AP Inter 1st Year

For systematic learning, students can include AP Inter 1st Year English Study Material and Odd Sounds in English Inter 1st Year in their daily study routine.

Inter 1st Year English Odd Sounds with Answers

A set of three words are given. One or two letters which are common in all the three words are underlined. The underlined letter stands for one sound in two of the given words and for a different sound in the other word. The word with a different sound is to be written as the response.

Exercise For Pracitce

Exercises

Question 1.
Find the odd words in the given words keeping the underlined parts of the words.

Question 2.
Identify the odd words from the following words in respect of their underlined sounds and write them.

Question 3.
Identify the odd words from the following words in respect of their underlined sounds and write them.

Referring to the AP Inter 1st Year Maths Study Material Chapter 8 Sequences and Series Exercise 8d Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sequences and Series Solutions Exercise 8d

Find the sum to infinity in each of the following Geometric Progression.

Question 1.
1, \(\frac{1}{3}, \frac{1}{9}\), …………..
Solution:
Let S = 1, \(\frac{1}{3}, \frac{1}{9}\), …………..; Here, a = 1 and r = \(\frac{1}{3}\)
So, S_∞ = \(\frac{a}{1-r}=\frac{1}{1-\frac{1}{3}}=\frac{1}{2 / 3}=\frac{3}{2}\) = 1.5

Question 2.
6, 1.2, 0.24, …………………
Solution:
Let S = 6 + 1.2 + 0.24 + ………………….; Here, a = 6 and r = 0.2
So, S_∞ = \(\frac{a}{1-r}=\frac{6}{1-(0.2)}=\frac{6}{0.8}\) = 7.5

Question 3.
5, \(\frac{20}{7}, \frac{80}{49}\) ………..
Solution:
Let S = 5, \(\frac{20}{7}, \frac{80}{49}\), ………….; Here a = 5, r = \(\frac{4}{7}\)
So, S_∞ = \(\frac{a}{1-r}=\frac{5}{1-\frac{4}{7}}=\frac{35}{3}\)

Question 4.
\(\frac{-3}{4}, \frac{3}{16}, \frac{-3}{64}\) ………….
Solution:
Let S = \(\frac{-3}{4}, \frac{3}{16}, \frac{-3}{64}\) ………….; Here, a = \(\frac{-3}{4}\) and r = \(\frac{-1}{4}\)

S_∞ = \(\frac{a}{1-r}=\frac{-3 / 4}{1-\left(\frac{-1}{4}\right)}=\frac{-3 / 4}{1+\frac{1}{4}}=\frac{-3}{4} \times \frac{4}{5}=\frac{-3}{5}\)

Question 5.
Prove that 3^{\(\frac{1}{2}\)} × 3^{\(\frac{1}{4}\)} × 3^{\(\frac{1}{8}\)} × ……. = 3
Solution:
LHS = 3^{\(\frac{1}{2}\)} × 3^{\(\frac{1}{4}\)} × 3^{\(\frac{1}{8}\)} × ……. = 3^{\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots \ldots .\)};
Here a = \(\frac{1}{2}\) and r = \(\frac{1}{2}\)
∴ S_∞ = 3^{\(\left(\frac{a}{1-r}\right)\)}
= 3^{\(\left(\frac{1 / 2}{1-1 / 2}\right)\)}
= 3^{\(\left(\frac{1 / 2}{1 / 2}\right)\)}
= 3¹
= 3 = R.H.S. [using S_∞ = \(\frac{a}{1-r}\)]

Question 6.
Let x = 1 + a + a² + …….. and y = 1 + b + b² + ….. where |a| < 1 and |b| < 1.
Prove that 1 + ab + a²b² + ….. = \(\frac{x y}{x+y-1}\)
Solution:
Here, x = 1 + a + a² + ……………… = \(\frac{1}{1-a}\) [using S_∞ = \(\frac{a}{1-r}\)]

And y = 1 + b + b² + ………… = \(\frac{1}{1-b}\)

LHS = 1 + ab + a²b² + ……. = \(\frac{1}{1-ab}\) …………(1)

RHS = \(\frac{x y}{x+y-1}=\frac{\left(\frac{1}{1-a}\right)\left(\frac{1}{1-b}\right)}{\left(\frac{1}{1-a}\right)+\left(\frac{1}{1-b}\right)-1}=\frac{\frac{1}{(1-a)(1-b)}}{\frac{1-b+1-a-(1-a)(1-b)}{(1-a)(1-b)}}\)

= \(\frac{1}{(1-a)(1-b)} \times \frac{(1-a)(1-b)}{2-a-b-1+a+b-a b}=\frac{1}{1-a b}\) ……. (2)

From (1) and (2), 1 + ab + a²b² + ………… = \(\frac{x y}{x+y-1}\)

Referring to the AP Inter 1st Year Maths Study Material Chapter 8 Sequences and Series Exercise 8c Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sequences and Series Solutions Exercise 8c

I.

Question 1.
Find the 20^th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.
Solution:
Given 2 × 4 + 4 × 6 + 6 × 8 + … + n terms
Therefore the n^th term, a_n = 2n (2n + 2) = 4n² + 4n
a₂₀ = 4(20)² + 4(20)
= 4(400) + 80
= 1600 + 80
= 1680
Therefore, 1680 is the 20^th term.

Question 2.
The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms.
Solution:
Let the sum of n terms of the G.P be 315. It is known that S_n = \(\frac{1}{2}\)
It is given that a = 5 and common ratio r = 2
Hence, 315 = \(\frac{1}{2}\)
⇒ 2ⁿ – 1 = 63
⇒ 2ⁿ = 64 = (2)⁶
⇒ n = 6
Therefore, last term of the G.P. is 6th term
a₆ = ar^6-1 = ar⁵
= 5 × (2)⁵
= 5 × 32
= 160

Question 3.
The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
Solution:
Let a and r be the first term and common ratio of the G.P respectively.
Hence, a = 1 ⇒ a₃ = ar² = r²
⇒ a₅ = ar⁴ = r⁴
Therefore, r² + r⁴ = 90
r² + r⁴ – 90 = 0
⇒ r² = \(\frac{-1 \pm \sqrt{1+360}}{2}\)
= \(\frac{-1 \pm \sqrt{361}}{2}=\frac{-1 \pm 19}{2}\) = -10 or 9
r² = 9 [Taking real roots]
r = ±3
Thus, the common ratio of the G.P is ±3.

II.

Question 1.
The sum of three consecutive terms in a G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
Solution:
Let the three numbers in G.P. be a, ar and ar².
From the given condition, a + ar + ar² = 56
⇒ a(1 + r + r²) = 56 ………….. (1)
Now, (a – 1), (ar – 7), (ar² – 21) forms an A.P.
Therefore, (ar – 7) – (a – 1) = (ar² – 21) – (ar – 7)
⇒ ar – a – 6 = ar² – ar – 14
⇒ ar² – 2ar + a = 8
⇒ a(r² + 1 r 2r) = 8
⇒ 7a(r² – 2r + 1) = 56 ………….(2)
From (1) and (2), we get
⇒ 7(r² – 2r + 1) – (1 + r + r²)
⇒ 7r² – 14r + 7 – 1 – r – r² = 0
⇒ 6r² -15r + 6 = 0
⇒ 2r² – 5r + 2 = 0
⇒ 2r² -4r – r + 2 = 0
⇒ 2r(r – 2) – 1 (r – 2) = 0
⇒ (2r – 1) (r – 2) = 0
⇒ r = \(\frac{1}{2}\), 2

Case I :
Substituting r = 2 in (1), we obtain a(1 + 2 + 22) = 56
⇒ 7a = 56
⇒ a = 8
Hence, the three numbers in G.P are 8, 16 and 32.

Case II :
Substituting r = \(\frac{1}{2}\) in (1), we obtain a(1 + \(\frac{1}{2}\) + \(\left(\frac{1}{2}\right)^2\)) = 56
⇒ \(\frac{1}{2}\)a = 56
a = 32
Hence, the three numbers in G.P are 32, 16 and 8.
Therefore, from case (i) and case (ii), the three required numbers are 8, 16 and 32.

Question 2.
A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
Solution:
Let the G.P be T₁, T₂, T₃, T₄, … T_2n. According to the question,
⇒ T₁ + T₂ + T₃ + T₄ + … + T_2n = 5[T₁ + T₃ + T₅ +… + T_2n-1]
⇒ T₁ + T₂ + T₃ + T₄ +… + T_2n – 5 [T₁ + T₃ + T₅ +… + T_2n-1] = 0
⇒ T₂ + T₄ + ………. + T_2n = 4[T₁ + T₃ + ………. + T_2n-1]
Let the G.P be a, ar, ar²,
\(\frac{\operatorname{ar}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}=\frac{4 \times \mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\)
⇒ ar = 4a
⇒ r = 4
Thus, the common ratio of the G.P is 4.

Question 3.
If \(\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}\) (x ≠ 0), then show that a, b, c and d are in G.P.
Solution:
It is given that = \(\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}\) (x ≠ 0)
⇒ \(\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}\) = (a + bx) (b – cx) = (b + cx) (a – bx)
⇒ ab – acx + b²x – bcx² = ab – b²x + acx — bcx²
⇒ 2b²x = 2acx
⇒ b² = ac
⇒ \(\frac{b}{a}=\frac{c}{b}\) (1)
Also, ⇒ \(\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}\)
⇒ (b + cx)(c – dx) = (b – cx)(c + dx)
⇒ bc – bdx + c²x – cdx² = bc + bdx – c²x – cdx²
⇒ 2c²x = 2bdx
⇒ c² = bd
⇒ \(\frac{c}{b}=\frac{d}{c}\) …(2)
From Equations (1) and (2), we obtain \(\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\)
Thus, a, b, c and d are in G.P.

Question 4.
Let S be the sum, P be the product and R be the sum of reciprocals of n terms in a G.P. Prove that P²Rⁿ = Sⁿ.
Solution:
Let the G.P be a, ar, ar², ar³ … ar^n-1
According to the given information; Sum of the n terms, S = \(\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\)
Product of the n terms P = aⁿ × r^{1 + 2 + 3 + …. n-1}
= aⁿr^{\(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\)} [∵ Sum of first n natural numbers = n\(\frac{(\mathrm{n}+1)}{2}\)

Sum of reciprocals of n terms R = \(\frac{1}{2}\)
\(\frac{r^{n-1}+r^{n-2}+\ldots+r+1}{a r^{n-1}}=\frac{1\left(r^n-1\right)}{(r-1)} \times \frac{1}{a r^{n-1}}=\frac{r^n-1}{a r^{n-1}(r-1)}\) [∵ 1, r, ……….. r^n-1 forms a G.P]
Therefore, P²Rⁿ = a²ⁿr^n(n-1)\(\frac{\left(r^n-1\right)^n}{a^n r^{n(n-1)}(r-1)^n}=\frac{a^n\left(r^n-1\right)^n}{(r-1)^n}=\left[\frac{a\left(r^n-1\right)}{(r-1)}\right]^n\) = Sⁿ
Hence, P²Rⁿ = Sⁿ proved.

Question 5.
If a, b, c, d are in G.P., prove that (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P.
Solution:
It is given that a, b, c, d are in G.P.
Therefore, b² = ac ………. (1)
c² = bd … (2)
ad = bc … (3)
We need to prove (an + bn), (bn + cn), (cn + dn) are in G.P i.e.,
(bⁿ + cⁿ)² = (aⁿ + bⁿ) . (cn + dn)
Consider (bⁿ + cⁿ)² = b²n + 2bncn + c²n
= (b²)ⁿ + 2bⁿcⁿ + (c²)ⁿ = (ac)ⁿ + 2bⁿcⁿ + (bd)ⁿ [Using (1) and (2)]
= cⁿ (aⁿ + bⁿ) + dⁿ (aⁿ + bⁿ) = (aⁿ + bⁿ) (cⁿ + dⁿ)
Hence, (bⁿ + cⁿ)² = (aⁿ + bⁿ) . (cⁿ + dⁿ)
Thus, (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P.

Question 6.
If a and b are the roots of x² – 3x + p * 0 and c, d are the roots of x² – 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17 : 15.
Solution:
It is given that a and b are the roots of x² – 3x + p = 0
Therefore, a + b = 3, ab = p
Also, c and d are the roots of x² – 12x + q = 0; c + d = 12, cd = q
It is given that a, b, c, d form a G.P.
Let a = x, b = xr, c = xr², d = xr³
From (1) and (2), we obtain, ⇒ x + xr = 3
⇒ x(1 + r) = 3
⇒ xr² + xr³ = 12
⇒ xr² (1 + r) = 12 … (4)

On dividing (4) by (3), we obtain \(\frac{\mathrm{xr}^2(1+\mathrm{r})}{\mathrm{x}(1+\mathrm{r})}=\frac{12}{3}\)
⇒ r² = 4 ⇒ r = ±2

Case I:
When r = 2, x = 1 ⇒ ab = x²r = 2 & cd = x²r⁵ = 32
Therefore, ⇒ \(\frac{q+p}{q-p}=\frac{32+2}{32-2}=\frac{34}{30}=\frac{17}{15}\)
⇒ (q + p) : (q – p) = 17 : 15

Case II: when r = -2, x = -3
⇒ ab = x²r = -18 & cd = x²r⁵ = -288
Therefore, ⇒ \(\frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-306}{-270}=\frac{17}{15}\)
Thus, (q + p) : (q – p) = 17 : 15 is proved.

Question 7.
The ratio of the A.M and G.M. of two positive numbers a and h, is m : n. Show that a : b = (m +\(\sqrt{m^2-n^2}\)):(m – \(\sqrt{m^2-n^2}\)).
Solution:
Let the two numbers be a and b. A.M = \(\frac{a+b}{2}\), G.M = \(\sqrt{a b}\)
Adding (1) and (2), we obtain \(\frac{a+b}{2 \sqrt{a b}}=\frac{m}{n} \Rightarrow \frac{(a+b)^2}{4 a b}=\frac{m^2}{n^2}\)
⇒ (a + b)² = \(\frac{4 \mathrm{abm}^2}{\mathrm{n}^2}\)
⇒ (a + b) = \(\frac{2 \mathrm{~m} \sqrt{\mathrm{ab}}}{\mathrm{n}}\) .. (1)

Using this in the identity (a – b)² = (a + b)² – 4ab, we obtain

Substituting the value of a in (1), we obtain

III.

Question 1.
Find the sum of the following series upto n terms.
(i) 5 + 55 + 555 + …
Solution:
Given that 5 + 55 + 555 + …
Let S_n = 5 + 55 + 555 + … n terms
S_n = \(\frac{5}{9}\)[9 + 99 + 999 + …n terms] = \(\frac{5}{9}\) (10 – 1) + (10² – 1) + (10³ – 1) + …nterms
= \(\frac{5}{9}\)[(10 + 10² + 10³ + ……………. n terms) – (1 + 1 + ….. n terms)]
= \(\frac{5}{9}\)[\(\frac{10\left(10^{\mathrm{n}}-1\right)}{10-1}\) – n]

= \(\frac{5}{9}\)[\(\frac{10\left(10^{\mathrm{n}}-1\right)}{9}\) – n]

= \(\frac{50}{81}\)(10ⁿ – 1) – \(\frac{5n}{9}\)

(ii) .6 + .66 + .666 + …
Solution:
S_n = .6 + .66 + .666 + … n terms
= 6[.1 + .11 + .111 + …… upto nterms] = \(\frac{6}{9}\)[0.9 + 0.99 + 0.999 + upto n terms]

Therefore, the sum of n terms of the given series is \(\frac{2}{3}\) n – \(\frac{2}{27}\)(1 – 10^-n)

Question 2.
If f is a function satisfying f(x + y) = f(x) . f(y) for all x, y ∈ N such that f(1) = 3 and \(\sum_{x=1}^n\) f(x) = 120. And the value of n.
Solution:
f(x + y) = f(x) . f(y) for all x, y ∈ N and f(1) = 3
Taking x = y = 1 in the given function, we obtain f(x + y) = f(x) . f(y) ⇒ f(l + 1) = f(1) f(1) f(2) = 3 x 3 = 9
Similarly, f(x + y) = f(x) . f(y) ⇒ f(1 + 2) = f(1) f(2) ⇒ f(3) = 3 x 9 = 27
Also, f(x + y) = f(x) . f(y) ⇒ f(1 + 3) = f(1) f(3) ⇒ f(4) = 3 x 27 = 81
Therefore, f(1), f(2), f(3) i.e., 3, 9, 27, 81 forms a G.P. with both the first term and common ratio equal to 3.
It is known that S_n = \(\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\). It is given that \(\sum_{x=1}^n\) f(x) = 120

⇒ 120 = \(\frac{3\left(3^{\mathrm{n}}-1\right)}{3-1}\)
⇒ 120 = \(\frac{3}{2}\)(3ⁿ – 1)
⇒ 3ⁿ – 1 = 80
⇒ 3ⁿ = 3⁴
⇒ n = 4
Thus, the value of n = 4.

Question 3.
A fanner buys a used tractor for ₹ 12000. He pays ₹ 6000 cash and agrees to pay the balance in annual installments of ₹ 500 plus 12% interest on the unpaid amount. How much will the tractor cost him ?
Solution:
Given, the farmer pays ₹ 6000 in cash.
So, the unpaid amount = ₹ 12000 – ₹ 6000 = ₹ 6000
From the question, the interest paid annually will be 12% of 6000, 12% of 5500, 12% of 5000, … , 12% of 500
Hence, the total interest to be paid = 12% of 6000 + 12% of 5500 + 12% of 5000 + … + 12% of 500 = 12% of (6000 + 5500 + 5000 + … + 500)
= 12% of (500 + 1000 + 1500 + … + 6000)
It’s seen that the series 500, 1000, 1500 … 6000 is an A.P. with the first term and common difference both are equal to 500.
Let’s take the number of terms of the A.P to be n.
So, 6000 = 500 + (n – 1) 500
⇒ 1 + (n – 1) = 12
⇒ n = 12
Now, The sum of the series in A.P = \(\frac{12}{2}\) [2(500) + (12 – 1) (500)]
= 6[1000 + 5500]
= 6(6500)
= 39000
Hence, the total interest to be paid = 12% of (500 + 1000 + 1500 +
= 12% of 39000 = ₹4680
Therefore, the tractor will cost the farmer = (₹ 12000 + ₹ 4680) = ₹ 16680

Question 4.
Shamshad Ali buys a scooter for ₹ 22000. He pays ₹ 4000 cash and agrees to pay the balance in annual installment of? 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him ?
Solution:
It is given that Shamshad Ali buys a scooter for ₹ 22000 and pays ₹ 4000 in cash. Hence, unpaid amount ₹ = 22000 – 4000 = 18000
According to the given condition, the interest paid annually will be (10% of 18000), (10% of 17000), (10% of 16000), … (10% of 1000)
Thus, total interest to be paid = (10% of 18000) + (10% of 17000) + … + (10% of 1000) = 10% of (18000 + 17000 + … + 1000) = 10% of (1000 + 2000 + … + 18000)
Now, the series 1000 + 2000 + … + 18000 forms an A.P. with a and d both equal to 1000.
Let the number of terms of the A.P. be n.
⇒ a_n = a + (n – 1) d
⇒ 18000 = 1000 + (n – 1) (1000)
⇒ 18000 – 1000 = 1000 (n – 1)
⇒ n – 1 = \(\frac{17000}{1000}\)
⇒ n = 17 + 1
⇒ n = 18
Now, total cost of scooter in ₹ = 22000 + 17100 = 39100
Thus, the scooter will cost him ₹ 39100.

Question 5.
A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter, find the amount spent on the postage when the 8th set of letters is mailed.
Solution:
The number of letters mailed forms a G.P : 4, 4², ……….. , 4⁸; Here a = 4, n = 8
Hence, S = \(\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\)
S8 = \(\frac{4\left(4^8-1\right)}{4-1}=\frac{4(65536-1)}{3}=\frac{4 \times 65535}{3}\)
= 4 × 21845
= 87380
It is given that the cost to mail one letter is 50 paisa.
Therefore, cost in ₹ for mailing 87380 letters = \(\frac{50}{100}\) × 87380 = 43690
Thus, ₹ 43690 is the amount on the postage when 8th set of letters is mailed.

Question 6.
A man deposited ₹ 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount, also calculate the total amount after 20 years.
Solution:
It is given that the man deposited ₹ 10000 in a bank at the rate of 5% simple interest annually. Hence, annual interest received in ₹ = \(\frac{50}{100}\) × 10000 = 500
Amount in ₹ at the end of first year = 10000 + 500 = 10500
Amount in ₹ at the end of second year = 10500 + 500 = 11000
Amount in ₹ at the end of third year = 11000 + 500 = 11500
Therefore, 10000, 10500, 11000, … is an A.P.
Hence, the amount in the 15^th year,
a₁₅ = 10000 + (15 – 1) (500) = 10000 + 7000 = 17000
Amount after 20 years
a₂₁ = 10000 + (21 – 1) (500) = 10000 + 10000 = 20000
Thus, the amount in 15th year is ₹ 17000 and after 20 years amount is ₹ 20000.

Question 7.
A manufacturer reckons that the value of a machine, which costs him ₹ 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
Solution:
Cost of a machine is ₹ 15625. Machine depreciates by 20% every year.
∴ Its value after every year is 80% of the original cost i.e., 4/5 of the original cost.
Value at the end of 5 years =

= 15628 × \(\left(\frac{4}{5}\right)^5\)
= 5 × 1024
= 5120
Thus, value of the machine at the end of 5 years is ₹ 5120.

Question 8.
150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
Solution:
Let x be the number of days in which 150 workers finish the work.
According to the given information, 150x = 150 + 146 + 142 + … (x + 8) terms
Here, 150 + 146 + 142 + … (x + 8) terms forms an A.P. with a = 150, d = – 4, n = (x + 8)
Therefore 150x = \(\frac{(x+8)}{2}\)[2(150) + (x + 8 -1)(-4)]
⇒ 150x – (x + 8) [150 + (x + 7) (-2)]
⇒ 150x = (x + 8) (150 – 2x – 14)
⇒ 150x = (x + 8) (136 – 2x)
⇒ 75x – (x + 8) (68 – x)
⇒ 75x – 68x – x² + 544 – 8x
⇒ x² + 15x – 544 = 0
⇒ x² + 32x – 17x – 544 = 0
⇒ x(x + 32) – 17(x + 32) = 0
⇒ (x – 17) (x + 32) = 0
⇒ x = 17, – 32
However, x cannot be negative, hence, x = 17
∴ The number of days in which the work was completed is (17 + 8) = 25 days. Thus, the work was completed in 25 days.

For systematic learning, students can include AP Inter 1st Year English Study Material and Phonetic Transcription in English Inter 1st Year in their daily study routine.

Inter 1st Year English Phonetic Transcription with Answers

Meaningful speech sounds are the basic raw material for any language. These sounds are represented by symbols in writing. We refer to these symbols as the alphabet.

Most of the Indian languages have a fixed relationship between the sounds and their symbols. In other words, one symbol always stands for one sound and vice versa. Therefore, we find absolutely no problem while reading or writing Indian languages once we learn the alphabet of that language.

The same is not the case with English. One letter may stand for many sounds. Example c as / k / in car, as / s/ in century as /t ʃ/ in change. One sound is also represented by various letters.
Ex: / f / is represented by ph’ in philosophy, by ‘f in fan, by gh’ in rough. This results in a lot of problems for the learners, particularly for foreign learners in writing the spelling and pronouncing the written words.

A way out of this probelm is a set of forty-four symbols, called phonetic symbols. Each of these symbols stands for one sound only. Learning these symbols arms us with the necessary weapons to war against the problems in pronunciation and spelling.

Study the following forty four phonetic symbols carefully and learn to identify and use them. Phonetic symbols are always placed between two Slant lines.

Vowels

1) / i: / as in see
2) /ɪ/ as in sit
3) /e/ as in ten
4) / æ / as in bat
5) /ɑː/ as in arm
6) / ɒ/ as in got
7) /ɔː/ as in saw
8) /u/ as in put
9) /u:/ as in too
10) /ʌ/ as in cup
11) /ɜː/ as in fur
12) /ə/ as in ago
13) /eɪ/ as in page
14) /əʊ/ as in home
15) /aɪ/ as in five
16) /aʊ/ as in cow
17) /ɔɪ/ as in boil
18) /ɪə/ as in near
19) /eə/ as in hair
20) /ʊə/ as in poor

Consonants

1) / p / as in pen
2) / b / as in bin
3) /t / as in tea
4) / d / as in dog
5) / k/ as in cat
6) / g / as in goat
7) /t ʃ/ as in change
8) / d ʒ / as in July
9) / f / as in fall
10) /v/ as in voice
11) /θ/ as in thin
12) /ð/ as in then
13) / s / as in so
14) / z / as in zoo
15) / ʃ / as in she
16) / ʒ / as in vision
17) / h / as in how
18) / n / as in may
19) / n / as in now
20) / ŋ / as in sing
21) /l/ as in long
22) / r / as in ring
23) /j/ as in yes
24) /w / as in wall

Writing the symbols that represent the sounds in a word is called phonetic transcript. In many examinations phonetic transcription is given and the examinee is asked to write the spelling of those words.

Look at the following words and their phonetic transcription and you will certainly have an idea of it:

Word – Transcription

1. almond – /ɑːmənd/
2. asthma – /ˈæsmə/
3. business – /ˈbɪznɪs/
4. bury – /ˈber.i/
5. cache – /kæʃ/
6. colonel – /ˈkɜːnl/
7. coupon – /ˈkuːpɒn/
8. chassis – /ˈʃæsɪ/
9. chaos – /ˈkeɪɒs/
10. choir – /ˈkwaɪər/
11. cucumber – /ˈkjuːkʌm.bər/
12. data – /ˈdeɪ.tə/
13. dengue – /ˈdɛŋɡi/
14. determine – /dɪˈtɜːmɪn/
15. debris – /ˈdeɪb.riː/
16. discretion – /dɪˈskrɛʃᵊn/
17. epitome – /ɪˈpɪtəmi/
18. extempore – /eksˈtempərɪ/
19. etc./ et cetera – /ɪtˈsɛt.ər.ə/
20. flour – /flaʊə/
21. forte – /ˈfɔːteɪ/
22. gauge – /ɡeɪʤ/
23. genre – /ˈʒɒn.rə/
24. iron – /aɪən/
25. itinerary – /aɪˈtɪnərəri/
26. jewellery – /ˈʤuːəlri/
27. mischievous – /ˈmɪsʧɪvəs/
28. photography – /fəˈtɒɡrəfi/
29. picture – /ˈpɪkʧər/
30. plumber – /ˈplʌm.ər/
31. poem – /ˈpəʊɪm/
32. police – /pəˈliːs/
33. pizza – /ˈpiːt.sə/
34. receipt – /rɪˈsiːt/
35. saloon – /səˈluːn/
36. sour – /ˈsaʊər/
37. status – /ˈstaeɪ.təs/
38. subtle – /ˈsʌtl/
39. suite – /swiːt/
40. vehicle – /ˈviə.kəl/
41. vegetable – /ˈveʤɪtəbəl/
42. violence – /ˈvaɪə.ləns/
43. vineyard – /ˈvɪnjɑːd/
44. Wednesday – /ˈwɛnz.deɪ/
45. women – /ˈwɪm.ɪn/

Exercises

Study the following phonetic transcriptions and write them in ordinary spelling.

1. Exercise from UNIT – 1

Question 1.
/kəˈmɛns/
Answer:
commence

Question 2.
/ˈfænsi/
Answer:
fancy

Question 3.
/ˈeɪlmənt/
Answer:
ailment

Question 4.
/dɪˈspeə(r)/
Answer:
despair

Question 5.
/ɡlɑːns/
Answer:
glance

Question 6.
/plʌnʤ/
Answer:
plunge

Question 7.
/fɪəs/
Answer:
fierce

Question 8.
/ˈblʌʤənɪŋ/
Answer:
bludgeoning

Question 9.
/ˈhæmpə(r)/
Answer:
hamper

Question 10.
/ˈsɜːkəmstɑːns/
Answer:
circumstance

2. Exercise from UNIT – 2

Question 1.
/pʌɡˈnæsəti/
Answer:
pugnacity

Question 2.
/ɪˈlæbərəɪt/
Answer:
elaborate

Question 3.
/ˈʤɒsl/
Answer:
jostle

Question 4.
/ˈɜːrʧɪn/
Answer:
urchin

Question 5.
/skɜːʤ/
Answer:
scourge

Question 6.
/ˈmʌŋɡrəl/
Answer:
mongrel

Question 7.
/ˈtrɛbəl/
Answer:
treble

Question 8.
/ˈmjuːtɪleɪt/
Answer:
mutilate

Question 9.
/slæk/
Answer:
Slack

Question 10.
/ˈpɛrɪʃ/
Answer:
perish

3. Exercise from UNIT – 3

Question 1.
/ˈslɔːtər/
Answer:
slaughter

Question 2.
/ˈθɪkɪt/
Answer:
thicket

Question 3.
/ˈvænɪʃ/
Answer:
vanish

Question 4.
/ˈænsɛstə(r)/
Answer:
ancestor

Question 5.
/ˈæŋɡwɪʃ/
Answer:
anguish

Question 6.
/ˈpænɪk/
Answer:
panic

Question 7.
/rɪˈtriːvd/
Answer:
retrieved

Question 8.
/ˈsɛntəns/
Answer:
sentence

Question 9.
/ˈprəʊtɛst/
Answer:
protest

Question 10.
/vɔːlt/
Answer:
vault

4. Study the following phonetic transcriptions and write ANY FIVE of them in their ordinary spelling.

Question 1.
/ˈʤiːniəli/
Answer:
genially

Question 2.
/rɛtrɪˈbjuːʃən/
Answer:
retribution

Question 3.
/ʌnˈkɒŋkərəbəl/
Answer:
unconquerable

Question 4.
/ˈkæpʃən ɒv maɪ səʊl/
Answer:
caption of my soul

Question 5.
/səʊl/
Answer:
soul

Question 6.
/kənˈtɛmpt/
Answer:
contempt

Question 7.
/ˈpreəri/
Answer:
prairie

Question 8.
/bɪˈfɔːl/
Answer:
befall

Question 9.
/kəˈlɪɡrəfi/
Answer:
calligraphy

Question 10.
/’endəskəʊp/
Answer:
endoscope

5. Study the following phonetic transcriptions and write ANY FIVE of them in their ordinary spelling.

Question 1.
/əˈblaɪʤ/
Answer:
oblige

Question 2.
/skɜː(r)ʤ/
Answer:
scourge

Question 3.
/ˈeɪlmənt/
Answer:
ailment

Question 4.
/ɪnˈdjuːs/
Answer:
induce

Question 5.
/spɑːrs/
Answer:
sparse

Question 6.
/ˈʤɒsəl/
Answer:
jostle

Question 7.
/ˈɪnəsəns/
Answer:
innocence

Question 8.
/ˈbɜːnɪŋ/
Answer:
burning

Question 9.
/kənˈvʌlʃən/
Answer:
convulsion

Question 10.
/ˈθrɒtl/
Answer:
throttle

For systematic learning, students can include AP Inter 1st Year English Study Material and Parts of Speech in English Inter 1st Year in their daily study routine.

Inter 1st Year English Parts of Speech with Answers

There are eight parts of speech. They are

1. Noun,
2. Pronoun,
3. Verb,
4. Adjective,
5. Adverb,
6. Preposition,
7. Conjunction and
8. Interjection.

Noun

A noun is a word for a person, place, thing, or idea. Nouns are often used with an article (the, a, an), but not always. Proper nouns always start with a capital letter; common nouns do not. Nouns can be singular or plural, concrete or abstract. Nouns show possession by adding’s. Nouns can function in different roles within a sentence; for example, a noun can be a subject, direct object, indirect object, subject complement, or object of a preposition.

నామవాచకం అనేది ఒక వ్యక్తి, స్థలం, వస్తువు లేదా ఆలోచనను సూచించే పదం. నామవాచకాలు తరచుగా ఒక article (the, a, an)తో ఉపయోగించబడతాయి, కానీ ఎల్లప్పుడూ కాదు. సరైన నామవాచకాలు ఎల్లప్పుడూ capital అక్షరంతో ప్రారంభమవుతాయి. సాధారణ నామవాచకాలు అలా చేయవు. నామవాచకాలు ఏకవచనం లేదా బహువచనం, కాంక్రీటు లేదా వియుక్తంగా ఉండవచ్చు. నామవాచకాలు ‘s ని జోడించడం ద్వారా possession ని చూపుతాయి. నామవాచకాలు ఒక వాక్యంలో వేర్వేరు పాత్రలలో పనిచేయగలవు; ఉదాహరణకు, ఒక నామవాచకం ఒక కర్త, ప్రత్యక్ష వస్తువు, పరోక్ష వస్తువు, కర్త యొక్క వస్తువు కావచ్చు.

Common Nouns : man, mountain, state, ocean, country, building, cat, airline, etc.

Proper Nouns: Himalayas, Andhra Pradesh, Atlantic Ocean, Australia, Vijayawada, Moon, Sun, Ram, Raghu, Sita, etc. Abstract Nouns: temperature, wealth, happiness, pride, fear, religion, belief, history, communication, fever, headache, heat, etc.

Concrete Nouns: stone, water, wood, iron, glass, grass, bird, photograph, banana, hands, dog, suitcase, flowers, etc. Countable Nouns: coin, bed, cat, movie, train, country, book, phone, match, speaker, clock, pen, pencil, violin, etc.

Uncountable Nouns : money, milk, rice, snow, rain, water, food, music, oil, sugar, hair, etc.

Compound Nouns : headmaster, handkerchief, tablecloth, eyeglasses, seabird, classmate, daughter-in-law, pigtails, sunlight, snowfall, playground, doorbell, etc.

Collective Nouns: bunch, audience, flock, team, group, family, band, bundle, socks, trousers, herd, flock, group, procession, staff, jury, etc.

Pronoun

A pronoun is a word used in place of a noun. A pronoun is usually substituted for a specific noun. Pronouns are further defined by type: personal pronouns refer to specific persons or things; possessive pronouns indicate ownership; reflexive pronouns are used to emphasize another noun or pronoun; relative pronouns introduce a subordinate clause; and demonstrative pronouns identify, point to, or refer to nouns.

సర్వనామం అనేది నామవాచకం స్థానంలో ఉపయోగించే పదం. సర్వనామం సాధారణంగా ఒక నిర్దిష్ట నామవాచకానికి ప్రత్యామ్నాయంగా ఉంటుంది, దీనిని, ఆమె అనే సర్వనామం యొక్క పూర్వపదం అమ్మాయి. వ్యక్తిగత సర్వనామాలు నిర్దిష్ట వ్యక్తులు లేదా వస్తువులను సూచిస్తాయి.

Personal Pronouns:

Demonstrative Pronouns : This, that, these, those.

• This is a book.
• That is a book.
• These are my books.
• Those are your books.

Indefinite Pronouns : Anyone, something, all, some, both. Anyone can do it.

• Anybody can do it.
• Someone can do it.
• Somebody can do it.
• All can do it.
• Both are my sisters.

Interrogative Pronouns: What, whom, whose, who, which. What happened?

• With whom did you come ?
• Whose are these books?
• Who do you want?
• Which is your pen?

Distributive Pronouns : Each, everyone, either, neither. Each one of you will get a chance.

• Everyone of you will get a chance.
• Either of you can get a chance.
• Neither of you will get a chance.

Verb

The verb in a sentence expresses action or being. There is a main verb and sometimes one or more helping verbs. (“She can sing.” Sing is the main verb; can is the helping verb.) A verb must agree with its subject in number (both are singular or both are plural). Verbs also take different forms to express tense.

ఒక వాక్యంలోని క్రియ చర్య లేదా ఉనికిని వ్యక్తపరుస్తుంది. ఒక ప్రధాన క్రియ మరియు కొన్నిసార్లు ఒకటి లేదా అంతకంటే ఎక్కువ సహాయ క్రియలు ఉంటాయి. (“ఆమె పాడగలదు.” sing అనేది ప్రధాన క్రియ; can అనేది సహాయ క్రియ.) ఒక క్రియ దాని కర్తతో ఏకీభవించాలి (రెండూ ఏకవచనం లేదా రెండూ బహువచనం). క్రియలు కూడా కాలాన్ని వ్యక్తీకరించడానికి వేర్వేరు రూపాలను తీసుకుంటాయి.

• Action verbs : go, come, sit, stand, walk, jump, take, drive, do, drink, clean, cut, drive, eat, fly, go, live, make, play, read, run, shower, sleep, smile, stop, sweep, swim, think, throw, trip, wash, work, write etc. (There are thousands of these verbs.)
• Being verbs : am, is, are, was, were (be, been).
• Possessive verbs : have, has, had.
• Helping verbs : Am, is, are, was, were, have, has, had, will, would, shall, should, can, could, may, might, need, dare, ought, used, and must.

Adjective

An adjective is a word used to modify or describe a noun or a pronoun. It usually answers the question of which one, what kind, or how many. (Articles [a, an, the] are usually classified as adjectives.)
Examples : homeless, adorable, delightful, homely, adventurous, horrible, aggressive, determined, hungry, real, agreeable, etc.

విశేషణం అనేది నామవాచకం లేదా సర్వనామాన్ని సవరించడానికి లేదా వివరించడానికి ఉపయోగించే పదం. ఇది సాధారణంగా ఏది, ఏ రకం లేదా ఎన్ని అనే ప్రశ్నకు సమాధానం ఇస్తుంది. (articles [a, an, the] ను సాధారణంగా విశేషణాలుగా వర్గీకరిస్తారు.)

Kinds of adjectives:

a. Quantitative Adjectives : Quantitative adjectives provide information about the quantity of the nouns / pronouns. This type belongs to the question category of ‘how much’ and ‘how many.’
Examples:

I have just 200 rupees in my purse. (How much)
They have three children. (How many)
You should have completed the whole task. (How much)
There are a few students in the class. (How many)

b. Descriptive Adjectives : Descriptive adjectives are words which describe nouns and pronouns. Most of the adjectives belong to this type. These adjectives provide information and attribute to the nouns/pronouns they modify or describe. Descriptive adjectives are also called qualitative adjectives.

Examples:

• I have a fast car which can be used in races.
• I am hungry so let us go to any hotel.
• The mischievous children are making a lot of noise.
• I saw a flying Eagle.
• He bought an expensive old silver mirror.
• He gave me four fresh yellow roses.
• I want to buy a diamond ring.

c. Proper Adjectives: Proper adjectives are the adjective form of proper nouns. When proper nouns modify or describe other nouns/pronouns, they become proper adjectives. ‘Proper’ means ‘specific.’
Example:

• American cars are comparatively very strong.
• Chinese people are hard workers but their leaders are cunning.
• KFC restaurants are found only in big cities.
• Marxist philosophers don’t like capitalism.
• He belongs to the Communist party of India.
• Indian culture is getting diluted.

d. Demonstrative Adjectives: Demonstrative adjectives directly refer to something or someone. Demonstrative adjectives include the words: this, that, these, those.

Examples:

• These books are borrowed from our college library.
• This house is ours.
• That problem needs a very good solution.
• Those bridges were built by the British.

e. Possessive adjectives : Possessive adjectives indicate possession or ownership. It suggests the belongingness of something to someone/something. Some of the most used possessive adjectives are my, his, her, our, their, your.

Examples:

• Your cycle was stolen yesterday.
• We all liked his performance.
• The students of class submitted their assignments.
• I could not understand her intentions.
• My mind stops working when I think about the problem.
• Our intention is not to hurt anyone’s feelings.

f. Interrogative Adjectives : Interrogative adjectives ask a question. An interrogative adjective must be followed by a noun or a pronoun. The interrogative adjectives are: which, what, whose. These words will not be considered as adjectives if a noun does not follow right after them. ‘Whose’ also belongs to the possessive adjective type.

Examples:

• Whose book was that?
• Which pen do you like more?
• What books are you buying today?
• Which team scored the higher in the last match?
• What curry do you like on Sundays?
• Which song are you listening to?

g. Indefinite Adjectives : Indefinite adjective describes or modifies a noun unspecifically. They provide indefinite/ unspecific information about the noun. The common indefinite adjectives are few, many, much, most, all, any, each, every, either, nobody, several, some, etc.

Examples:

• I gave some ice cream to my friend.
• I want a few moments to take a decision.
• Several writers wrote about the recent incidents.
• Each student will have to submit homework tomorrow.

Adverb

An adverb describes or modifies a verb, an adjective, or another adverb, but never a noun. It usually answers the questions of when, where, how, why, under what conditions, or to what degree. Adverbs often end in ‘ly’.

ఒక క్రియా విశేషణం ఒక క్రియ, విశేషణం లేదా మరొక క్రియా విశేషణాన్ని వర్ణిస్తుంది లేదా మారుస్తుంది, కానీ ఎప్పుడూ నామవాచకంగా ఉండదు. ఇది సాధారణంగా ఎప్పుడు, ఎందుకు, ఎలా, ఎందుకు, ఏ పరిస్థులలో, లేదా ఎంతవరకు అనే ప్రశ్నలకు సమాధానం ఇస్తుంది. క్రియా విశేషణాలు తరుచుగా ly తో ముగుస్తాయి.

Adverbs of questions : How, when, where, why.
How did they do it ?
When did they do it ?
Where did they do it ?

Adverbs of Time : today, yesterday, later, now, then, last year, all day, for a while, sometimes,

Adverbs of Frequency: hourly, daily, weekly, monthly, yearly, every, second, once a minute, twice a year, once, twice, once or twice, three times, always, constantly, ever, frequently, generally, never, normally, occasionally, often, rarely, regularly, seldom, sometimes, usually, annually, daily, fortnightly, hourly, monthly, nightly, quarterly, weekly, yearly, yes, no, not, so, too, very, etc.

Preposition

A preposition is a word placed before a noun or pronoun to form a phrase modifying another word in the sentence. Therefore a preposition is always part of a prepositional phrase. The prepositional phrase almost always functions as an adjective or as an adverb.

సర్వనామం ముందు ఉంచిన పదం, వాక్యంలోని మరొక పదాన్ని సవరించి విభక్తి ప్రత్యయము ఒక పదబంధాన్ని ఏర్పరుస్తుంది. అందువల్ల, విభక్తి ప్రత్యయము రెండు పదాల మధ్య బంధాన్ని ఏర్పరుస్తుంది.

Examples: aboard, about, above, across, after, against, ahead of, along, amid, amidst, among, around, as, as far as, as of, aside from, at, athwart, atop barring, before, behind, below, beneath, beside, besides, between, beyond, but, by, despite, down, during, except, excluding, for, from, in, minus, near, next to, of, off, on, opposite, out, outside, over, since, through, till, to, toward, towards, under, underneath, unlike, until, up, upon, with, within, without, etc.

Conjunction

A conjunction joins words, phrases, or clauses, and indicates the relationship between the elements joined.

Coordinating conjunctions connect grammatically equal elements: and, but, or, nor, for, so, yet.

Subordinating conjunctions connect clauses that are not equal: because, although, while, since, etc.

There are other types of conjunctions as well.

ఒక సంయోగం పదాలు,పదబంధాలు లేదా నిబంధనలను కలుపుతుంది మరియు జతచేయబడిన elements మధ్య సంబంధాన్ని సూచిస్తుంది. సమన్వయ సంయోగాలు వ్యాకరణపరంగా సమాన అంశాలను కలుపుతాయి: and, but, or, nor, for, so, yet. సబార్డినేటింగ్ సంయోగాలు సమానం కానీ నిబంధనలు కలుపుతాయి: because, although, while, since, etc. ఇతర రకాల సంయోగాలు కూడా ఉన్నాయి.

Subordinating conjunctions: though, although, even though, even if, as, since, because, for, after, as if, as long as, as much as, as soon, as though, before, even if, if, lest, than, that, till, unless, until, when, where, how, what, why, whenever, wherever, while etc.

Coordinating conjunctions: but, yet, still, so, therefore, that’s why, and or, otherwise, then, either………… or, neither………… nor, not only……… but also etc.

Interjection

An interjection is a word used to express emotion. It is often followed by an exclamation point (!).

భావోద్వేగానికి వ్యక్తీకరించడానికి ఉపయోగించే పదం దాని తర్వాత తరచుగా ఆశ్చర్యార్థంకంగా గుర్తు (!) ఉంటుంది.

Example: Aha! Ahh! Ahoy! Alas! Boo! Bravo! Congratulations! Eh! Eureka! Goodbye! Goodness! Hey! Hurray! Oh! Oops! Ouch! Phew! uhhuh!

Wahoo! Wow! Yeah!, etc.

Examples

Question 1.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 2.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

(Note : In sentence 5, ‘out’ modifies the verb ‘went’ by indicating direction or movement away from a place, which is typical of an adverb of place.)

Exercises For Practice

Question 1.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 2.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 3.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 4.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 5.
Here’s a day from the life of Padmini. Can you identify at least 15 nouns from the given paragraph ?

Last weekend, Padmini visited the museum with her friends. They admired the paintings and marvelled at the sculptures in the art gallery. Later, they had lunch at a cozy cafe, where Padmini ordered a delicious sandwich and a cup of coffee. The group talked about their plans for the upcoming holiday and shared their ideas for a fun trip. Everyone enjoyed the afternoon, and they decided to visit the museum again next month.
Answers:

1. weekend
2. museum
3. friend
4. paintings
5. sculptures
6. gallery
7. lunch
8. cafe
9. sandwich
10. cup
11. coffee
12. group
13. plans
14. holiday
15. ideas
16. trip
17. afternoon
18. month

Question 6.
Fill in the blanks with correct plural form of the nouns given in the brackets.

1. The park was full of playful ___________. (child)
Answer:
children

2. We ran several ___________. (diagnosis)
Answer:
diagnoses

3. He explained the ___________ (focus) of the study.
Answer:
foci

4. My ___________ (foot) were sore after the run.
Answer:
feet

5. My dentist says I have healthy ___________. (tooth)
Answer:
teeth

6. A flock of ___________ (goose) flew across the sky.
Answer:
geese

7. All my ___________ (brother-in-law) are teachers.
Answer:
brothers-in-law

Question 7.
Fill in the blanks with appropriate Pronouns to complete the story.

One day, a young boy named Sam went to the park with ___________ (1) sister. ___________ (2) both loved playing on the swings and running around. While Sam was climbing a tree, ___________ (3)
saw a bird’s nest. ___________ (4) called out to ___________ (5) sister, “Look! There’s a nest up here!” ___________ (6) sister looked up and said. “Be careful, Sam. Don’t scare the birds!” Sam carefully climbed down and joined ___________ (7) on the bench. Together, ___________ (8) decided to draw the birds in ___________ (9) notebooks. When ___________ (10) finished, ___________ (11) showed the drawings to ___________ (12) parents, who were very impressed.
Answers:
1. his
2. They
3. he
4. He
5. his
6. His
7. her
8. they
9. their
10. they
11. they
12. their

Question 8.
Read each sentence carefully and underline or highlight the Adjectives.

1. The beautiful garden was filled with vibrant flowers.
Answer:
The beautiful garden was filled with vibrant flowers.

2. She wore a red dress and carried a small bag.
Answer:
She wore a red dress and carried a small bag.

3. The tall building cast a long shadow on the street.
Answer:
The tall building cast a long shadow on the street.

4. We had a delicious meal at the new restaurant.
Answer:
We had a delicious meal at the new restaurant.

5. His quick decision helped to avoid a serious mistake.
Answer:
His quick decision helped to avoid a serious mistake.

6. The young boy found a golden key in the ancient chest.
Answer:
The young boy found a golden key in the ancient chest.

7. It was a cold night with a bright moon in the sky.
Answer:
It was a cold night with a bright moon in the sky.

8. The teacher gave us a challenging yet interesting assignment.
Answer:
The teacher gave us a challenging yet interesting assignment.

9. I love my cozy, little home in the peaceful countryside.
Answer:
I love my cozy, little home in the peaceful countryside.

10. The angry crowd demanded immediate action from the authorities:
Answer:
The angry crowd demanded immediate action from the authorities.

Question 9.
Identify the Adverbs in the following sentences.

1. She sings melodiously during every performance.
Answer:
melodiously

2. The train arrived late because of the storm.
Answer:
late

3. He quickly finished his homework before heading out.
Answer:
quickly

4. The team played extremely well in the finals.
Answer:
extremely, well

5. They are always kind to their neighbours.
Answer:
always

6. The baby slept peacefully in her crib.
Answer:
peacefully

7. We will visit the museum tomorrow.
Answer:
tomorrow

8. He spoke very softly during the meeting.
Answer:
very, softly

9. The cat crept silently across the room.
Answer:
silently

10. I have never seen such a beautiful sunset.
Answer:
never

Question 10.
Underline the Conjunctions in the following sentences.

1. She wanted to go to the party, but she was feeling unwell.
Answer:
She wanted to go to the party, but she was feeling unwell.

2. I will call you when I reach home.
Answer:
I will call you when I reach home.

3. He likes both cats and dogs, but he cannot decide which to adopt.
Answer:
He likes both cats and dogs, but he cannot decide which to adopt.

4. They left early because they had a train to catch.
Answer:
They left early because they had a train to catch.

5. You can stay here or you can go with them.
Answer:
You can stay here or you can go with them.

Exercises

Question 1.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 2.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 3.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 4.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 5.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 6.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 7.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 8.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 9.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 10.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 11.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Question 12.
Match the underlined words of the sentences under Section A with their respective Parts of Speech under Section B.

Answer:

Referring to the AP Inter 1st Year Maths Study Material Chapter 8 Sequences and Series Exercise 8e Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sequences and Series Solutions Exercise 8e

I. Find the sum to n terms of each of the series in Exercises 1 to 7.

Question 1.
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
Solution:
The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … n terms
Hence, a_n = n(n + 1)
Therefore, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) a_k = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k(k + l) = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k² + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k

= \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{\mathrm{n}(\mathrm{n}+1)}{2}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left(\frac{2 \mathrm{n}+1}{3}+1\right)\)

= \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left(\frac{2 \mathrm{n}+4}{3}\right)\)

= \(\frac{\mathrm{n}}{3}\)(n + 1)(n + 2)

Question 2.
1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
Solution:
The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + … n terms
Hence, a_n = n(n + 1) (n + 2) = (n² + n) (n + 2) = n³ + 3n² + 2n
Therefore, S_n = \(\sum_{k=1}^n\) ak = \(\sum_{k=1}^n\)(k³ + 3k² + 2k) = \(\sum_{k=1}^n\) k³ + 3 \(\sum_{k=1}^n\) k² + 2 \(\sum_{k=1}^n\) k

Question 3.
3 × 1² + 5 × 2² + 7 × 3² + …
Solution:
The given series is 3 × 1² + 5 × 2² + 7 × 3² + …. n terms
Hence, a_n = (2n + 1) n² = 2n³ + n²
Therefore, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) a_k = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) (2k³ + k²) = 2\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k³ + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k²

= 2\(\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]^2\) + \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\)

= \(\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{2}+\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\)

= \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left(\mathrm{n}(\mathrm{n}+1)+\frac{2 \mathrm{n}+1}{3}\right)\)

= \(\frac{n(n+1)}{2}\left[\frac{3 n^2+3 n+2 n+1}{3}\right]\)

= \(\frac{n(n+1)}{2}\left[\frac{3 n^2+5 n+1}{3}\right]\)

= \(\frac{n}{6}\)(n + 1)(3n + 5n + 1)

Question 4.
\(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}\) + ………..
Solution:
The given series is \(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}\) + ……….. n terms
Hence, a_n = \(\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\)
∴ a₁ = \(\frac{1}{1}-\frac{1}{2}\), a₂ = \(\frac{1}{2}-\frac{1}{3}\), a₃ = \(\frac{1}{3}-\frac{1}{4}\), …………….., a_n = \(\frac{1}{n}-\frac{1}{n+1}\),

Adding the above terms column wise, we obtain
a₂ + a₂ + …… + a_n = \(\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{\mathrm{n}}\right]-\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{\mathrm{n}+1}\right]\)

Thus, S_n = 1 – \(\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1}\)

Question 5.
5² + 6² + 7² + ………. +20²
Solution:
The given series is 5² + 6² + 7² + ………. +20²
Hence, a_n = (n + 4)² = n² + 8n + 16

Therefore, S_n = \(\sum_{k=1}^n\)a_k = \(\sum_{k=1}^n\)(k² + 8k + 16)

S_n = \(\sum_{k=1}^n\)k² + 8\(\sum_{k=1}^n\)k + \(\sum_{k=1}^n\) 16

= \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{8 \mathrm{n}(\mathrm{n}+1)}{2}\) + 16n

Since, 16^th term is (16 + 4)² = (20)²

Then, S_n = \(\frac{16(16+1)(2(16)+1)}{6}+\frac{8(16)(16+1)}{2}\)+ 16(16)

= \(\frac{(16)(17)(33)}{6}+\frac{(128)(17)}{2}\) + 256

= 1496 + 1088 + 256 = 2840

Thus, 5² + 6² + 7² + ………. +20² = 2840

Question 6.
3 × 8 + 6 × 11 + 9 × 14 + …………..
Solution:
The given series is 3 × 8 + 6 × 11 + 9 × 14 + ………….. n terms
Hence, a_n = (n^th term of 3, 6, 9, ………..) × (n^th term of 8, 11, 14 ……)
= (3n)(3n + 5)
= 9n² + 15n

Therefore, S_n = \(\sum_{k=1}^n\) ak = \(\sum_{k=1}^n\) (9k² + 15k)

= 9 \(\sum_{k=1}^n\) k² + 15\(\sum_{k=1}^n\)k

= 9 × \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\) + 15 × \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

= \(\frac{3 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{2}+\frac{15 \mathrm{n}(\mathrm{n}+1)}{2}\)

= \(\frac{3 \mathrm{n}(\mathrm{n}+1)}{2}\)(2n + 1 + 5)

= \(\frac{3 \mathrm{n}(\mathrm{n}+1)}{2}\)(2n + 6)

= 3n(n + 1)(n + 3)

Question 7.
1² + (1² + 2²) + (1² + 2² + 3²) + ………..
Solution:
The given series is 1² + (1² + 2²) + (1² + 2² + 3²) + ……….. n terms

Hence, a_n = (1² + 2² + 3² + …… + n²) = \(\frac{n}{2}\)(n + 1)(2n + 1)
= \(\frac{n}{6}\) (2n² + 3n + 1)
= \(\frac{1}{3}\)n³ + \(\frac{1}{2}\)n² + \(\frac{1}{6}\)n

Therefore,

II. Find the sum to n terms of the series in Exercises 1 to 3 whose nth terms is given by

Question 1.
n(n + 1) (n + 4).
Solution:
The given n* term is a_n = n(n + 1) (n + 4)
Hence, a_n = n(n + 1) (n + 4) = n(n² + 5n + 4) = n³ + 5n² + 4n
Therefore, S_n = \(\sum_{k=1}^n\) a_k = \(\sum_{k=1}^n\)k³ + 5 \(\sum_{k=1}^n\) k² + 4 \(\sum_{k=1}^n\) k

Question 2.
n² + 2n
Solution:
The given n^th term is an = n² + 2ⁿ
Hence, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k² + 2^k

= \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k² + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)2^k …………..(1)
Consider \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 2^k = 2¹ + 2² + 2³ +…

The above series 2¹ + 2² + 2³ + … is a G.P with both the first term and common ratio equal to 2.
\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 2^k = \(\frac{1}{2}\) = 2(2ⁿ – 1) …………..(2)

From (1) and (2) we obtain
S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k² + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 2^k
= \(\frac{n}{6}\)(n + 1)(2n + 1) + 2(2ⁿ – 1)

Question 3.
(2n – 1)²
Solution:
The given n^th term is a_n = (2n – 1)²
Hence, a_n = (2n – 1)² = 4n² – 4n + 1

Therefore, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) a = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) (4k² – 4k + 1)

= 4\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k2 – 4\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 1

= \(\frac{4 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\frac{4 \mathrm{n}(\mathrm{n}+1)}{2}\) + n

= \(\frac{2 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{3}\) – 2n(n + 1) + n

= n[\(\frac{2\left(2 n^2+3 n+1\right)}{3}\) – 2(n + 1) + 1]

= n\(\left[\frac{4 n^2+6 n+2-6 n-6+3}{3}\right]\) = n\(\left[\frac{4 \mathrm{n}^2-1}{3}\right]\)

= \(\frac{\mathrm{n}}{3}\) (2n + 1)(2n – 1)

Referring to the AP Inter 1st Year Maths Study Material Chapter 7 Binomial Theorem Exercise 7a Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Binomial Theorem Solutions Exercise 7a

I. Expand each of the expressions in Exercises 1 to 8.

Question 1.
(1 – 2x)⁵
Solution:
By using Binomial theorem, the expression (1 – 2x)⁵ can be expanded as
⁵C₀(1)⁵(-2x)⁰ + ⁵C₁(1)⁴(-2x)¹ + ⁵C₂ (1)³ (-2x)² + ⁵C₃(1)²(-2x)³ + ⁵C₄(1)¹ (-2x)⁴ + ⁵C₅(1)° (-2x)⁵
= 1 – 5(2x) + 10(4x²) – 10(8x³) + 5(16x⁴) – (32x⁵)
= 1 – 10x + 40x² – 80x³ + 80x⁴ – 32x⁵.

Question 2.
\(\left(\frac{2}{x}-\frac{x}{2}\right)^5\)
Solution:
By using Binomial theorem, the expression \(\left(\frac{2}{x}-\frac{x}{2}\right)^5\) can be expanded as

Question 3.
(2x – 3)⁶
Solution:
By using Binomial theorem, the expression (2x – 3)⁶ can be expanded as ⁶C₀ (2x)⁶ (-3)° + ⁶C₁ (2x)⁵ (-3)¹ + ⁶C₂ (2x)⁴ (-3)² + ⁶C₃ (2x)³ (-3)³ + ⁶C₄ (2x)² (-3)⁴ + ⁶C₅ (2x)¹ (-3)⁵ + ⁶C₆ (2x)⁰ (-3)⁶
= 64x⁶ – 6(32x⁵) (3) + 15(16x⁴)(9) – 20(8x³)(27) + 15(4x²)(81) – 6(2x)(243) + 729
= 64x⁶ – 576x⁵ + 2160x⁴ – 4320x³ + 4860x² – 2916x + 729

Question 4.
\(\left(\frac{x}{3}+\frac{1}{x}\right)^5\)
Solution:
By using Binomial theorem, the expression \(\left(\frac{x}{3}+\frac{1}{x}\right)^5\) can be expanded as

Question 5.
\(\left(x+\frac{1}{x}\right)^6\)
Solution:
By using Binomial theorem, the expression \(\left(x+\frac{1}{x}\right)^6\) can be expanded as

Question 6.
(4x + 5y)⁷
Solution:
By using Binomial theorem,
(4x + 5y)⁷ = ⁷C₀ (4x)⁷ (5y)⁰ + ⁷C₁ (4x)⁶ (5y)¹ + ⁷C₂ (4x)⁵ (5y)² + ⁷C₃ (4x)⁴ (5y)³ + ⁷C₄ (4x)³ (5y)⁴ + ⁷C₅ (4x)² (5y)⁵ + ⁷C₆ (4x)¹ (5y)⁶ + ⁷C₇ (4x)° (5y)⁷
= \(\sum_{r=0}^7\) ⁷C_r (4x)^7-r (5y)^r

Question 7.
\(\left(\frac{2}{3} x+\frac{7}{4} y\right)^5\)
Solution:
By using Binomial theorem, we have

Question 8.
\(\left(\frac{2 p}{5}-\frac{3 q}{7}\right)^6\)
Solution:

Question 9.
Write down and simplify.
i) 6^th term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\)
Solution:
Given that \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\)
The general term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\) is
T_r+1 = ⁹C_r \(\left(\frac{2 x}{3}\right)^{9-r}\left(\frac{3 y}{2}\right)^r\)
If r = 5
T₆ = ⁹C₅\(\frac{1}{2}\)
= ⁹C₅\(\frac{1}{2}\)
= ⁹C₄\(\frac{1}{2}\) x⁴y⁵
= \(\frac{1}{2}\). x⁴y⁵
= 189 x⁴y⁵

ii) 7^th term in (3x – 4y)¹⁰.
Solution:
Given that (3x – 4y)¹⁰
The general term in (3x – 4y)¹⁰ is T_r+1 = (-1)^{r 10}C_r (3x)^10-r (4y)^r
If r = 6 ⇒ T₇ = (-1)^{6 10}C₆ (3x)⁴ (4y)⁶
= ₂C₄3₄4₆x₄y₆
= \(\frac{10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4}\). 3₄.4₆. x₄. y₆
= 280 (12)₅x₄y₆ (∵ _nC_r = _nC_n-r)

iii) 10^th term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\)
Solution:
Given that \(\left(\frac{3 p}{4}-5 q\right)^{14}\)
The general term in \(\left(\frac{3 p}{4}-5 q\right)^{14}\) is T_r+1 = (-1)^{r 14}C_r \(\left(\frac{3 \mathrm{p}}{4}\right)^{14-\mathrm{r}}\) (5q)^r
If r = 9
T₁₀ = (-1)^{9 14}C₉\(\left(\frac{3 p}{4}\right)^5\) (5q)⁹
= – ¹⁴C₅\(\left(\frac{3}{4}\right)^5\) 5⁹.p⁵.q⁹ (∵ _nC_r = _nC_n-r)
= \(-\frac{14 \times 13 \times 12 \times 11 \times 10}{1 \times 2 \times 3 \times 4 \times 5} \cdot \frac{3^5}{4^5}\).5⁹.p⁵.q⁹
= \(\frac{-(2002) 3^5 \cdot 5^9}{4^5}\)p⁵.q⁹

iv) r^th term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) (1 ≤ r ≤ 8)
Solution:
Given that \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\)
The general term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) is T_r+1 = ⁸C_r \(\left(\frac{3 a}{5}\right)^{8-r}\left(\frac{5 b}{7}\right)^r\)
Replace r by r – 1, then
T_r = ⁸C_(r-1)\(\left(\frac{3 a}{5}\right)^{8-r+1}\left(\frac{5 b}{7}\right)^{r-1}\)
= ⁸C_(r-1)\(\left(\frac{3 a}{5}\right)^{9-r}\left(\frac{5 b}{7}\right)^{r-1}\) (1 ≤ r ≤ 8)

v) Middle term in \(\left(3-\frac{x^3}{6}\right)^7\)
Solution:
Given that \(\left(3-\frac{x^3}{6}\right)^7\)

The expression contains 8 terms. Thus T₄ and T₅ are middle terms.
T₄ = T₃₊₁ = ⁷C₃ (3)^7-3\(\left(-\frac{x^3}{6}\right)^3=\frac{35 \times 3^4(-x)^9}{6^3}=-\frac{105}{8}\)x⁹.

T₅ = T₄₊₁ = ⁷C₄ (3)^7-4\(\left(-\frac{x^3}{6}\right)^4=\frac{35 \times 3^3 x^{12}}{6^4}=\frac{35}{48}\)x¹²

Thus, the middle terms in the expansion of \(\left(3-\frac{x^3}{6}\right)^7\) are \(-\frac{105}{8}\) x⁹ and \(\frac{35}{48}\)x¹²

vi) Middle term in \(\left(\frac{x}{3}+9 y\right)^{10}\)
Solution:
Given that \(\left(\frac{x}{3}+9 y\right)^{10}\)
The number of terms in the given expansion is 11. Thus T₆ is the middle term.
T₆ = T₅₊₁ = ¹⁰C₅ \(\frac{1}{2}\) (9y)⁵
= \(\frac{1}{2}\) 3¹⁰.x⁵y⁵
= 252 × 3⁵ x⁵.y⁵
= 61236 x⁵y⁵

vii) 4th term in (x – 2y)¹²
Solution:
Given that (x – 2y)¹²
General term in the expansion of (x + y)ⁿ is Tr+ x = nCr Xn_r (Y)r
Here, X = x, Y = -2y, and n = 12; T_r+1 = ¹²C_r (x)^12-r (-2y)^r
Putting r = 3, ,
T₃₊₁ = ¹²C₃x^12-3 (-2y)³
T₄ = \(\frac{12 \times 11 \times 10}{6}\) x⁹(-2)³y³
= 2 × 11 × 10x⁹(-8)y³
= 220 × (-8)x⁹y³
= -1760 x⁹y³

viii) 13^th term in \(\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}\), x ≠ 0
Solution:
The general term in the expansion of (X + Y)ⁿ is T_r+1 = ⁿC_r X^n-rY^r

Here X = 9x, Y = –\(\frac{1}{3 \sqrt{x}}\), and n = 18

The General term in the expansion of \(\frac{1}{2}\) is T_r = ¹⁸C_r (9x)^18-r\(\left(\frac{-1}{3 \sqrt{x}}\right)^r\)
Putting r = 12
T₁₂₊₁ = ¹⁸C₁₂ (9x)⁶ \(\left(\frac{-1}{3 \sqrt{x}}\right)^{12}\)
= ¹⁸C₁₂ 9⁶ × x⁶\(\frac{1}{3^{12}(\sqrt{x})^{12}}\) (∵ _nC_r = _nC_n-r)
= \(\frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \times \frac{3^{12} \times x^6}{3^{12} \times x^6}\)
= 18564

Question 10.
Find the number of terms in the expansion of
i) \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\)
Solution:
The number of terms in (x + a)ⁿ is (n + 1), where n is a positive integer.
Hence number of terms in \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\) are 9 + 1 = 10

ii) (3p + 4q)¹⁴
Solution:
The number of terms in (x + a)ⁿ is (n + 1), where n is a positive integer.
Hence number of terms in (3p + 4q)¹⁴ are 14 + 1 = 15

II. Using binomial theorem, evaluate each of the following.

Question 1.
(96)³
Solution:
96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial theorem can be applied, it can be written that, 96 = 100 – 4
Therefore, (96)³ = (100 – 4)³
= ³C₀ (100)³ (-4)⁰ + ³C₁ (100)² (-4)¹ + ³C₂ (100)¹ (-4)² + ³C₃ (100)° (-4)³
= (100)³ – 3(100)²(4) + 3(100)(4)² – (4)³
= 1000000 – 120000 + 4800 – 64
= 884736

Question 2.
(102)⁵
Solution:
102 can be expressed as the sum or difference of two numbers whose powers are
easier to calculate and then, Binomial theorem can be applied.
It can be written that, 102 = 100 + 2.
Therefore, (102)⁵ = (100 + 2)⁵
= ⁵C₀ (100)⁵ (2)° + ⁵C₁ (100)⁴ (2)¹ + ⁵C₂ (100)³ (2)² + ⁵C₃ (100)² (2)³ + ⁵C₄ (100)¹ (2)⁴ + ⁵C₅ (100)° (2)⁵
= (100)⁵ + 5(100)⁴ (2) + 10(100)³(2)² + 10(100)²(2)³ + 5(100) × (2)⁴ + (2)⁵
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32
= 11040808032

Question 3.
(101)⁴
Solution:
101 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial theorem can be applied.
It can be written that, 101 = 100+ 1.
(101)⁴ = (100 + 1)⁴ = ⁴C₀ (100)⁴(1)°+ ⁴C₁ (100)³ (1)¹ + ⁴C₂ (100)² (1)² + ⁴C₃ (100)¹ (1)³ + ⁴C₄ (100)° (1)⁴
= (100)⁴ + 4(100)³ + 6(100)² + 4(100)¹ + 1
= 100000000 + 4000000 + 60000 + 400 + 1
= 104060401

Question 4.
(99)⁵
Solution:
99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial theorem can be applied.
It can be written that 99 = 100 – 1
=> (99)⁵ = (100 – 1)⁵
= ⁵C₀(100)⁵(-1)° + ⁵C₁(100)⁴(-1)¹ + ⁵C₂(100)³(-1)² + ⁵C₃(100)²(-1)³ + ⁵C₄(100)¹(-1)⁴ + ⁵C₅(100)°(-1)⁵
= (100)⁵ – 5(100)⁴ + 10(100)³ – 10(100)² + 5(100) – 1
= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1
= 10010000500 – 500100001
= 9509900499

Question 5.
Using Binomial theorem, indicate which number is larger (1.1)¹⁰⁰⁰⁰ 0r 1000.
Solution:
By splitting 1.1 and then applying Binomial theorem, the first few terms of (1.1)10000 can be obtained as (1.1)¹⁰⁰⁰⁰
= (1 + 0.1)¹⁰⁰⁰⁰
= ¹⁰⁰⁰⁰C₀ + ¹⁰⁰⁰⁰C₁(0.1)¹ + other positive terms
= 1 + 10000 × 0.1 + other positive terms
= 1001 + other positive terms > 1000
Hence, (1.1)¹⁰⁰⁰⁰ > 1000

Question 6.
Find (a + b)⁴ – (a – b)⁴, hence evaluate (√3 + √2)⁴ – (√3 – √2)⁴
Solution:
Using Binomial theorem, the expression, (a + b)4 – (a – b)4, can be expanded as (a + b)⁴ – (a – b)⁴
= ⁴C₀a⁴b° + ⁴C₁b¹ + ⁴C₂a²b² + ⁴C₃b³ + ⁴C₄a°b⁴ – [⁴C₀a⁴b° – ⁴C₁a³b¹ + ⁴C₂a²b² – ⁴C₃a¹b³ + ⁴C₄a°b⁴]
= 2[⁴C₁a³b¹ + ⁴C₃a¹b³]
= 2[4a³b + 4ab³]
= 8ab(a² + b²)

Now putting a = √3 and b = √2 , we have
(√3 + √2)⁴ – (√3 – √2)⁴ = 8(√3)(√2) [(√3)² +(√2)²]
= 8√6(3 + 2)
= 40√6

Question 7.
Find (x + 1)⁶ + (x – 1)⁶. Hence or otherwise evaluate (√2 + 1) + (√2 – 1)
Solution:
Using Binomial theorem, the expression, (x + 1)⁶ + (x – 1)⁶, can be expanded as (x + 1)⁶ + (x – 1)⁶
= ⁶C₀x⁶1⁰ + ⁶C₁⁵1¹ + ⁶C₂x⁴1² + ⁶C₃x³1³ + ⁶C₄x²1⁴ + ⁶C₅x¹1⁵ + ⁶C₆x°1⁶ + ⁶C₀x⁶1°- ⁶C₁x⁵1¹ + ⁶C₂x⁴1² – ⁶C₃x³1³ + ⁶C₄x²1⁴ – ⁶C₅x¹1⁵ + ⁶C₆x°1⁶
= 2[⁶C₀x⁶ + ⁶C₂x⁴ + ⁶C₄x² + ⁶C₆x⁰]
= 2(x⁶ + 15x⁴ + 15x² + 1)
Now putting x = √2 , we have (√2)⁶ + (√2)⁶
= 2[(√2)⁶ + 15(√2)⁴ + 15(√2)² + 1]
= 2(8 + 60 + 30 + 1)
= 198

Question 8.
Show that 9n +1 – 8n – 9 is divisible by 64, whenever n is a positive integer.
Solution:
Using Binomial theorem,
(1 + x)ⁿ⁺¹ = ⁿ⁺¹C₀ x° + ⁿ⁺¹C₁ x¹ + ⁿ⁺¹C₂ x² + …………. + ⁿ⁺¹C_n+1 xⁿ⁺¹
Now putting x = 8, we have
(1+8)ⁿ⁺¹ = ⁿ⁺¹C₀ 8° + ⁿ⁺¹C₁ 8¹ + ⁿ⁺¹C₂ 8² + ⁿ⁺¹C_n+1 8ⁿ⁺¹
⇒ 9ⁿ⁺¹ = 1 + (n + 1)(8) + 8²[ⁿ⁺¹C₂ + …….. + ⁿ⁺¹C_n+1 8^n-1]
⇒ 9ⁿ⁺¹ – 1 – 8(n+1) = 64[ⁿ⁺¹C₂ + + ⁿ⁺¹C_n+1 8^n-1]
⇒ 9ⁿ⁺¹ – 8n – 9 = 64 [ⁿ⁺¹C₂ + …. + ⁿ⁺¹C_n+1 8^n-1]
Now, RHS is divisible by 64 for all positive integers n. So, LHS is also divisible by 64. Hence, 9ⁿ⁺¹ – 8n – 9 is divisible by 64, whenever n is a positive interger.

Question 9.
Prove that \(\sum_{r=0}^n\)3^rnC_r = 4ⁿ.
Solution:
LHS = \(\sum_{r=0}^n\)3^rnC_r = 3⁰ⁿC₀ + 3¹ⁿC₁ + 3²ⁿC₂ + …………… 3^{n-1 n}C_n-1 + ⁿC_n+1. 3ⁿ

= ⁿC₀ 3° + ⁿC₁ 3¹ + ⁿC₂ 3² + …. + 3^n-1 + ⁿC_n.3ⁿ
= (1 + 3)ⁿ = 4ⁿ = RHS

Question 10.
Using the binomial theorem, prove that 50” – 49n – 1 is divisible by 492 for all positive integers n.
Solution:
Given 50ⁿ – 49n – 1 = (49 + 1)ⁿ – 49n – 1
= ⁿC₀(49)ⁿ + ⁿC₁(49)^n-1 + ⁿC₂(49)^n-2 + … + ⁿC_n 2(49)² + ⁿC_n-1(49) + ⁿC_n(49)° – 49n – 1
= (49)ⁿ + ⁿC₁(49)^n-1 + ⁿC₂(49)^n-2 + … + ⁿC_n-2(49)² + (n) (49) + 1 – 49n – 1
= (49)² [(49)^n-2 + ⁿC₂ (49)^n-3 + ⁿC₂(49)^n-4 + … + ⁿC_n-2]
= (49)² [a positive integer]
∴ 50ⁿ – 49n – 1 is divisible by 492 for all positive integers of n.

Practice AP Inter 1st Year Maths Study Material Chapter 8 Sequences and Series MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Sequences and Series MCQ

Question 1.
Which of the following is finite sequence ?
1) 48, 24, 12, ………..
2) 1, 2, 3, …….
3) 2, 4, 6, 8, 10 ………..
4) 2, 3, 5, 7, 11, 13, …….
Answer:
3) 2, 4, 6, 8, 10 ………..

Explanation:
2, 4, 6, 8, 10 is a finite sequence.

Question 2.
Which of the following relation gives Fibonacci sequence ?
1) a_n = a_n-1 + a_n-2, n > 2
2) a_n-1 = a_n + a_n-2, n > 2
3) a_n-2 = a_n + a_n-1, n> 2
4) a_n = a_n+1 + a_n-2, n > 2
Answer:
1) a_n = a_n-1 + a_n-2, n > 2

Explanation:
Relation of fibonacci sequence is a_n = a_n-1 + a_n-2, n > 2

Question 3.
The first term of Fibonacci sequence is
1) 0
2) 1
3) 2
4) 3
Answer:
2) 1

Explanation:
The first term of fibonacci sequence is “1”.

Question 4.
If a_n = 4n + 6, then 15^th term of the sequence is
1) 6
2) 10
3) 60
4) 66
Answer:
4) 66

Explanation:
a_n = 4n + 6
a₁₅ = 4(15) + 6
= 60 + 6 = 66

Question 5.
A series can also be denoted by the symbol …………
1) πa_n
2) Σa_n
3) Φa_n
4) θa_n
Answer:
2) Σa_n

Explanation:
Series denoted by Σa_n

Question 6.
The sum of first five terms of series 2 + 4 + 6 + ….. is
1) 14
2) 16
3) 20
4) 30
Answer:
4) 30

Explanation:
Given series 2 + + 6 + ………..
Sum of even number S_n = n(n + 1)
S₅ = 5(5 + 1) = 5.(6) = 30

Question 7.
\(\sum_{n=1}^4\) 2n + 3 = ……….
1) 5
2) 12
3) 21
4) 32
Answer:
4) 32

Explanation:
\(\sum_{n=1}^4\) 2n + 3, If n = 1, 2, 3, 4; Series is 5, 7, 9, 11 …… ; S₄ = 5 + 7 + 9 + 11 = 32

Question 8.
If a_n+1 = a_n. r then the sequence is called
1) Arithmetic progression
2) Geometric progression
3) Harmonic progression
4) Special progression
Answer:
2) Geometric progression

Explanation:
a_n+1 = a_n. r is G.P.

Question 9.
The nth term of G.P. is ( 3 )
1) a_n = a + (n – 1) d
2) a_n = a + nd
3) a_n = a . r^n-1
4) a_n = arⁿ
Answer:
3) a_n = a . r^n-1

Explanation:
n^th term of G.P. is a_n = a . r^n-1

Question 10.
If r = 1 in G.P. then the sum of first n terms is
1) na
2) a/n
3) (n – 1) a
4) (n + 1) a
Answer:
1) na

Explanation:
If r = 1, the sum of first n terms = na

Question 11.
If 4, 8, 16, 32 are in G.P. then the sum upto 5th term is
1) 16
2) 64
3) 128
4) 124
Answer:
4) 124

Explanation:
4, 8, 16, 32, are in G.P; a = 4, r = 2
S₅ = \(\frac{4\left(2^5-1\right)}{2-1}=\frac{4(32-1)}{1}\) = 4(31) = 124

Question 12.
The Sum of series 1 + \(\frac{1}{2}+\frac{1}{4}\) + …….. upto 6 terms is
1) \(\frac{63}{32}\)
2) \(\frac{32}{63}\)
3) \(\frac{26}{53}\)
4) \(\frac{53}{26}\)
Answer:
1) \(\frac{63}{32}\)

Explanation:
Series 1 + \(\frac{1}{2}+\frac{1}{4}\) + …….. upto 6 terms
Sum of 6 terms = \(\frac{\left(1-\left(\frac{1}{2}\right)^6\right)}{1-\frac{1}{2}}\)
= 2(1 – \(\frac{1}{64}\))
= 2 – \(\frac{1}{32}=\frac{64-1}{32}=\frac{63}{32}\)

Question 13.
The geometric mean of 3 and 12 is ………….
1) 4
2) 6
3) 9
4) 12
Answer:
2) 6

Explanation:
Geometric mean of 3 and 12 = \(\sqrt{3 \times 12}=\sqrt{36}\)
= 6

Question 14.
If A.M of two numbers is \(\frac{15}{2}\) and their G.M is 6 then the two numbers are
1) 6 and 8
2) 12 and 3
3) 24 and 6
4) 27 and 3
Answer:
2) 12 and 3

Explanation:
Let the two numbers be and b
A.M = \(\frac{15}{2} \Rightarrow \frac{a+b}{2}=\frac{15}{2}\)
⇒ a + b = 15 ……….(1)

G. M = 6 ⇒ \(\sqrt{\mathrm{ab}}\)
⇒ ab = 36

a – b = \(\sqrt{(a+b)^2-4 a b}=\sqrt{225-144}=\sqrt{81}\)
a – b = 9 …(2)
(1) + (2) ⇒ a = 12, b = 3

Question 15.
If A means arithmetic mean and G means geometric mean then which of the following is true ?
1) A > G
2) A ≥ G
3) G > A
4) G ≥ A
Answer:
2) A ≥ G

Explanation:
A ≥ G

Question 16.
The ratio of A.M. and G.M of two positive numbers a and b is 5 : 3. Then the ratio of a to b is ( 1 )
1) 9 : 1
2) 3 : 5
3) 1 : 9
4) 3 : 1
Answer:
\(\frac{A \cdot M}{G \cdot M}=\frac{5}{3} \Rightarrow \frac{a+b}{2 \sqrt{a b}}=\frac{5}{3}\)
⇒ 9(a + b)² = 25 (4ab)
⇒ 9a² + 18ab + 9b² = 100 ab
⇒ 9a² – 82ab + 9b² = 0
⇒ (9a – b) (a – 9b) = 0
9a-b = 0
9a = b
\(\frac{a}{b}=\frac{1}{9}\)

a – 9b = 0
a = 9b
\(\frac{a}{b}=\frac{9}{1}\)
∴ a : b = 9 : 1 (∵ a > b)

Referring to the AP Inter 1st Year Maths Study Material Chapter 7 Binomial Theorem Exercise 7b Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Binomial Theorem Solutions Exercise 7b

I.

Question 1.
If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer. [Hint : Write aⁿ = (a – b + b)ⁿ and expand]
Solution:
Here, aⁿ = (a – b + b)ⁿ = [(a – b) + b)ⁿ
By using Binomial theorem, the expression [(a – b) + b]ⁿ can be expanded as
ⁿC₀(a- b)ⁿ b⁰ + ⁿC₁(a- b)^n-1 b¹ + ⁿC₂(a- b)^n-2 b² + …. + ⁿC_n-1(a – b)b^n-1 + ⁿC_n(a- b)⁰bⁿ
= (a – b)ⁿ + ⁿC₁(a – b)^n-1 b + ⁿC₂(a – b)^n-2 bⁿ + ………….. + ⁿC_n-1(a – b)b^n-1] + bⁿ
⇒ aⁿ = (a – b)[(a – b)^n-1 + ⁿC₁(a – b)^n-2 b + ⁿC₂(a – b)^n-3 b² + ⁿC_n-1b^n-1] + bⁿ
⇒ aⁿ – bⁿ = (a – b)[(a – b)^n-1 + ⁿC₁(a – b)^n-2 b + ⁿC₂(a – b)^n-3 b² + ……. + ⁿC_n-1b^n-1]
now, a – b is a factor of R.H.S, so, a – b is also a factor of L.H.S hence, a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer.

Question 2.
Evaluate (√3 + √2)⁶ – (√3 – √2)⁶
Solution:
(√3 + √2)⁶ – (√3 – √2)⁶
= ⁶C₀(√3)⁶ + ⁶C₁(√3)⁵(√2) + ⁶C₂(√3)⁴(√2)² + ⁶C₃(√3)³(√2)³ + ⁶C₄(√3)²(√2)⁴
+ ⁶C₅(√3)¹(√2)⁵ + ⁶C₆(√2)⁶ – [⁶C₀(√3)6 – ⁶C₁(√3 )⁵(√2) + ⁶C₂(√3)⁴(√2)² – ⁶C₃(√3)³(√2)³ + ⁶C₄(√3)²(√2)⁴ – ⁶C₅(√3)¹(√2 )⁵ + ⁶C₆(√2)⁶]
= 2[⁶C₁(√3)⁵(√2) + ⁶C₃(√3)³(√2)³ + ⁶C₂(√3)¹(√2)⁵]
= 2[6 × 9√6 + 20 × 6√6 + 6 × 4√5]
= 2[54 + 120 + 24]√6
= 396√6

Question 3.
Find the value of (a² + \(\sqrt{a^2-1}\))⁴ + (a² – \(\sqrt{a^2-1}\))⁴
Solution:

= 2(a⁸ + 6a⁴(a² – 1) + (a² – 1)²]
= 2[a⁸ + 6a⁶ – 6a⁴ + a⁴ + 1 – 2a²]
= 2[a⁸ + 6a⁶ – 5a⁴ – 2a² + 1]
= 2a⁸ + 12a⁶ – 10a⁴ – 4a² + 2

II.

Question 1.
Find an approximation of (0.99)⁵ using the first three terms of its expansion.
Solution:
0.99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then Binomial theorem can be applied. It can be written that, 0.99 = 1 – 0.01
Therefore, (0.99)⁵ = (1 – 0.01)⁵
= ⁵C₀(-0.01)° + ⁵C₁(-0.01)¹ + ⁵C₂ (-0.01)² + ………………..
= 1 – 5(0.01) + 10(0.0001) + …………
= 1 – 0.05 + 0.001
= 0.951

Question 2.
Expand (1 + \(\frac{x}{2}-\frac{2}{x}\))⁴ using Binomial theorem,
Solution:
Here (1 + \(\frac{x}{2}-\frac{2}{x}\))⁴ = [(1 + \(\frac{x}{2}\)) – \(\frac{2}{x}\)]⁴
Using Binomial theorem, the expression (1 + \(\frac{x}{2}-\frac{2}{x}\))⁴ can be expressed as

Again applying Binomial theorem, we have

Now putting the value of (1 + \(\frac{x}{2}\))⁴ and (1 + \(\frac{x}{3}\))³ in (1) we have

Question 3.
Find the expansion of (3x² – 2ax + 3a²)³ using binomial theorem.
Solution:
Here, (3x² – 2ax + 3a²)³ = ((3x² – 2ax) + 3a²)³
Using Binomial theorem, the expression [(3x² – 2ax) + 3a²)³ can be expressed as
³C0(3x² – 2ax)³ + ³Cj(3x² – 2ax)² (3a²) + ³C2(3x² – 2ax)¹ (3a²)² + ³C3(3a²)³ = (3x² – 2ax)³ + 3(9x⁴ + 4a²x² – 12ax³)(3a²) + 3(3x² – 2ax) (9a⁴) + (27a⁶)
= (3x² – 2ax)³ + 81a²x⁴ + 36a⁴x² – 108a³x³ + 81a⁴x² – 54a⁵x + 27a⁶

Again applying Binomial theorem, we have
(3x² – 2ax)³ = ³C₀(3x²)³ + ³C₁(3x²)² (-2ax)¹ + ³C₂(3x²) (-2ax)² + ³C₃(-2ax)³
= 27x⁶ – 3 × 18ax⁵ + 3 × 12a²x⁴ – 8a³x³
= 27x⁶ – 54ax⁵ + 36a²x⁴ – 8a³x³

Now putting the value of (3x² – 2ax)³, we have
= 27x⁶ – 54ax⁵ + 36a²x⁴ – 8a³x³ + 81a²x⁴ + 36a⁴x² – 108a³x³ + 81a⁴x² – 54a⁵x + 27a⁶
= 27x⁶ – 54ax⁵ + 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶

Question 4.
Prove that C₀ + 3.C₁ + 5.C₂ + … + (2n + 1). C_n = (2n + 2). 2^n-1
Solution:
Let S = C₀ + 3.C₁ + 5.C₂ + … + (2n + 1).C_n ……….. (1)
By writing the terms in (1) in the reverse order we get
S = (2n + 1). C_n + (2n – 1) . C_n-1 + (2n – 3).C_n-2 + …….. + 3.C₁ + 1.C₀
= (2n + 1).C₀ + (2n – 1).C₁ + (2n – 3).C₂ + ……… + 3.C_n-1 + 1.C_n …………. (2)
On adding (1) and (2), we get
2S = (2n + 2).C₀ + (2n + 2).C₁ + … + (2n + 2).C_n-1 + (2n + 2) . C_n
= (2n + 2) (C₀ + C₁ + C₂ + … + C_n) = (2n + 2).2n Therefore, S = (2n + 2) 2^n-1

Question 5.
Prove that, for any real numbers a, d,
a.C₀ + (a + d) . C₂ + (a + 2d) . C₂ + … + (a + nd) . C_n = (2a + nd) . 2^n-1
Solution:
Let S = a.C₀ + (a + d) . C₂ + (a + 2d) . C₂ + … + (a + nd) . C_n ….. (1)
∴ S = (a + nd) . C_n + [a + (n – 1) d] . C_n-1 + … aC₀
⇒ S = (a + nd) . C₀ + [a + (n – 1)d]. + [a + (n – 2)d].C₂ + … + a . C_n ……….. (2)
(1) + (2) => 2S = (2a + nd) C₀ + (2a + nd) C₁ + (2a + nd) C₂ + … + (2a + nd) C_n
= (2a + nd) (C₀ + C₁ + C₂ + … + C_n)
= (2a + nd) 2n ⇒ S = (2a + nd) 2_n – 1.

Question 6.
If n is a positive integer, then prove that
C₀ + \(\frac{C_1}{2}+\frac{C_2}{3}\) + …… + \(\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\)
Solution:
Let S = C₀ + \(\frac{C_1}{2}+\frac{C_2}{3}\) + …… + \(\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\)
Now multiplying on both sides with (n + 1), we get
(n + 1) S = (n + 1)ⁿC₀ + \(\frac{1}{2}\) (n + 1)ⁿC₁ + —(n + 1)ⁿC₂ + … + ⁿC_n
= (n +1) + \(\frac{(\mathrm{n}+1) \mathrm{n}}{1 \cdot 2}+\frac{(\mathrm{n}+1) \mathrm{n}(\mathrm{n}-1)}{1 \cdot 2 \cdot 3}\) + … + 1
= ⁽ⁿ⁺¹⁾C₁ + ⁽ⁿ⁺¹⁾C₂ + ⁽ⁿ⁺¹⁾C₃ + ……… + ⁽ⁿ⁺¹⁾C_n+1
= ⁽ⁿ⁺¹⁾C₁ + ⁽ⁿ⁺¹⁾C₁ + ⁽ⁿ⁺¹⁾C₂ + ⁽ⁿ⁺²⁾C₃ ……… + ⁽ⁿ⁺¹⁾C_n+1 – ⁽ⁿ⁺¹⁾C₀
= 2ⁿ⁺¹ -1
⇒ S = \(\frac{2^{n+1}-1}{n+1}\)
C₀ + \(\frac{C_1}{2}+\frac{C_2}{3}\) + …… + \(\frac{C_n}{n+1}=\frac{2^{n+1}-1}{n+1}\)

Question 7.
If n is a positive integer and x is any non-zero real number, then prove that
C₀ + C₁.\(\frac{x}{2}\) + C₂.\(\frac{x^2}{3}\) + C₃.\(\frac{x^3}{4}\) + ……….. + C_n.\(\frac{x^n}{n+1}=\frac{(1+x)^{n+1}-1}{(n+1) x}\)
Solution:
Let S ≡ C₀ + C₁.\(\frac{x}{2}\) + C₂.\(\frac{x^2}{3}\) + C₃.\(\frac{x^3}{4}\) + ……….. + C_n.\(\frac{x^n}{n+1}\)

Multiplying both sides with be ‘x’

Question 8.
Prove that 2.C₀ + 5.C₁ + 8.C₂ + ……….. + (3n + 2) C_n = (3n + 4)2^n-1
Solution:
Let S = 2.C₀ + 5.C₁ + 8.C₂ + ……….. + (3n – 1) C_n-1 + (3n + 2) C_n ……(1)
S = (3n + 2) C₀ + (3n – 1)C₁ + (3n – 4) C₂ + + 5C_n-1 + 2.C_n … (2)
(∵ C_n = C₀.C_n-1 = Ci…)
Adding (1) 86 (2) we get 2S = (3n + 4)C0 + (3n + 4)Cj + (3n + 4)C2 + … + (3n + 4) Cn = (3n + 4) (C0 + Cj + C2 + … + Cn) = (3n + 4)2n
S = (3n + 4) . 2^n-1

Question 9.
Prove tnat C₀ +2C₁ + 4.C₂ + 8.C₃ + ……….. + 2ⁿ.C_n = 3ⁿ
Solution:
We know that,
(1 + x)ⁿ = C₀ + C₁x + C₂x² + C₃x² + ………… + C_nxⁿ
Put x = 2 in equation (1) we get,
(1 + 2)ⁿ = C₀ + C₁(2) + C₂(2)² + C₃(2)³ + ………. + C_n(2)ⁿ
∴ C₀ +2C₁ + 4.C₂ + 8.C₃ + ……….. + 2ⁿ.C_n = 3ⁿ

Question 10.
If (1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + ………… + a_2nx²ⁿ, then prove that
i) a₀ + a₁ + a₂ + ………… + a_2n = 3ⁿ
Solution:
(1 + x + x²)ⁿ = a₀ + a₁x + a₂x² + ………… + a_2nx²ⁿ
Put x = 1 ⇒ a₀ + a₁ + a₂ + ………… + a_2n = (1 + 1 + 1)ⁿ = 3ⁿ …….(1)
Put x = -1 ⇒ a₀ – a₁ + a₂ + ………. + a_2n = (1 – 1 + 1)ⁿ = 1 ………. (2)

a₀ + a₁ + a₂ + ………… + a_2n = 3ⁿ

ii) a₀ + a₂ + a₄ + ………… + a_2n = \(\frac{3^n+1}{2}\)
Solution:
(1) + (2) ⇒ 2(a₀ + a₂ + a₄ + ………… + a_2n) = 3ⁿ + 1
⇒ ∴ a₀ + a₂ + a₄ + ………… + a_2n = \(\frac{3^n+1}{2}\)

iii) a₁ + a₃ + a₅ + ….. + a_2n-1 = \(\frac{3^n-1}{2}\)
Solution:
(1) – (2) ⇒ 2(a₁ + a₃ + a₅ + ….. + a_2n-1) = 3ⁿ – 1
⇒ ∴ a₁ + a₃ + a₅ + ….. + a_2n-1 = \(\frac{3^n-1}{2}\)

iv) a₀ + a₃ + a₆ + a₉ + ………… + a_2n = 3^n-1
Solution:
Put x = 1
⇒ a₀ + a₃ + a₆ + a₉ + ………… + a_2n = 3ⁿ …(a)
Put x = ω
a₀ + a₁ω + a₂ω² + a₃ω³ + ………… + a_2nω²ⁿ = 0
Hint: 1 + ω + ω² = 0; ω³ = 1 ……………. (b)
Put x = ω²
⇒ a₀ + a₁ω² + a₂ω⁴ + a₃ω⁶ + ………… + a_2nω⁴ⁿ = 0 …..(c)
Adding (a), (b), (c)
∴ 3₀ + a₁(1 + ω + ω²) + a₂(1 + ω³ + ω⁶) + ……… + a_2n(1 + ω²ⁿ + ω⁴ⁿ) = 3ⁿ
⇒ 3a₀ + a₁(0) + a₂(0) + 3a₃ + ……. = 3ⁿ
∴ a₀ + a₃ + a₆ + ……… = \(\frac{3^{\mathrm{n}}}{3}\) = 3^n-1

Question 11.
If P and Q Me the sum of odd terms and the sum of even terms respectively in the expansion of (x + a)ⁿ then prove that
i) P² – Q² = (x² – a²)ⁿ
ii) 4PQ = (x + a)²ⁿ – (x – a)²ⁿ
Solution:
We know that (x + a)ⁿ = ⁿC₀xⁿ + ⁿC₁x^n-1a + ⁿC₂ x^n-2a² + ………………
(x + a)ⁿ = T₁ + T₂ + T₃ + T₄ + T₅ + T₆ + …………….
Given that P = (T₁ + T₃ + T₅ + T₇ + …) and Q = (T₂ + T₄ + T₆ + ………….)
⇒ (x + a)ⁿ = P + Q …………(1) and (x – a)ⁿ = P – Q ………….(2)

i) P² – Q² = (P + Q) (P – Q) = (x + a)ⁿ (x – a)ⁿ = [(x + a) (x – a)]ⁿ = (x² – a²)ⁿ

ii) 4PQ = (P + Q)² – (P – Q)²
= [(x + a)ⁿ]² – [(x – a)ⁿ]²
= (x + a)²ⁿ – (x – a)²ⁿ

Question 12.
If (1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + … + a₂₀x²⁰, then prove that
i) a₀ + a₁ + a₂ + … + a₂₀ = 2¹⁰
ii) a₀ – a₁ + a₂ – a₃ + … + a₂₀ = 4¹⁰
Solution:
(1 + 3x – 2x²)¹⁰ = a₀ + a₁x + a₂x² + … + a₂₀x²⁰ ………….. (1)

i) Put x = 1 in eq. (1)
(1 + 3 – 2)¹⁰ = a₀ + a₁ + a₂ + … + a₂₀
∴ a₀ + a₁ + a₂ + … + a₂₀ = 2¹⁰

ii) Put x = -1 in eq (1)
(1 – 3 – 2)¹⁰ = a₀ – a₁ + a₂ – a₃ + … + a₂₀
∴ a₀ – a₁ + a₂ – a₃ + … + a₂₀ = (-4)¹⁰ = 4¹⁰

Question 13.
Show that the middle term in the expansion of (1 + x)²ⁿ is \(\frac{1 \cdot 3 \cdot 5 \ldots(2 n-1)}{n!}\) 2ⁿ . xⁿ, where n is a positive integer.
Solution:
As 2n is even, the middle term of the expansion (1 + x)²ⁿ is i.e., (n + 1)^th term
Now T_n+1 = ²ⁿC_n (1)^2n-n (x)ⁿ = ²ⁿC_n xⁿ = \(\frac{(2 \mathrm{n})!}{\mathrm{n}!\mathrm{n}!}\)xⁿ

Question 14.
The second, third and fourth terms in the binomial expansion (x + a)ⁿ are 240, 720, 1080 respectively. Find the value of x, a and n.
Solution:
Given expression is (a + x)ⁿ. General term T_r+1 = ⁿC_r a^n-r x^r
T₂ = T₁₊₁ = ⁿC₁ a^n-1 x¹ = 240 … (1)
T₃ = T₂₊₁ = ⁿC₂ a^n-2 x² = 720 … (2)
T₄ = T₃₊₁ = ⁿC₃ a^n-3 x³ = 1080 … (3)

⇒ 4n – 8 = 3n – 3
⇒ 4n – 3n = -3 + 8
∴ n = 5

Substitute n = 5 in (4) ⇒ 4x = 6a ⇒ x = 3a/2
Substitute x, n values in (1)
⇒ 5 a⁴\(\left(\frac{3 \mathrm{a}}{2}\right)\) = 240
⇒ a⁵ = \(\frac{240 \times 2}{15}\) = 32
⇒ a⁵ = 2⁵
⇒ a = 2

Substitute the values of n = 5 and a = 2 in (6)
⇒ x = \(\frac{3(2)}{2}\) = 3
∴ a = 2, x = 3, n = 5

Regular practice with AP Inter 1st Year Physics Study Material Chapter 2 Motion in a Straight Line Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Physics 2nd Lesson Motion in a Straight Line Questions and Answers

I. Multiple Choice Questions

Question 1.
The velocity of a body depends on time according to the equation V = 10 + 2t². The body undergoing
(1) uniform acceleration
(2) uniform retardation
(3) non-uniform acceleration
(4) zero acceleration
Answer:
(3) non-uniform acceleration

Question 2.
Which of the following is true?
(1) Speed is always equal to magnitude of velocity.
(2) Speed is always greater than the magnitude of velocity.
(3) Velocity is always greater than the speed.
(4) Speed in general is either equal to or greater than the magnitude of velocity.
Answer:
(1) Speed is always equal to magnitude of velocity.

Question 3.
A bullet is dropped from the same time another bullet is fired horizontally. They will hit the ground
(1) one after the other
(2) simultaneously
(3) depends on the observer
(4) depends on the velocity of the fired bullet
Answer:
(2) simultaneously

Question 4.
If a car is moving with a uniform velocity of 60 km/h. Then its acceleration is
(1) 0km / h²
(2) 10km / h²
(3) 20km / h²
(4) 30km / h²
Answer:
(1) 0km / h²

Question 5.
A ball is thrown vertically upward with an initial velocity of 50 m/s. Its maximum height is (take g = 10/s²)
(1) 125m
(2) 250 m
(c) 500 m
(d) 1000 m
Answer:
(1) 125m

Question 6.
A particle moves in a straight line with an initial velocity of 10m/s and a constant acceleration of 2m/s². Its displacement after 5 seconds will be
(1) 50 m
(2) 75 m
(3) 150 m
(4) 200 m
Answer:
A = ut + \(\frac{1}{2}\)at² (or) x = v₀t + \(\frac{1}{2}[latex]at², ∴ x = (10 × 5) + [latex]\frac{1}{2}\)(2) × 5² = 50 + 25 = 75 m

Question 7.
A body starts from rest and moves with a uniform acceleration of 2m / s². Its velocity after 10 seconds will be
(1) 5 m/s
(2) 10 m/s.
(3) 15 m/s
(4) 20 m/s
Answer:
(4) 20 m/s
V = u + at = 0 + (2 × 10) = 20m/s

Question 8.
For the motion with uniform velocity, the slope of the velocity-time graph is equal to
(1) 1 m/s
(2) zero
(3) initial velocity
(4) final velocity
Answer:
(2) zero

Question 9.
If a cyclist takes two minutes to complete half revolution on a circular path of 120 m radius, what is the average speed?
(1) 1 m/s
(2) 2 m/s
(3) 3.14 m/s
(4) 4.14 m/s
Answer:
(3) 3.14 m/s
Average Speed = \(\frac{\text { Total distance }}{\text { Total Time }}=\frac{\pi \mathrm{r}}{2 \times 60}=\frac{3.14 \times 120}{2 \times 60}\) = 3.14 m/s

Question 10.
In a particular time, the net displacement travelled by the body is equal to
(1) the area which v -1 graph encloses with displacement axis
(2) the area which x -1 graph encloses with the time axis
(3) the area which v -1 graph encloses with time axis
(4) the area which a -1 graph encloses with displacement axis
Answer:
(3) the area which v -1 graph encloses with time axis

Question 11.
A particle moves along a straight line. At a time t, the distance x of the particle from the origin given by x = 20 + 12t – t³ where x is in meters and t in sec, the time taken for the particle fo coming to rest is
(1) 1 s
(2) 2 s
(3) 4 s
(4) 6 s
Answer:
(2) 2 s
x = 20 + 12t – t³, \(\frac{d x}{d t}\) = 12 – 3t² = V, 12 – 3t² = 0 ⇒ 3t² = 12 ⇒ t² = 4, t = 2s

Question 12.
A car travels half of the distance with speed v and the remaining distance with speed 2v. Its average speed is
(1) \(\frac{4 v}{3}\)
(2) \(\frac{3 v}{3}\)
(3) \(\frac{v}{3}\)
(4) \(\frac{2 v}{3}\)
Answer:
(1) \(\frac{4 v}{3}\)
Average Speed = \(\frac{2 u v}{u+v}=\frac{2 \times(v)(2 v)}{v+2 v}=\frac{2 \times v \times 2}{3}=\frac{4}{3} v\)

Question 13.
A car runs at constant speed on a circular track of radius 100 m, taking 62.8 seconds for every circular lap. The average velocity and average speed of each circular lap respectively is
(1) 10 m/s, 0 m/s
(2) 0m/s. 0 m/s
(3) 0 m/s, 10 m/s
(4) 10 m/s, 10 m/s
Answer:
(3) 0 m/s, 10 m/s
Hint: Average Velocity = 0
Average Speed = \(\frac{2 \pi \mathrm{r}}{62.8}=\frac{2 \times 3.14 \times 100}{62.8}=\frac{6.28 \times 100}{62.8}\) = 10m/s

Question 14.
The motion of the particle is given by the equation S = (2t³ + 5t² + 16t + 8)m. The value of acceleration of the particle at t = 1 sec is
(1) 20 m/s²
(2) 22 m/s²
(3) 24 m/s²
(4) 26 m/s²
Answer:
(2) 22 m/s²
S = 2t³ + 5t² + 16t + 8, v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 6t² + 10t + 16; a = \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 12t + 10
at t = 1, a = 12 + 10 = 22m/s²

Question 15.
The ratio of distances travelled by a freely falling body in the 1^st, 2^nd, 3^rd, 4^th second
(1) 1:2:3:4
(2) 1:4:9:16
(3) 1:3:5:7
(4) 1:1:1:1
Answer:
(3) 1:3:5:7
S_n = u + \(\frac{\mathrm{a}}{2}\)(2n – 1), S₁ = \(\frac{\mathrm{g}}{2}\)(2(1) – 1) = \(\)
S₂ = \(\frac{\mathrm{g}}{2}\)(2(2)- 1) = \(\frac{\mathrm{3g}}{2}\), S₃ = \(\frac{\mathrm{g}}{2}\)(2(3) – 1) = \(\frac{\mathrm{5g}}{2}\) ⇒ S₁ : S₂: S₃ : S₄ = 1 : 3 : 5 : 7

Question 16.
A stone falls freely under gravity. It covers distances h₁, h₂ and h₃ in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h₁, h₂ and h₃ is
(1) h₂ = 3h₁ and h₃ = 3h₂
(2) h₁ = h₂ = h₃
(3) h₁ = 2h₂ = 3 h₃
(4) h₁ = \(\frac{\mathrm{h}_2}{3}=\frac{\mathrm{h}_3}{5}\)
Answer:
(4) h₁ = \(\frac{\mathrm{h}_2}{3}=\frac{\mathrm{h}_3}{5}\)
h₁ = \(\frac{1}{2}\)g(5²) = \(\frac{1}{2}\)g(25), x₁ = \(\frac{1}{2}\)g(10²) = \(\frac{1}{2}\)g(100)
h₂ = x₁ – h₁ = \(\frac{1}{2}\)g (100 – 25) = \(\frac{1}{2}\)g(75), x₂ = \(\frac{1}{2}\)g (15²) = \(\frac{1}{2}\)g(225)
h₃ = x₂ – x₁ = \(\frac{1}{2}\)g(225 – 100) = \(\frac{1}{2}\)g(125), h, = \(\frac{\mathrm{h}_2}{3}=\frac{\mathrm{h}_3}{5}\) (or) 1:3:5:7

II. Fill in the Blanks

Question 1.
________ deals with the study of motion of objects without taking into account, factors which cause the motion.
Answer:
Kinematics

Question 2.
The shortest path travelled by a moving object in a given time from initial to final position is called _______.
Answer:
displacement

Question 3.
The distance covered by a body along a semicircular path of radius r is _______ .
Answer:
πr

Question 4.
The slope of velocity-time graph gives _______.
Answer:
acceleration

Question 5.
The slope of displacement-time graph gives _______.
Answer:
velocity

Question 6.
The area under acceleration-time graph gives _____.
Answer:
change in velocity

Question 7.
The acceleration of a body moving with uniform velocity is ________.
Answer:
zero

Question 8.
The velocity of a body at any instant of time is called _______.
Answer:
instantaneous velocity

Question 9.
The value of an acceleration of a freely falling body is ________.
Answer:
g = 9.8 m/s²]

Question 10.
The ratio of distances travelled by a freely falling body from the rest in the first three seconds is ______.
Answer:
1 : 3 : 5

III. One Word Answer Questions

Question 1.
What is uniform motion ?
Answer:
If an object moving along a straight line covers equal displacement in equal intervals of time, said to be in uniform motion along a straight line.

Question 2.
What is acceleration ?
Answer:
The rate of change of velocity is called acceleration.

Question 3.
What is average speed ?
Answer:
The ratio of total distance travelled by a body to total time is called average speed.

Question 4.
What is instantaneous velocity ?
Answer:
The velocity of a particle at a particular instant is called instantaneous velocity.

Question 5.
Draw position – time graph for an object at rest.
Answer:

Question 6.
What represents the area under the cure of velocity – time graph ?
Answer:
Area = v × t = displacement.

Question 7.
What is the maximum height reached by a body when it is vertically projected up with an initial velocity ‘u’ ?
Answer:
H_max = \(\frac{u^2}{2 g}\)

Question 8.
Mention one similarity between speed and velocity.
Answer:
Having same units and dimensions.

Question 9.
Draw velocity-time graph for a particle moving with uniform velocity.
Answer:

Question 10.
What is the time taken by the body, when it is freely dropped from a height “h” ?
Answer:
h = \(\frac{1}{2}\)gt²; t² = \(\frac{2 h}{g} \Rightarrow t=\sqrt{\frac{2 h}{g}}\)

IV. Very Short Answer Questions

Question 1.
The states of motion and rest are relative. Explain. [dlarch – 2026]
Answer:
REST: If the position of a body does not change with respect to surroundings, it is said to be at “Rest”.

MOTION : If the position of a body changes with respect to surroundings, it is said to be in “Motion”. By definitions, rest and motion are relative with respect to surroundings.

Question 2.
Is it necessary that the direction of acceleration and velocity of a body should be the same? Give an example.
Answer:
No. For a decelerating body velocity is in forward direction and acceleration will be in opposite direction.

Question 3.
Is it possible for a body to have acceleration but zero velocity? Give an example.
Answer:
In case of vertically projected body at maximum height its velocity v = 0. But acceleration due to gravity ‘g’ is not zero.
So even though velocity v = 0 ⇒ acceleration is not zero.

Question 4.
Which is greater, a velocity of 72kmph or 25m/s ?
Answer:
V₁ = 72 × \(\frac{5}{18}\) = 20m/s; V₂ = 25m/s ∴ 25m/s is greater than 72kmph.

Question 5.
Give an example of one-dimensional motion where a particle moving along a straight line comes to rest and moves backward.
Answer:

1. A body projected vertically upwards.
2. When the length of pendulum is high and amplitude is less then its motion is along a straight line. The pendulum will come to a stop at extreme position and moves back in forward direction (‘x’ + ve) periodically.

Question 6.
A vehicle travels half the distance L with speed Vj and the other half with speed v₂. What is the average speed ?
Answer:
The average speed of a vehicle for the two equal parts \(\frac{\mathrm{L}}{2}, \frac{\mathrm{~L}}{2}\) is
V_avg = \(\frac{\text { Total distance }}{\text { Total time }}=\frac{\frac{\mathrm{L}}{2}+\frac{\mathrm{L}}{2}}{\mathrm{t}_1+\mathrm{t}_2}\)
V_avg = \(\frac{\mathrm{L}}{\left(\frac{\frac{\mathrm{~L}}{2}}{\mathrm{~V}_1}+\frac{\frac{\mathrm{L}}{2}}{\mathrm{~V}_2}\right)}=\frac{\mathrm{L}}{\frac{\mathrm{~L}}{2}\left(\frac{1}{\mathrm{~V}_1}+\frac{1}{\mathrm{~V}_2}\right)}=\frac{1}{\frac{1}{2}\left(\frac{\mathrm{~V}_1+\mathrm{V}_2}{\mathrm{~V}_1 \mathrm{~V}_2}\right)}\) (∵ t = \(\frac{\mathrm{L} / 2}{\mathrm{~V}}\)) ⇒ V_avg = \(\frac{2 V_1 V_2}{V_1+V_2}\)

Question 7.
A body covers a distance of 5m along a semicircular path from its one end to another end. What is the ratio of its distance covered to its displacement ?
Answer:
πr = 5; r = 5/π

\(\frac{\text { Distance }}{\text { Displacement }}=\frac{\pi \mathrm{r}}{2 \mathrm{r}}=\frac{\pi}{2}=\frac{22}{7 \times 2}=\frac{11}{7}\)

Question 8.
Displacement travelled by a body is given by S = 2t + 5t² m. What is the acceleration of that body ?
Answer:
S = 2t + 5t²; V = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 2 + 10t; a= \(\frac{\mathrm{dv}}{\mathrm{dt}}\) = 10m/s²

Question 9.
An object falling through a fluid is observed to have an acceleration given by a = g – bv where g is the gravitational acceleration and b is a constant. After a long time it is observed to fall with a constant velocity. What would be the value of this constant velocity ?
Answer:
Acceleration, a- = g – bv when moving with constant velocity, a = 0 ⇒ 0 = g – bv
∴ Constant velocity, v = \(\frac{\mathrm{g}}{\mathrm{~b}}\) m/sec.

Question 10.
Under what circumstances is the relationship s = vt valid ?
Answer:
Body moving with uniform velocity (a = 0).

V. Short Answer Questions

Question 1.
Can the equations of kinematics be used when the acceleration varies with time ? If not, what form would these equations take ?
Answer:
The equations of motion are 1) v =u + at 2) s = ut + 1/2at² and 3) v² – u² = 2as. All these three equations applicable to a body moves with uniform acceleration ‘a’.

No, the equations are not applicable when the acceleration varies with time.

Question 2.
A particle moves in a straight line with uniform acceleration. Its velocity at time t = 0 is V₁ and at time t = t is v₂. The average velocity of the particle in this time interval is (v₁ + v₂) / 2. Is this correct ? Substantiate your answer.
Answer:
t₁ = 0 ⇒ u = v₁ ; t₂ = t ⇒ v = v₂; Average velocity = \(\frac{u+v}{2}=\frac{v_1+v_2}{2}\)

Question 3.
Can the velocity of an object be in a direction other than the direction of acceleration of the object ? If so, give an example.
Answer:
Yes. Velocity of a body and its acceleration may be in different directions.
Ex :
1) In case of vertically projected body ⇒ velocity of body is in the upward direction and acceleration is in downward direction.
2) When brakes are applied the velocity of body before coming to rest is opposite to retarding acceleration.

Question 4.
Obtain the equation of motion x = ut + \(\frac{1}{2}\)at² for constant acceleration using graphical method.
Answer:
The velocity – time graph of a body moving with initial velocity ‘u’ and with uniform acceleration ‘a’ as shown. Let V be the velocity of the body after a time t.
In v – t graph, area of velocity-time graph = total displacement travelled by it. Area under velocity – time graph = area of OABCD
∴ Area of Rectangular part OACD = Area of OACD + Area of ABC.
A₁ = OA × OD = ut …..(1)
Area of triangle ABC = A₂
A₂ = \(\frac{1}{2}\) × Base × height = \(\frac{1}{2}\) × AC × (BC) = \(\frac{1}{2}\)t(u – v)
But v – u = at; ∴ A₂ = \(\frac{1}{2}\)t. at = \(\frac{1}{2}\)at²
∴ Total area under graph = s = A₁+ A₂; s(n) = ut + \(\frac{1}{2}\)at².
∴ s = ut + \(\frac{1}{2}\)at² is graphically proved.

Question 5.
Explain the terms the average velocity and instantaneous velocity. When are they equal?
Answer:
Average velocity : It is the ratio of total displacement to total time taken. It is independent of path of the body.
∴ Average velocity = \(\frac{\mathrm{s}_2-\mathrm{s}_1}{\mathrm{t}_2-\mathrm{t}_1}\) (Or) V_avg = \(\frac{\mathrm{ds}}{\mathrm{dt}}\)
Velocity of a particle at a particular instant of time is known as instantaneous velocity. Here time interval is very small. V = \(\frac{\mathrm{dx}}{\mathrm{dt}}\)
Only in uniform motion, instantaneous velocity = average velocity. For all other cases instantaneous velocity may differ from average velocity.

Problems

Question 1.
A man walks on a straight road from his home to a market 2.5 km away with a speed of km h^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^-1. What is the (a) magnitude of average velocity mid (b) average speed of the man over the time interval 0 to 50 minutes ?
Solution:
Time taken by man to go from his home to market, t₁ = \(\frac{\text { distance }}{\text { speed }}=\frac{2.5}{5}=\frac{1}{2} \mathrm{~h}\)
Time taken by man to go from market to his home, t₁ = \(\frac{2.5}{7.5}=\frac{1}{3} h\)
∴ Total time taken = t₁ + t₂ = \(\frac{1}{2}+\frac{1}{3}=\frac{5}{6} h\) = 50min.
In time interval 0 to 50 min,
Total distance travelled = 2.5 + 2.5 = 5 km.; Total displacement = zero.

1. Average velocity = \(\frac{\text { displacement }}{\text { time }}\) = 0
2. Average speed = \(\frac{\text { distance }}{\text { time }}=\frac{5}{5 / 6}\) = 6 km/h

Question 2.
A car travels the first third of a distance with a speed of 10 kmph, the second third at 20 kmph and the last third at 60 kmph. What is its mean speed over the entire distance ?
Solution:
Total distance = s; distance travelled, S₁ = \(\frac{s}{3}\); velocity, v₁ = 10 kmph ; distance, s₂ = \(\frac{s}{3}\) velocity, v₂ = 20 kmph; distance, s₃ = \(\frac{s}{3}\); velocity, v₃ = 60 kmph
Average velocity = \(\frac{\text { total distance }}{\text { total time }}\)

Question 3.
A bullet moving with a speed of 150 m s^-1 strikes a tree and penetrates 3.5 cm before stopping. What is the magnitude of its retardation in the tree and the time taken for it to stop after striking the tree?
Solution:
Velocity of bullet, u = 150 m/s; Final velocity, v = 0 ; Distance travelled, s = 3.5 cm = 3.5 × 10^-2 m,
a) Acceleration, a = \(\frac{v^2-u^2}{2 s}=\frac{0^2-150^2}{2 \times 3.50 \times 10^{-2}}=\frac{-22500}{7 \times 10^{-2}}\)
= -3.214 × 10⁵ m/sec² (-ve sign for retardation)

b) Time taken to stop, t = \(\frac{v-u}{a}=\frac{-150}{-3.214 \times 10^5}\) = 4.67 × 10^-4 sec.

Question 4.
A motorist drives north for 30 min at 85 km/h and then stops for 15 min. He continues travelling north and covers 130 km in 2 hours. What is his total displacement and average velocity ?
Solution:
In first part:
Velocity, v₁ = 85 kmph ; Time, t₁ = 30 min ; Distance travelled, s₁ = v₁ t₁ = 85 × \(\frac{30}{60}\) = 42.5 km

In second part:
Distance travelled, s₂ = 0 ; Time, t₂ = 15.0 min.

In third part:
Distance travelled, s₃ = 130 km ; Time, t₃ = 2 hours = 120 min.
a) Total distance of the motorist, s = s₁ + s₂ + s₃ = 42.5 + 0 + 130 = 172.5 km

b) Total time travelled,
t = t₁ + t₂ + t₃ = 30 + 15 + 120 = 165 minutes = 2 hrs 45 minutes = 2\(\frac{3}{4}\) hrs. = \(\frac{11}{4}\) hrs.
∴ Average velocity, v_avg = \(\frac{\text { total displacement }}{\text { total time }}=\frac{172.5}{(11 / 4)}\) = 62.7 kmph.

Question 5.
A ball A is dropped from the top of a building and at the same time an identical ball B is thrown vertically upward from the ground. When the balls collide the speed of A is twice that of B. At what fraction of the height of the building did the collision occur ?
Solution:
Given at time of collision velocity of A = V_A = 2 × V_B (velocity of B)
Let the body be dropped from a height ‘h’. Let the two stones collide at x from the ground.
For the body dropped, s = h – x = \(\frac{1}{2}\)gt² → (1); For the body thrown up, x = ut – \(\frac{1}{2}\)gt² → (2)
For the body dropped, v = u + at ⇒ V_A = gt → (3)
For the body thrown up, v = u – gt ⇒ V_B = u – gt → (4)
Given V_A = 2V_B ⇒ gt = 2 (u – gt) or u = \(\frac{3gt}{2}\) → (5)
Divide equation (1) with equation (2)

∴ Fraction of height of collision = \(\frac{2}{3}\)

Question 6.
Drops of water fall at regular intervals from the roof of a building of height 16 m. The first drop strikes the ground at the same moment as the fifth drop leaves the roof. Find the distances between successive drops.
Solution:
Height of building, h = 16 m
∴ Time taken to reach the ground, t = \(\sqrt{\frac{2 h}{g}}\)
∴ Time taken to fall, t = \(\sqrt{\frac{2 \times 16}{9.8}}=\sqrt{\frac{32}{9.8}}=\sqrt{3.26}\) = 1.80

Number of drops, n = 5
∴ Number of intervals = n – 1 = 5 – 1 = 4
Time interval between drops = \(\frac{1.8}{4}\) = 0.45 sec
Time of travel of 1st drop, t₁ = 4 × 0.45 = 1.8
∴ Distance travelled by
1st drop, S₁ = \(\frac{1}{2}\)gt₁² = \(\frac{1}{2}\) × 9.8 × 1.8 × 1.8 = 16 m

For 2nd drop, t₂ = 3 × 0.45 = 1.35 sec., S₂ = \(\frac{1}{2}\)gt₂² ;
∴ S₂ = \(\frac{1}{2}\) × 9.8 × 1.35² = 4.9 × 1.822 = 8.9278 ≅ 9m

For 3rd drop, t₃ = 2 × 0.45 = 0.9 sec.
Distance, S₃ = \(\frac{1}{2}\)gt₃² = \(\frac{1}{2}\) × 9.8 × 0.9² = 3.97 ≅ 4m

For 4th drop, t₄ = 1 × 0.45 = 0.45 sec.
Distance travelled, S₄ = \(\frac{1}{2}\)gt₄² = \(\frac{1}{2}\) × 9.8 × (0.45)² ≅ 1

For 5 th drop, t₅ = 0 ⇒ S₅ = 0
Distance between 1^st and 2^nd drop S_{1, 2} = S₁ – S₂ = 16 – 9 = 7m
Distance between 2^nd and 3^rd drop S_{2, 3} = S₂ – S₃ = 9 – 4 = 5 m
Distance between 3^rd and 4^th drop S_{3, 4} = S₃ – S₄ = 4 – 1 = 3m
Distance between 4^th and 5^th drop S_{4, 5} = S₄ – S₅ = 1 – 0 = 1 m
∴ Distances between successive drops are 7m, 5m, 3m and 1m.

Question 7.
A food packet is dropped from an aeroplane, moving with a speed of 360 kmph in a horizontal direction, from a height of 500m. Find (i) its time of descent (ii) the horizontal distance between the point at which the food packet reaches the ground and the point above which it was dropped.
Solution:
Velocity of plane, V = 360 kmph = 360 × \(\frac{5}{18}\) = 100 m/s.
Height above the ground, h = 500 m; g = 10 m/s²

i) Time of descent, t = \(\) = 10sec.

ii) Horizontal distance between point of dropping and point where it reaches the ground = Range R.
∴ R = u \(\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\) = 100 \(\sqrt{\frac{2 \times 500}{10}}\) = 100 x 10 = 1000 m.

Motion in a Straight Line Class 11 NCERT Solutions

Question 1.
In which of the following examples of motion, can the body be considered approximately point object.
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on the top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
Answer:
(a) A railway carriage can be considered as a point object because the size of carriage is much smaller than the distance between the two stations.

(b) A monkey sitting on the top of the cyclist can be considered as a point object if the cyclist describes a circular path of radius which is sufficiently larger than the size of monkey.

(c) A spinning cricket ball cannot be considered as a point object because the size of ball is not negligible as compared to the distance through which the ball may turn on hitting the ground.
Remark: The spinning motion of cricket ball is due to its size. A point mass cannot have spin motion.

(d) A beaker slipping off the edge of a table cannot be considered as a point object because the size of beaker is not negligible as compared to the height of the table.

Question 2.
The position – time (x – t) graphs for two children A and B returning from their school O to their homes P and Q respectively as the shown in Fig.
Choose the correct entries in the brackets below :

(a) (A/B) lives closer to the school than (B/A).
(b) (A/B) starts from the school earlier than (B/A).
(c) (A/B) walks faster than (B/A).
(d) A and B reach home at the (same/different) time.
(e) (A/B) overtakes (B/A) on the road (once/twice).
Answer:
(a) As OP < OQ ; A lives closer to school than B.
(b) For A, x = 0 at t = 0 and for B, x = 0 at t = t₀ (let)
This implies that A starts from the school earlier than B.
(c) The speed = slope of x – t graph. As slope of B is greater than the slope of A, therefore, B walks faster than Answer:
(d) From graph, it is clear that the positions P and Q are attained at the same time, this implies that both A and B reach home at the same time.
(e) x – t graphs for A and B intersect each other only once, so B overtakes A on the road once.

Question 3.
A woman starts from her home at 9.00 am, walks with a speed of 5 kmh^-1 on a straight road upto her office 2.5 km away, stays at the office upto 5.00 pm and returns home by an auto with a speed of 25 kmh^-1. Choose suitable scales and plot the x-t graph of her motion.
Answer:
For motion from home to office :
Time to which woman leaves the office = 09.00 am
Speed = 5 km/h, distance traversed, s = 2.5 km
Time taken, t₁ = \(\frac{\text { Distance traversed }}{\text { Speed }}=\frac{2.5}{5}\) = 0.5 h = 30 min.
Time at which woman reaches office
= 9.00 am + 0.30 = 9.30 am
Stay at office : 9.30 am – 5.00 pm (position fixed)

For motion from office to home :
Time at which she leaves office = 5.00 pm
Speed of auto = 25 km/h, distance = 2.5 km
Time spent, t₂ = \(\frac{\text { Distance }}{\text { Speed }}=\frac{2.5}{25}\) = 0.1 h = 0.1 × 60 min = 6 min
Time at which she reaches home = 5.06 pm
The position-time (x – t) graph of motion of woman is as shown in the Fig.

Question 4.
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x – t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Answer:
The drunkard takes 5 steps forward and 3 steps backward. Each step is of 1 m length and takes 1 second. Therefore, the displacement of drunkard as a function of time is as shown below.

The position – time graph of motion of drunkard is as shown in the Fig.
From graph, it is clear that the drunkard will fall into the pit 37 seconds after the beginning.

Question 5.
A car moving along a straight highway with a speed of 126 kmir1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform); and how long does it take for the car to stop ?
Answer:
Initial speed of car, u = 126 km/h
1 km/h = \(\frac{1000 \mathrm{~m}}{3600 \mathrm{~s}}=\frac{5}{18}\) m/s ; u = 126 × \(\frac{5}{18}\) = 35 m/s
Final speed of car, v = 0, Distance traversed, s = 200 m
Using the formula v² = u² + 2as, we get
0 = (35)² + 2a × 200 ⇒ -400a = (35)² ⇒ a = \(\frac{(35)^2}{400}\) = -3.06 ms^-2
Retardation of Car = 3.06 ms^-2
If t is the time taken, then formula v = u + at gives
0 = 35 – 3.06 t ⇒ 3.06t = 35 ⇒ t = \(\frac{35}{3.06}\) = 11.4 s

Question 6.
A player throws a ball upwards with an initial speed of 29.4 ms1.
(a) What is the direction of acceleration during the upward motion of the ball ?
(b) What are the velocity and acceleration of the ball at the highest point of its motion ?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis and give the signs of position, velocity and acceleration of the ball during its upward and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands ?[Take g = 9.8 ms^-2 and neglect air resistance].
Answer:
(a) The ball moves under gravity throughout its motion. The acceleration due to gravity is always directed vertically downward. So, the direction of acceleration during the upward motion of the ball is vertically downward.
(b) At the highest point of motion, the velocity of ball is zero and the acceleration is 9.8 ms^-2 directed vertically downward.
(c) As the ball remains between highest point and earth, so during upward and downward motion of ball, x > 0 (or x is positive).
During vertically upward direction v is negative (v < 0) and during vertically downward direction v is positive (v > 0).
Acceleration is always g = + 9.8 ms^-2 (positive).
(d) Using formula v² = u² + 2as, we have
a = + g = 9.8 ms^-2, s = – H, v = 0 (at highest point)
u = – 29.4 ms^-1 (given)
0 = (-29.4)² – 2 × 9.8 × H ⇒ H = \(\frac{(29.4)^2}{2 \times 9.8}\) m = 44.1 m
Also from relation, s = ut + \(\frac{1}{2}\)at²,
we have s = 0, u = -29.4 m/s, a = 98m/s²
0 = -29.4 t + \(\frac{1}{2}\) × 9.8t² or \(\frac{1}{2}\) 9.8t² = 29.4 t ⇒ t = \(\frac{29.4}{4.9}\) = 6s.
∴ Height, (H) = 44.1m and time taken (t) = 6s.

Question 7.
Read each statement below carefully and state with reasons and examples, if it is true or false.
A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant.
(b) with zero speed may have non-zero velocity,
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.
Answer:
(a) True; at highest point of particle thrown in vertically upward direction, the speed is zero but acceleration is g = 9.8 ms^-2 vertically downward.
(b) False; because velocity = speed + direction. As speed is zero, velocity must be zero.
(c) True; as v = constant; acceleration a = \(\frac{d v}{d t}\) = 0
(d) False, when a body is thrown upward its speed decreases, but gravitational acceleration is downward (assumed positive direction).
However, if body continues to move along a straight line on a smooth horizontal surface, then positive acceleration will necessarily imply speeding up.

Question 8.
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one-tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Answer:
Let t₁ be the time taken by the ball to hit the floor.
Given, u = 0, h = 90 m, g = 9.8 m/s²
Formula, h = ut + \(\frac{1}{2}\)gt² gives
90 = 0 + \(\frac{1}{2}\)9.8t₁² ⇒ t₁² = \(\frac{90}{4.9}=\frac{900}{49}\) ⇒ t₁ = \(\frac{30}{7}\) = 4.3 s.
The speed of the ball when it hits the floor first time is given by
v₁ = u + gt₁ = 0 + 9.8 × \(\frac{30}{7}\) = 42 m/s
Now the ball bounces upward after collision. As the ball loses one-tenth of speed during collision, therefore, upward speed of ball just after collision
v₂ = \(\frac{9}{10}\) × 42 = 37.8 m/s
The ball now starts moving upward with retardation g = 9.8 m/s². The speed of ball at any instant t will be (u¹ = v₂ = 37.8 m/s).
∴ v = u¹ – gt = 37.8 – 9.8 t
The speed of ball becomes zero at highest point after time t₂¹ is given by
0 = 37.8 – 9.8 t₂¹ ⇒ t₂¹ = \(\frac{37.8}{9.8}\) = 3.85 s
Now the ball begins to fall downward and attains a speed of 37.8 m/s just before collision with the floor. As time of ascent = time of descent.

∴ Time taken by ball to hit the floor
t₂¹¹ = 3.85 s
Total time upto this instant
T = t₂ + t₂¹ + t₂¹¹ = 4.3 + 3.85 + 3.85 = 12 s
The speed of ball just after second collision
v₃ = \(\frac{9}{10}\) × 37.8 ≅ 34m/s
The graph of speed and time is as shown in the Fig.
Remark: Speed is never negative.

Question 9.
Explain clearly, with examples, the distinction between
a) magnitude of displacement (sometimes called distance) over an interval of time and the total length of the path covered by a particle over the same interval.
b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total paths length divided by the time interval].
Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only].
Answer:
For simplicity we suppose one-dimensional motion, where an object traverses from A to B and then returns back in total time t where AB = s.
(a) Displacement = 0 and total path length = AB + AB = 2s.
(b) Average velocity = \(\frac{\text { Displacement }}{\text { Time }}=\frac{0}{t}\) = 0
Average speed = \(\frac{\text { Total path length }}{\text { Time }}=\frac{2 \mathrm{~s}}{\mathrm{t}}\)

In both the cases (a) and (b), the second quantity (i.e., total path length and speed) turns out to be greater than the first (i.e., magnitude of displacement and velocity respectively). However, if the motion takes place along a straight line without change in direction, then equality sign will be true.
That is then, magnitude of displacement = Total path length
(i.e., motion along a straight line in one direction) and magnitude of velocity = Speed.

Question 10.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 kmh^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 kmh ‘. What is the
(a) Magnitude of average velocity, and
(b) The average speed of the man over the interval of time
(i) 0 to 30 min,
(ii) 0 to 50 min,
(iii) 0 to 40 min ?
Answer:
Distance of market from home = 2.5 km; Speed of man from home to market = 5 km/h
Time taken by man to reach market, t =Speed of man, while returning from the market = 7.5 km/h
∴ Time taken by man to return the home, t₁ = \(\frac{\text { Distance }}{\text { Speed }}=\frac{2.5}{5}=\frac{1}{2} \mathrm{~h}\) = 20 min
Speed of man, while returning from the market = 7.5 km/h
Time taken by man to return the home, t₂ = \(\frac{2.5}{7.5}=\frac{1}{3} h\) = 20 min
Total time of man returning to his home = 30 min + 20 min = 50 min

(i) Over the interval 0 to 30 min :
During this time, man goes from home to market; therefore, displacement, s = 2.5 km
Average velocity = \(\frac{\text { Displacement }}{\text { Time }}=\frac{2.5}{1 / 2}\) = 5km/h
Average speed = \(\frac{\text { Distance }}{\text { Time }}=\frac{2.5}{1 / 2}\) = 5 km/h
Clearly, during motion along a straight line (without turn), the average speed and average velocity are equal in magnitude.

(ii) Over the interval 0 to 50 min :
In 50 min the man goes from home to market and returns back to home.
Net displacement = zero
Average velocity = \(\frac{\text { Net displacement }}{\text { Total time }}=\frac{0}{50}\)
Average speed = \(\frac{\text { Total distance }}{\text { Total time }}=\frac{2.5 \mathrm{~km}+2.5 \mathrm{~km}}{(50 \mathrm{~min})}+\frac{5 \mathrm{~km}}{50 \times \frac{1}{60} \mathrm{~h}}\) = 6 km/h.

(iii) Over the interval 0 to 40 min :
During first 30 min man goes from home to market covering a distance of 2.5 km; in next 10 min the man is in path from market to home and covers a distance
= (75 km/h) × \(\left(\frac{10}{60} h\right)\) = 1.25 km
Displacement = 2.5 – 1.25 = 1.25 km
Distance = 2.5 + 1.25 = 3.75 km
Average velocity = \(\frac{\text { Displacement }}{\text { Time }}=\frac{1.25}{40 \times \frac{1}{60} \mathrm{~h}}\) = 1.875 km/h
Average speed = \(\frac{\text { Distance }}{\text { Time }}=\frac{3.75 \mathrm{~km}}{40 \times \frac{1}{60} \mathrm{~h}}\) = 5.625 km/h

Question 11.
In problems 9 and 10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why ?
Answer:
Instantaneous velocity, v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\)
Where ds is infinitesimal displacement in infinitesimal time interval.
For infinitesimal interval of time there is no change in direction so the magnitude of displacement is equal to the length of the path; hence magnitude of instantaneous velocity = instantaneous speed.

Question 12.
Look at the graphs (a) to (d) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.

Answer:
(a) It does not represent one-dimensional motion because according to this graph particle has two positions at the same time, which is impossible.

(b) It does not represent one dimensional motion because according to this graph the particle have equal positive and negative velocity at the same time, which is impossible.

(c) It cannot represent one-dimensional motion because according to graph speed is negative at certain instants, which is impossible (speed can never be negative).

(d) It cannot represent one-dimensional motion because according to this graph after a certain time the total path length begins to decrease with increase of time, which is impossible (total path length will always increase with increase of time).

Question 13.
Fig. shows the x – t graph of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0 ? If not suggest a suitable physical context for this graph.
Answer:
For t < 0, the particle is at rest; since, its position remains at x = 0. For t > 0, x – t graph is parabolic, it is possible if motion of particle is uniformly accelerated since x = ut + \(\frac{1}{2}\)at²
If u = 0, x = \(\frac{1}{2}\)at² or x ∝ t² (parabola)

Thus, it is not correct to say that the particle moves in a straight line for t < 0, because during t < 0, the particle does not move at all. However, for t > 0, the trajectory is parabolic which is possible if particle moves along a straight line with uniform acceleration. It is again not correct to say that particle moves on parabolic path for t > 0.

Physical context: The graph may represent the motion of a particle released from the top of a hill or tower.

Question 14.
Suggest a suitable physical situation for each of the following graphs.

Answer:
(a) The given x – t graph shows that initially x = 0, then ball gains constant velocity which then reverses in direction and finally its velocity becomes zero.
Physical situation : The ball is initially at rest on a smooth floor. It is kicked and given a constant velocity. Then, it strikes a wall, rebounds with reduced speed and strikes the opposite wall which stops it.

(b) The given velocity – time graph (v – t) shows that the velocity changes sign again and again with the passage of time and each time, it loses some speed.
Physical situation : A ball thrown up with some initial velocity rebounding from the floor with reduced speed after each collision.

(c) The given a – t graph shows that initially acceleration of object is zero. It gains acceleration for a very short duration and immediately it is decelerated and again, its acceleration is zero.
Physical situation: A cricket ball moving with a uniform speed is hit back with a bat for a very short duration.

Question 15.
Fig. gives the x – t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s and – 1.2 s.
Answer:
Position is positive when particle is above t – axis and negative if particle is below t – axis. Velocity is positive, when x increases with increase of time and velocity is negative when x decreases with increase of time. Acceleration is positive when the slope of x – t curve increases and negative when slope of x – t curve decreases.

At 0.3 s, x is negative, v is negative and a is positive.
At 1.2 s, x is positive, v is positive and a is negative.
At – 1.2 s, x is negative, v is positive and a is positive.

Question 16.
Fig. shows x – t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval.

Answer:
The average speed inx-t graph is equal to the slope of line joining the initial and final points in that interval. From the graph the slope is greatest in interval 3 and least in interval 2. Therefore, average speed is greatest in interval 3 and average speed is least in interval 2. Slope is positive in intervals 1 and 2 and negative in interval 3, so the average velocity is positive during intervals 1 and 2 and negative during interval 3.

Question 17.
Fig. shows a speed – time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which intervals is the average acceleration greatest in magnitude ? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ?

Answer:
The average acceleration is given by the slope of line joining the two points onv-f graph for given interval of time. From the graph, out of three intervals the slope is maximum in interval (2) but it is negative. Therefore, the magnitude of average acceleration is greatest in interval 2.
Average speed in a given interval = \(\frac{v_1+v_2}{2}\)
From the graph, it is clear that the magnitudes of v₁ and v₂ are maximum in interval 3, so average speed is maximum in interval 3.

Signs of v: The speed v is always positive. Therefore, it is positive in all the three intervals.
Signs of a: In interval T slope at v -1 graph is positive, therefore, acceleration is positive. In interval 2, the slope of v – t graph is negative, therefore, the acceleration is negative. In interval 3, the slope of v – t graph is positive, therefore, the acceleration is positive.

The instantaneous acceleration is equal to the slope of tangent on v – t graph at given instant. The tangents drawn on given v – t curve at points A, B, C and D are parallel to time axis, so slope is zero.
Hence, acceleration at points A, B, C and D are zero.

Question 18.
A police van moving on a highway with a speed of 30 km/h fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km/h, if the muzzle speed of the bullet is 150 m/s, with what speed does the bullet hit the thief’s car ?
(Note : Obtain that speed which is relevant for damaging thief’s car ?)
Answer:
Let v_p be the velocity of police van, v_T be the velocity of thief’s car and u_BP the muzzle velocity of bullet. Muzzle velocity means the velocity with respect to gun (or police van).
Given, v_P = 30 km/h = 30 × \(\frac{1100 \mathrm{~m}}{(60 \times 60) \mathrm{s}}=\frac{25}{3} \mathrm{~m} / \mathrm{s}\)
v_T = 192 km/h = 192 x \(\frac{1100 \mathrm{~m}}{(60 \times 60) \mathrm{s}}=\frac{160}{3} \mathrm{~m} / \mathrm{s}\)
u_B = 150 m/s
All velocities are in same direction, therefore, if u_B is the velocity of bullet with respect to earth, the velocity of bullet with respect to police van
u_BP = u_B + v_P = 150 + \(\frac{25}{3}=\frac{475}{3}\)m/s
Relative velocity of bullet with respect to thief’s car
u_BT = u_B – v_T = \(\frac{475}{3}-\frac{160}{3}=\frac{315}{3}\)= 105 m/s.

Regular practice with AP Inter 1st Year Chemistry Study Material Chapter 6 Equilibrium Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Chemistry 6th Lesson Equilibrium Questions and Answers

I. Multiple Choice Questions

Question 1.
The pH of a solution of hydrochloric acid is 4. The molarity of the solution is :
1) 4.0 M
2) 0.4 M
3) 0.0001 M
4) 0.04 M
Answer:
3) 0.0001 M
If [H⁺] = 0.0001M (Molarity), then
pH = -log₁₀ [H⁺] = -log 0.0001 = -log 10^-4 = 4.

Question 2.
The strong conjugate base in the following is
1) NO₃^–
2) Cl^–
3) SO₄^2-
4) CH₃COO^–
Answer:
4) CH₃COO^–
Strong conjugate base is CH₃COO^–, which is obtaining from weak acid.

Question 3.
When NH₄Cl is added to NH₄OH solution the dissociation of ammonium hydroxide is reduced. It is due to
1) common ion effect
2) hydrolysis
3) oxidation
4) reduction
Answer:
1) common ion effect
Due to common ion effect dissociation of NH₄OH is reduced.

Question 4.
The following reaction is at equilibrium.
Fe³⁺_(aq) + SCN^–_(aq) ⇌ [Fe(SCN)]²⁺_(aq)
In the above reaction, colour intensity of red colour can be increased by
l) addition of oxalic acid which reacts with Fe³⁺ ions.
2) addition of Hg²⁺ ions which reacts with SCN^– ions.
3) addition of KSCN.
4) red colour intensity cannot be changed.
Answer:
3) addition of KSCN.
\(\mathrm{Fe}_{(\mathrm{aq})}^{+3}+\mathrm{SCN}_{(\mathrm{aq})}^{-} \rightleftharpoons[\mathrm{Fe}(\mathrm{SCN})]_{(\mathrm{aq})}^{+2}\)
on adding KSCN colour intensity of red colour increased.

Question 5.
K_sp of a sparingly soluble salt AB₂ is 4 × 10^-12. The solubility of the salt is
1) 2 × 10^-6M
2) 4 × 10^-4M
3) 1 × 10^-12
4) 1 × 10^-4M
Solution:
4) 1 × 10^-4M
For AB₂ type of salt, K_sp = 4s³ ⇒ s² = \(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\)
⇒ s = \(\sqrt[3]{\frac{\mathrm{K}_{\mathrm{sp}}}{4}}=\sqrt[3]{\frac{4 \times 10^{-12}}{4}}\) = 10^-4m

Question 6.
An aqueous solution of ammonia consists of
1) H⁺ only
2) OH^– only
3) NH₄⁺ only
4) NH₄⁺ and OH^–
Answer:
Aqueous solution of ammonia contains NH₄⁺ and OH^– ions.
NH₄OH_(aq) → NH₄⁺ + OH^–

Question 7.
The pH of a 0.05 M of H₂SO₄ solution is
1) 0.05
2) 1
3) -1
4) 1.3010
Answer:
2) 1
For 0.05M of H₂SO₄
[H⁺] = 2 × 0.05(Di basic H₂SO₄) = 0.1
pH = -log₁₀[H⁺] = -log 0.1 = -log 10^-1 = 1

Question 8.
If the molar solubility of Zirconium phosphate Zr₃(PO₄)₄ is ‘S’. Solubility product K_sp of Zirconium phosphate is
1) 6912 (S)⁵
2) 6912 (S)⁷
3) 108 (S)⁷
4) 27 (S)⁵
Answer:
2) 6912 (S)⁷

Question 9.
For a reversible reaction, if the concentration of reactants is increased, the equilibrium constant of the reaction
1) Increases
2) Remains constant
3) Decreases
4) Depends on amount of reactant
Answer:
Equilibrium constant remains constant on increasing the concentration of reactants.

Question 10.
The pK_a of acetic acid and pK_b of ammonium hydroxide are 4.76 and 4.75 respectively.
1) 6.098
2) 7.005
3) 8
4) 5.0
Answer:
2) 7.005
CH₃COONH₄ is the salt formed by weak acid and weak base
∴ pH = 7 + — (PK_a – PK_b) = 7 + \(\frac{1}{2}\) (4.76 – 4.78) = 7 × \(\frac{1}{2}\) × 0.01 = 7.005

Question 11.
Which among the following salt solutions is basic in nature?
1) Ammonium chloride
2) Ammonium sulphate
3) Ammonium nitrate
4) Sodium acetate
Answer:
4) Sodium acetate
Sodium acetate is the salt obtained from sodium hydroxide (Strong base) and acetic acid (weak acid). So aq. CH₃COONa is in basic nature.

Question 12.
Observe the following equilibrium at T (Kelvin)
H_2(g) + I_2(g) ⇌ HI_(g)
Which one of the following does not disturb the above equilibrium at constant volume?
1) Addition of H_2(g)
2) Removal of HI_(g)
3) Addition of I_2(g)
4) Addition of He_(g)
Answer:
4) Addition of He_(g)
He is a Noble gas, it is chemically inert, so by adding inertgas the equilibrium does not disturbed.

Question 13.
Solution of 0.1 N NH₄OH and 0.1 N NH₄Cl has pH 9.25. Find out pK_b of NH₄OH.
1) 9.25
2) 4.75
3) 3.75
4) 8.25
Solution:
2) 4.75
P^OH = PK_b + log \(\frac{[\text { salt }]}{[\text { base }]}\) (basic buffer)
pH + P^OH = 14 ⇒ P^OH = 14 – 9.25 = 4.75
4.75 = PK_b + log \(\frac{0.1}{0.1}\) ⇒ PK_b = 4.75

Question 14.
For the following process, the expression for the K_c
Ni_(s) + 4CO_(g) ⇌ Ni(CO)_4(g)
1) K_c = \(\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]^2}{[\mathrm{Ni}][\mathrm{CO}]^4}\)
2) K_c = \(\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]}{[\mathrm{CO}]^4}\)
3) K_c = \(\frac{[\mathrm{Ni}][\mathrm{CO}]^4}{\left[\mathrm{Ni}(\mathrm{CO})_4\right]^{2-}}\)
4) K_c = \(\frac{[\mathrm{CO}]^4}{\left[\mathrm{Ni}(\mathrm{CO})_4\right]^2}\)
Answer:
2) K_c = \(\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]}{[\mathrm{CO}]^4}\)
K_c = \(\frac{\text { Molar concentrations of products }(\mathrm{g})}{\text { Molar concentrations of reactants }(\mathrm{g})}\) ⇒ K_c = \(\frac{\left[\mathrm{Ni}(\mathrm{CO})_4\right]}{[\mathrm{CO}]^4}\)

Question 15.
K₁ and K₂ are equilibrium constants for reactions (i) & (ii). Then the relation between
K₁ and K₂
(i) H_2(g) + I_2(g) ⇌ 2HI(g)
(ii) HI_(g) ⇌ 1/2 H_2(g) + 1/2 I_2(g)
1) K₁ = K₂²
2) K₁ = \(\frac{1}{\mathrm{~K}_2}\)
3) K₁ = K₂⁰
4) K₁ = \(\frac{1}{K_2^2}\)
Solution:
4) K₁ = \(\frac{1}{K_2^2}\)
H_2(g) + I_2(g) ⇌ 2HI(g) …………… (1)
HI_(g) ⇌ \(\frac{1}{2}\) H_2(g) + \(\frac{1}{2}\) I_2(g) ………… (2)
K₁ = Equilibrium constant
K₂ = Equilibrium constant
K₂ = \(\frac{1}{\sqrt{\mathrm{~K}_1}}\) [eq.(2) is reverse reaction and halved]
Squaring on both sides, \(\mathrm{K}_2^2=\frac{1}{\mathrm{~K}_1}\)
⇒ K₁ = \(\frac{1}{K_2^2}\)

II. Fill in the Blanks

Question 1.
Acidity of BF₃ can be explained on the basis of __________ theory of acids and bases.
Answer:
Lewis acid base theory

Question 2.
The nature of aqueous solution of ammonium chloride (NH₄Cl) is ______.
Answer:
NH₄Cl is the salt of weak base (NH₄OH) and strong acid (HCl).
So aq. solution of NH₄Cl is acidic.

Question 3.
K_p for the following equation: ______.
CaCO_3(s) ⇌ CaO_(s) + CO_2(g)
Answer:
K_p = P_CO2

Question 4.
The molarity of pure water is ______.
Answer:
Molarity of pure water is = 55.5.M
M = \(\frac{\mathrm{W}}{\mathrm{GMW}} \times \frac{1000}{\mathrm{~V}(\mathrm{ml})}=\frac{1000}{18} \times \frac{1000}{1000}\) = 55.5 M

Question 5.
If K is in the range of ________ appreciable concentrations of both reactants and products are present.
Answer:
10^-3 to 10³

III. One Word Answer Questions

Question 1.
What is a bronsted base?
Answer:
The substance which can accept a proton is said to be bronsted base. Eg : Cl^–, HCO₃^–.

Question 2.
What is Lewis acid?
Answer:
Lewis acid is the substance which can accept a pair of electrons and form a coordinate covalent bond. Eg : BF₃, H⁺.

Question 3.
What is the value of K_w at 25°C? What are its units ?
Answer:
Ionic product of water K_w = 1 × 10^-14 ; Units are mole²/lit².

Question 4.
What is conjugate acid-base pair?
Answer:
An acid-base pair which differ by proton (H⁺) is called a conjugate acid base pair.
Eg : HCl and Cl^–.

Question 5.
Write the relation between Gibbs energy and equilibrium constant.
Answer:
∆G° : – 2.303 RT log K_c
R = gas constant ; T= temperature ; K_c = Equilibrium constant

IV. Very Short Answer Questions

Question 1.
State law of chemical equilibrium.
Answer:
The ratio of product of molar concentrations of products to the product of molar concentrations of reactants is a constant value. This constant is called equilibrium constant and this is called law of chemical equilibrium.
Eg : aA + bB ⇌ cc + dD (or) Equilibrium constant (K_c) = \(\frac{[\mathrm{C}]^{\mathrm{c}}[\mathrm{D}]^{\mathrm{d}}}{[\mathrm{~A}]^{\mathrm{a}}[\mathrm{~B}]^{\mathrm{b}}}\)
Eg : H_2(g) + I_2(g) ⇌ (K_c) = \(\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}\)

Question 2.
What is homogeneous equilibrium? Give two examples.
Answer:
The equilibrium in which all the substances are present in the same phase is known as homogeneous equilibrium.
Eg: N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
CH₃COOH (l) + C₂H₅OH (l) ⇌ CH₃COO C₂H₅ (l) + H₂O (l)

Question 3.
What is heterogeneous equilibrium? Give two examples.
Answer:
The equilibrium in which the substances involved are present in different phases is called heterogeneous equilibrium.
CaCO₃ (s) ⇌ CO(s) + CO₂ (g) and H₂O (l) ⇌ H₂O (g)

Question 4.
What is meant by dynamic equilibrium?
Answer:
A state in a reversible reaction where the forward and backward reaction continue to takes place with same rate is known as dynamic equilibrium.

Question 5.
Give two chemical equilibrium reactions for which K_p > K_c
Answer:
In order that K_p > K_c, ∆n should be +ve
1) PCl₅ (g) ⇌ PCl₃ (g) + Cl₂(g); ∆n = (2 – 1)
2) N₂O₄(g) ⇌ 2NO₂ (g) ; ∆n = (2 – 1)

Question 6.
Give two chemical equilibrium reactions for which K_p <K_c.
Answer:
In order that K_p <K_c, ∆n should be -ve
1) N₂ (g) + 3H₂ (g) ⇌ 2NH₃(g) ; ∆n = (2 – 4)
2) 2SO₂ (g) + O₂ (g) ⇌ 2SO₃(g) ; ∆n = (2 – 3)

Question 7.
Write the equations for the conversion of K_c to K_p for the following reaction
CO_(g) + H₂O_(g) ⇌ CO_2(g) + H_2(g)
Answer:
CO_(g) + H₂O_(g) ⇌ CO_2(g) + H_2(g)
K_p = K_c (RT)^∆n
∆n = (1+1) – (1 + 1) = O
K_p = K_p (RT)
K_p = K_c

Question 8.
What are the factors which influence the chemical equilibrium?
Answer:
Molar concentration, temperature and pressure influence the equilibrium.

Question 9.
What is meant by ionic product of water?
Answer:
The product of concentration of H⁺ and OH^– ions in aqueous solution (or) in pure water is called ionic product of water.

Question 10.
Give two examples of salts whose aqueous solutions are acidic.
Answer:
NH₄Cl (Ammonium Chloride); Cu₂SO₄ (copper sulphate) and (NH₄)₂SO₄ (Ammonium sulphate) .
These aqueous solutions are acidic in nature.

V. Short Answer Questions

Question 1.
Derive the relation between K_p and K_c for the equilibrium reaction.
N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
Answer:
N_2(g) + 3H_2(g) ⇌ 2NH_3(g)
K_c = \(=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}\) ; K_p = \(\frac{P_{\mathrm{NH}_3}^2}{P_{\mathrm{N}_2} \times \mathrm{P}_{\mathrm{H}_2}^3}\)
Partial pressures of P_N2, P_H2 and P_NH3 can be written as
P_N2 = \(\frac{\mathrm{n}_{\mathrm{N}_2} \mathrm{RT}}{\mathrm{~V}}\) = C_N2RT ; P_H2 = \(\frac{\mathrm{n}_{\mathrm{H}_2} \mathrm{RT}}{\mathrm{~V}}\) = C_H2RT and P_NH3 = \(\frac{\mathrm{n}_{\mathrm{NH}_3} \mathrm{RT}}{\mathrm{~V}}\) = C_NH3
Substituting these values in K_p equation, K_p = \(\frac{\left(\mathrm{C}_{\mathrm{NH}_3}\right)^2}{\left(\mathrm{C}_{\mathrm{N}_2}\right)\left(\mathrm{C}_{\mathrm{H}_2}\right)^3}\) × [RT]^-2
∴ K_p = K_c[RT]^-2 [∵ \(\frac{\mathrm{C}_{\mathrm{NH}_3}^2}{\left(\mathrm{C}_{\mathrm{N}_2}\right)\left(\mathrm{C}_{\mathrm{H}_2}\right)^3}\) = K_c]

Question 2.
State Le Chateliers’s Principle. Discuss the application of Le Chatelier’s principle for the industrial synthesis of Ammonia by Haber’s process.
Answer:
Le Chatelier’s principle : If a chemical reaction at equilibrium is subjected to a change in temperature, pressure or concentration that govern the chemical equilibrium, the equilibrium position shifts in the direction in which the change is reduced or nullified.
Application of Le Chatelier’s principle to by Haber’s process :
N_2(g) + 3H_2(g) ⇌ 2NH_3(g); ∆H = -92 kJ

Effect of pressure :
Increase of external pressure on the reaction at equilibrium favours the reaction in the direction in which the volume or the number of molecules decreases.
In the above reaction of formation of ammonia forward reaction involves decrease in volume or number of molecules. So high pressure is favourable for the formation of ammonia. Hence ammonia is manufactured at 200 atm pressure.

Effect of temperature :
In a reversible reaction, if the forward reaction is exothermic, the backward reaction will be endothermic.
High temperature favours the endothermic reaction while low temperature favours the exothermic reaction.

• Since the formation of ammonia is exothermic reaction, low temperature is favourable for the manufacture of ammonia.
• But at low temperature, the reaction is slow. So an optimum temperature of about 725 – 775 K is used.

Effect of catalyst:

• To increase the rate of formation of ammonia iron powder mixed with molybdenum powder is used as catalyst.
• Molybdenum acts as promoter to iron catalyst.

Effect of concentration :

• Increase in concentrations of reactants increases the rate of forward reaction.
• Removal of the products from reaction mixture decreases the rate of backward reaction.
• So more ammonia can be obtained by increasing the concentration of N₂ and H₂ and removing the NH₃ formed at regular intervals.
• Optimum conditions for the synthesis of ammonia are
  Pressure = 200 atm
  Temperature = 725 – 775 K
  Catalyst = Fe
  Promoter = Mo

Question 3.
State Le Chateliers’s Principle. Discuss the application of Le Chatelier’s principle for the manufacture of SO₃ by contact process.
Answer:
Le Chatelier’s principle : If a chemical reaction at equilibrium is subjected to a change in temperature, pressure or concentration that govern the chemical equilibrium, the equilibrium position shifts in the direction in which the change is reduced or nullified.

Application of Le Chatelier’s principle by contact process :
2SO_2(g) + O_2(g) ⇌ 2SO_3(g) + Heat; ∆H = -189 kJ

Effect of pressure :

• Increase of external pressure on the reaction at equilibrium favours the reaction in the direction in which the volume or the number of molecules decreases.
• Formation of SO₃ in the above reaction involves decrease in volume.
• According to Le Chatelier’s principle high pressure is favourable for the formation of SO₃.
• So, SO₃ is prepared at a pressure of 1.5 to 1.7 atmospheres.

Effect of temperature :

• In a reversible reaction if the forward reaction is exothermic, the backward reaction is endothermic.
• High temperatures are favourable for the endothermic reaction, while the low temperatures are favourable for the exothermic reactions.
• Since the formation of SO₃ is exothermic reaction, low temperatures are favourable. So, SO₃ is prepared at a temperature of about 673 K.
  Effect of catalyst: To increase the rate of formation of SO₃ catalyst like V₂O₅ or platinised asbestos is used.

Effect of concentration :

• Increase in the concentration of the reactants, increase the rate of forward reaction.
• Removal of products from reaction mixture decreases the rate of backward reaction.
• So more SO₃ can be obtained by increasing the concentrations of SO₂ and O₂ and by removing, the SO₃ is formed at regular intervals.
  Optimum conditions for the synthesis of ammonia are
  Pressure : 1.5 – 1.7 atmospheres Temperature : 673 K
  Catalyst: V₂O₅ (Vanadium pentoxide) or Platinised asbestos.

Question 4.
The species H₂O, HCO₃^–, HSO₄^– and NH₃ can act as both bronsted acids and bases. Give the corresponding conjugate acid and base for each of them.
Answer:

Question 5.
Write notes on
(i) Common ion effect.
(ii) Solubility product
Answer:
i) Common ion Effect : The decrease in the dissociation of the aq. solution of weak electrolyte by adding another strong electrolyte having one common ion is called common ion effect. Eg : To the saturated solution of AgCl, electrolyte NaCl is added to decrease the dissociation.
ii) Solubility Product (K_sp) : The solubility product is the product of molar equilibrium concentrations of cation and anion at a particular temperature in a saturated solution of sparingly soluble salt.
Applying the concept at equilibrium, A_xB_y(s) ⇌ xA^y+ + yB^x-
Applying the concept at equilibrium, K_sp = [A^y+]^x [B^x-]^y
AgCl ⇌ Ag⁺ + Cl^–
K_sp = [Ag⁺] [Cl^–]
⇒ K_sp = s × s [where s is the solubility]
⇒ K_sp = s²
Solubility s = \(\sqrt{\mathrm{K}_{\mathrm{sp}}}\)

Question 6.
Explain the concept of Bronsted-Lowry acids and bases. Illustrate with suitable examples.
Answer:
Bronsted – Lowry theory : Any substance that can lose a proton or protons is an acid.
Eg : HCl, H₂SO₄, H₃PO₄, CH₃COOH.

• Any substance that can gain a proton or protons is a base. Eg : NH₃, H₂O, OH^– etc.
• When an acid react with base neutralization takes place. As per Bronsted – Lowry theory proton transfer from an acid to a base is called neutralization.
  Eg : HCl + H₂O ⇌ H₃O⁺ + Cl^–
• The above reaction is reversible. In this reaction,
  1. HCl donates proton to water. So it is a bronsted acid.
  2. H₂O gains the proton. So it is a bronsted base.
  3. H₂O⁺ donates a proton to Cl^–. So it is bronsted acid.
  4. Cl^– can gain a proton. So it is a bronsted base.
• The acid base pair which differ by a single proton is said to be conjugate acid-base pair.
• In the above reaction HCl & Cl^– and H₃O⁺ and H₂O are conjugate acid-base pairs.
• Every bronsted acid has its conjugate base and every bronsted base has a conjugate protonic acid.
• According to bronsted – Lowry theory greater the tendency to donate proton stronger is the acid. Higher the tendency of the base to accept protons, stronger is the base.
• In the above reaction, HCl gives proton easily than H₃O⁺. So HCl is stronger acid than H20. H20 accepts the proton more easily than Cl^–. So H₂O is a stronger base than NH₃
• In all acid – base reactions, the reaction takes place in the direction of formation of weaker acid and a weaker base.
• A stronger acid has a weaker conjugate base and a stronger base has a weaker conjugate acid.

Advantages of Bronsted – Lowry theory :

• This theory explains the behaviour of acids and bases in both aqueous and non-aqueous solvents.
• It explains the behaviour of NH₃, CaO, etc. as bases and CO₂, SO₂, etc. as acids.

Drawbacks of Bronsted – Lowry theory :

• Proton donation or acceptance happens only in the presence of other substances.
• It could not explain the behaviour of electron deficient molecules like AlCl₃, BCl₃, etc. as acids.

Question 7.
Explain Lewis acids and bases with suitable examples. Classify the following species into Lewis acids and Lewis bases.
a) OH^–
b) F^–
c) H⁺
d) BCl₃.
Answer:
Lewis acid : The substance which can accept an electron pair to form a coordinate covalent bond is called Lewis acid.
Types of Lewis acids :
1) All +vely charged metal ions or cations act as Lewis acids.
Eg : Ag⁺, CO⁺³, Cu⁺², Fe⁺³, etc.
2) Electron deficient compounds whose central atom has an incomplete octet, possessing an empty orbital acts as Lewis acid. Eg : BF₃, BCl₃, AlCl₃, FeCl₃.
3) Compounds in which central atom has vacant’d’ orbitals and which can expand its octet. Eg : SiF₄, SnCl₄, SF₄, TeF₄, FeCl₃.
4) Molecules having multiple bonds between atoms of different electronegativities.
Eg : CO₂, SO₂, SO₃, NO₂, Cl₂O₇, P₄O₁₀.
5) Elements with an electron sextet. Eg : S, O.

Lewis base : The substance which donates electron pair to form a coordinate covalent bond is called Lewis base.
Types of Lewis bases :
1) All anions will act as Lewis bases. Eg : Cl^–, OH^–, CN^–, NH₂^–, F^–, SCN^–.
2) Molecules with one or two lone pairs on the central atom.
Eg : H₂O, NH₃, R – OH, R – NH₂,
R-O-R, C₅H₅N.
3) Molecules with multiple bonds. Eg : CO, NO, HC ☰ CH, H₂C ☰ CH₂
a) OH^– is Lewis base, b) F^– is Lewis base, c) H⁺ is Lewis acid, d) BCl₃ is Lewis acid.

Question 8.
Calculate the pH of
a. 10^-3 M of HCl
b. 10^-3 M of H₂SO₄ [log 2 = 0.3010]
Solution:
a) 10^-3M of HCl [H⁺] = 10^-3M
pH = – log [H⁺] = – Log 10^-3 = 3 log 10 = 3

b) 10^-3 M of H₂SO₄
[H⁺] = 2 × 10^-3
pH = -log [H⁺] = -log 2 × 10^-3 = – log 2 – log 10^-3
= – log 2 + 3 log 10 = -log 2 + 3
= 3 – log 2
= 3 – 0.3010 = 2.6990

Question 9.
Calculate the pH for
a. 0.001 M of NaOH
b. 0.01 M of Ca(OH)₂
Solution:
a) 0.001 M of NaOH
[OH⁺] = 0.001M
P^OH = -log [OH^–] = – log 10^-3 = 3 log 10 = 3
pH + POH =14
pH = 14 – 3 = 11.
∴ pH = 11

b) 0.01 M of Ca(OH)₂
[OH^–] = 0.01 × 2 = 2 × 10^-2
P^OH = – log [OH^–]
= – log 2 × 10^-2
= – log 2 – log 10^-2 = – log2 + 2 log 10
= 2 – log2 = 2 – 0.3010 = 1.6990
pH + P^OH = 14
pH = 14 – 1.699 = 12.3010

Question 10.
The pH of 0.1M solution of weak mono protic acid is 4.0. Calculate its [H⁺] and K_a.
Solution:
HA ⇌ H⁺ + A^–
pH = 4 ; [H⁺] = 10^-4
K_a = \(\frac{c \alpha^2}{1-\alpha}=\frac{(0.1)\left(10^{-4}\right)^2}{1}\) [Neglecting α in the denominator]
K_a = 10^-7

Question 11.
One litre of buffer solution contains 0.1 mole of acetic acid and 1 mole of sodium acetate. Find its pH if pK_a of CH₃COOH is 4.8.
Solution:
Henderson’s Equation, pH = PK_a + log \(\frac{[\text { Salt }]}{[\text { Acid }]}\)
Given [Salt] = 1M; [Acid] = 0.1M; PK_a = 4.8
= 4.8 + log \(\frac{1}{0.1}\)
= 4.8 + log 10
= 4.8 +1 = 5.8

VI. Long Answer Questions

Question 1.
What are equilibrium processes? Explain equilibrium in Physical and Chemical processes with examples.
Answer:

• In a reversible process, a state will be attained such that the rate of forward reaction is equal to the rate of backward reaction.
• Such a state is known as equilibrium state.
• Equilibrium processes are of two types. They are

1. Physical equilibrium process : In such a process, equilibrium occurs in a phase transformation process.
Ex: Solid ⇌ Liquid
Solid ⇌ Vapour
Vapour ⇌ Liquid

2. Chemical equilibrium process: If there exists an equilibrium in between reactants and products, such reactions are said to be chemical equilibrium process.
1) N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)
2) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

Question 2.
What is meant by dynamic equilibrium? Explain with suitable examples.
Answer:

• In a reversible process, a state will be attained such that the rate of forward reaction is equal to the rate of backward reaction. Such a state is known as equilibrium state.
• At equilibrium we find no change in the concentration of either reactants (or) products then we may feel that the reaction has stopped. But actually the reaction will be going on. Such a state of reaction is called dynamic equilibrium.

Equilibrium processes are of two types. They are :
1. Physical equilibrium process : In such a process, equilibrium occurs in a phase transformation process.
Ex : Solid ⇌ liquid
Solid ⇌ vapour
Vapour ⇌ liquid

2. Chemical equilibrium process: If there exists an equilibrium in between reactants and products, such reactions are said to be chemical equilibrium process.
1) N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)
2) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

The forward and backward reactions of a reversible reaction continuously takes place with equal rates simultaneously at the equilibrium stage also. Hence the equilibrium is called dynamic equilibrium.

At this stage, the reaction appears to be stopped.

All chemical equilibrium processes are dynamic equilibriums in nature.

Question 3.
What are the important features of equilibrium constant? Discuss any two applications of equilibrium constant.
Answer:

• The equilibrium constant has a definite value for every chemical reaction at a particular temperature. The value of equilibrium constant is independent of initial concentrations of reacting species.
• The value of equilibrium constant K changes with change in temperature.
• For a reversible reaction, the equilibrium constant for the backward reaction is inverse of the equilibrium constant for the forward reaction.
• The equilibrium constant is independent of the presence of catalyst.
• The equilibrium constant is expressed in terms of concentration, it has different units for different reactions.
• When addition of two equilibria leads to another equilibrium then the product of their equilibrium constants gives the equilibrium constant of the resultant equilibrium.
  Eg : N₂ + O₂ ⇌ 2NO ; K₁
  2NO + O₂ + O₂ ⇌ 2NO₂; K₂
  N₂ + 2O₂ ⇌ 2NO₂; K₃
  K = K₁ × K₂ × K₃

Applications of equilibrium constant:
i) Predicting of extent of reaction: The magnitude of equilibrium constant tells us about the extent to which the reactants are converted into the products before the equilibrium is attained.

ii) Predicting the direction of the reaction : From the knowledge of equilibrium con¬stant it is possible to predict the direction in which the net reaction is taking place at a given concentrations or partial pressures of reactants and products.

• If Q_c > K_c the reaction taking place in backward direction is more.
• If Q_c < K_c the reaction taking place in forward direction is more.
• If Q_c = K_c the reaction is at equilibrium. No reaction taking place.

Question 4.
What is Le Chatelier’s principle? Discuss briefly the factors which can influence the equilibrium.
Answer:
Le Chatelier’s principle : If a chemical reaction at equilibrium is subjected to a change in temperature, pressure or concentration that govern the chemical equilibrium, the equilibrium position shifts in the direction in which the change is reduced or nullified.
1) Effect of pressure : ,

• Pressure is inversely proportional to volume.
• When pressure is increased the equilibrium shifts in such a direction where the decrease in volume is observed.
• Similarly when pressure is decreased, the equilibrium shifts in such a direction where increase in volume is observed. This is applicable exclusively for gaseous reactions.

2) Effect of temperature :

• When temperature is increased, equilibrium shift in such a direction where the temperature change is nullified or equilibrium shifts towards the direction in which temperature is decreased or heat is absorbed. That is endothermic reaction is favoured.
• When temperature is decreased equilibrium shifts in such a direction where temperature is increased or heat is liberated. That is exothermic reaction is favoured.

3) Effect of concentration :

• Increase in concentration of the reactants favours forward reaction and decrease in concentration of the reactants favours the backward reaction.
• When products are added to the equilibrium reaction the concentration of the products increases and reaction proceeds in backward direction.
• Removal of products leads to decrease in concentration of the products and leads to forward reaction.

Question 5.
Describe the effect of:
a) addition of H₂
b) addition of CH₃OH
c) removal of CO
d) removal of CH₃OH on the equilibrium of the reaction 2H₂ (g) + CO (g) ⇌ CH₃OH (g)
Answer:
a) According to Le Chatelier’s principle increase in the concentration of reactant increase the forward reaction.
So the addition of H₂ results in the formation of more amount of CH₃OH.

b) According to Le Chatelier’s principle increase in the concentration of the product, shift the equilibrium in the backward direction. So the addition of CH₃OH (product) results in the decrease in concentration and CH₃OH and increase in the concentration of H₂ and CO.

c) According to Le Chatelier’s principle the decrease in the concentration of reactant equilibrium shift in the backward direction. So removal of CO (reactant) the equilibrium shift in the backward direction. So formation of CH₃OH decreases.

d) According to Le Chatelier’s principle the decrease in concentration of product shifts the equilibrium in the forward direction. So the removal of CH₃OH (product) shift the equilibrium in the forward direction. Thus more amount of CH₃OH is formed.

Question 6.
What is degree of ionization in respect of weak acids and weak bases? Derive the relationship between degree of ionization (α) and ionization constant (K_a) for the weak acid HX.
Answer:
Degree of ionization : It can be defined as the fraction of acid (a) base that undergoes ionization per one mole of acid or base. For strong acids or bases degree of ionization (a) is equal to ‘1’.
Let’s consider a weak acid HX with concentration ‘c’. It undergoes partial ionization.

where α is degree of ionization
Ionization constant K_a = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{X}^{-}\right]}{[\mathrm{HX}]}\)
⇒ K_a = \(\frac{(\mathrm{c} \alpha)(\mathrm{c} \alpha)}{\mathrm{c}(1-\alpha)}\) for weak acids α < < 1, so α is negligible.
⇒ K_a = \(\frac{c^2 \alpha^2}{c}\) ⇒ α = \(\sqrt{\frac{K_a}{c}}\)

Question 7.
Define pH. What is buffer solution? Derive Henderson – Hasselbalch equation for calculating the pH of an acid buffer solution.
Answer:
The pH of a solution may be defined as negative logarithm of the activity of hydrogen ion [H⁺]
pH = – log [H⁺]
Buffer solution : A buffer solution is the solution which can resist the change in pH on
addition of small amount of acid or base. The ability of a solution to resist change in pH on
addition of small amount of acid or base is called buffer solution.
Henderson – Hasselbalch equation :
HA + H₂ ⇌ H₃O⁺ + A^– Partial ionisation.
NaA → Na⁺ + A^– Complete ionisation
K_a = \(\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)
Since HA is weakly dissociated, the conc. HA ion solution can be taken equal to the initial conc. of the acid.
Since NaA is completely dissociated, conc. of A- can be taken equal to the conc. of NaA (salt)
K_a = \(\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right][\text {Salt }]}{[\text { Acid }]}\)
H₃O⁺ = \(\frac{\mathrm{K}_{\mathrm{a}} \text { [Acid] }}{[\text { Salt }]}\)
-log [H₃O⁺] = – logK_a – log\(\frac{\text { [Acid] }}{[\text { Salt }]}\)
pH = pKa + log \(\frac{\text { [Salt] }}{[\text { Acid }]}\)