursued further innovations In atomic structure exploration.
3. Thomson won the 1906 Nobel Prize in Physics, among many accolades.
4. He died on August 30, 1940.

(c) Ernest Rutherford:

1. Chemist and Physicist Ernest Rutherford was born on August 30, 1871, n Spring Grove, New Zealand.
2. A pioneer of Nuclear Physics and the first to split the atom, Rutherford was awarded the 1908 Nobel Prize in Chemistry for his theory of atomic structure.
3. He was the fourth of 12 children and second son. His father, James, had little education and struggled to support the large family on a flax-miller’s income.
4. Ernest’s Mother, Martha, worked as a school teacher. She believed that knowledge was power, and placed a strong emphasis on her children’s education.
5. Together, Rutherford and Thomson studied the effects of X-rays on the conductivity of gases, resulting Is a paper about dividing atoms and molecules into ions.

6. Famous as the “Father of the Nuclear Age”, Rutherford died in Cambridge, England, on October 19, 1937.

(d) Neils Bohr:

1. (1) Born on October 7, 1885, In Copen Hagan, Denmark, Nells Bohr went on to become an accomplished physicist who came up with a revolutionary theory on atomic structures and radiation emission.
2. (2) He won the 1922 Nobel Prize in physics for his ideas and years later, after working on the Manhattan project in the United States, called for responsible and peaceful applications of atomic energy across the world.
3. (3) In 1957, Bohr received the “Atoms for Peace Award” for his trail-blazing theories and effects to use atomic energy wisely.
4. (4) After having a stroke, he died on November 18, 1962, in Copen hagen. Bohr’s son Aage shared with two others the 1975 Nobel Prize in Physics for his research on motion in atomic nuclei.

Question 3.
Make the s, p and d – orbital models.
Answer:
Material required: 10 iron spokes, 9 round-shaped wooden pieces, and dumbbell-shaped beads having holes along its length.

Procedure:

1. Cut the each iron spoke into three pieces In which one of them Is longer than the other two.
2. Make a hole at the middle of the wooden piece and insert the long spoke into that hole.
3. Fix the other two spokes to the longer spoke horizontally such that the three spokes are perpendicular to each other.
4. Now the three spokes represent X, Y and Z – axis.
5. Make nine of these types of models, one for s-orbital, three for p-orbitals and five for d-orbitals.
6. Fix the dumb-bell-shaped beeds to the Iron spokes as shown in the fig – to make three (P_x, P_y, P₂) P – orbitals.
7. Fix the dumb-bell-shaped beads to the iron spokes as shown in the fig. to make five d-orbitaIs. (d_XY,
   d_YZ,d_ZX,d_X² – y² and d_X²)

TS 10th Class Physical Science Structure of Atom Intext Questions

Page 106

Question 1.
How do these sub-atomic particles coexist in an electrically neutral atom?
Answer:
Protons and neutrons are present at the centre of the atom called nucleus, and electrons revolve around this nucleus in a circular manner.
As the number of protons and the number of electrons in a neutral atrn1s the same, the +ve charges and -ve charges being equal, they coexist.

Question 2.
Do all atoms have the same sub-atomic particles?
Answer:
Yes, all atoms have same sub-atomic particles but differ in number.

Question 3.
Why is an atom of one element different from the atoms of other elements?
Answer:
Because they have different numbers of sub-atomic particles.

Question 4.
How are the electrons distributed In the space of an atom?
Answer:
Electrons are distributed around the nucleus In subshells of orbits.

Page 107

Question 5.
How many colour sare there in a rainbow?
Answer:
There are seven colours namely violet, Indigo, blue, green, yellow, orange and red (VIBGYOR) in a rainbow.

Question 6.
How do the vibrating electric and magnetic fields around the charge become a wave that travels through space?
Answer:
A vibrating electric charge creates a change In the electric field. The changing electric field creates a changing magnetic field. This process continues, with both the created fields being perpendicular to each other and at right angles to the direction of propagation of the wave.

Question 7.
What are the characteristics of electromagnetic waves?
Answer:
The electromagnetic wave is characterized by wavelength ( λ), frequency (v), velocity (c) and amplitude.

Page 108

Question 8.
Can we apply C=vλ equation to a sound wave?
Answer:
Yes. It is a universal relationship and applies to all waves. As the frequency increases, the wavelength becomes smaller.

Question 9.
Are there any other wavelengths of light other than visible spectrum?
Answer:
Yes. Cosmic Rays, Gamma Rays, X-rays, U.V. Ray5, I.R. Rays, Micro Waves, Radio Waves are other waves besides visible spectrum.

Page 109

Question 10.
What happens when you heat an iron rod on a flame? Do you find any change In colour on heating an Iron rod?
Answer:
When we heat an Iron rod, some of the heat energy is emitted as light. First It turns red (lower energy corresponding to higher wavelength) and as the temperature rises It glows in orange, yellow, blue (higher energy and of lower wavelength) or even white (all visible wavelengths) colour If the temperature Is high enough.

Question 11.
Do you observe any other colour at the same time when one colour is emitted?
Answer:
When the temperature Is high enough, other colours will also be emitted, but due to higher Intensity of one particular emitted colour (e.g., red), others cannot be observed.

Question 12.
Do you enjoy Deepavall fireworks?
Answer:
Yes, I enjoy because variety of colours are emitted from fireworks.

Question 13.
How do these colours come from fireworks?
Answer:
Inside each handmade crackers, there are small packets filled with special chemicals, mainly metal salts and metal oxides, which react to produce an array of colours. When heated, the atoms of each element in the mixture absorbs energy, causing its electrons to rearrange from their lowest energy state to a higher ‘excited” state. As the electrons plummet back down to their lower energy state, the excess energy gets emitted as light. Each element releases a different amount of energy, and this energy is what determines the colour or wavelength of the light that is emitted.

Page 110

Take a pinch of cupric chloride in a watch glass and make a paste with concentrated hydrochloric acid. Take this paste on a platinum loop and introduce it into a non-luminous flame.

Question 14.
Do you observe yellow light In street lamps?
Answer:
Yes, Sodium vapours produce yellow light in street lamps.

Question 15.
Why do different elements emit different flame colours when heated by the same non-luminous flame?
Answer:
Each element emits Its own characteristic colour when they come back from excited state to ground state. These colours correspond to certain discrete wavelengths of light.

Question 16.
What does a line spectrum tell us about the structure of an atom?
Answer:
We can identify the atom as to what element It belongs. Each element gives a unique light spectrum. That is how spectroscopy works for identifying the elements that make up a certain substance.

Question 17.
What happens when an electron gains energy?
Answer:
The electron moves to a higher energy level that is, the excited state.

Page 111

Question 18.
Does the electron retain the energy forever?
Answer:
The electron loses the energy and comes back to its ground state. The energy emitted by the electron is seen in the form of electromagnetic energy and when the wavelength is in the visible region it is visible as an emission line.

Question 19.
Did Bohr’s model account for splitting of line spectra of a hydrogen atom Intofiner lines?
Answer:
Bohr’s model failed to account for splitting of line spectra.

Question 20.
Why is the electron in an atom restricted to revolve around the nucleus at certain fixed distances?
Answer:
Because the angular momentum of an electron in an orbit is quantized.

Page 112

Question 21.
Do the electrons follow defined paths around the nudeus?
Answer:
If the electron revolves around the nucleus in definite paths or orbits, the exact position of the electron at various times will be known.

Question 22.
What is the velocity of the electron?
Answer:
3 x 10⁸ m/s

Question 23.
Is It possible to find the exact position of the electron?
Answer:
We can find the probable position of the electron alone but not position and momentum simultaneously.

Question 24.
What do we call the region of space where the electron might be, at a given time?
Answer:
The region or space around the nucleus where the probability of finding the electron Is maximum Is called an atomic orbital.

Page 113

Question 25.
What Information do the quantum numbers provide?
Answer:
The quantum numbers describe the space around the nucleus where the electrons can be found and also their energies. In fact, they give us the address of an electron.

Question 26.
What does each quantum number signify?
Answer:

• The principal quantum number is related to the size and energy of the main shell.
• The azimuthal quantum number gives the shape of a particular sub-shell in the space around the nucleus.
• The magnetic quantum number describes the orientation of the orbital in space relative to the other orbitals in the atom.
• The spin quantum number refers to the two possible orientations of the spin of an electron, one clockwise and the other anticlockwise spin. These are represented by +1/2 and -1/2. If both are positive values, then the spins are parallel otherwise the spins are non-parallel.

Page 114

Question 27.
What is the maximum value of ‘I’ for n=4?
Answer:
Three (Recall, l = n-1)

Question 28.
How many values can ‘I’ have for n=4?
Answer:
9 (2l+1=2×4+1=9)

Question 29.
Do these three p-orbitals have the same energy?
Answer:
The three p-orbitals have same energy and hence these are called degenerate orbitals.

Page 116

Question 30.
How do electrons In an atom occupy shells, sub-shells and orbitals?
Answer:
Electrons occupy shells, sub-shells and orbitals according to Hund’s rule, Pauli’s rule and Aufbau’s principle.

Question 31.
Helium (Z=2) atom has two electrons. How are these two electrons arranged?
Answer:
Helium atom has two electrons. The first electron occupies ‘is’ orbital. The second electron joins the first In the 1s-orbital, and so the electron configuration of the ground state of He’ is 1s². According to Pauli Exclusion Principle, no two electrons of the same atom can have all four quantum numbers the same.

If n, l, and m_l, are same for two electrons then m_s, must be different. In the helium atom the spins must be opposite. Electrons with paired spins are denoted by ‘↓↑’. One electron has m = + 1/2, the other has m_s = -1/2. They have non-parallel spins.

Page 117

Question 32.
What are the spins of these two electrons?
Answer:
According to Paull’s Exclusion Principle, no two electrons of the same atom can have all four quantum numbers the same. If n, l, and m are same for two electrons then m must be different. In the helium atom, the spins must be paired. Electrons with paired spins are denoted by ‘↓↑’ one electron has m_s=+½, th other has m_s= -½. They have anti-parallel spins.

Question 33.
How many electrons can occupy an orbital?
Answer:
An orbital can hold only two electrons and they must have opposite spins.

Page 118

Question 34.
For carbon (C) atom (Z=6), where does the 6th electron go?
Answer:
In carbon, 6th electron enters into P_y orbital but not P_x orbital.

Question 35.
Whether the electron pairs up In the same p-orbItal or will it go to the next p-orbital?
Answer:
In carbon 6th electron enters into P_y orbital but not P_x orbital because according to Hunds rule electron pairing does not take place until each degenerate orbital is filled with one electron each.

TS 10th Class Physical Science Structure of Atom Activities

Activity 1

Question 1.
Explain the wave nature of light.
Answer:

1. Light is an electromagnetic wave.
2. Electromagnetic waves are produced when an electric field and magnetic field are perpendicular to each other.
3. This vibrating electric charge creates a change in the electric field. The change in electric field creates a change In magnetic field.
4. This process continues, with both the created fields being perpendicular to each other and at right angles to the direction of propagation of the wave.
5. The electromagnetic wave is produced.

Activity 2

Question 2.
Write an activity which shows metals produce different colours In flame.
Answer:
(A)

1. Take a pinch of cupric chloride in a watch glass and make a paste with concentrated hydrochloric acid.
2. Take this paste on a platinum loop and introduce it into a non-luminous flame.
3. Cuprlc chloride produces a green colour flame.

(B)

1. Take a pinch of strontium chloride in a watch glass and make a paste with concentrated hydrochloric acid.
2. Take this paste on a platinum loop and Introduce It into a non-luminous flame.
3. Strontium chloride produces a crimson-red flame.

Activity 3

Question 3.
Complete the electronic configuration of the following elements.
Answer:

Students can practice TS Class 10 Maths Solutions Chapter 8 Similar Triangles InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 8 Similar Triangles InText Questions

Do This

Question 1.
Fill in the blanks with similar / not similar. (AS₃) (Page No : 194)
i) All squares are …………..
Solution:
similar

ii) All equilateral triangles are …………….
Solution:
similar

iii) All isosceles triangles are ……………
Solution:
similar

iv) Two polygons with same number of sides are ……………. if their corresponding angles are equal and corresponding sides are equal.
Solution:
similar

v) Reduced and Enlarged photographs of an object are ……………
Solution:
similar

vi) Rhombus and squares are …………….. to each other.
Solution:
not similar

Question 2.
Write the True / False for the following statements. (AS₃) (Page No : 194)
i) Any two similar figures are congruent.
Solution:
False

ii) Any two congruent figures are similar.
Solution:
True

iii) Two polygons are similar if their corresponding angles are equal.
Solution:
False

Question 3.
Give two different examples of pair of (AS₃) (Page No : 194)
i) Similar figures
ii) Non-similar figures
Solution:
i) Similar figures :
a) Any two circles
b) Any two squares
c) Any two equilateral triangles

ii) Non – similar figures :
a) A square and a rhombus
b) A square and a rectangle

Question 4.
What value(s) of x will make DE//AB, in the given figure ?
AD = 8x + 9, CD = x + 3, (AS₁)
BE = 3x + 4, CE = x (Page No : 200) (A.P. Mar. ’16, ’15)

Solution:
Given : In ∆ABC, DE // AB
AD = 8x + 9, CD = x + 3,
BE = 3x + 4, CE = x
By Basic Proportional theorem,
If DE // AB then we should have
\(\frac{\mathrm{CD}}{\mathrm{DA}}\) = \(\frac{\mathrm{CE}}{\mathrm{EB}}\)
∴ \(\frac{x+3}{8 x+9}\) = \(\frac{x}{3 x+4}\)
⇒ (x + 3) (3x + 4) = x(8x + 9)
⇒ x(3x + 4) + 3(3x + 4) = 8x² + 9x
⇒ 3x² + 4x + 9x + 12 = 8x² + 9x
⇒ 8x² + 9x – 3x² – 4x – 9x – 12 = 0
⇒ 5x² – 4x – 12 = 0
⇒ 5x² – 10x + 6x – 12 = 0
⇒ 5x(x – 2) + 6(x – 2) = 0
⇒ (5x + 6) (x – 2) = 0
⇒ 5x + 6 = 0 (or) x – 2 = 0 – 6
⇒ x = \(\frac{-6}{5}\) (or) x = 2
∴ The value of x = 2 will make DE // AB.

Question 5.
In ∆ABC, DE // BC. AD = x, DB = x – 2, AE = x + 2 and EC = x – 1. Find the value of x. (AS₁) (Page No : 200)

Solution:
Given : In ∆ABC, DE // BC
By Basic Proportionality theorem,
we have
\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\)
⇒ \(\frac{x}{x-2}\) = \(\frac{x+2}{x-1}\)
⇒ x(x – 1) = (x + 2) (x – 2)
⇒ x² – x = x² – 4
⇒ – x = – 4
∴ x = 4

Try This

Question 1.
E and F are points on the sides PQ and PR respectively of ∆PQR. For each of the fol-lowing, state whether. EF || QR or not ? (AS₂) (Page No : 197)
i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
Solution:

Hence, EF is not parallel to QR.

ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
Solution:
Here, \(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{4}{4.5}\) = \(\frac{0.8}{0.9}\) = \(\frac{8}{9}\)
\(\frac{\mathrm{PE}}{\mathrm{RF}}\) = \(\frac{8}{9}\)
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PE}}{\mathrm{RF}}\)
∴ EF // QR
Hence, EF is parallel to QR.

iii) PQ = 1.28 cm, PR = 2.56 cm, PE =1.8 cm and PF = 3.6 cm
Solution:

Given, PQ = 1.28 cm, PE = 1.8 cm
⇒ EQ = PE – PQ = 1.8 – 1.28
⇒ EQ = 0.52 cm
Also, PR = 2.56 cm, PE = 3.6 cm
FR = PF – PR = 3.6 cm – 2.56 cm
FR = 1.04 cm
Now
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{1.8}{0.52}\) = \(\frac{0.9}{0.26}\)
\(\frac{\mathrm{PF}}{\mathrm{FR}}\) = \(\frac{3.6}{1.04}\) = \(\frac{0.9}{0.26}\)
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PF}}{\mathrm{FR}}\)
∴ EF // QR (By converse of Basic proportionality theorem)
Hence, EF is parallel to QR.

Question 2.
In the following figures DE || BC.
i) Find EC (AS₁) (Page No : 198)

Solution:
\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\)
⇒ \(\frac{1.5}{3}\) = \(\frac{\mathrm{1}}{\mathrm{EC}}\)
∴ EC = \(\frac{3}{1.5}\) = 2 cm

ii) Find AD

Solution:
\(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\)
⇒ \(\frac{\mathrm{AD}}{7.2}\) = \(\frac{1.8}{5.4}\)
∴ AD = \(\frac{1.8}{5.4}\) × 7.2 = latex]\frac{12.96}{5.4}[/latex] = 2.4 cm.

Think – Discuss

Question 1.
Can you give some more examples from your daily life where scale factor is used ? (AS₃) (Page No : 192)
Solution:
Scale factor is used in drawing maps, designing machines and in sculpture, etc.

Question 2.
Can you say that a square and a rhombus are similar ? Discuss with your friends. Write why the conditions are not sufficient. (AS₂) (Page No : 193)
Solution:
A square ☐ ABCD and a rhombus
▱ PQRS are not similar.
Though the ratio of their corresponding sides are equal, the corresponding angles are not equal.

But ∠A ≠ ∠P; ∠B ≠ ∠Q, ∠C ≠ ∠R; ∠D ≠ ∠S

Try This

Question 1.
Are the triangles formed in each figure similar ? If so, name the criterion of similarity. Write the similarity relation in symbolic form. (AS₃) (Page No : 207)
i)

!! ∠G = ∠I alt.int.angles for the ∠F = ∠K parallel lines GF // KI
Solution:
∆GHF and ∆lKH are similar by ASA similarity rule.
∆GHF ~ ∆lKH

ii)

Solution:
∆PQR and ∆LMN
\(\frac{\mathrm{LM}}{\mathrm{PQ}}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
\(\frac{\mathrm{MN}}{\mathrm{QR}}\) = \(\frac{4}{10}\) = \(\frac{2}{5}\)
\(\frac{\mathrm{LM}}{\mathrm{PQ}}\) ≠ \(\frac{\mathrm{MN}}{\mathrm{QR}}\)
∆PQR is not similar to ∆LMN

iii)

Solution:
∠A = ∠A (Common)
\(\frac{\mathrm{AX}}{\mathrm{XB}}\) = \(\frac{2}{3}\) ; \(\frac{\mathrm{AY}}{\mathrm{YC}}\) = \(\frac{2}{3}\)
\(\frac{\mathrm{AX}}{\mathrm{XB}}\) = \(\frac{\mathrm{AY}}{\mathrm{YC}}\)
∆ABC and ∆AXY are similar by SAS similarity condition.
∆ABC ~ ∆AXY

iv)

Solution:
\(\frac{\mathrm{AP}}{\mathrm{PB}}\) = \(\frac{3}{5}\) ; \(\frac{\mathrm{AJ}}{\mathrm{JC}}\) = \(\frac{2}{3 \frac{1}{3}}\)
= \(\frac{2}{10 / 3}=\frac{2 \times 3}{10}=\frac{6}{10}=\frac{3}{5}\)
\(\frac{\mathrm{AP}}{\mathrm{PB}}\) = \(\frac{\mathrm{AJ}}{\mathrm{JC}}\)
∆ABC and ∆APJ are similar.

v)

Solution:
∆AOQ and ∆BOP are similar by AA criterion. ∆AOQ ~ ∆BOP

vi)

Solution:
∆ABC and ∆PQR are similar AAA Similarity condition
∆ABC ~ ∆QPR

vii)

Solution:

∆ABC and ∆PQR are not similar.

viii)

Solution:
∆ABC and ∆PQR are not similar.

Question 2.
If pairs of the triangles are similar and then find the value of x. (AS₁, AS₂) (Page No : 207)
i)

Solution:
Gwen : In ∆PQR and ∆LTS
∠Q = ∠T
∠R = ∠S
∴ ∠P = ∠T
(by angle sum property of triangles)
Hence, ∆PQR ~ ∆LTS [∵ AAA]
Then \(\frac{\mathrm{PQ}}{\mathrm{LT}}\) = \(\frac{\mathrm{QR}}{\mathrm{TS}}\) = \(\frac{\mathrm{PR}}{\mathrm{LS}}\)
∴ \(\frac{3}{4.5}\) = \(\frac{5}{\mathrm{x}}\)
3x = 5 × 4.5
⇒ x = \(\frac{5 \times 4.5}{3}\) = 5 × 1.5 = 7.5

ii)

Solution:
Given : In ∆ABC and ∆PQC
∠B = ∠Q
[∵ ∠PQC = 180° – 110° = 70° – linear pair of angles]
∠C = ∠C [∵ common]
∴ ∠A = ∠P
[ ∵ Angle sum property of triangles]
∴ ∆ABC ~ ∆PQC (by AAA similarity condition).
Then the ratio of their corresponding sides are equal.
∴ \(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{BC}}{\mathrm{QC}}\) = \(\frac{\mathrm{AC}}{\mathrm{PC}}\)
\(\frac{5}{\mathrm{x}}\) = \(\frac{6}{3}\)
⇒ x = \(\frac{5 \times 3}{6}\) = 2.5

ii)

Solution:
Given : In ∆ABC and ∆ECD
∠A = ∠E
∠ACB = ∠ECD [∵ Vertically opposite angles]
∴ ∠B = ∠D [∵ Angles um property of triangles]
∴ ∆ABC ~ ∆ECD by AAA rule.
Hence, \(\frac{\mathrm{AB}}{\mathrm{ED}}\) = \(\frac{\mathrm{BC}}{\mathrm{CD}}\) = \(\frac{\mathrm{AC}}{\mathrm{EC}}\)
\(\frac{24}{14}\) = \(\frac{22}{\mathrm{x}}\)
⇒ x = \(\frac{22 \times 14}{24}\) = 12.83

iv)

Solution:
Given : In ∆RAB and ∆RST
∠R = ∠R (common)
∠A = ∠S and ∠B = ∠T [∵ Pair of corresponding angles for AB // ST]
∴ ∆RAB ~ ∆RST [∵ AAA similarity]
∴ \(\frac{\mathrm{RA}}{\mathrm{RS}}\) = \(\frac{\mathrm{AB}}{\mathrm{ST}}\) = \(\frac{\mathrm{RB}}{\mathrm{RT}}\)
\(\frac{6}{8}\) = \(\frac{9}{\mathrm{x}}\)
[from the figure, RS = 6 + 2 = 8]
⇒ x = \(\frac{9 \times 8}{6}\) = 12

v)

Solution:
Given: In ∆PQR and ∆PMN
∠P = ∠P (∵ Common)
∠Q = ∠M
(∵ Pair of corresponding angles for MN // OR]
∠R = ∠N
∴ ∆PQR ~ ∆PMN [∵ AAA similarity]
∴ \(\frac{\mathrm{PQ}}{\mathrm{PM}}\) = \(\frac{\mathrm{QR}}{\mathrm{MN}}\) = \(\frac{\mathrm{PR}}{\mathrm{PN}}\)
\(\frac{15}{5}\) = \(\frac{4+x}{4}\)
[From the figure, PR = 4 + x]
⇒ 3 × 4 = 4 + x
∴ x = 12 – 4 = 8

vi)

Solution:
Given : In ∆XYZ and ∆XBA
∠X = ∠X (∵ Common)
∠B = ∠Y [∵ Pair of corresponding angles for AB // ZY]
∴ ∆XYZ ~ ∆XBA [∵ AAA similarity]
Hence, \(\frac{\mathrm{XY}}{\mathrm{XB}}\) = \(\frac{\mathrm{YZ}}{\mathrm{BA}}\) = \(\frac{\mathrm{XZ}}{\mathrm{XA}}\)
\(\frac{18}{12}\) = \(\frac{7.5+x}{x}\)
[from the figure, XZ = 7.5 + X]
\(\frac{3}{2}\) = \(\frac{7.5+x}{x}\)
3x = 15 + 2x
3x – 2x = 15
x = 15

vii)

Solution:
Given : With the given conditions, we can’t find the value of x.
Note: If it is given that ∠A = ∠E then we can say that
∆ABC ~ ∆ EDC (by A.A.A rule)
and \(\frac{\mathrm{AB}}{\mathrm{ED}}\) = \(\frac{\mathrm{BC}}{\mathrm{CD}}\) = \(\frac{\mathrm{AC}}{\mathrm{EC}}\)
⇒ \(\frac{1.6}{\mathrm{x}}\) = \(\frac{1.5}{15}\)
⇒ x = \(\frac{1.6 \times 15}{1.5}\) = 16

viii)

Solution:
Given : Data not sufficient.

Think – Discuss

Question 1.
Discuss with your friends that In what way similarity of triangles is different from similarity of other polygons? (AS₃) (Page No. 203)
Answer:
In two triangles if the corresponding angles are equal then they are similar, whereas in two polygons if the corresponding angles are equal, they may not be similar.
i.e., In triangles,
(Pairs of corresponding angles are equal) ⇔ (Ratio of corresponding sides are equal).
But this is not so with respect to polygons.

Do This

Question 1.
In ∆ACB, ∠C = 90° and CD ⊥ AB. Prove that \(\frac{\mathrm{BC}^2}{\mathrm{AC}^2}\) = \(\frac{\mathrm{BC}}{\mathrm{AC}}\) (AS₂) (Page No. 218)

Solution:
Proof : ∆ADC and ∆CDB are similar
∴ \(\frac{\Delta \mathrm{ADC}}{\Delta \mathrm{CDB}}\) = \(\frac{\frac{1}{2} \times \mathrm{AD} \times \mathrm{DC}}{\frac{1}{2} \times \mathrm{BD} \times \mathrm{DC}}\) = \(\frac{\mathrm{AD}}{\mathrm{BD}}\)
⇒ \(\frac{\Delta \mathrm{CDB}}{\Delta \mathrm{ADC}}\) = \(\frac{\mathrm{BC}}{\mathrm{AD}}\) ……………. (1)
(or) \(\frac{\Delta \mathrm{CDB}}{\Delta \mathrm{ADC}}\) = \(\frac{\mathrm{BC}^2}{\mathrm{AC}^2}\) ………….. (2)
[Ratio of areas of similar triangles is equal to the ratio of squares of their corresponding sides.]
From (1) and (2),
\(\frac{\mathrm{BC}}{\mathrm{AD}}\) = \(\frac{\mathrm{BC}^2}{\mathrm{AC}^2}\) (Q.E.D)

Question 2.
A ladder 15m long reaches a window which is 9m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to other side of the street to reach a window 12m high. Find the width of the street. (AS₄) (Page No : 218)
Solution:

Let A and D be the windows on the either sides of the street.
From Pythagoras theorem,
AC² = AB² + BC²
15² = 9² + BC²
BC² = 225 – 81
BC = \(\sqrt{144}\) = 12 ……………… (1)
Also,
CD² = DE² + CE²
15² = 12² + CE²
CE² = 225 – 144
CE = \(\sqrt{81}\) = 9
∴ BE = BC + CE
= 12 + 9 = 21
∴ Width of the street = 21 m

Question 3.
In the given fig. if AD ⊥ BC, prove that AB² + CD² = BD² + AC². (AS₂) (Page No : 219)

Solution:
Given : In ∆ABC, AD ⊥ BC
R.T.P. : AB² + CD² = BD² + AC²
Proof : ∆ABD is a right angled triangle.
AB² – BD² = AD² …………….. (1)
∆ACD is a right angle triangle.
AC² – CD² = AD² ……………. (2)
From (1) and (2)
AB² – BD² = AC² – CD²
AB² + CD² = BD² + AC²

Think – Discuss

Question 1.
For a right angled triangle with integer sides atleast one of its measurements must be an even number. Why ? Discuss with your friends and teachers. (AS₃) (Page No : 215)
Solution:
Given : A right triangle with integer sides.
To discuss : At least one measurement must be an even number.
Case – (i) : If the measurements of the sides are 3, 4, 5 – Primitive Pythagorean tripplet then one measurement ‘4’ is even and the given statement is true.

Case – (ii) : If the measurements are an integer multiple of primitive i.e., 3n, 4n and 5n then also ‘4n’ is even [irrespective of n-even / odd]
∴ Given statement is true.

Case – (iii) : If one measurement ‘n’ is odd, then n, \(\frac{\mathrm{n}^2+1}{2}\) and \(\frac{\mathrm{n}^2-1}{2}\) are the measurements of the sides, then also \frac{\mathrm{n}^2-1}{2} is even. [∵ n = 2k + 1]
n² = (2k + 1)² = 4k² + 4k + 1
n² – 1 = 4k² + 4k + 1 – 1
= 4(k² + k)
= 2(2k² + 2k) [even]
∴ In any case the given statement is true.