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)

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

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.

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 \(\frac{\text { [Salt] }}{[\text { Acid }]}\)

Question 8.
Write notes on
(i) Common ion effect.
(ii) The relation between K_sp and solubility (S) of a sparingly soluble salt BaSO₄.
Answer:
i) Common ion effect: The solubility of an electrolyte in water decreases by the addition of another electrolyte which has one ion common with the electrolyte.

• For example the solubility of NaCl decreases on addition of HCl to the solution of NaCl in water which is known as salting out of water.
• Acetic acid is weak acid and it undergoes partial ionization as follows.
  CH₃COOH ⇌ CH₃COO^– + H⁺
• In case of such electrolytes there exists an equilibrium in between ions and molecular form. In such a condition, the addition of electrolyte with common ions shifts the equilibrium in backward direction which causes the decrease in solubility.

Significance :
1) Common ion effect is an important phenomenon in qualitative analysis.
2) The common ion effect principle is also used in controlling the H⁺ ion concentration in buffer solution.

ii) The relation between K and solubility (S) of a sparingly soluble salt BaSO₄ : The sparingly soluble salt like BaSO₄, when added to water form saturated solution. There exists an equilibrium between undissolved solid and the ions in saturated solution.
BaSO₄ ⇌ Ba²⁺ + SO₄^2-
The equilibrium constant, K = \(\frac{\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]}{\left[\mathrm{BaSO}_4\right]}\)
K = \(\frac{\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]}{\mathrm{K}_1}\) (∴ Activity of solid BaSO₄ is constant)

K × K₁ = [Ba²⁺][SO^2-] or K_sp = [Ba²⁺][SO₄^2-]
The new constant K_sp is called solubility product.

If S moles lit^-1 is the solubility of BaSO₄.

K_sp = s × s = s²
or solubility of BaSO₄ is s = \(\sqrt{\mathrm{K}_{\mathrm{sp}}}\) .

Equilibrium Class 11 NCERT Solutions

Question 1.
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature.
The volume of the container is suddenly increased.
i) What is the initial effect of the change on vapour pressure ?
ii) How do rates of evaporation and condensation change initially ?
iii) What happens when equilibrium is restored finally and what will be the final vapour pressure ?
Answer:
i) The vapour pressure will decrease due to increase in volume.
ii) Since, the temperature is fixed, the rate of evaporation will remain constant. However, the rate of condensation will be low initially due to lesser number of molecules per unit volume in the vapour phase.
iii) When equilibrium is restored, the final vapour pressure will be the same as original vapour pressure because at equilibrium the rates of evaporation and condensation are equal.

Question 2.
W hat is K_c for the following equilibrium when the equilibrium concentration of each substance is : [SO₂] = 0.60 M, [O₂] = 0.82 M and [SO₃] = 1.90 M ?
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Answer:
K_c = \(\frac{\left[\mathrm{SO}_3(\mathrm{~g})\right]^2}{\left[\mathrm{SO}_2(\mathrm{~g})\right]^2\left[\mathrm{O}_2(\mathrm{~g})\right]}=\frac{(1.90)^2}{(0.60)^2 \times(0.82)}\) = 12.229 L mol^-1

Question 3.
At a certain temperature and a total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms.
I_2(g) ⇌ 2I_(g)
Calculate K_p, for the equilibrium.
Answer:
Partial pressure of I atom (pI₁) = \(\frac{40}{100}\) × 10⁵
Partial pressure of I₂ (pI₂) = \(\frac{(100-40)}{100}\) × 10⁵
∴ K_p = \(\frac{\mathrm{p}_{\mathrm{I}}^2}{\mathrm{p}_{\mathrm{I}_2}}=\frac{\left(4.0 \times 10^4\right)^2}{6.0 \times 10^4}\)
= 2.67 × 10⁴ Pa

Question 4.
Write the expression for the equilibrium constant, K_c for each of the following reactions.
i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g)
ii) 2Cu(NO₃)₂(s) ⇌ 2Cu0(s) + 4NO₂(g) + O₂(g)
iii) CH₃COOC₂H₅(aq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(aq)
iv) Fe³⁺(aq) + 3OH^–(aq) ⇌ Fe(OH)₃(s)
v) I₂(s) + 5F₂(g) ⇌ 2IF₅(g)
Answer:
i) K_c = \(\frac{[\mathrm{NO}(\mathrm{~g})]^2\left[\mathrm{Cl}_2(\mathrm{~g})\right]}{[\mathrm{NOCl}(\mathrm{~g})]^2}\)
ii) K_c = [NO₂(g)]⁴[O₂(g)] {∵ [CuO(s) = [Cu(NO₃)₂(s)] = 1}
iii) K_c = \(\frac{\left[\mathrm{CH}_3 \mathrm{COOH}(\mathrm{aq})\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(\mathrm{aq})\right]}{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(\mathrm{aq})\right]}\) (∵ [H₂O(l)] = 1)
iv) K_c = \(\frac{1}{\left[\mathrm{Fe}^{3+}(\mathrm{aq})\right]\left[\mathrm{OH}^{-}(\mathrm{aq})\right]^3}\) (∵ [Fe(OH)₃(s)] = 1)
v) K_c = \(\frac{\left[\mathrm{IF}_5\right]^2}{\left[\mathrm{~F}_2\right]^5}\) (∵ [I₂(s) = 1)

Question 5.
Find out the value of Kc for each of the following equilibria from the value of K_p.
i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) ;K_p = 1.8 × 10^-2 at 500 K
ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); K_p = 167 at 1073 K
Answer:
i) For the reaction, 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) ; ∆n = (2 + 1) – 2 = 1
K_p = K_c (RT)^∆n
∴ K_c = \(\frac{\mathrm{K}_{\mathrm{p}}}{(\mathrm{RT})^{\Delta \mathrm{n}}}=\frac{1.8 \times 10^{-2}}{(0.0831 \times 500)}\) = 4.33 × 10^-4 (∵ R = 0.0831 L bar mol^-1K^-1)

ii) For the reaction, CaCO₃(s) ⇌ CaO(s) + CO₂(g) ; ∆n = 1 – 0 = 1
∵ K_c = \(\frac{\mathrm{K}_{\mathrm{p}}}{(\mathrm{RT})^{\mathrm{An}}}=\frac{167}{(0.0831 \times 1073)}\) = 1.873

Question 6.
For the following equilibrium, K_c = 6.3 × 10¹⁴ at 1000 K.
NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reactions ?
Answer:
K_c(reverse) = \(\frac{1}{\mathrm{~K}_{\mathrm{c}(\text { forward })}}\) = \(\frac{1}{3.3 \times 10^{14}}\) = 1.59 × 10^-15

Question 7.
Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression ?
Answer:
[Pure solid] or [Pure liquid] = \(\frac{\text { No.of moles }}{\text { Volume (in } \mathrm{L})}\) = \(\frac{\text { Mass } / \text { molecular mass }}{\text { Volume }}\)
= \(\frac{\text { Density }}{\text { Molecular mass }}\) = Constant
(The density of a pure solid or a pure liquid is constant at constant temperature)

Question 8.
Reaction between N₂ and O₂ takes place as follows.
2N₂(g) + O₂(g) ⇌ 2N₂O(g)
If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a 10L reaction vessel and allowed to form N₂O at a temperature for which K_c = 2.0 × 10^-37, determine the composition of equilibrium mixture.
Answer:
Suppose x moles of N₂ react till equilibrium is attained.

Since, the value of Kc for the reaction (K_c = 2.0 × 10^-37) is very small, the value of x must be very small. Therefore,
[N₂(g)] = \(=\frac{0.482-x}{10}=\frac{0.482}{10}\) = 0.0482 mol L^-1
[O₂(g)] = \(\frac{0.933-x / 2}{10}=\frac{0.933}{10}\frac{0.933-x / 2}{10}=\frac{0.933}{10}\) = 0.0933 mol L^-1 and [N₂O(g)] = \(\frac{\mathrm{x}}{10}\)
From K_c = \(\frac{\left[\mathrm{N}_2 \mathrm{O}(\mathrm{~g})\right]^2}{\left[\mathrm{~N}_2(\mathrm{~g})\right]^2\left[\mathrm{O}_2(\mathrm{~g})\right]}\)
We have, 2.0 × 10^-37 = \(\frac{(x / 10)^2}{(0.0482)^2 \times 0.0933}\)
Which gives, x = 6.6 × 10^-20
∴ [N₂O(g)] = \(\frac{x}{10}=\frac{6.6 \times 10^{-20}}{10}\) = 6.6 × 10^-21

Question 9.
Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below.
2NO(g) + Br₂(g) ⇌ 2NOBr(g)
When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br₂.
Answer:

Given that x = 0.0518
No. of moles of NO(g) at equilibrium = 0.087 – x = 0.087 – 0.518 = 0.0352
and number of moles of Br₂(g) at equilibrium = 0.0437 – \(\frac{\mathrm{x}}{2}\) = 0.0437 – \(\frac{0.0518}{2}\) = 0.0178

Question 10.
At 450 K, K_p = 2.0 × 10¹⁰ bar for the given reaction at equilibrium.
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
What is K_c at this temperature ?
Answer:
For the given reaction, ∆n = 2 – (2 + 1) = -1
K_p = K_c(RT)^∆n
⇒ K_c = \(\frac{\mathrm{K}_{\mathrm{p}}}{(\mathrm{RT})^{\Delta \mathrm{n}}}=\frac{2.0 \times 10^{10}}{(0.0831 \times 450)^{-1}}\) L mol^-1, (∵ R = 0.0831 L bar K^-1 mol^-1)
= 2.0 × 10¹⁰ × 0.831 × 450 L mol^-1
= 7.48 × 10¹¹ L mol^-1

Question 11.
A sample of HI(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is K_p for the given equilibrium ?
2HI(g) ⇌ H₂(g) + I₂(g)
Answer:

[∵ (0.2 – 0.04) is the decrease in the pressure of HI]
K_p = \(\frac{\mathrm{p}_{\mathrm{H}_2} \times \mathrm{p}_{\mathrm{I}_2}}{\mathrm{p}_{\mathrm{HI}}^2}=\frac{0.08 \times 0.08}{(0.04)^2}\) = 4.0

Question 12.
A mixture of 1.57 mol of N₂, 1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 1.7 × 10². Is the reaction mixture at equilibrium ? If not, what is the direction of the net reaction ?
Answer:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
Q_c = \(\frac{\left[\mathrm{NH}_3(\mathrm{~g})\right]^2}{\left[\mathrm{~N}_2(\mathrm{~g})\right]\left[\mathrm{H}_2(\mathrm{~g})\right]^3}=\frac{(8.13 / 20)^2}{\left(\frac{1.57}{20}\right) \times\left(\frac{1.92}{20}\right)^3}\) = 2.38 × 10³ (V = 20 L)
Given that K_c = 1.7 × 10²
Obviously, Q_c ≠ K_c
Therefore, the reaction is not in equilibrium because Q_c ≠ K_c. It will proceed in the reverse direction.

Question 13.
The equilibrium constant expression for a gas reaction is, K_c = \(\frac{\left[\mathrm{NH}_3\right]^4\left[\mathrm{O}_2\right]^5}{[\mathrm{NO}]^4\left[\mathrm{H}_2 \mathrm{O}\right]^6}\)
Write the balanced chemical equation corresponding to this expression.
Answer:
4NO(g) + 6H₂O(g) ⇌ 4NH₃(g) + 5O₂(g)

Question 14.
One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with CO according to the equation,
H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)
Calculate the equilibrium constant for the reaction.
Answer:
The given reaction is H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)
Initial number of moles of H₂O(g) = 1
Number of moles of H₂O(g) (reacted) = \(\frac{1 \times 40}{100}\) = 0.4
∴ Number of moles of H₂O(g) left (unreacted) = 1 – 0.4 = 0.6
From the given equation it is clear that 0.4 moles of H₂O(g) react with 0.4 moles of CO(g) to form 0.4 moles of H₂(g) and 0.4 moles of CO₂(g). The total volume is 10 litres.
∴ At equilibrium,
[H₂O(g)] = \(\frac{1-0.4}{10}\) = \(\frac{0.6}{10}\) = 0.06 mol L^-1 ; [CO(g)] = \(\frac{1-0.4}{10}\) = \(\frac{0.6}{10}\) = 0.06 mol L^-1
[H₂(g)] = \(\frac{0.4}{10}\) = 0.04 mol L^-1 ; [CO₂(g)] = \(\frac{0.4}{10}\) = 0.04 mol L^-1
Hence, K = \(\frac{\left[\mathrm{H}_2(\mathrm{~g})\right]\left[\mathrm{CO}_2(\mathrm{~g})\right]}{\left[\mathrm{H}_2 \mathrm{O}(\mathrm{~g})\right][\mathrm{CO}(\mathrm{~g})]}=\frac{0.04 \times 0.04}{0.06 \times 0.06}\) = 0 44

Question 15.
At 700 K, equilibrium constant for the reaction :
H₂(g) + I₂(g) ⇌ 2HI(g)
is 54.8. If 0.5 mol L_1 of HI(g) js present at equilibrium at 700 K, what are the concentration of H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K ?
Answer:
The given reaction can be written as 2HI(g) ⇌ H₂(g) + I₂(g)
For this reaction, K = \(\frac{1}{54.8}\) = 1.82 × 10^-2 (∵ because for the reverse reaction, K = 54.8)
As the number of moles of H₂ and N₂ are equal, their concentrations at equilibrium will also be equal.
Suppose [H₂(g)] = [I₂(g)] = x mol L^-1
Given that [HI(g)] = 0.5 mol L^-1
K = \(\frac{\left[\mathrm{H}_2(\mathrm{~g})\right]\left[\mathrm{I}_2(\mathrm{~g})\right]}{[\mathrm{HI}(\mathrm{~g})]^2}\) ⇒ 1.82 × 10^-2 = \(\frac{x \times x}{(0.5)^2}\) ⇒ x = [1.82 × 10^-2 × (0.5)²]^1/2 = 0.068 mol L^-1
Hence, at equilibrium, [H₂(g)] = [I₂(g)] = 0.068 mol L^-1

Question 16.
What is the equilibrium concentration of each of the substances in the equilibrium when the initial, concentration of ICl was 0.78 M ?
2ICl(g) ⇌ I₂(g) + Cl₂(g); K_c = 0.14
Answer:
Suppose at equilibrium, [I₂(g)] = [Cl₂(g>] = x mol L^-1

Given that, K_c = 0.14
K_c = \(\frac{\left[\mathrm{I}_2(\mathrm{~g})\right]\left[\mathrm{Cl}_2(\mathrm{~g})\right]}{[\mathrm{ICl}(\mathrm{~g})]^2}\)
⇒ 0.14 = \(\frac{x \times x}{(0.78-2 x)^2}\) ⇒ x² = 0.14(0.78 -2x)²
⇒ x = \(\sqrt{0.14}\) (0.78 – 2x) = 0.374(0.78 – 2x)
⇒ x = 0.292 – 0.748 x
⇒ 1.748 x = 0.292
x = \(\frac{0.292}{1.748}\) = 0.167
Hence, equilibrium concentrations will be as follows.
[I₂(g)] – [Cl₂(g)] = x = 0.167 M
[ICl(g)] = 0.78 – 2x = 0.78 – (2 × 0.167) = 0.446 M

Question 17.
K_p = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
C₂H₆(g) ⇌ C₂H₄(g) + H₂(g)
Answer:
Suppose the partial pressure of C₂H₆ decreases by p atm till equilibrium is attained.

K_p = \(\frac{\mathrm{p}_{\mathrm{C}_2 \mathrm{H}_4} \cdot \mathrm{p}_{\mathrm{H}_2}}{\mathrm{p}_{\mathrm{C}_2 \mathrm{H}_6}}\) = \(\frac{\mathrm{p} \times \mathrm{p}}{4.0-\mathrm{p}}\) ⇒ 0.04 = \(\frac{\mathrm{p}^2}{4.0-\mathrm{p}}\) = 0.16 – 0.04 p ⇒ p² + 0.04 p – 0.16 = 0
p = \(\frac{-0.04 \pm \sqrt{(0.04)^2-4 \times 1 \times(-0.16)}}{2 \times 1}=\frac{-0.04 \pm 0.801}{2}\)
Taking positive value, p = 0.38
∴ [C₂H₆(g)]_eq = 4.0 – p = 4.0 – 0.38 = 3.62 atm

Question 18.
Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as,
CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)
i) Write the concentration ratio (reaction quotient), Q_c, for this reaction (note : water is not in excess and is not a solvent in this reaction)
ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
Answer:
i) Q_c = \(\frac{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(l)\right]\left[\mathrm{H}_2 \mathrm{O}(l)\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}(l)\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)\right]}\)

ii)

If the volume of the reaction mixture is V L, we have at equilibrium,
[CH₃COOH(l)] = \(\frac{0.829}{\mathrm{V}}\) mol L^-1 ; [C₂H₅OH(l)] = \(\frac{0.009}{\mathrm{V}}\) mol L^-1
[CH₃COOC₂H₅(l)] = \(\frac{0.171}{\mathrm{V}}\) mol L^-1 ; [H₂O(l)] = \(\frac{0.171}{\mathrm{V}}\) mol L^-1
K_c = \(\frac{\left[\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5(l)\right]\left[\mathrm{H}_2 \mathrm{O}(l)\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}(l)\right]\left[\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}(l)\right]}=\frac{\frac{0.171}{\mathrm{~V}} \times \frac{0.171}{\mathrm{~V}}}{\frac{0.829}{\mathrm{~V}} \times \frac{0.009}{\mathrm{~V}}}=\frac{0.171 \times 0.171}{0.829 \times 0.009}\) = 3.92

iii)

∴ Q_c = \(\frac{\frac{0.214}{V} \times \frac{0.214}{V}}{\frac{0.786}{V} \times \frac{0.286}{V}}\) = 0.204
Since, Q_c < K_c, equilibrium has not been reached.

Question 19.
A sample of pure PCl₅, was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5 × 10^-1 mol L^-1. If value of K_c is 8.3 × 10^-5 what are the concentrations of PCl₃ and Cl₂ at equilibrium ?
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
Answer:
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
Since, the number of moles of PCl₃(g) and Cl₂(g) are equal, their concentrations at equilibrium will also be equal.
Suppose, at equilibrium, [PCl₃(g)] = [Cl₂(g)] = x mol L^-1
Given that at equilibrium, [PCl₅] = 0.5 × 10^-1 mol L^-1
∴ K_c = \(\frac{\left[\mathrm{PCl}_3(\mathrm{~g})\right]\left[\mathrm{Cl}_2(\mathrm{~g})\right]}{\left[\mathrm{PCl}_5(\mathrm{~g})\right]}\) ⇒ 8.3 × 10^-3 = \(\frac{\mathrm{x} \times \mathrm{x}}{0.5 \times 10^{-1}}\)
⇒ x = (8.3 × 10^-3 × 0.5 × 10^-1)^1/2 = 0.02
Hence, [PCl₃(g)] = [Cl₂(g)] = 0.02 mol L^-1.

Question 20.
One of the reaction that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO₂.
FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g); K_p = 0.265 atm at 1050 K
What are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial partial pressures are : P _CO = 1.4 atm and P_CO2 = 0.80 atm?
Answer:

Q_p = \(\frac{\mathrm{P}_{\mathrm{CO}_2}}{\mathrm{P}_{\mathrm{CO}}}=\frac{0.80}{1.4}\) = 0.571
Since, Q_p > K_p, the reaction will proceed in the reverse direction. On account of this partial pressure of CO₂ will decrease while that of CO will increase.
Suppose, the decrease in the partial pressure of CO₂ is p atm.
∴ At equilibrium, P_CO2 = (0.80 – p) atm and P_CO = (1.4 + p) atm
∵ K_p = \(\frac{\mathrm{P}_{\mathrm{CO}_2}}{\mathrm{P}_{\mathrm{CO}}}\), we have
0.265 = \(\frac{0.80-\mathrm{p}}{1.4+\mathrm{p}}\) ⇒ 0.265 (1.4 + p) = 0.80 – p
⇒ 0.371 + 0.265 p = 0.80 – p
⇒ 1.265 p = 0.429 ⇒ p = \(\frac{0.429}{1.265}\) = 0.339 atm
∴ P _CO at equilibrium = 1.4 + 0.339 = 1.739 atm.
and P_CO2 at equilibrium = 0.80 – 0.339 = 0.461 atm.

Question 21.
Equilibrium constant, K_c for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mole L^-1 N₂(g) 2.0 mol L^-1 H₂ and 0.5 mol L^-1 NH₃. Is the reaction at equilibrium ? If not in which direction does the reaction tend to proceed to reach equilibrium ?
Answer:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ; K_c = 0.061
For the given set of conditions, Q_c = \(\frac{\left[\mathrm{NH}_3(\mathrm{~g})\right]^2}{\left[\mathrm{~N}_2(\mathrm{~g})\right]\left[\mathrm{H}_2(\mathrm{~g})\right]^3}=\frac{(0.5)^2}{(3.00) \times(2.00)^3}\) = 0.0104
Since, Q_c < K_c, the reaction is not at equilibrium. It will proceed in the forward direction.

Question 22.
Bromine monochioride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:
2BrCl(g) ⇌ Br₂(g) + Cl₂(g)
for which K_c = 32 at 500 K. If initially pure BrCl is present at a concentration 3.3 × 10^-3 mol L^-1, what is its molar concentration in the mixture at equilibrium?
Answer:

K_c = \(\frac{\left[\mathrm{Br}_2(\mathrm{~g})\right]\left[\mathrm{C} l_2(\mathrm{~g})\right]}{[\mathrm{BrCl}(\mathrm{~g})]^2}\) ⇒ 32 = \(\frac{\frac{x}{2} \times \frac{x}{2}}{\left(3.30 \times 10^{-3}-x\right)^2}\)
Taking square root on both sides, we get
5.66 = \(\frac{\frac{x}{2}}{3.30 \times 10^{-3}-x}=\frac{x}{2\left(3.30 \times 10^{-3}-x\right)}\)
⇒ 0.037 – 11.32 x = x ⇒ (1 + 11.32)x = 0.037 ⇒ x = \(\frac{0.037}{1+11.32}\) = 3.0 × 10^-3
∴ Equilibrium centration of BrCl = 3.30 × 10^-3 – x
= 3.30 × 10^-3 – 3.0 × 10^-3
= 3.0 × 10^-4 mol L^-1

Question 23.
At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass, C(s) + CO₂(g) ⇌ 2CO(g)
Calculate K_c for this reaction at the above temperature.
Answer:
If the total mass of the gaseous mixture at equilibrium is 100 g then
mass of CO(g) = 90.55 g
and mass of CO₂(g) = 100 – 90.55 = 9.45 g
Number of moles of COg) = \(\frac{90.55}{28}\) = 3.234,
(∵ molar mass of CO = 28 g/mol, molar mass of CO₂ = 44 g/mol)
and number of moles of CO₂(g) = \(\frac{9.45}{44}\) = 0.215
Total number of gaseous moles = 3.234 + 0.2 15 = 3.449
∴ P_CO = \(\frac{3.234}{3.449}\) × 1 = 0.938 atm (∵ Partial pressure = \(\frac{\text { No.of moles }}{\text { Total no.of moles }}\) × Total pressure)
and P_CO2 = \(\frac{0.215}{3.449}\) × 1 = 0.062 atm
Hence, for the reaction C(s) + CO₂(g) ⇌ 2CO(g)
K_p = \(\frac{\mathrm{p}{\mathrm{CO}}^2}{\mathrm{p}{\mathrm{CO}_2}}=\frac{(0.938)^2}{0.062}\) = 14.19
For the above reaction, ∆n = 2 – 1 = 1
K_p = K_p(RT)^∆n ⇒ K_c = \(\frac{\mathrm{K}_{\mathrm{p}}}{(\mathrm{RT})^{\mathrm{An}}}=\frac{14.19}{(0.0821 \times 1127)^1}\) = 0.153

Question 24.
Calculate (i) ∆G° and (ii) the equilibrium constant for the formation of NO₂ from NO and O₂ at 298
NO(g) + 1/2O₂(g) ⇌ NO₂(g)
where ∆_fG°(NO₂) = 52.0 kJ/mol ; ∆_fG°(NO) = 87,0 kJ/mol ; ∆_fG°(O₂) = 0 kJ/mol
Answer:
i) ∆_rG° = ∆_fG° (products) – ∆_fG° (reactants)
or = ∆_rG° [NO₂] – {∆_fG° [NO] + ½∆_fG° [O₂])
or ∆_r = 52.0 – (87.0 + 0) = -35.0 kJ mol ^-1

ii) ∆_rG° = -2.303 RT log10K
⇒ log₁₀K = –\(\frac{\Delta_{\mathrm{r}} G^{\circ}}{2.303 \mathrm{RT}}\) = \(\frac{(-35.0)}{-2.303 \times 8.314 \times 10^{-3} \times 298}\) = 6.1341
(∵ R = 8.314 × 10^-3 kJ K^-1 mol^-1)
⇒ K = antilog₁₀6.134l = 1.362 × 10⁶

Question 25.
Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease In pressure by increasing the volume?
i) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
ii) CaO(s) + CO₂(g) ⇌ CaCO₃(s)
iii) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)
Answer:
According to Le Chateliers principle, if the pressure on the system is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more.
i) No.of moles of products will increase on decreasing pressure as the forward reaction will be favoured.
ii) No.of moles of products will decrease as the backward reaction will be favoured.
iii) No.of moles of products will remain the same as the number of moles of gaseous substances is the same on both the sides.

Question 26.
Which of the following reactions will get affected by increasing the pressure ? Also, mention whether change will cause the reaction to go into forward or backward direction.
i) COCl₂(g) ⇌ CO(g) + Cl₂(g)
ii) CH₄(g) + 2S₂(g) ⇌ CS₂(g) + 2H₂S(g)
iii) CO₂(g) + C(s) ⇌ 2C0(g)
iv) 2H₂(g) + CO(g) ⇌ CH₃OH(g)
v) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)
Answer:
i) Reaction will go in the backward direction as n_P > n_R
ii) Reaction will remain unaffected as n_P = n_R.
iii) Reaction will go in the backward direction as n_P > n_R
iv) Reaction will go in the forward direction as n_P < n_R
v) Reaction will go in the backward direction as n_P > n_R
vi) Reaction will go in the backward direction as n_P > n_R.

Question 27
The equilibrium constant for the following reaction is 1.6 × 10⁵ at 1024 K
H₂(g) + Br₂(g) ⇌ 2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.
Answer:
For the reaction

For the above reaction, K = \(\frac{1}{1.6 \times 10^5}\) = 6.25 × 10^-6
Further, K = K = 6.25 × 10^-6 ⇒ K_p = \(\frac{\mathrm{p}{\mathrm{H}_2} \cdot \mathrm{p}{\mathrm{Br}2}}{\mathbf{p}{\mathrm{HBr}}^2}\)
Hence, 6.25 × 10^-6 = \(\frac{\frac{\mathrm{p}}{2} \cdot \frac{\mathrm{p}}{2}}{(10.0-\mathrm{p})^2}\)
Taking square mot on both sides, we have, 2.5 × 10^-3 = \(\frac{\frac{p}{2}}{10.0-p}=\frac{p}{2(10.0-p)}\)
⇒ 0.05 – 5.0 × 10^-3 p ⇒ p = p(1 +5.0 × 10^-3)= 0.05 ⇒ p = \(\frac{0.05}{1+5.0 \times 10^{-3}}\) = 4.98 × 10^-2 bar
Hence, at equilibrium, P_H2 = P_Br2 = \(\frac{p}{2}=\frac{4.98 \times 10^{-2}}{2}\) = 2.49 × 10^-2 bar
and P_HBr = (10.0 – p) = 10.0 – 4.98 × 10^-2 = 9.95 bar

Question 28.
Dihydrogen gas ¡s obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)
i) Write an expression for K_p for the above reaction.
ii) How will the values of K_p and composition of equilibrium mixture be affected by
a) increasing the pressure b) increasing the temperature c) using a catalyst?
Answer:
i) K_p = \(\frac{\mathrm{p}{\mathrm{CO}} \times \mathrm{p}{\mathrm{H}2}^3}{\mathrm{p}{\mathrm{CH}4} \times \mathrm{p}{\mathrm{H}_2 \mathrm{O}}}\)
ii) a) Equilibrium will shift in the backward direction as n_p > n_R.

b) Equilibrium will shift in the forward direction because the reaction is endothermic.

c) Composition of equilibrium mixture will remain unchanged. However, the equilibrium will be attained earlier,
The value of K_p will remain unchanged in (a) and (C). K_p will increase in (b).

Question 29.
Describe the effect of:
i) addition of H₂
ii) addition of CH₃OH
iii) removal of CO
iv) removal of CH₃OH
on the equilibrium of the reaction:
2H₂(g) + CO(g) ⇌ CH₃OH(g)
Answer:
i) Equilibrium will shift in the forward direction.
ii) Equilibrium will shift in the backward direction.
iii) Equilibrium will shift in the backward direction.
iv) Equilibrium will shift in the forward direction.

Question 30.
At 473 K, equilibrium constant K_c for decomposition of phosphorus pentachioride, PCl₅ is 8.3 × 10^-3. If decomposition is depicted as,
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); ∆_rH° = 124.0 kJ mol^-1
i) Write an expression for K_c for the reaction.
ii) What is the value of K_c for the reverse reaction at the same temperature?
iii) What would be the effect on K_c if (a) more PCl₅ is added (b) pressure is increased (c) the temperature is increased?
Answer:
i) K_c = \(\frac{\left[\mathrm{PCl}_3(\mathrm{~g})\right]\left[\mathrm{Cl}_2(\mathrm{~g})\right]}{\left[\mathrm{PCl}_5(\mathrm{~g})\right]}\)
ii) K’ = \(\frac{1}{K_c}=\frac{1}{8.3 \times 10^{-3}}\) = 120.48
iii) a) M constan temperature, the value of K_c is not affected by addition of reactants.
Therefore, K_c will remain unchanged.
b) K_c will remain unchanged.
c) Since, K_c = \(\frac{K_1}{K_2}\), an increase in temperature will favour the forward reaction as the reaction is endothermic. Thus, K₁ will Increase. Hence, the value of K_c will increase or increasing temperature.

Question 31.
Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)
If a reaction vessel at 4000 C is charged with an equimolar mixture of CO and steam such that p_CO = p_H2O = 4.0 bar, what will be the partial pressure of H₂ at equilibrium? K_p = 10.1 at 400°C.
Answer:
Suppose the partial pressure of H₂ at equilibrium is p bar.

∴ K_p = \(\frac{\mathrm{p}_{\mathrm{CO}_2} \cdot \mathrm{p}_{\mathrm{H}_2}}{\mathrm{p}_{\mathrm{CO}} \cdot \mathrm{p}_{\mathrm{H}_2 \mathrm{O}}}\)
⇒ 10.1 = \(\frac{p \times p}{(4.0-p) \times(4.0-p)}=\left(\frac{p}{4.0-p}\right)^2\) ⇒ \(\frac{p}{4.0-p}\) = \(\sqrt{10.1}\) = 3.178
⇒ p = 3.178 (40 – p) ⇒ p = 12.712 – 3.178 p ⇒ 3.178 + p = 12.712 bar
∴ p = 12.712/4.178 bar
p = 3.04 bar

Question 32.
Predict which of the following reaction will have appreciable concentration of reactants and products.
i) Cl₂(g) ⇌ 2Cl(g) ; K_c = 5 × 10^-39
ii) Cl₂(g) + 2NO(g) ⇌ 2NOCl(g) ; K_c = 3.7 × 10⁸
iii) Cl₂(g) + 2NO₂(g) ⇌ 2NO₂Cl(g) ; K_c = 1.8
Answer:
i) For reaction (a), K_c is very low. Therefore, the concentration of products will be very low.
ii) For reaction (b), K_c is quite high. Therefore, the reaction will proceed almost to completion and there will be no appreciable concentrations of reactants.
iii) The reaction (c) has a moderate value of K_c and will have appreciable concentrations of both reactants and products.

Question 33.
The value of K_c for the reaction 3O₂(g) ⇌ 2O₃(g) is 2.0 × 10^-50 at 25°C. If the equilibrium concentration of O₂ in air at 25°C is 1.6 × 10^-2, what is the concentration of O₃ ?
Answer:
K_c = \(\frac{\left[\mathrm{O}_3(\mathrm{~g})\right]^2}{\left[\mathrm{O}_2(\mathrm{~g})\right]^3}\) ⇒ 2.0 × 10^-30 = \(\frac{\left[\mathrm{O}_3(\mathrm{~g})\right]^2}{\left(1.6 \times 10^{-2}\right)^3}\)
⇒ [O₃(g)]² = 2.0 × 10^-50 × (1.6 × 10^-2)³ = 8.192 × 10-56
∴ [O₃(g)] = (8.192 × 10-56)¹⁰ = 2.86 × 10-28 M

Question 34.
The reaction, CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g) is at equilibrium at 1300 K in a 1 L flask. It also contain 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant, K_c for the reaction at the given temperature is 3.90.
Answer:
K_c = \(\frac{\left[\mathrm{CH}_4(\mathrm{~g})\right]\left[\mathrm{H}_2 \mathrm{O}(\mathrm{~g})\right]}{[\mathrm{CO}(\mathrm{~g})]\left[\mathrm{H}_2(\mathrm{~g})\right]^3}\)
∴ 3.90 = \(\frac{\left[\mathrm{CH}_4(\mathrm{~g})\right] \times 0.02}{0.30 \times(0.10)^3}\) (V = 1 L)
⇒ [CH₄(g)] = \(\frac{3.90 \times 0.30 \times(0.10)^3}{0.02}\) = 0.0585 = 5.85 × 10^-2 mol L^-1

Question 35.
What is meant by the conjugate acid-base pair ? Find the conjugate acid/base for the following species.
HNO₂, CN^–, HClO₄, F^–, OH^–, \(\mathrm{CO}_3^2\) and S^2-
Answer:
A conjugate acid-base pair that differs only by one proton. Conjugate pairs of the given species are NO₂^–, HCN, ClO₄^–, HF, H₂O, HCO₃^– and HS^– respectively.

Question 36.
Which of the followings are Lewis acids ? H₂O, BF₃, H⁺, and \(\mathrm{NH}_4^{+}\)?
Answer:
BF₃, H⁺ and \(\mathrm{NH}_4^{+}\).

Question 37.
What will be the conjugate bases for the Bronsted acids : HF, H₂SO₄ and \(\mathrm{HCO}_3^{-}\) ?
Answer:
F^–, HSO₄^– and CO₃^2-
(Conjugate base = Conjugate acid – H⁺)

Question 38.
Write the conjugate acids for the following Bronsted bases : NH₂^–, NH₃ and HCO₃^–.
Answer:
NH₃, NH₄⁺, HCOOH (Conjugate acid = Conjugate base + H⁺)

Question 39.
The species : H₂O, HCO₃^–, HS₄^– and NH₃ can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base.
Answer:

Question 40.
Classify the following species into Lewis acids and Lewis bases and show these act as Lewis acid/base :
(a) OH^–
(b) F^–
(c) H⁺
(d) BCl₃.
Answer:
a) OH^– is a Lewis base because it can donate an electron pair.
b) F^– is a Lewis base because it can donate electron pair.
c) H⁺ is a Lewis acid because it can accept an electron pair.
d) BCl₃ is a Lewis acid because it is an electron deficient molecule and can accept an electron pair.

Question 41.
The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^-3 M. What is its pH?
Answer:
pH = -log₁₀(H⁺) = -log₁₀(3.8 × 10^-3) = 2.42

Question 42.
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Answer:
log₁₀[H⁺] = -3.76
[H⁺] = antilog₁₀(-3.76) = antilog₁₀\(\overline{4} .24\) = 1.74 × 10^-4 M

Question 43.
The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 × 10^-4, 1.8 × 10^-4 and 4.8 × 10^-9 respectively. Calculate the ionization constants of the corresponding conjugate base.
Answer:
i) The conjugate base of HF is F^–. It will react with water as follows.
F^– + H₂O ⇌ HF + OH^–
Obviously, this is the equation of hydrolysis of a salt of HF with a strong base. Hence,
K_F = K_h = \(\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{~K}_{\mathrm{a}}}=\frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}}\) = 1.5 × 10^-11

ii) Similarly, for HCOOH, K_HCOO^– = K_h = \(\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{~K}_{\mathrm{a}}}=\frac{1.0 \times 10^{-14}}{1.8 \times 10^{-9}}\) = 5.6 × 10^-11

iii) For HCN, K_CN^– = K_h = \(\frac{K_w}{K_a}=\frac{1.0 \times 10^{-14}}{4.8 \times 10^{-9}}\) = 2.1 × 10^-6

Question 44.
The ionization constant of phenol is 1.0 × 10^-10. What is the concentration of phenolate ion is 0.05 M solution of phenol ? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate ?
Answer:

∴ K_a = \(\frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{O}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\right]}=\frac{\mathrm{x} . \mathrm{x}}{0.05-\mathrm{x}}\) = 1.0 × 10^-10(given)
⇒ \(\frac{x^2}{0.05-x}\) = 1.0 × 10^-10
Since, phenol is not much dissociated, 0.05 – x ≈ 0.05.
Hence, \(\frac{x^2}{0.05}\) = 1.0 × 10^-10 ⇒ x = (0.05 × 1.0 × 10^-10)^1/2 = 2.24 × 10^-6
Hence, [C₆H₅O^–] in solution = x = 2.24 × 10^-6 M

In the presence of 0.01 M of C₆H₅ONa :
Suppose y is the amount of phenol dissociated. Therefore, at equilibrium
[C₆H₅OH] – 0.05 – y, [C₆H₅O^–] = 0.01 + y and [H⁺] = y
∴ K_a = \(\frac{(0.01+\mathrm{y})(\mathrm{y})}{(0.05-\mathrm{y})}\) = 1.0 × 10^-10 (given)
Since, in the presence of sodium phenolate, the dissociation of phenol will get further suppressed we can take
0.01 + y ≈ 0.01 and 0.05 – y ≈ 0.05
Hence, \(\frac{0.01+\mathrm{y}}{0.05}\) = 1.0 × 10^-10 ⇒ y = \(\frac{1.0 \times 10^{-10} \times 0.05}{0.01}\) = 5.0 × 10^-10
Hence, degree of dissociation of phenol
α = \(\frac{\text { Number of moles dissociated }}{\text { Total no.of moles initially taken }}=\frac{5.0 \times 10^{-10}}{0.05}\) = 1.0 × 10^-10

Question 45.
The first ionization constant of H₂S is 9.1 × 10^-8. Calculate the concentration of HS^– ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H₂S is 1.2 × 10^-13, calculate the concentration of Sounder both conditions.
Answer:
Calculation of [HS^–]
H₂S ionises in two stages. The first stage of ionisation is as follows.

K_a1 = 9.1 × 10^-8 (given)
According to the law of equilibrium, K_a1 = \(\)
⇒ 9.1 × 10^-8 = \(\frac{x \cdot x}{0.1-x}\) = \(\frac{x^2}{0.1}\) (as x is very small so 0.1 – x = 0.1)
⇒ x = (9.1 × 10^-8 × 0.1)^1/2 = 9.54 × 10^-5
Hence, [HS^–] in 0.1 M solution = x = 9.54 × 10^-5.
In the presence of 0.1 M HCl, suppose H₂S dissociates to an extent of y M.
Therefore, [H₂S] = 0.1 – y ≈ 0.1M, [H⁺] = y + 0.1 ≈ 0.1M and [HS^–] = y M
(Since, in the presence of HCl, the dissociation of H₂S will further get suppressed)
∴ K_a1 = \(\frac{0.1 \times \mathrm{y}}{0.1}\) = 9.1 × 10^-8 ⇒ y = 9.1 × 10^-8M
Hence, in the presence of 0.1 M of HCl,
[HS^–] = y = 9.1 × 10^-8M
Calculation of [S^2-]

The overall dissociation can be expressed as follows.
H₂S ⇌ 2H⁺ + S^2-
K_a = \(\frac{\left[\mathrm{H}^{+}\right]^2\left[\mathrm{~S}^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{~S}\right]}\)
K_a = K_a1 × K_a2 = 9.1 × 10^-8 × 1.2 × 10^-13 = 1.1 × 10^-20

In the absence of 0.1 M of HCl:

⇒ x’ = \(\left(\frac{1.1 \times 10^{-20} \times 0.1}{4}\right)^{1 / 3}\) = 6.5 × 10^-8 M
Hence, [S^2-] in the absence of 0.1 M of HCl = 6.5 × 10^-8 M

In the presence of 0.1 M of HCl:
Suppose H₂S dissociates to an extent of y’ M
∴ [H₂S] = 0.1M – y’ ≈ 0.1 M, [H⁺] = 2y’ + 0.1 ≈ 0.1 M and [S^2-] = y
∴ K_a = \(\frac{(0.1)^2 \times y^{\prime}}{0.1}\) = 1.1 × 10^-20 ⇒ y’ = \(\frac{1.1 \times 10^{-20}}{0.1}\) = 1.1 × 10^-19 M
Hence, [S^2-] in the presence of 0.1 M of HCl = 1.1 × 10^-19 M

Question 46.
The ionization constant of acetic acid is 1.74 × 10^-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.
Answer:

According to the law of equilibrium,
⇒ K = \(\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}\) ⇒ 1.74 × 10^-5 = \(\frac{x . x}{0.05-x}=\frac{x^2}{0.05}\) (∵ x is very small)
⇒ x = (1.74 × 10^-5 × 0.05)^1/2 = 9.33 × 10^-4 M
[CH3COO^–] = x = 9.33 × 10^-4 mol L^-1
[H⁺] = x = 9.33 × 10^-4 mol L^-1
∴ pH = -log₁₀[H⁺] = -log₁₀(9.33 × 10^-4) = 3.03

Question 47.
It has been found that the pH of a 0.01 M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK_a.
Answer:
Suppose the organic acid is HA. HA ⇌ H⁺ + A^–
pH = -log₁₀[H⁺]
4.15 = -log₁₀[H⁺]
∴ [H⁺] = antilog10(-4.15) = 7.08 × 10^-5 M
[A] = [H⁺] = 7.08 × 10^-5 M
∴ K_a = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)
K_a = \(\frac{7.08 \times 10^{-5} \times 7.08 \times 10^{-5}}{0.01}\) = 5.01 × 10^-7 (∵ Dissociation of the acid is very low)
pK_a = -log₁₀ K_a = -log₁₀(5.01 × 10^-7) = 6.3002

Question 48.
Assuming complete dissociation, calculate the pH of the following solutions :
i) 0.003 M HCl
ii) 0.005 M NaOH
iii) 0.002 M HBr
iv) 0.002 M KOH
Answer:
i) [H₃O⁺] = [HCl] – 0.003 mol L^-1
∴ pH = -log₁₀[H₃O⁺] = -log₁₀ (0.003) = 2.52

ii) [OH^–] = [NaOH] = 0.005
∴ [H₃O⁺] = \(\frac{\mathrm{K}_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{0.005}\) = 20 × 10^-12
pH = -log₁₀[H₃O⁺] = -log₁₀(2.0 × 10^-12) = 11.7

iii) [H₃O⁺] = [HBr] = 0.002 mol L^-1
∴ pH = -log₁₀[H₃O⁺] = -log₁₀(0.002) = 2.70

iv) [OH^–] = [KOH] = 0.002 mol L^-1
∴ [H₃O⁺] = \(\frac{\mathrm{K}_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{0.002}\) = 5.0 × 10^-12
pH = -log₁₀[H₃O⁺] = -log₁₀(5.0 × 10^-12) = 11.3

Question 49.
Calculate the pH of the following solutions.
i) 2 g of TlOH dissolved in water to give 2 litres of solution.
ii) 0.3 g of Ca(OH)₂ dissolved in water to give 500 mL of solution.
iii) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
iv) 1 mL of 13.6 M HCl is diluted with water to give 1 litre of solution,
Answer:
i)

iv) Suppose the molarity of the diluted solution is M’.
M₁V₁ = M’V’ ⇒ 13.6 × 1 = M’ × 1000
⇒ M’ = \(\frac{13.6 \times 1}{1000}\) = 1.36 × 10^-2
∴ [H₃O⁺] = 1.36 × 10^-2 M
pH = -log₁₀[H₃O⁺] = -log₁₀(1.36 × 10^-2) = 1.87

Question 50.
The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the PK. of bromoacetic acid.
Answer:

∴ K_a = \(\frac{\left[\mathrm{BrCH}_2 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{BrCH}_2 \mathrm{COOH}\right]}=\frac{(0.1 \times 0.132) \times(0.1 \times 0.132)}{0.1(1-0.132)}\) = 2.01 × 10^-3
pK_a = -log₁₀K_a = -log₁₀(2.01 × 10^-3) = 2.70
[H⁺] = 0.1 × 0.132 = 0.0132 = 1.32 × 10^-2
pH = -log₁₀[H⁺] = – log₁₀(1.32 × 10^-2) = 1.88

Question 51.
The pH Of 0.005 M codeine (C₁₈H₂₁NO₃) solution is 9.95. Calculate its ionization constant as pK_b.
Answer:
Codeine is a base, therefore, it is expected to ionise in solution as follows.
Codeine + H₂ ⇌ Codeine H⁺ + OH^–
Given, pH = 9.95
∴ POH = 14 – 9.95 = 4.05
i.e., -log₁₀[OH^–] = 4.05
[OH^–] = antilog₁₀(-4.05) = 8.91 × 10^-5
[Codeine H⁺] = [OH^–] = 8.91 × 10^-5
[Codeine] = 0.005 M
K_b = \(\frac{\left[\text { Codeine } \mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\text {Codeine }]}=\frac{\left(8.91 \times 10^{-5}\right) \times\left(8.91 \times 10^{-5}\right.}{0.005}\) = 1.587 × 10^-5
[Codeine W] [OH ]
and pK_b = -log₁₀K_b = -log₁₀(1.587 × 10^-6) = 5.8

Question 52.
What is the pH of 0.001 M aniline solution ? The ionization constant of aniline can be taken as 4.27 × 10^-10. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Answer:
K_b for aniline = 4.27 × 10^-10
C₆H₅NH₂ + H₃O⁺ ⇌ C₆H₅NH₃⁺ + OH^–
K_b = \(\frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_3^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\right]}\)
Since, [C₆H₅NH₃⁺] = [OH^–] and dissociation of aniline is not much, we have
and
[OH^–] = {K_b[C₆H₅NH₂]}^1/2 = (4.27 × 10^-10 × 0.00)^1/2 = 6.53 × 10^-7 M
∴ POH = -log₁₀(6.53 × 10^-7) = 6.185
and pH = 14 – 6.185 = 7.815

∴ K_b = \(\frac{0.001 \alpha \times 0.001 \alpha}{0.001(1-\alpha)}=\frac{0.001 \alpha^2}{1-\alpha}\) = 0.001 α² (Since, α is very small, 1 – α = 1)
Thus, 0.001 α² = 4.27 × 10^-10 ⇒ α = \(\left(\frac{4.27 \times 10^{-10}}{0.001}\right)^{1 / 2}\) = 6.53 × 10^-4
For a pair of conjugate acid and base,
pK_b + pK_a = 14 ⇒ pK_a = 14 – pK_b = 14 – (-log₁₀ 4.27 × 10^-10) = 14 – 9.37 = 4.63
Hence, the ionisation constant of the conjugate acid.
K_a = antilog₁₀(-pK_a) (∵ pK_a = -log10K_a)
= antilog₁₀ (-4.63) = 2.34 × 10^-5

Question 53.
Calculate the degree of ionization of 0.05 M acetic acid if its pK_a value is 4.74. How is the degree of dissociation affected when its solution also contains
i) 0.01 M
ii) 0.1 M of HCl?
Answer:
pK_a = -log₁₀K_a = 4.74
∴ K_a = antilog₁₀(-4.74) = 1.82 × 10^-5
For a weak electrolyte, α = \(\sqrt{\frac{K_a}{c}}\); For acetic acid, α = \(\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{c}}}=\sqrt{\frac{1.82 \times 10^{-5}}{0.05}}\) = 1.908 × 10^-2
In the presence of HCl, the dissociation of acetic acid will be suppressed due to common ion effect.

i) When the solution contains 0.01 M of HCl also : Suppose the concentration of CH₃COOH decreases by x.

K_a = \(\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}\) ⇒ 1.80 × 10^-5 = \(\frac{x(x+0.01)}{(0.05-x)}=\frac{0.01 x}{0.05}\)
(∵ Since, x is very small, x + 0.01 = 0.01 and 0.05 – x = 0.05)
⇒ x = \(\frac{1.82 \times 10^{-5} \times 0.05}{0.01}\) = 9.1 × 10^-5
Degree of dissociation (α) = \(\frac{\text { Amount dissociated }}{\text { Initial amount }}=\frac{9.1 \times 10^{-5}}{0.05}\) = 1.85 × 10^-5

ii) When the solution contains 0.1 M of HCl:
Proceeding as above, we have, α’ = \(\frac{1.82 \times 10^{-5} \times 0.05}{0.1}\) = 9.1 × 10^-4 and
α’ = \(\frac{9.1 \times 10^{-6}}{0.05}\) = 1.82 × 10^-4

Question 54.
The ionization constant of dimethylamine is 5.4 × 10^-4. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.11 in NaOH?
Answer:
Dimethylamine is weak base. For a weak base, α = \(\sqrt{\frac{K_b}{c}}\)
Therefore, for dimethylamine, α = \(\sqrt{\frac{5.4 \times 10^{-4}}{0.02}}\) = 0.164
In presence of 0.1 M of NaOH, its dissociation will get suppressed.

∴ K_a = \(\frac{x(x+0.1)}{0.02-x}=\frac{0.1 x}{0.02}\) (∵ x is very small, x + 0.1 ≈ 0.1 and 0.02 – x ≈ 0.02)
⇒ 5.4 × 10^-4 = \(\frac{0.1 x}{0.02}\) ⇒ x = \(\frac{5.4 \times 10^{-4} \times 0.02}{0.1}\) = 1.08 × 10^-4
∴ Percent ionisation of dimethylamine = \(\frac{\text { Amount dissociated }}{\text { Total amount }}\) × 100
= \(\frac{x \times 100}{0.02}=\frac{1.08 \times 10^{-4}}{0.02}\) × 100 = 0.57%

Question 55.
Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below.
i) Human muscle-fluid, 6.83
iii) Human blood, 7.38
ii) Human stomach fluid, 1.2
iv) Human saliva, 6.4
Answer:
Given, pH = -log₁₀[H₃O⁺]
[H₃O⁺] = antilog₁₀(-pH)
i) [H₃O⁺] in human muscle fluid = antilog₁₀(-6.83) = 1.47 × 10^-7 mol L^-1
ii) [H₃O⁺] in stomach fluid = antilog₁₀(-1.2) = 6.3 × 10^-2 mol L^-1
iii) [H₃O⁺] in human blood = antilog₁₀(-7.38) = 4.1 × 10^-8 mol L^-1
iv) [H₃O⁺] in human saliva = antilog₁₀(-6.4) = 3.98 × 10^-7 mol L^-1

Question 56.
The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate the corresponding hydrogen ion concentration in each.
Answer:
[H⁺] = [H₃O⁺]
= = antilog₁₀(-pH) = antilog₁₀(-6.8) = 1.58 × 10^-7 mol L^-1
[H₃O⁺] in milk = antilog₁₀(-5.0) = 1.0 × 10^-5 mol L^-1
[H₃O⁺] in black coffee = antilog₁₀(-4.2) =1.0 × 10^-5 mol L^-1
[H₃O⁺] in tomato juice = antilog₁₀(-2.2) = 6.31 × 10^-5 mol L^-1
[H₃O⁺] in lemon juice = antilog₁₀(-7.8) = 6.31 × 10^-3 mol L^-1
[H₃O⁺] in egg white = antilog₁₀(-7.8) = 1.58 × 10^-8 mol L^-1

Question 57.
If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH ?
Answer:
Molarity of a solution is given by w = \(\frac{\mathrm{M} \times \mathrm{M}^{\prime} \times \mathrm{V}}{1000}\)
∴ 0.561 = \(\frac{\mathrm{M} \times 56 \times 200}{1000}\) (mol. mass of KOH = 56)
⇒ M = \(\frac{0.561 \times 1000}{56 \times 200}\) = 0.05 M
Since, KOH dissociates completely in the solution, KOH → K⁺ + OH^–
∴ [K⁺] = [OH^–] = Molarity of KOH = 0.05 M
and [H⁺] = \(\frac{K_w}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{0.05}\) = 2.0 × 10^-13 M
Now, pH = -log₁₀[H⁺] = -log₁₀(2 × 10^-13) = -log₁₀2 + 13 log 10 = -0.3010 + 13 = 12.70

Question 58.
The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Answer:
Molecular mass of Sr(OH)₂ = 87.6 + 2 × (16 + 1) = 121.6
Solubility of Sr(OH)₂ = 19.23 g/L = \(\frac{19.23}{121.6}\) mol/L^-1 = 0.1581 mol L^-1
Sr(OH)₂ ⇌ Sr²⁺ + 2OH^–
If Sr(OH)₂ is assumed to be completely ionised, we have
[Sr²⁺] = 0.1581 mol L^-1
and [OH^–] = 2 × 0.1581 = 0.3162 mol L^-1
∴ [H₃O⁺] = \(\) = 3.16 × 10^-14
and pH = -log₁₀[H₃O⁺] = -log₁₀(3.16 × 10^-14) = 13.50

Question 59.
The ionization constant of propanoic acid is 1.32 × 10^-5. Calculate the degree of ionization of the acid in its 0.05 M solution and also its pH. What will be its degree of ionization if the solution is 0.01 M in HCl also ?
Answer:

∴ K_a = \(\frac{\left[\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}\right]}\)
or 1.32 × 10^-5 = \(\frac{0.05 \alpha \times 0.05 \alpha}{0.05(1-\alpha)} \approx \frac{(0.05 \alpha)^2}{0.05}=\frac{0.05 \alpha^2}{1}\)
α = 1.63 × 10^-2
Then, [H₃O⁺] = 0.05α = 0.05 × 1.63 × 10^-2 = 8.15 × 10^-4 M
∴ pH = 3.09
In the presence of 0.1 M of HCl, let ‘α’ be the degree of ionization.
Then, [H₃O⁺] = 0.01; [A^–] = 0.05 α’ ; [H_A] = 0.05
K_a = \(\frac{0.01 \times 0.05 \alpha^{\prime}}{0.05}\) ⇒ 1.32 × 10^-5 = 0.01 α’
⇒ α’ = 1.32 × 10^-3

Question 60.
The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Answer:
Suppose the degree of ionisation of HCNO is α.

Given that pH of the solution is 2.34.
∴ -log₁₀(0.1 × α) = 2.34
⇒ log₁₀(0.1 × α) = -2.34 ⇒ 0.1 × α = antilog₁₀(-2.34) = 0.00457 = \(\frac{0.00457}{0.1}\) = 0.0457
K_a = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{CNO}^{-}\right]}{[\mathrm{HCNO}]}=\frac{(0.1 \times \alpha) \times(0.1 \times \alpha)}{0.1(1-\alpha)}\)
= \(\frac{(0.1 \times \alpha)^2}{0.1}=\frac{(0.1 \times 0.0457)^2}{0.1}\) = 2.09 × 10^-4

Question 61.
The ionization constant of nitrous acid is 4.5 × 10^-4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
Answer:
Sodium nitrite is a salt of a strong base and a weak acid. For such a salt,
pH = \(\frac{1}{2}\)pK_w + \(\frac{1}{2}\)pK_a + \(\frac{1}{2}\)log₁₀c
= \(\frac{1}{2}\) × (-log₁₀1.0 × 10^-14) + \(\frac{1}{2}\) × (-log₁₀4.5 × 10^-4) + \(\frac{1}{2}\) × log₁₀0.04
= 7.0 + 1.673 – 0.698 = 7.975
The degree of hydrolysis of such a salt is given by
h = \(\sqrt{\frac{K_w}{K_a \cdot c}}=\sqrt{\frac{1.0 \times 10^{-14}}{4.5 \times 10^{-4} \times 0.04}}\) = 2,36 × 10^-5

Question 62.
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Answer:
Pyridinium hydrochloride is the salt of a weak base (pyridine) and a strong acid (HCl). For the salt of a weak base and strong acid.
pH = \(\frac{1}{2}\)pK_w + \(\frac{1}{2}\)pK_a + \(\frac{1}{2}\)log₁₀c
Substituting the values, we have
3.44 = [-\(\frac{1}{2}\)log₁₀1.0 × 10^-14 – \(\frac{1}{2}\) × (- log₁₀ K_b) – \(\frac{1}{2}\) × log₁₀ 0.02
⇒ 3.44 = – \(\frac{1}{2}\) × (-14) + \(\frac{1}{2}\)log₁₀ K_b – \(\frac{1}{2}\) × (-1.699)
⇒ 3.44 = 7 + \(\frac{1}{2}\)log₁₀K_b + 0.849
log₁₀K_b = (3.44 – 7 – 0.849) × 2 = -8.82
⇒ K_b = anti log₁₀(-8.82) = 1.5 × 10^-9

Question 63.
Predict if the solutions of the following salts are neutral, acidic or basic.
NaCl, KBr, NaCN, NH₄NO₃, NaNO₂ and KF
Answer:
Neutral solutions : NaCl, KBr.
Acidic solutions : NH₄NO₃.
Basic solutions : NaCN, NaNO₂, KF.

Question 64.
The ionization constant of chloroacetic acid is 1.35 × 10^-3. What will be the pH of 0.1 M acid and its 0.1 M of sodium salt solution ?
Answer:
pH of 0.1 M acid :
Suppose the degree of ionisation of chloroacetic acid is α.

∴ K_a = \(\frac{\left[\mathrm{Cl} \mathrm{CH}_2 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{ClCH}{ }_2 \mathrm{COOH}\right]}\)
⇒ 1.35 × 10^-3 = \(\frac{0.1 \alpha \times 0.1 \alpha}{0.1(1-\alpha)}\) ≈ \(\frac{(0.1 \alpha)^2}{0.1}\) ⇒ α = \(\left(\frac{1.35 \times 10^{-3} \times 0.1}{(0.1)^2}\right)^{1 / 2}\) = 0.116
∴ [H⁺] = 0.1 × α = 0.1 × 0.116 = 0.0116
and pH = -log₁₀[H⁺] = -log₁₀(0.0116) = 1.94
pH of 0.1 M of sodium salt:
The sodium salt of chloroacetic acid is the salt of a weak acid and a strong base.
For such as salt
pH = \(\frac{1}{2}\)pK_w + \(\frac{1}{2}\)pK_a + \(\frac{1}{2}\)log₁₀c
= \(\frac{1}{2}\)(-log₁₀1.0 × 10^-14) + \(\frac{1}{2}\)(-log₁₀1.35 × 10^-3) + \(\frac{1}{2}\) × log₁₀0.1
= 7.0 + 1.435 + (-0.5) = 7.94

Question 65.
Ionic product of water at 310 K is 2.7 × 10^-14. What is the pH of neutral water and temperature?
Answer:
In neutral water, [H₃O⁺] = [OH^–]
K_w = [H₃O⁺] [OH^–] = [H₃O⁺]²
⇒ [H₃O⁺]² = K_w = 2.7 × 10^-14
[H₃O⁺] = (2.7 × 10^-14)12 = 1.64 × 10^-7
Hence, pH = -log₁₀[H₃O⁺] = -log₁₀(1.64 × 10^-7) = 6.78

Question 66.
Calculate the pH of the resultant mixtures.
i) 10 mL Of 0.2 M of Ca(OH)₂ + 25 mL of 0.1 M of HCl
ii) 10 mL Of 0.01 M of H₂SO₄ + 10 mL of 0.01 M of Ca(OH)₂
iii) 10 mL of 0.1 M of H₂SO₄ + 10 mL of 0.1 M of KOH
Answer:
i) Normality of Ca(OH)₂ solution = 0.2 × 2 = 0.4 N
and normality of HCl solution = 0.1 × 1 = 0.1 N
Gram equivalents of is taken as = \(\frac{\mathrm{N} \times \mathrm{V}}{1000}=\frac{0.4 \times 10}{1000}\) = 0.004
Gram equivalents of HCl is taken as = \(\frac{N \times V}{1000}=\frac{0.1 \times 25}{1000}\) = 0.0025
0.0025 gram eq. of HCl would neutralise 0.0025 gram eq. of Ca(OH)₂.
∴ Amount of left unreacted = 0.004 – 0.0025 = 0.0015 gram equivalents
Total volume of the mixture = 10 + 25 = 35 mL
Suppose the normality of the resultant solution is N’.
No.of gram eq. = \(\frac{\mathrm{N}^{\prime} \times \mathrm{V}}{1000}\)
⇒ 0.0015 = \(\frac{\mathrm{N}^{\prime} \times 35}{1000}\)
⇒ N’ = \(\frac{0.0015 \times 1000}{35}\) = 0.0429
∴ [OH^–] = N’ = 0.042
∴ [H₃O⁺] = \(\frac{\mathrm{K}_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{0.0429}\) = 2.331 × 10^-13
pH = -log₁₀[H₃O⁺] = -log₁₀2.331 × 10^-13 = 12.63

ii) 10 mL of 0.01 M of H₂SO₄ would exactly neutralise 10 mL of 0.01 M Ca(OH)₂ and resultant solution would be neutral.
Hence, pH of the resultant mixture = 7.00

iii) Normality of H₂SO₄ = 0.1 × 2 = 0.2 N and Normality of KOH = 0.1 × 1 = 0.1 N
Gram equivalents of H₂SO₄ = \(\frac{\mathrm{N} \times \mathrm{V}}{1000}=\frac{0.2 \times 10}{1000}\) = 0.002
Gram equivalents of NaOH = \(\frac{\mathrm{N} \times \mathrm{V}}{1000}=\frac{0.1 \times 10}{1000}\) = 0.001
0.001 gram equation of NaOH would neutralise 0.001 gram equation of H₂SO₄.
Total volume of the mixture = 10 + 10 = 20 mL. If the normality of the resultant solution is N’.
0.001 = \(\frac{\mathrm{N}^{\prime} \times 20}{1000}\) ⇒ \(\frac{0.001 \times 1000}{20}\) = 0.05
∴ [H₃O⁺] = N’ = 0.05
pH = -log₁₀[H₃O⁺] = -log₁₀0.05 = 1.30

Question 67.
Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility product constants given. Determine also the molarities of individual ions.
Given, K_sp Ag₂CrO₄ = 1.1 × 10^-12, K_sp BaCrO₄ = 1.2 × 10^-10
K_sp = 1.0 × 10^-38, K_sp PbCl₂ = 1.6 × 10^-5,
K_sp Hg₂I₂ = 4.5 × 10^-29
Answer:
i) For silver chromate (Ag₂CrO₄) :
1
s = \(\) = \(\) = 2.24 × 10^-10 M
and [Hg₂²⁺] = s = 2.2 × 10^-10 M and [I^–] = 2s = 2.24 × 10^-10 × 2 = 4.48 × 10^-10 M

Question 68.
The solubility product constant of Ag₂CrO₄ and AgBr are 1.1 × 10^-2 and 5.0 × 10^-15 respectively. Calculate the ratio of the molarities of their saturated solutions.
Answer:
For Ag₂CrO₄ (a ternary salt), s = \(\left(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\right)^{1 / 3}\) = \(\left(\frac{1.1 \times 10^{-12}}{4}\right)^{1 / 3}\) = 6.5 × 10^-5 M
For AgBr (a binary salt), s’ = \(\sqrt{\mathrm{K}_{\mathrm{sp}}}\) = \(\sqrt{5.0 \times 10^{-13}}\) = 7.1 × 10^-7 M
Ratio of molarities of their saturated solutions = \(\frac{\mathrm{s}}{\mathrm{~s}^{\prime}}\) = \(\frac{6.5 \times 10^{-5}}{7.1 \times 10^{-7}}\) = 91.5

Question 69.
Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate K_sp = 7.4 × 10^-8).
Answer:
[I\(\mathrm{O}_3^{-}\) = 0.002 M and [Cu²⁺] = 0.002 M
Cu²⁺ + 2I\(\mathrm{O}_3^{-}\) → Cu(IO₃)₂
K,sub.sp Cu(IO₃)₂ = [Cu²⁺][I\(\mathrm{O}_3^{-}\)]² = 7.4 × 10^-8 (given)
On mixing equal volumes, the concentration gets halved.
[Cu²⁺][I\(\mathrm{O}_3^{-}\)]₂ = \(\left(\frac{0.002}{2}\right)\) × \(\left(\frac{0.002}{2}\right)^2\) = 1.0 × 10^-9
Since, the ionic product is less than the solubility product, precipitation will not take place.

Question 70.
The ionization constant of benzoic acid is 6.46 × 10^-5 and K_sp for silver benzoate is 2.5 × 10^-13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared
to its solubility in pure water?
Answer:
C₆H₅COOAg \(\text { ⇌ }\) C₆H₅COO^– + Ag⁺
Solubility in water : s = \(\sqrt{\mathrm{K}_{\mathrm{sp}}}\) = \(\sqrt{2.5 \times 10^{-13}}\) = 5.0 × 10^-7 mol L^-7
Solubility in a buffer of pH = 3.19:
In a buffer of pH = 3.19, -log₁₀[H⁺] = 3.19
∴ [H⁺] = antilog₁₀(-3.19) = 6.457 × 10^-4 mol L^-1.
The H⁺ ions present in buffer solution will combine with C₆H₅COO^– ions to form C₆H₅COOH but (H⁺) in the solution will remain unchanged because the solution is a buffer solution.
C₆H₅COOH \(\text { ⇌ }\) C₆H₅COO^– + H⁺
∴ K_a = \(\frac{\left[\mathrm{C}6 \mathrm{H}_5 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right]}\) ⇒ \(\frac{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\right]}{\left[\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}\right]}\) = \(\frac{\left[\mathrm{H}^{+}\right]}{\mathrm{K}{\mathrm{a}}}\) = \(\frac{6.457 \times 10^{-4}}{6.46 \times 10^{-5}}\) = 9.995 ….. (i)
Suppose the solubility of silver benzoate in the buffer is s’ mol L^-1.
Since, almost all the C₆H₅COO^– ions get converted to benzoic acid which remains almost unionised, we have
s’ = [Ag⁺] = [C₆H₅COO-] + [C₆H₅COOH]
= [C₆H₅COO^–] + 9.995 × [C₆H₅COO^–] (From eq. (i))
∴ [C₆H₅COO-] = \(\frac{\mathrm{s}^{\prime}}{10.995}\)
K_sp = [C₆H₅COO^–][Ag⁺]
⇒ 2.5 × 10^-13 = \(\frac{\mathrm{s}^{\prime}}{10.995}\) × s’ = s^‘2 = 2.5 × 10^-13 × 10.995
⇒ s’ = \(\sqrt{2.5 \times 10^{-13} \times 10.995}\) = 1.657 × 10^-6 mol L^-1
∴ \(\frac{s^{\prime}}{s}\) = \(\frac{1.657 \times 10^{-6}}{5.0 \times 10^{-7}}\) = 3.314

Question 71.
What is the maximum concentration of equlmolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For an iron sulphide, K_sp = 6.3 × 10^-18).
Answer:
Suppose the required concentration is x M. On mixing equal volumes, concentration gets halved. Therefore, [Fe²⁺] = \(\frac{\mathbf{x}}{2}\) and [S^2-] = \(\frac{\mathbf{x}}{2}\)
For the state of no precipitation, [Fe²⁺] [S^2-] = K_sp
⇒ \(\frac{x}{2} \cdot \frac{x}{2}[latex] 6.3 × 10^-18 ⇒ x = (6.3 × 10^-18 × 4)^1/2 = 5.02 × 10^-9 mol L^-1

Question 72.
What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, K_sp is 9.1 × 10^-8). V
Answer:
For a binary salt, s = [latex]\sqrt{K_{s p}}\)
∴ For CaSO₄, S = 1\(\sqrt{9.1 \times 10^{-6}}\) = 3.0 × 10^-3 mol L^-1
Solubility of CaSO₄ in g/L = 3.0 × 10^-3 × 136 = 0.41 gL^-1
(Mol. mass of CaSO₄ = 40 + 32 + 64 = 136 g/mol)
Thus, a saturated solution of CaSO₄ contains 0.41 g of the salt in 1 L.
Hence, the volume of water required to dissolve 1 g of CaSO₄ = \(\frac{1}{0.41}\) = 2.44 L

Question 73.
The concentration of suiphide ion in 0.1 M HCl solution saturated with hydrogen sulphide is 1.0 × 10^-19M. If 10 ml. of this is added to 5 mL of 0.04 M solution of the following FeSO₄, MnCl₂, ZnCl₂ and CdCl₂. In which of these solutions precipitation will take place?
Answer:
Precipitation takes only when the ionic product exceeds the solubility product.
Since 10 mL of the solution containing S^2- ion is mixed with 5 mL of the different metal salt solutions, volume of the mixture after mixing in each case = 10 + 5 = 15 mL.
∴ [S^2-] = 1.0 × 10^-19 × \(\frac{10}{15}\) = 6.67 × 10^-20M
and M²⁺ = [Fe²⁺] = [Mn²⁺] = [Zn²⁺] = [Cd²⁺] = 0.04 × \(\frac{5}{15}\) = 1.33 × 10^-2 M
Therefore, in each case
Ionic product = [M²⁺][S²⁺] = (1.33 × 10^-2) × (6.67 × 10^-20) = 8.87 × 10^-22
This ionic product is greater than the solubility product of ZnS and CdS. Hence, precipitation will take place only in ZnCl₂ and CdCl₂ solutions.

Inter 1st Year English Dialogue Writing Skills with Answers

Conversation forms the basis of communication skills. Regular practice is required to equip oneself with this essential skill. Greetings, courtesies, polite way of agreement or otherwise, etc. form an important component of conversation. Learning to use them properly is, therefore, of great help.

In the Public Examination, an incomplete dialogue usually with four blanks is given. Four appropriate words or phrases are also provided. The student is required to understand the given part and respond to fill in the blanks suitably. Careful observation of the part given before and after the blank and a little bit of common sense will help the student answer this question correctly. There may be more than one possible way of filling in the given blank. One should not worry about this as long as the completed dialogue sounds meaningful. Study the following dialogues.

Examples

1. Read the dialogue given below regarding a complaint over a phone call and All the blanks with the suitable dialogues given below.

Making a Complaint – a faulty Television Set

Shopkeeper : ___________ (1) ___________
Customer : Yes please! I have a complaint regarding with television set.
Shopkeeper : ___________ (2)___________
Customer : Oh, it s SM004 – NEW. I bought this television here about three months ago, but the sound and picture quality are not so good.
Shopkeeper : Can you explain in more detail ?
Customer : ___________ (3) ___________ there’s a dark line down the left – hand side of the screen.
Shopkeeper : Do you have an outdoor antenna ?
Customer : Yes, I do.
Shopkeeper : ___________ (4) ___________
Customer : Several times.
Shopkeeper : ___________ (5) ___________
Customer : I am not so happy with the television set. I want a refund.
Shopkeeper : I’m afraid it isn’t our policy to give refunds, sir.
Customer : ___________ (6) ___________

Choose the correct dialogue from the following to make the above conversation meaningful.

a) Well then, I want to talk to the manager.
b) I’ll get our engineers to have a look at it.
c) Have you tried adjusting the antenna ?
d) Good afternoon, can 1 halp you ?
e) May I know the model number of it ?
f) Oh, yes, of course, the picture is always flickering and there’s an annoying hissing sound in the background.
Answers:
1) Good afternoon, can I help you ?
2) May I know the model number of it ?
3) Oh, yes, of course, the picture is always flickering and there is an annoying hissing sound in the background.
4) Have you tried adjusting the antenna ?
5) I’ll get our engineers to have a look at it.
6) Well then, I want to talk to the manager.

2. Choose the appropriate dialogue from the given options to make the conversation below meaningful.
(Sita dropped a glass on the floor. Naturally, her mother asks for an explanation.)

Mother : What was that terrible noise ? ___________ (1) ___________ anything ?
Sita : I’m sorry, mom. ___________ (2) ___________ and it broke.
Mother : Were you hurt ?
Sita : No, but I’m really sorry.
Mother : I’m glad that you aren’t hurt ! It was ___________ (3) ___________ on your 1st birthday!
Sita : I feel bad for breaking the glass.
Mother : Haven’t I told you to handle glassware carefully ?
Sita : The glass was too slippery.
Mother : It’s okay, but you’ve ___________ (4) ___________.
Sita : Sure, Mom. ___________ (5) ___________ again.

a) I dropped a glass on the floor.
b) a gift from your aunt Meena
c) got to be cautious next time!
d) I’ll not let it happen
e) Did you drop
Answers:
a) Did you drop
b) I dropped a glass on the floor
c) a gift from your aunt
d) Meena got to be cautious next time!
e) I’ll not let it happen

Exercises For Practice

I. what happened, mind, see, slowly recuperating

Mahesh : Hi, Suresh. Why didn’t you turn up to college yesterday?
Suresh : Hello, Mahesh. I wasn’t feeling well yesterday.
Mahesh : ___________ (1) ___________ ?
Suresh : I had a severe stomach ache.
Mahesh : How are you now? Are you okay ?
Suresh : Yes, I’m ___________ (2) ___________ now.
Mahesh : Did you ___________ (3) ___________ the doctor ?
Suresh : Yes, I have seen the doctor. He told me to avoid fast food and have my meals on time.
Mahesh : I’ve seen you skip meals for a plate of pani poori many times.
Suresh : I love pani poori. But now, I’ve made up my mind not to have them.
Mahesh : Thank God. That must have been a terrible pain. Good that you changed your ___________ (4) ___________.
Answers:
1. What happened?
2. slowly recuperating
3. see
4. mind

II. sure, hungry, else you need, replace

Victor : Excuse me, I ordered pasta, but this is noodles.
Waiter : I’m terribly sorry, sir. Let me ___________ (1) ___________ this immediately.
Victor : Thank you. Please ensure it’s vegetarian, as I requested earlier.
Waiter : ___________ (2) ___________, sir. I’ll double-check your order and bring it back soon.
Victor : I hope it doesn’t take too long. I’m quite ___________ (3) ___________.
Waiter : It won’t take long, sir. Thank you for your patience.
Victor : I appreciate your quick response.
Waiter : You’re welcome. Please let me know if there’s anything ___________ (4) ___________.
Answers:
1. replace
2. Sure
3. hungry
4. else you need

III. resolved, sure, hours, phone number

Rahul : Excuse me, my internet hasn’t been working for two days now.
Support Agent : I’m sorry to hear that, sir. Could you please provide your ___________(1)___________?
Rahul : Sure, it’s 9876543210. Can you fix this by today ?
Support Agent : I’ll escalate the issue right away. You should have it resolved within a few ___________ (2) ___________.
Rahul : Thank you. I really need it for an important online meeting.
Support Agent : I understand, sir. We’ll prioritize your request.
Rahul : Will you notify me once the issue is resolved ?
Support Agent : ___________ (3) ___________, sir. You’ll receive a confirmation message once it’s ___________ (4) ___________.
Answers:
1. phone number
2. hours
3. Sure
4. resolved

Exercises

Read each set of dialogues below and complete the blanks with appropriate words or phrases from the word bank provided.

I. expensive / iPhones / budget phones / tough

Ramu : I’m thinking of buying a mobile phone. I really like the iPhone.
Charan : ___________ (1) ___________ are amazing, no doubt!
Ramu : Many of my school friends have bought one.
Charan : Yeah, some of my friends have it too. But………… do you have the money ? It’s pretty expensive.
Ramu : It is ___________ (2) ___________. I know. That’s why I’m planning to ask my mom.
Charan : You know, the price of an iPhone is like two months of your mom’s salary. It might be tough for her.
Ramu : Hmm, you’re right. It is ___________ (3) ___________ for her. I didn’t think of her difficulties. Now, I think a budget phone will be enough for me. I’ll get a budget phone.
Charan : Exactly! These days, ___________ (4) ___________ have almost all the features we need. It’s better not to trouble our parents unnecessarily, especially just to follow what others are doing.
Answers:
1. iPhones
2. expensive
3. tough
4. budget phones

II. homemade food / for both of us / I’m hungry / let’s go

Sowmya : I’m famished. Shall we go out to eat something?
Navya : ___________ (1) ___________ too, but I want homemade food.
Sowmya : Homemade food isn’t tasty. At least I want something spicy and tasty.
Navya : Sure, but ___________ (2) ___________ is healthy. And we can cook our
Sowmya : But we have neither the energy nor the patience, not even the skill to cook.
Navya : You are correct.
Sowmya : I will pay for both of us. Shall we go out?
Navya : Sure, but you will pay ___________ (3) ___________
Sowmya : Let’s go out!
Navya : ___________ (4) ___________
Answers:
1. I’m hungry
2. homemade food
3. for both of us
4. Let’s go.

III. having fun / photos / social media / memories

Tinnu : Why are you always clicking ___________ (1) ___________ ? We are here to enjoy the trip, not to take photos.
Bannu : But I want to make memories. Lots of memories.
Tinnu : ___________ (2) ___________ of taking photos? If you don’t really enjoy, you can’t make real memories. First, let’s have some fun.
Bannu : Fun ? I’m all in our ___________ (3) ___________. But don’t you think others need to know that we are having fun ?
Tinnu : We need to share our happiness, not our photos. By sharing our photos, we are just making others jealous.
Bannu : I think we are just inspiring them. It depends on how you look at it. I don’t want to blame social media.
Tinnu : Even I won’t blame ___________ (4) __________. But social media just increases our fears. So please keep your camera away.
Bannu : Sure. Come here, one more selfie.
Answers:
1. photos
2. Memories
3. having fun
4. social media

IV. deep breath /nothing / do you / shout

Rahim : Hey, I didn’t understand what our teacher said. I have a doubt.
Raju : __________ (1) __________ ? Then ask him. He’ll explain.
Rahim : But, I’m afraid. What if he shouts?
Raju : Why will he __________ (2) __________ at you? He won’t do such things. He will explain. Trust me.
Rahim : But what if others laugh at me? What if they think I know nothing?
Raju : Knowing __________ (3) __________ is not a problem. Not learning is a problem. Even others will learn something through your question. Take a deep breath.
Rahim : Ok, I’ll ask the doubt myself. Let me take a __________ (4) __________.
Raju : All the best. I know you’ll rock!
Answers:
1. Do you
2. shout
3. nothing
4. deep breath

V. our parents / greedy / everything / exams

Gita : Why do you always study so much? You are too greedy.
Madhu : I’m not __________ (1) __________. I just want to learn as much as I can. I don’t want to leave anything to chance.
Gita : But that’s too much. If I get some 50%, that’s enough for me. If I get less than that, I have to face the anger of my parents.
Madhu : I don’t think we should read for __________(2) __________. I read it because I love it.
Gita : But those questions won’t be asked in the exam.
Madhu : We shouldn’t read for the sake of __________ (3) __________. Imagine you go to a doctor with a strange problem, and the doctor says that your problem wasn’t asked in her exam. How would you feel?
Gita : I don’t go to such doctors. I want my doctor to know everything.
Madhu : It’s good thinking. But isn’t it fair? You expect others to know __________ (4) __________ but you don’t learn everything? W’e all should do our best.
Answers:
1. Thank you
2. cold and cough
3. That’s very nice
4. 200 rupees

VI. the train to Delhi / you ‘re welcome / on time / excuse me

Doctor : Please take your seat.
Patient : __________ (1) __________
Doctor : Please tell me, how can I help you?
Patient : Sir, I have been suffering from fever for the last two days.
Doctor : Ok. Do you have a cold, cough, or anything else?
Patient : I have __________ (2) __________. Please give me medicine, doctor. I don’t want an injection.
Doctor : Don’t worry. I’m not giving you an injection. These medicines are enough.
Patient : __________ (3) __________ of you. How much is the fee?
Doctor : The fee is __________ (4) __________. Please pay it at the counter.
Answers:
1. Thank you
2. cold and cough
3. That’s very nice
4. 200

VII. the train to Delhi / you ‘re welcome / on time / excuse me

Customer : ___________, on which platform will the train to Delhi arrive?
Enquiry : __________ (2) __________ will arrive on platform number four.
Customer : When will it arrive? Is it on time?
Enquiry : The train will arrive in fifteen minutes, and yes, it is __________ (3) __________
Customer : Thank you for the information.
Enquiry : __________ (4) __________. Do you need any more information?
Customer : No, that’s all. Thank you for the help once again.
Answers:
1. Excuse me
2. The train to Delhi
3. on times
4. You’re welcome

VIII. thank you / good morning/hear that / studies

Principal : Good morning, Mr Anil. Welcome to the parent-teacher meeting.
Mr Anil : __________ (1) __________. Sir.
Principal : Please take a seat.
Mr Anil : __________ (2) __________. Sir. I am eager to know about my son’s progress.
Principal : Of course, your son is doing well in __________ (3) __________.
Mr Anil : It’s nice to __________ (4) __________. The whole credit goes to your staff.
Principal : It’s our duty to bring the best out of the students. I thank you for visiting.
Answers:
1. Good morning
2. Thank you
3. studies
4. hear that

IX. neglecting / just an exam / a movie / completing

Arjun : Hi! I am planning to go for a movie. Will you accompany me?
Karthik : __________ (1) __________! Are you even thinking? Tomorrow we have an exam.
Arjun : It’s __________ (2) __________ I’m very eager to see the first day first show.
Karthik : Bro, it’s not just an exam. You are __________ (3) __________ your studies. It’s not good on your part.
Karthik : Exams can’t be written the other day. But you can go to the movie after the exam. Even heroes don’t skip their schedules. movie after __________ (4) __________ the exam.
Karthik : That’s a good idea.
Answers:
1. A movie
2. just an exam
3. neglecting
4. completing

X. be completed / harm / microplastics / playing

Mom : Will you please stop playing games on your mobile phone?
Son : Mom, I’m not __________ (1) __________ games. I’m watching a video.
Mom : Whatever, it’s not good for your eyes to keep looking at the screen for hours together.
Son : Mom, I’ve got a project to __________ (2) __________ by this weekend. I’m watching a video related to that.
Mom : What’s the project about?
Son : It’s about microplastics.
Mom : __________ (3) __________. What is it?
Son : Just like plastic, there are very tiny particles that can cause harm to the Earth.
Mom : How do they cause __________ (4) __________ to the Earth?
Son : The usage of plastic has become a part of our life. When we dispose of any plastic object, it doesn’t degrade in the soil. Moreover, it pollutes our air, soil, and water too.
Mom : Interesting. Tell me more about microplastics once you complete your project.
Son : Sure, Mom.
Answers:
1. Playing
2. be completed
3. Microplastics
4. harm

XI. Read the following dialogue and fill the blanks with the words given below.

for you / thank you very much /your bill/ please wait

Sophia : Excuse me, I believe there’s been a msitake in my bill.
Manager : I’m sorry to hear that. Could you please show me __________ (1) __________ ?
Sophia : Here it is. I was charged for two desserts, but I only had one.
Manager : Let me check this for you. __________ (2) __________ while I verify it.
Sophia : Sure, I’ll wait. Thank you for looking into it.
Manager : You’re right. There was an error. I’ll fix it and issue a new bill right away.
Sophia : __________ (3) __________. Mistakes happen, but it’s great to see such prompt service.
Manager : Thank you for your understanding. Here’s your corrected bill. Please let us know if there’s anything else we can do __________ (4) __________.
Sophia : Thank you. Everything else was perfect. I’ll definitely visit again.
Answers:
1. your bill
2. Please wait
3. Thank you very much
4. for you

XII. Read the following dialogue and fill the blanks with the words given below.

have / hoped / advance / thank you very much

Elsa : Hi, Mia. I’m so sorry I couldn’t come to your party last night.
Mia : It’s okay, Elsa. But I really __________ (1) __________ you would be there.
Elsa : I know, and I feel terrible about it. Something urgent came up at home.
Mia : I understand. But next time, please let me know in __________ (2) __________.
Elsa : Absolutely. I’ll make it up to you with a coffee treat soon!
Mia : That sounds good ! But next time, you really __________ (3) __________ to make it.
Elsa : I promise. Your party pictures looked amazing!
Mia : __________(4) __________! It was a great evening. I hope to
see you at the next one.
Answers:
1. hoped
2. advance
3. have
4. Thank you very much

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

Find the coefficient of

Question 1.
x⁵ in (x + 3)⁸
Solution:
The general term T_r+1 of the binomial expansion (a + b)n is given by T_r+1 = ⁿC_r a^n-rb^r
Here x5 is the T_r+1 term so a = x, b = 3 and n = 8
T_r+1 = 8C_rx^8-r3^r ………..(1)
To find out the coefficient of x⁵, We have to equate x⁵ = x^8-r =>r = 3
Putting the value of r in (1), we get T₃₊₁ = ⁸C₃ × x^8-3 3³
T₄ = \(\frac{8!}{3!5!}\) × x⁵ × 27 = 1512 x⁵
Hence the coefficient of x⁵ = 1512

Question 2.
a⁵b⁷ in (a – 2b)¹².
Solution:
The general term Tr+ t of the binomial expansion (a + b)ⁿ is given by T_r+1 = ⁿC_r a^n-rb^r
Here a = a, b = -2b & n = 12
Substituting above values, we get T_r+1 = ¹²C_r a^12-r (-2b)^r … (1)
To find out the coefficient of a⁵, We equate a^12-r = a⁵, r = 7
Putting r = 7 in (1) ⇒ Tg = ¹²C₇ a⁵(-2b)⁷
T₈ = \(\frac{12!}{7!5!}\) × a⁵ × (-2)⁷b⁷
= -101376 a⁵b⁷
Hence, the coefficient of a⁵b⁷ = -101376.

Write the general term in the expansion of

Question 3.
(x³ – y)⁶
Solution:
The general term T_r+1 of the binomial expansion (a + b)n is given by T_r+1 = nCra”-b’… (1)
Here, a = x², n = 6 and b = – y
Putting above values in (1)
T_r+1 = ⁶C_r x^2(6-r) (-1)^r y^r
= ⁶C_r x^12-2r x (-1)^r X y^r
= (-1)^{r 6}C_r x^12-2r y^r

Question 4.
(x² – yx)¹², x ≠ 0.
Solution:
The general term T_r+x of the binomial expansion (a + b)n is given by T_r+1 = nC_r a^n-r b^r
Here n = 12, a = x² and b = -y x
Substituting the above values, we get
T_r+1 = ¹²C_r x^2(12-r) (-1)^r y^rx^r = (-1)r ¹²C_r x^24-2r y^r x^r
= (-1)^{r 12}C_r x^24-r. y^r

Question 5.
In the expansion of (1 + a)^m+n, prove that the coefficients of am and an are equal.
Solution:
We know that the general term T_r+1 of the binomial expansion (a + b)ⁿ is given by
T_r+1 = ⁿC_r a^n-r b^r
Here n = m + n, a = 1 and b = a
Substituting the values in the general term
T_r+1 = ^m+nC. 1^m+n-ra^r = ^m+nC_r a^r ………..(1)
Now, we have the general term for the expression is, T_r+1 = ^m+nC_r a^r
Now, for coefficient of a^m, T_m+1 = ^m+nC_m a^m
Hence, for the coefficient of m, the value of r = m
So, the coefficient is ^m+nC_m
Similarly, the coefficient of an is ^m+nC_n
^m+nC_m = \(\frac{(\mathrm{m}+\mathrm{n})!}{\mathrm{m}!\mathrm{n}!}\) And also, ^m+nC_n = \(\frac{(\mathrm{m}+\mathrm{n})!}{\mathrm{m}!\mathrm{n}!}\)
The coefficient of am and an are same that is \(\frac{(\mathrm{m}+\mathrm{n})!}{\mathrm{m}!\mathrm{n}!}\)

Question 6.
The coefficients of the (r – 2)^th, r^th and (r + 2)^th terms in the expansion of (x + 1)ⁿ are in the ratio 1:3:5. Find n and r.
Solution:
The general term T_r+1 of the binomial expansion (a + b)n is given by T_r+1 = ⁿC_r a^n-r b^r
Here, the binomial is (1 + x)ⁿ where a = 1, b = x and n = n

The (r + 1)^th term is given by
T_r+1 = ⁿC_r 1^n-r x_r ⇒ T_{(r + 1)} = ⁿC_r x^r
The coefficient of (r + 1)^th term is ⁿC_r

The rth term is given by (r – 1)^th term
T_{(r – 1 – 1)} = ⁿC_r-1
⇒ T_r = ⁿC_{r – 1}x^r-1
∴ The coefficeint of (r – 1)^th term is ⁿC_{r – 1}

For (r – 1)^th term, we will take (r – 2)^th term
T_{r – 2 + 1} = ⁿC_r-2 x^r-2
T_r-1 = ⁿC_r-2x^r-2
∴ The coefficeint of (r – 1)^th term is ⁿC_{r – 2}

Given that the coefficient of (r – 1)^th, r^th and r + 1^th term are in ratio 1: 3 : 5
Therefore,
\(\frac{\text { Coefficient of }(\mathrm{r}-1)^{\text {th }} \text { term }}{\text { Coefficient of } \mathrm{r}^{\text {th }} \text { term }}=\frac{1}{3}\)

On rearrangement, we get
\(\frac{n!}{(r-2)!(n-r+2)!} \times \frac{(r-1)!(n-r+1)!}{n!}=\frac{1}{3}\)

By multiplying
⇒ \(\frac{(\mathrm{r}-1)(\mathrm{r}-2)!(\mathrm{n}-\mathrm{r}+1)!}{(\mathrm{r}-2)!(\mathrm{n}-\mathrm{r}+2)!}=\frac{1}{3} \Rightarrow \frac{(\mathrm{r}-1)(\mathrm{n}-\mathrm{r}+1)!}{(\mathrm{n}-\mathrm{r}+2)(\mathrm{n}-\mathrm{r}+1)!}=\frac{1}{3}\)

On simplifying we get,
⇒ \(\frac{(\mathrm{r}-1)}{(\mathrm{n}-\mathrm{r}+2)}=\frac{1}{3}\)
⇒ 3r – 3 = n – r + 2
⇒ n – 4r + 5 = 0 …. (1)

Also, \(\frac{\text { the coefficient of } r^{\text {th }} \text { term }}{\text { coefficient of } r+1^{\text {th }} \text { term }}=\frac{3}{5} \Rightarrow \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{3}{5}\)

On rearranging we get,
⇒ \(\frac{n!}{(r-1)!(n-r+1)!} \times \frac{r!(n-r)!}{n!}=\frac{3}{5}\)

By multiplying
⇒ \(\frac{r(r-1)!(n-r)!}{r-1)!(n-r+1)!}=\frac{3}{5} \Rightarrow \frac{r(n-r)!}{(n-r+1)!}=\frac{3}{5} \Rightarrow \frac{r(n-r)!}{(n-r+1)(n-r)!}=\frac{3}{5}\)

On simplifying we get,
⇒ \(\frac{r}{(n-r+1)}=\frac{3}{5}\)
⇒ 5r = 3n – 3r + 3
⇒ 8r – 3n – 3 = 0 … (2)
We have 1 and 2 as
n – 4r + 5 = 0 …(1)
8r – 3n – 3 = 0 …(2)
Multiplying equation (1) by number (2)
2n – 8r + 10 = 0 …(3)
Adding equations (2) and (3)
2n – 8r + 10 = 0
⇒ -3n – 8r – 3 = 0
⇒ -n = -7
⇒ n = 7 and r = 3

Question 7.
Prove that the coefficient of xn in the expansion of (1 + x)²ⁿ is twice the of xⁿ in the expansion of (1 + x)^2n-1.
Solution:
The general term T_r+1, of the binomial expansion (a + b)ⁿ is given by T_r+1 = ⁿC_r+1 a^n-rb^r
The general term for binomial (1 + x)²ⁿ is
T_(r+1) = ²ⁿC_rx^r …(1)
To find the coefficient of xⁿ, r = n
T_n+1 = ²ⁿC_nxⁿ
The coefficeint of xⁿ = ²ⁿC_n

The general term for binomial (1 + x)^2n-1 is
T_r+1 = ^2n-1C_r x^r

To find the co-efficient of xⁿ
Putting r = n
T_r+1 = ^2n-1C_r x^r
The coefficient of xⁿ = ^2n-1C_n
We have to prove
Coefficient of xⁿ in (1 + x)²ⁿ = 2 coefficient of xⁿ in (1 + x)^2n-1
Consider LHS = ²ⁿC_n = \(\frac{2 n!}{n!(2 n-n)!}=\frac{2 n!}{n!(n)!}\)
Again consider RHS = 2 × ^2n-1C_n
= 2 × \(\frac{(2 n-1)!}{n!(2 n-1-n)!}=2 \times \frac{(2 n-1)!}{n!(n-1)!}\)
Now multiplying and dividing by n we get, 2 × \(\frac{(2 n-1)!}{n!(n-1)!} \times \frac{n}{n}=\frac{2 n(2 n-1)!}{n!n(n-1)!}=\frac{2 n!}{n!n!}\)
From above equation LHS = RHS.
Hence proved.

Question 8.
Find a positive value of m for which the coefficient of x² in the expansion (1 + x)^m is 6.
Solution:
The general term T_r+1 of the binomial expansion (a + b)ⁿ is given by T_r+1 = ⁿC_r a^n-r b^r
Here, a = 1, b = x and n = m
Putting above values, T_r+1 = ^mC_r 1^n-r x^r = ^mC_r x^r
We need the coefficient of x²
Putting r = 2, T₂₊₁ = ^mC₂ x²
The coefficient of x² = ^mC₂
Given that coefficient of x² = ^mC₂ = 6
⇒ \(\frac{\mathrm{m}!}{2!(\mathrm{m}-2)!}\) = 6
⇒ \(\frac{\mathrm{m}(\mathrm{~m}-1)(\mathrm{m}-2)!}{2 \times 1 \times(\mathrm{m}-2)!}\) = 6
⇒ m(m – 1) = 12
⇒ m² – m – 12 = 0
⇒ m² – 4m + 3m – 12 = 0
⇒ m(m – 4) + 3(m – 4) = 0
⇒ (m + 3) (m – 4) = 0
⇒ m = -3, 4
We need the positive value of m, so m = 4.

AP Inter 1st Year English Common Errors with Answers

(A) USAGE OF ARTICLES (A, An, The) :

Rule 1 : Don’t use the’ before material nouns used in general sense.
(పదార్ధాల పేర్ల ముందు ‘the‘ నుపయోగింపరాదు.)
e.g. : 1. The gold is a precious metal. (Wrong)
Gold is a precious metal. (Right)

Rule 2 : No article is used before the branches of knowledge.
(విద్యలేక జ్ఞాన సంబంధ వాక్యాలముందు ఏ విధమైన Article నుపయోగింపరాదు.)
2. The Geology is an interesting subject. (W)
Geology is an interesting subject. (R)

Rule 3 : No article is used before the names of days, months and seasons.
(దినముల, నెలల, ఋతువుల పేర్ల ముందు ఏ Article నుపయోగింపరాదు.)
e.g. : 3. Sunday is a holiday. (not the Sunday)
4. If winter comes, can spring be far behind. (not ‘the winter’, not the spring)
5. May is the hottest month. (not the May)

Rule 4 : Don’t use any article before nouns in certain phrases.
(కొన్ని phrases లోని నామవాచకాల ముందు ఏ విధమైన Article నుపయోగింపరాదు.)
e.g. : 6. by air, by sea, over hill and dale etc.

Rule 5 : Don’t use ‘the’ before the names of games.
(ఆటల పేర్ల ముందు, ‘the’ నుపయోగింపరాదు.)
e.g. : 7. I play hockey. (not ‘the hockey’ )
8. Chess is a fine game. (not ‘the chess’)

Rule 6 : No article is used before a Common Noun used in the vocative case.
(సంబోధనాత్మక సాధారణ నమాచకాల ముందు ఎట్టి Article నుపయోగింపరాదు.)
e.g. : 9. Friend, I have news to tell you. (not ‘the friend’)
10. Can you help me, brother ? (not ‘the brother’)

Rule 7 : Article is omitted when a noun indicates rank, titles, occupation etc.
(బిరుదు, వృత్తి, హూదాలను తెలిపేటపుడు Article నుపయోగింపరాదు.)
e.g. : 11. King Arthur. (not ‘the King Arthur’)
12. General Kaul. (not ‘the General Kaul’)

Rule 8 : No article is used before Collective Nouns like mankind, people, society, company etc.
(సామూహిక నామవాచకాల ముందు ఏ Article నుపయోగింపరాదు.)
e.g. : 13. We are members of society, (not ‘the society’)

Rule 9 : In certain phrases consisting of a preposition, no article is placed.
(కొన్ని phrases లో విభక్తి ప్రత్యయముల ముందు ఏ విధమైన Article నుపయోగింపరాదు.)
e.g. : 14. on demand (not ’on the demand’)
15. at home (not ‘at the home’ or ‘at a home’)

Rule 10 : Don’t put any article before names of meal when used in general sense.
(భోజన పదాలముందు ఏ Article నుపయోగింపరాదు.)
e.g.: 16. The lunch is at eleven o’ clock. (W)
Lunch is at eleven o’ clock. (R)

Rule 11 : ‘A’ is used before words beginning with a consonant sound.
e.g.: 17. Frank is an European. (W)
Frank is a European. (R)

18. Here is an one-rupee note. (W)
Here is a one-rupee note. (R)

Rule 12 : ‘An’ is used before words with a vowel sound.
e.g. : 19. Brutus is a honest man. (W)
Brutus is an honest man. (R)

Rule 13 : ‘The’ is used in the following cases :
e.g. : 20. Rajahmundry is on Godavari. (W)
Rajahmundry is on the Godavari. (R)

21. My sister lives in U.K. (W)
My sister lives in the U.K. (R)

22. The higher you climb colder you get. (W)
The higher you climb the colder you get. (R)

Rule 14 : No article is used before Proper Nouns.
(పేర్లకుముందు ఏ విధమైన Article నుపయోగింపరాదు.)
23. The Delhi is the capital of India. (W)
Delhi is the capital of India. (R)

(B) USAGE OF PREPOSITIONS:

Rule 15 : ‘Since’ denotes point of time. ‘For’ denotes period of time.
e.g. : 24. She has been typing since two hours. (W)
She has been typing for two hours. (R)

25. They have been living here for 1990. (W)
They have been living here since 1990. (R)

Rule 16 : ‘At’ is used for small places and villages. ‘In’ is used in the case of towns, cities and countries.
e.g. : 26. He was born in Pamarru. (W)
He was born at Pamarru. (R)

27. He was born at Sweden. (W)
He was born in Sweden. (R)

Rule 17 : ‘Beside’ means by ‘the side of nr ‘next to’.
‘Besides’ means ‘in addition to’:
e.g. : 28. His house is besides the temple. (W)
His house is beside the temple. (R)

29. Beside Rama, Krishna was also injured. (W)
Besides Rama, Krishna was also injured. (R)

(C) USAGE OF NOUNS :

Rule 18 : Some uncountable nouns are generally used in the Singular Number. They have no Plural form.
e.g. : information luggage news furniture
machinery poetry alphabet scenery
bedding clothing bread soap advice paper etc.

ఫై పదాలకు బహువచన రూపం ఉండదు. వాటిని ఎల్లప్పుడు ఏక వచనంగానే భావించాలి.వాటి ముందు a గాని an గాని, ఉపయోగింపరాదు.

furnitures (Wrong) ………………………… furniture (Right)
alphabets (Wrong) ………………….. alphabet (Right)

30. I know the alphabets. (W)
I know the alphabet. (R)

31. He gave me wrong advices. (W)
He gave me wrong advice. (R)

32. We heard a good news. (W)
We heard good news. (R)

Rule 19 : పైన పేర్కొనబడ్డ uncountable Nouns Subject గా వాడబడినప్పుడు క్రియాపదము ఏకవచనంలో ఉండాలి.
e.g. : 33. Where are my luggages? (W)
Where is my luggage? (R)

34. The news are true. (W)
The news is true. (R)

35. The news of the air crash are shocking. (W)
The news of the air crash is shocking. (R)

Rule 20 : Some nouns appear plural in form but they are used as singular.
(కొన్ని నామవాచకాలు బహువచనాలుగా కన్పించినా వాటిని ఏకవచనంగానే ప్రయోగించాలి.)
Branches of learning: Mathematics, Physics, Civics, Politics, Statistics, Mechanics etc.
Diseases : Measles, Mumps
Games : Billiards, Draughts, Gymnastics.
Other words : premises, innings, summons, wages, gallows etc.

e.g. : 36. Mathematics are his favourite subject. (W)
Mathematics is his favourite subject. (R)

37. Measles have broken out in the city. (W)
Measles has broken out in the city. (R)

38. Summons are served on him. (W)
Summons is served on him. (R)

39. Billiards are interesting. (W)
Billiards is interesting. (R)

40. The first innings are over. (W)
The first innings is over. (R)

Rule 21 : Some nouns are always used in plural.
(ఈ క్రింది నామవాచకాలను ఎల్లపుడు బహువచన రూపంలోనే వాడాలి)
Articles of dress : shoes, trousers etc.
Instruments: scissors, spectacles, bellows, tongs etc.
Other nouns : alms, ashes, annals, assets, thanks, proceeds, savings, spirits, riches, contents etc.
(ఈ పాదాలన్నింటిని బహువచనంలోనే వాడాలి. బహువచన క్రియనే ఉపయోగించాలి.)

e.g : 41. These scissors is very sharp. (W)
These scissors are very sharp. (R)

42. This is my spectacles. (W)
These are my spectacles. (R)

43. Father bought a bag of rices. (W)
Father bought a bag of rice. (R)

Exception: (మినహాయింపు)
A pair వాడబడినప్పుడు ఏకవచన క్రియనే వాడాలి.
e.g. : 44. A pair of shoes are all I want. (W)
A pair of shoes is all I want. (R)

Rule 22 : Plurals of compound words are formed by adding ‘s’ to the main word.
e.g. : 45. I have two brother-in-laws. (W)
I have two brothers-in-law. (R)

Rule 23 : The following words do not take plural when preceded by a numerical.
(ఈ క్రింది పదాల ముందు సంఖ్యారూపక పదమున్నపుడు ఆ పదాలను బహువచనంలో వ్రాయరాదు)
hundred, dozen, thousand, lakh, million etc.
e.g. : 46. I need five hundreds rupees. (W)
I need five hundred rupees. (R)

Rule 24 : The meaning of brother or sister is included in the word ’cousin’. So don’t add ‘brother’ or ‘sister’ after ‘cousin’.
(Cousin అను పదము తర్వాత brother గాని sister అను పదము గాని ప్రయోగింపరాదు)
e.g. : 47. Radha is my cousin sister. (W)
Radha is my cousin. (R)

Rule 25 : Running is not used with age.
(వయస్సు గురించి చెప్పేటప్పుడు running ప్రయోగించరాదు.)
e.g. : 48. Gopal is running sixteen. (W)
Gopal is sixteen. (R)

Rule 26 : The words ‘male’, ‘female’ are used as Adjectives but not as Nouns. ‘Men’ and ‘Women’ are used as Nouns.
(male, female అను పదాలను విశేషణాలుగా ప్రయోగిస్తారు. అవి నామవాచకాలు కావు. నామవాచకాలుగా male, female ప్రయోగించాలి.)

e.g. : 49. There are twenty females working here. (W)
There are twenty women working here. (R)

50. Five hundred males attended the meeting. (W)
Five hundred men attended the meeting. (R)

Rule 27 : When two nouns are in apposition ‘s’ is added to the second word.
e.g. : 51. It is my sister’s Kamala’s house. (W)
It is my sister Kamaia’s house. (R)

(D) USAGE OF PRONOUNS :

Rule 28 : To the word yours apostrophe is not added.
e.g. : 52. Your’s faithfully. (W)
Yours faithfully. (R)

53. This book is your’s. (W)
This book is yours. (R)

Rule 29 : The first person comes last. The second person is placed before the third.
e.g. : 54. I and Gopal went to the Zoo. (W)
Gopal and I went to the Zoo. (R)

55. Ravi invited me and you to dinner. (W)
Ravi invited you and me to dinner. (R)

Rule 30 : When a personal pronoun is used as the object of a verb, it is placed in the objective case.
e.g. : 56. I accused she of theft. (W)
I accused her of theft. (R)

57. How long did you wait for myself? (W)
How long did you wait for me? (R)

58. Besides he, many others voted for me. (W)
Besides him, many others voted for me. (R)

59. It is me that scored the goal. (W)
It is I that scored the goal. (R)

Rule 31 : If a pronoun stands for a collective noun singular verb is used. If it stands for separate Individuals plural verb is used.
(కొన్ని సామూహిక పద నామవాచకాలు గుంపును నిర్దేశిస్తే క్రియ ఏకవచనంలో ఉండాలి లేక ఆ నామవాచకం వేర్వేరు వ్యక్తులను నిర్దేశిస్తే క్రియ బహువచనంలో ఉండాలి.)
e.g. : 60. The fleet, after it had served, were withdrawn. (W)
The fleet, after it had served, was withdrawn. (R)

61. The committee was divided in their opinion. (W)
The committee were divided in their opinion. (R)

Rule 32 : The verbs ‘enjoy’, ‘avail’ take Reflexive pronouns.
e.g. : 62. He enjoyed during the holidays. (W)
He enjoyed himself during the holidays. (R)

63. We avail the opportunity. (W)
We avail ourselves of the opportunity. (R)
avail + …………. self + of + sth

Rule 33 : A relative pronoun agrees with its antecedent in gender, number and person.
e.g. : 64. It is I who is to blame. (W)
It is I who am to blame. (R)

65. He is one of the Directors who has served the company. (W)
He is one of the Directors who have served the company. (R)

Rule 34 : ‘Who’ is used for persons. ‘Which’ is used for animals and lifeless objects.
e.g. : 66. A triangle who has three sides equal is called an equilateral triangle. (W)
A triangle which has three sides equal is called an equilateral triangle. (R)

Rule 35 : Each, every, anybody, nobody, everybody, either, neither should always be treated as singular.
(Each, every, anybody, nobody మొదలగు పదాలను ఎల్లప్పుడూ ఏకవచన పదాలుగా పరిగణించాలి)
e.g. : 67. Each of these rooms are well ventilated. (W)
Each of these rooms is well ventilated. (R)

68. Neither John nor Henry are Indians. (W)
Neither John nor Henry is an Indian. (R)

69. Every boy are given a prize. (W)
Every boy is given a prize. (R)

70. Either Sekhar or his brother do the work. (W)
Either Sekhar or his brother does the work. (R)

71. None of these students are here. (W)
None of these students is here. (R)

Rule 36 : Use ‘each other’ for two, ‘one another’ for more than two.
e.g. : 72. The two girls hated one another. (W)
The two girls hated each other. (R)

73. The four thieves accused each other. (W)
The four thieves accused one another. (R)

(E) USAGE OF ADJECTIVES :

Rule 37 : Double Comparatives or double Superlatives are not used:
e.g. : 74. Sita is more prettier than Kamala. (W)
Sita is prettier than Kamala. (R)

75. Raju is the most tallest boy. (W)
Raju is the tallest boy. (R)

Rule 38 : Comparatives like prefer, preferable ………………… superior, inferior, junior, senior, prior, exterior, interior are followed by ‘to’ but not ‘than’.
eg. : 76. Sarala is junior than Kamala. (W)
Sarala is junior to Kamala. (R)

77. This cloth is inferior than that cloth. (W)
This cloth is inferior to that cloth. (R)

78. Coffee is preferable than tea.(W)
Coffee is preferable to tea. (R)

79. I prefer reading stories than seeing TV. (W)
I prefer reading stories to seeing TV. (R)

Rule 39 : ‘Some’ is used in the affirmative. ‘Any’ is used in the negative.
e.g. : 80. I am hungry, I want any food. (W)
I am hungry, I want some food. (R)

81. The plate does not have some fresh cake. (W)
The plate does not have any fresh cake. (R)

Much is used in case of uncountables. Many is used in case of countables.

82. There are much plans before me. (W)
There are many plans before me. (R)

83. There is many work for me to do. (W)
There is much work for me to do. (R)

Rule 40 : ‘Many’ is followed by plural verb. ‘Manya’ is followed by singular verb.

eg. : 84. Many a passenger are hurt. (W)
Many a passenger is hurt. (R)

85. Many passengers is hurt. (W)
Many passengers are hurt. (R)

Rule 41 : ‘Elder’ means the one above. ‘Older’ means more aged.
eg. : 86. Prakash is my older brother. (W)
Prakash is my elder brother. (R)

87. Prakash is elder than Raghu. (W)
Prakash is older than Raghu. (R)

Rule 42 : Superlative Degree cannot be used when the comparison is only between two.
e.g. : 88. This short story is the best of the two. (W)
This short story is the better of the two. (R)

Rule 43 : Some Adjectives have no degrees of comparison. They are perfect, unique, eternal, round etc.
e.g. : 89. It is the most unique work. (W)
It is unique work. (R)

Rule 44 : In certain comparisons ‘that of’ should be used.
e.g. : 90. The climate of Bangalore is better than Hyderabad. (W)
The climate of Bangalore is better than that of Hyderabad. (R)

Rule 45 : Farther = more distance; further = beyond.
e.g. : 91. He wants to study farther. (W)
He wants to study further. (R)

92. Let us not walk any further. (W)
Let us not walk any farther. (R)

Rule 46 : When ‘both’ is used as an adjective it should be followed by ‘the’.
e.g. : 93. We were present on both days. (W)
We were present on both the days. (R)

(F) USAGE OF ADVERBS :

e.g. : 94. It is a bitter cold today. (W)
It is a bitterly cold today. (R)

95. She is so rich. (W)
She is very rich. (R)

96. Drinking is quite harmful for health. (W)
Drinking is very harmful for health. (R)

97. He spoke fluent. (W)
He spoke fluently. (R)

98. It is so hot to go out. (W)
It is too hot to go out. (R)

99. He is too weak that he cannot walk. (W)
He is too weak to walk. (R)
(or)
He is so weak that he cannot walk. (R)

100. The boy sat alonely. (W)
The boy sat alone. (R)

101. He comes late often to school. (W)
He often comes late to school. (R)

102. He only married her for her money. (W)
He married her only for her money. (R)

103. Last night you returned lately. (W)
Last night you returned late. (R)

104. He is enough wise to understand the problem. (W)
He is wise enough to understand the problem. (R)

105. I met him three weeks before. (W)
I met him three weeks ago. (R)

106. His daughter is working hardly. (W)
His daughter is working hard. (R)

107. I came directly to my house. (W)
I came direct to my house. (R)

108. She bore cheerfully her miseries. (W)
She bore her miseries cheerfully. (R)

(G) USAGE OF VERB FORMS :

Rule 47 : Simple Present Tense is used to express a habit.
e.g. : 109. She is drinking milk daily. (W)
She drinks milk daily. (R)

Rule 48 : The Present Perfect Continuous Tense is used with ‘since’ and ‘for’.
e.g. : 110. They are working here for six years. (W)
They have been working here for six years. (R)

111. She is typing from 2 p.m. (W)
She has been typing since 2 p.m. (R)

Rule 49 : Some words like taste, hear, smell, feel, love, like, have, own, belong etc., are not used in the continuous tense form.
e.g. : 112. I am having two Maruthi cars. (W)
I have two Maruthi cars. (R)

113. This house is belonging to Ravi. (W)
This house belongs to Ravi. (R)

114. It is tasting good. (W)
It tastes good. (R)

Rule 50 : Past Tense is used after ‘It is time’ …………………
e.g. : 115. It is time you leave this place. (W)
It is time you left this place. (R)

Rule 51 : Simple Past is used to denote past action.
e.g. : 116. Last year my grandfather had died. (W)
Last year my grandfather died. (R)

Rule 52 : Present Perfect Tense is used to denote actions just completed.
e.g. : 117. The postman just came. (W)
The postman has just come. (R)

118. I just received the telegram. (W)
I have just received the telegram. (R)

Rule 53 : Use Simple Present to denote universal truths.
119. The earth was round. (W)
The earth is round. (R)

120. The sun rose in the east. (W)
The sun rises in the east. (R)

121. Honesty will be the best policy. (W)
Honesty is the best policy. (R)

When two past actions are mentioned, the earlier. (W)
I have just received the telegram. (R)

Rule 54 : When two past actions are mentioned, the earlier action should be in Past Perfect Tense :
(రెండు జరిగిన పనులను తెలిపేటప్పుడు ముందు జరిగిన పనిని Past Perfect Tense లో చెప్పాలి.)
122. The patient died before the doctor arrived. (W)
The patient had died before the doctor arrived.(R)

123. She took her supper before she went to bed. (W)
She had taken her supper before she went to bed. (R)

Rule 55 : Future Tense is not used after some conjunctions like until, when, before, after etc.
124. She will come when she will be ready. (W)
She will come when she is ready. (R)

125. I shall call on you when the doctor will be ready. (W)
I shall call on you when the doctor is ready. (R)

Rule 56 : Don’t use ‘will’ in First Person in questions.
126. Will I get you coffee ? (W)
Shall I get you coffee ? (R)

Rule 57 : ‘Could’ means capacity. ‘Was able’ denotes actual performance.
e.g. : 127. As he was a good batsman, he could score a century. (W)
As he was a good batsman, he was able to score a century. (R)

128. I knew he was able to do it. (W)
I knew he could do it. (R)

Rule 58 : ‘Lay’ means to place. ‘Lie’ means in a resting position.
e.g. : 129. You are tired. Lay here. (W)
You are tired. Lie here. (R)

130. She lies here books on the table. (W)
She lays her books on the table. (R)

Rule 59 : ‘Hanged’ means ‘Caused death’. Hung means ‘to display’.
e.g. : 131. She hanged her saree here. (W)
She hung her saree here. (R)

132. The murderer was hung. (W)
The murderer was hanged. (R)

Rule 60 : ‘Ing’ form is used after certain verbs and phrases like avoid, imagine, enjoy, finish, practise, with a view to, can’t help, look forward to etc.
e.g. : 133. I avoid to speak to her. (W)
I avoid speaking to her. (R)

134. She went to Mysore with a view to see Brindavan Gardens. (W)
She went to Mysore with a view to seeing Brindavan Gardens. (R)

135. She looks forward to help me. (W)
She looks forward to helping me. (R)

Rule 61 : ‘To’ infinitive is used after certain verbs like allow, order, command, wish, request, permit, encourage etc.
e.g. 136. He ordered me wait. (W)
He ordered me to wait. (R)

137. I allowed him read my essay. (W)
I allowed him to read my essay. (R)

Rule 62 : In Interrogative sentences the helping verb is placed before the subject.
e.g. : 138. Where you will go ? (W)
Where will you go ? (R)

Rule 63 : In a phrase like ‘Being a ……………’ it is placed before it.
e.g. : 139. Being a hot day, we stayed indoors. (W)
It being a hot day, we stayed indoors. (R)

Rule 64 : Lest is followed by should :
e.g. : 140. He ran away lest he might be caught. (W)
He ran away lest he should be caught. (R)

Rule 65 : The verb must agree with the subject in all aspects.
e.g. : 141. She wash her clothes. (W)
She washes her clothes. (R)

142. One of my friends have gone to the Andamans. (W)
One of my friends has gone to the Andamans. (R)

Rule 66 : Errors of Proximity:
e.g. : 143. The cost of all things have gone up. (W)
The cost of all things has gone up. (R)

144. The state of affairs are causing anxiety. (W)
The state of affairs is causing anxiety. (R)

Rule 67 : When the subject is the formal ‘There’, the verb agrees with the real subject that follows it.
e.g. : 145. There has been many difficulties. (W)
There have been many difficulties. (R)

146. There are nothing to fear. (W)
There is nothing to fear. (R)

Rule 68 : When two subjects are connected with ‘not only ……………… but also’ the verb agrees with the latter subject.
e.g. : 147. Not only the master but also his servants is at work. (W)
Not only the master but also his servants are at work. (R)

148. Not only the soldiers but also their captain have been captured. (W)
Not only the soldiers but also their captain has been captured. (R)

Rule 69 : In a sentence when two nouns as subjects refer to the same thing or person, the verb must be singular.
e.g. : 149. Rice and curry are his favourite dish. (W)
Rice and curry is his favourite dish. (R)

150. The poet and dramatist are honoured. (W)
The poet and dramatist is honoured. (R)

Rule 70 : When the subject consists of two nouns or pronouns joined with ‘with’, or ‘as well as’ the verb agrees with the first subject.
(ఒక వాక్యంలో రెండు నామవాచకాలు గాని సర్వనామాలు గాని కర్తలుగా ఉన్నట్లయితే ఆ వాక్యంలోని క్రియ మొదటి కర్త వచనం బట్టి ఉంటుంది)

e.g. : 151. Good students as well as good staff is necessary to show good results. (W)
Good students as well as good staff are necessary to show good results. (R)

152. The Minister with all the guests have arrived. (W)
The Minister with all the guests has arrived. (R)

Exercises

1. Read the following sentences, correct the underlined part of each sentence, and rewrite them correctly. The entire sentence must be written in exams.

Question 1.
Khyati is one of my best friend.
Answer:
Khyati is one of my best friends.

Question 2.
The sceneries of Araku Valiey are beautiful.
Answer:
The scenery of Araku Valley is beautiful.

Question 3.
The news are false.
Answer:
The news is false.

Question 4.
The audience were impressed by the magic show.
Answer:
The audience was impressed by the magic show.

Question 5.
The scissor is are very sharp.
Answer:
The scissors are very sharp.

Question 6.
He has two brother-in-laws.
Answer:
He has two brothers-in-law.

Question 7.
Myself is Gulshan from Kashmir.
Answer:
I am Gulshan from Kashmir.

Question 8.
We should avail the opportunity.
Answer:
We should avail ourselves of the opportunity.

Question 9.
I am loving my country.
Answer:
I love my country.

Question 10.
We have submitted our project an hour ago.
Answer:
We submitted our project an hour ago.

Question 11.
Chitra has a sweet voice, isn’t it ?
Answer:
Chitra has a sweet voice, doesn’t she?

Question 12.
It’s time you start preparation for the examinations.
Answer:
It’s time you started preparation for the examinations.

Question 13.
Before we reached the theatre, the movie began already.
Answer:
Before we reached the theatre, the movie had already begun.

Question 14.
I was studied in this college.
Answer:
I studied in this college. OR I was a student in this college.

Question 15.
The soldiers along with the major was wounded in the war.
Answer:
The soldiers along with the major were wounded in the war.

Question 16.
The two first chapters of the novel are very boring.
Answer:
The first two chapters of the novel are very boring.

Question 17.
Mihira wore a red Bengal cotton gorgeous sari.
Answer:
Mihira wore a gorgeous red Bengal cotton sari.

Question 18.
Vivek is too intelligent.
Answer:
Vivek is very intelligent.

Question 19.
Teja drives car very fastly.
Answer:
Teja drives car very fast.

Question 20.
He considers her as his guide.
Answer:
He considers her his guide.

Question 21.
He has been reading the novel since three hours.
Answer:
He has been reading the novel for three hours.

Question 22.
I met an European yesterday.
Answer:
I met a European yesterday.

Question 23.
Although he had a fever, but he attended the meeting.
Answer:
Although he had a fever, he attended the meeting.

Question 24.
This is the most perfect answer to the given question.
Answer:
This is the perfect answer to the given question.

Question 25.
She is the most tallest girl of the class.
Answer:
She is the tallest girl of the class.

Question 26.
This is the best of the two designs.
Answer:
This is the better of the two designs.

Question 27.
The climate of Visakhapatnam is better than Guntur.
Answer:
The climate of Visakhapatnam is better than that of Guntur.

Question 28.
Veeru is junior than Indraneel.
Answer:
Veeru is junior to Indraneel.

Question 29.
He is elder than all other boys in the class.
Answer:
He is older than all other boys in the class.

Question 30.
Teacher’s Day is celebrated on September 5th in India.
Answer:
Teacher’s Day is celebrated on the 5th of September in India.

Question 31.
That was a blunder mistake by him.
Answer:
That was a blunder by him. OR That was a mistake by him.

Question 32.
Ananya returned back from Bangalore.
Answer:
Ananya returned from Bangalore.

Question 33.
I was late why because I had missed the bus.
Answer:
I was late because I had missed the bus.

Question 34.
Meera is good in English.
Answer:
Meera is good at English.

Question 35.
Abhigna listens melodious songs.
Answer:
Abhigna listens to melodious songs.

Question 36.
We watched the film in TV.
Answer:
We watched the film on TV.

Question 37.
We should pray God.
Answer:
We should pray to God.

Question 38.
The little boy is suffering with fever.
Answer:
The little boy is suffering from fever.

Question 39.
They discussed about films.
Answer:
They discussed films.

Question 40.
We are awaiting for the results.
Answer:
We are awaiting the results.

Question 41.
Videep was elected as general secretary.
Answer:
Videep was elected general secretary.

Question 42.
Vihan is my cousin brother.
Answer:
Vihan is my cousin.

Question 43.
I go to the post office by walk.
Answer:
I go to the post office on foot.

Question 44.
We had better to wake up early.
Answer:
We had better wake up early.

Question 45.
She is running twenty-five.
Answer:
She is twenty-five years old.

Question 46.
The exam has been preponed to tomorrow.
Answer:
The exam has been postponed to tomorrow.

Question 47.
The students should pay their fees to the concerned clerk.
Answer:
The students should pay their fees to the clerk concerned.

Question 48.
Corrupted officers should be punished severely.
Answer:
Corrupt officers should be punished severely.

Question 49.
According to me, there is no short cut to success.
Answer:
In my opinion, there is no short cut to success.

Question 50.
May I know your good name ?
Answer:
May I know your name?

2. Rewrite the following sentences correcting the underlined part. The entire sentence must be written underlining the corrected part.

Question 1.
The rose is smelling sweet.
Answer:
The rose smells sweet.

Question 2.
Laahiri has seen him long ago.
Answer:
Laahiri saw him long ago.

Question 3.
Mathematics are hard subject.
Answer:
Mathematics is a hard subject.

Question 4.
Hardly he saw the tiger when he ran away.
Answer:
Hardly had he seen the tiger when he ran away.

Question 5.
He entered into the hall late.
Answer:
He entered the hall late.

3. Rewrite the following sentences correcting the underlined part. The entire sentence must be written underlining the corrected part.

Question 1.
Raghu has three brother-in-laws.
Answer:
Raghu has three brothers-in-law.

Question 2.
Her both sons are married.
Answer:
Both her sons are married.

Question 3.
Kittu has finished his homework yesterday.
Answer:
Kittu finished his homework yesterday.

Question 4.
We discussed about Grammar.
Answer:
We discussed Grammar.

Question 5.
I met an university professor in the meeting.
Answer:
I met a university professor in the meeting.

AP Inter 1st Year Physics 1st Lesson Units and Measurements Questions and Answers

I. Multiple Choice Questions

Question 1.
Pascal is the unit of
(1) power
(2) pressure
(3) potential energy
(4) pressure gradient
Answer:
(2) pressure

Question 2.
When the number 0.046508 is reduced to four significant figures then it becomes
(1) 0.04651
(2) 4650
(3) 4.651
(4) 4.650
Answer:
(1) 0.04651

Question 3.
Which of the following physical quantity has no dimensions?
(1) Angular Velocity
(2) Angular Acceleration
(3) Angular Displacement
(4) Stress
Answer:
(3) Angular Displacement

Question 4.
Mention same dimensions in the dimensional formula of torque and angular momentum.
(1) mass, time
(2) time, length
(3) mass, length
(4) mass
Answer:
(3) mass, length

Question 5.
The velocity of an object varies with time as v = At² + Bt + C. If v and t are expressed in SI,
what are the units of A, B and C ?
(1) LT^-1
(2) LT^-2
(3) LT^-3
(4) LT⁴
Answer:
(1) LT^-1
V = At² + Bt + C, From principal of Homogeneity. V = [LT^-1]

Question 6.
Which of the following equation is dimensionally correct?
(1) \(\frac{1}{2}\)mV² = mgh
(2) V = ut + at
(3) s = ut + at³
(4) t = s + uv
Answer:
(1) \(\frac{1}{2}\)mV² = mgh

Question 7.
The velocity (V) of sound with air pressure (P) and density (d) of air are related as V ∝ P^xd^y. The value of x and y are.
(1) 1, 1/2
(2) -1/2, 1/2
(3) 1/2, -1/2
(4) 1/2, 1/2
Answer:
(3) 1/2, -1/2
V = P^xd^y → LT^-1 = [ML^-1T^-2]^x [ML^-3]^y.
equality powers of M → x + y = 0 ……….(1), L → 1 = -x – 3y ….(2)
1 = y – 3y ⇒ -2y = 1 ⇒ y = -1/2, x = 1/2

Question 8.
The velocity of a body is given by V = At² + Bt + C. If V and t are expressed in SI units, then unit of A is
(1) m/s
(2) m/s²
(3) m/s³
(4) ms
Answer:
(3) m/s³
From principle of homogeneity, V = At² → A = \(\frac{\mathrm{v}}{\mathrm{t}^2}=\frac{\mathrm{LT}^{-1}}{\mathrm{~T}^2}\) = [LT³]

Question 9.
The dimensional formula for product of two physical quantities P and Q is ML² T^-2 and dimensional formula of P/ Q is MT^-2 then P is
(1) force
(2) displacement
(3) acceleration
(4) velocity
Answer:
(1) force
P × Q = ML²T^-2 ………(1), \(\frac{P}{Q}\) = mT² …(2)
(1) × (2), P² M²L²T^-2 × MT^-2 = M²L²T^-4 ⇒ P = MLT^-2 = Force.

Question 10.
E, M, L and G denote energy, mass, angular momentum and gravitational constant respectively. Then the dimension of EL²M^-5G^-2 is
(1) angle
(2) length
(3) mass
(4) time
Answer:
(1) angle

II. Fill in the Blanks

Question 1.
Plane angle is the ratio of length of arc to ________.
Answer:
Radius

Question 2.
The unit of plane angle is ________.
Answer:
Radian

Question 3.
The SI unit of luminous intensity is ______.
Answer:
Candela

Question 4.
The dimensional formula for the coefficient of viscosity is _______.
Answer:
ML^-1T^-1

Question 5.
The number of significant figures in 0.02308 is _____.
Answer:
4

Question 6.
If F, M and T are the force, mass and time, then FM^-1T² denotes _____.
Answer:
Length
FM^-1T² = MLT^-2 M^-1T² = [L]

Question 7.
A number 2.735 is rounded off to three significant figures is _______.
Answer:
2.74

Question 8.
Planck’s constant and ______ have same dimensions.
Answer:
angular momentum

Question 9.
The dimensional formula of angular impulse is ________.
Answer:
ML²T^-1

Question 10.
A force is represented by F = Ax² + Bt², where x = distance and t = time then the dimensions of ‘A’ ________.
Answer:
ML^-1T²
Principle of Homogeneity:
F = Ax² ⇒ A = \(\frac{\mathrm{F}}{\mathrm{x}^2}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}\), A = [ML^-1T^-2]

III. One Word Answer Questions

Question 1.
What is fundamental physical quantity?
Answer:
Fundamental Quantity : A fundamental quantity is one which is unique and freely existing. It does not depend on any other physical quantity.
Ex : Length (L), Time (T), Mass (M) etc.

Question 2.
What is derived physical quantity?
Answer:
Derived quantity: A derived quantity is produced by the combination of fundamental quanti¬ties (i.e., by division or by multiplication of fundamental quantities).
Ex: Velocity = \(\frac{\text { displacement }}{\text { time }}=\frac{\mathrm{L}}{\mathrm{~T}}\) or LT^-1
Acceleration = \(\frac{\text { change in velocity }}{\text { time }}=\frac{\mathrm{LT}^{-1}}{\mathrm{~T}}\) = LT^-2 etc.

Question 3.
What is the SI unit of solid angle?
Answer:
Steradian (SI)

Question 4.
Write any one physical constant that is having dimensions.
Answer:
Planck’s constant (D.F : M¹ L² T^-1), Gravitational Constant (G) : M^-1L³T^-2

Question 5.
Which physical quantity has negative dimensions in mass?
Answer:
Universal Gravitational Constant (G) = [M^-1 L³ T^-2]

Question 6.
Which physical quantity is represented by gR, where R is the radius of the Earth?
Answer:
\(\sqrt{\mathrm{gR}}=\sqrt{\mathrm{LT}^{-2} \times \mathrm{L}}\) = [LT^-1 ] = Velocity

Question 7.
Which physical quantity is represented by PV, where P is pressure and V is volume?
Answer:
PV = M¹L¹T^-2 × L³ = [ML²T^-2] = Work

Question 8.
Name atleast one physical quantity whose SI unit is Nm .
Answer:
Pressure = force/area \(\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}\)
= [M¹L¹T^-2]; Nm^-2 (or) Pascal

IV. Very Short Answer Questions

Question 1.
What are significant figures and what do they represent when reporting the result of a measurement ?
Answer:
Significant figures represent all practically measured digits plus one uncertain digit at the end. When a result is reported in this way we can know up to what extent the value is reliable and also the amount of uncertainty in that reported value.

Question 2.
Distinguish between fundamental units and derived units.
Answer:
The units of fundamental physical quantities are called fundamental units.
Ex : Kg, m, sec, etc.,
The units of derived physical quantities are called derived units.
Ex: m/sec, J, m/sec² etc.

Question 3.
What are the two fundamental physical quantities used to measure angle ?
Answer:
Plane angle (θ) = Radian.
Solid angle (ω) = Steradian.

Question 4.
Why do we have different units for the same physical quantity ?
Answer:
To measure the same physical quantity we use different units because sometimes the magnitude of the measured quantity value may be very big or very small.
Ex:
1. To measure astronomical distances, we will use light year.
1 light year = 9.468 × 10¹⁵m.
2. To measure atomic distances, we will use Angstrom A (or) Fermi.

Question 5.
Define plane angle.
Answer:
Plane angle is the ratio of length of arc ‘dS’ to the radius “r”.

Question 6.
What is meant by dimension of a physical quantity?
Answer:
Dimension : The power of a fundamental quantity in the given derived quantity is called dimension.
Ex : Force dimensional formula MLT^-2.
Here dimensions of Mass → 1, Length → 1, Time → 2.

Question 7.
What is meant by dimensional formula?
Answer:
Dimensional formula : It is a mathematical expression giving relation between various funda-mental quantities of a derived physical quantity.
Ex : Momentum (\(\overline{\mathrm{P}}\)) MLT^-1, Energy (E) ML²T^-2 etc.

Question 8.
Write the physical quantities that are having J m^-1 and J m^-2 as their units.
Answer:
Force and Surface Tension.

V. Short Answer Questions

Question 1.
In a system of units, the unit of force is 100N, unit of length is 10m and the unit of time is 100s. What is the unit of mass in this system ?
Answer:
Here, F = MLT^-2 = 100 N → (1) ; L = 10 m ; T = 100s
∴ From equation (1)
M × (10) × (100)^-2 = 100 ⇒ M × 10^-3 = 100 ⇒ M = \(\frac{100}{10^{-3}}\) ⇒ M = 10⁵ kg.

Question 2.
1 calorie = 4.2 J where 1J = 1 kg m² s^-2. Suppose we employ a system of units in which the unit of mass is α kg, the unit of length is β in and the unit of time γs, show that a calorie has a magnitude 4.2 α^-1 β^-2 γ² term of new units.
Answer:
Here, 1 calorie = 4.2 J = 4.2 kg m² / s² → (1)
As new unit of mass = α kg
∴ 1 kg = \(\frac{1}{\alpha}\) new unit of mass ⇒ α^-1 new unit of mass
Similarly lm = β^-1new unit of length and Is = γ^-1 new unit of time
Putting these values in (1) we get
1 calorie = 4.2 (α^-1 new unit of mass) (β^-1 new unit of length)² (γ^-1 new unit of time)^-2
= 4.2 α^-1β^-2γ² new unit of energy, which was proved.

Question 3.
The velocity of a body is given by v = At² + Bt + C. If v and t are expressed in SI, what are the units of A, B and C ?
Answer:
From principle of Homogeneity the terms At², Bt and C must have same dimensional formula of velocity ‘v’.
v = Velocity = LT^-1 ⇒ LT^-1 = A [ T² ]
∴ A = \(\frac{\mathrm{LT}^{-1}}{\mathrm{~T}^2}\) = LT^-3. So unit of A is m/sec³
LT^-1 = BT ⇒ B = LT^-2 So unit of B is m/sec²
LT^-1 = C So unit of C is m/sec.

Question 4.
What are fundamental quantities? State their units in SI system.
Answer:
(i) Fundamental unit: The units of the fundamental quantities are called the “fundamental units”.
Ex : Length → Meter (m), Mass → Kilogram (kg), Time → Second (s) etc.
Basic units or fundamental units of SI system :

Question 5.
Explain the uses of Dimensional methods with examples.
Answer:
Uses of dimensional methods :

1. To convert units from one system to another system.
2. To check the validity of given physical equations. For this purpose, we will use homogeneity of dimensions on L.H.S and on R.H.S.
3. To derive new relations between various physical quantities.

Question 6.
What are the rules for arithmetic operations with significant figures ?
Answer:
Rules for arithmetic operations with significant figures
1. In multiplication or division, the final result should retain as many significant figures as there in the original number with the least significant figures.
Ex: In the division \(\frac{4.237}{2.51}\) the significant figures are 4 and 3, so least significant figures are ‘3’.
∴ \(\frac{4.237}{2.51}\) = 1.69 i.e., final answer must have only ‘3’ significant digits.

2. In addition or subtraction, the final result should retain as many decimal places are there in the number with the least decimal places.
Ex : 436.26g + 227.2 g Here least number of significant figures after decimal point is one.
∴ 436.26 + 272.2 = 708.46 must be expressed as 708.5 (after rounding off the last digit).

Question 7.
What is the rounding off a number and what are the rules to be followed in it?
Answer:
Rounding off the uncertain digits
Rules for rounding off procedure : In rounding off the numbers to the required number of significant digits the following rules are followed.

1. The preceding significant digit is raised by one if the first non-significant digit is more than 5.
2. The preceding significant digit is left unchanged if the first non-significant digit is less than 5.
3. If the first non-significant figure is 5, then

• • If the preceding significant figure is an odd number, then add one to it.
  • If the preceding significant figure is an even number, then it is unchanged and 5 is discarded.

Problems

Question 1.
If the velocity of light (c). Planck’s constant (h) and the gravitational constant (G) are taken as fundamental quantities; then express mass, length and time in terms of dimensions of these quantities.
Solution:
Here, c = [L T^-1] ; h = [M L²T^-1] ; G = [M^-1L³T^-2]
∴ E = hv, h = \(\frac{\mathrm{E}}{v}\) ; G = \(\left(\frac{\mathrm{Fd}^2}{\mathrm{~m}_1 \mathrm{~m}_2}\right)\)
Let mass (M) = c^xh^yG^z → (1)
⇒ [M¹L⁰T⁰]=(LT^-1 )^x (M L²T^-1)^y (M^-1L³T^-2 )^z
⇒ [M¹L⁰T⁰] = M^y-zL^x+2y+3zT^-x-y-2z
Applying the principle of homogeneity of dimensions, we get
y – z = 1 → (2) ; x + 2y + 3z = 0 → (3) ; – x – y – 2z = 0 → (4)
Adding eq. (2), eq. (3) and eq. (4),
∴ From eq. (2) z = y – 1 = \(\frac{1}{2}\) – 1 = \(\frac{-1}{2}\), From eq. (4) x = -y – 2z = \(\frac{-1}{2}\) + 1 = \(\frac{1}{2}\)
Substituting the values of x, y & z in eq. (1), we get ⇒ M = c^1/2 h^1/2 G^1/2 ⇒ M = \(\sqrt{\frac{\mathrm{ch}}{\mathrm{G}}}\)
Proceeding as above we can show that L = \(\sqrt{\frac{\mathrm{hG}}{\mathrm{c}^3}} \text { and } \mathrm{T}=\sqrt{\frac{\mathrm{hG}}{\mathrm{c}^5}}\)

Question 2.
An artificial satellite is revolving around a planet of mass M and radius R. in a circular orbit of radius r. Using dimensional analysis show that the period of the satellite“where k is a dimensionless constant and g is acceleration due to gravity.
Solution:
Given that, T² ∝ r³ or T ∝ r^3/2. Also T is a function of g and R.
Let T ∝ r^3/2 g^a R^b where a, b are the dimensions of g and R. (or) T = k r^3/2 g^a R^b → (1)
where k is a dimensionless constant of proportionality
From equation (1) [M⁰L⁰T¹] = L^3/2 (LT^-2)^a (L)^b = M⁰L^{a+b+\(\frac{3}{2}\)}T^-2a Applying the principle of homogeneity of dimensions, we get
a + b + \(\frac{3}{2}\) = 0 → (2)
∴ -2a = 1 ⇒ a = \(\frac{-1}{2}\)
From eq (1), \(\frac{-1}{2}\) + b + \(\frac{3}{2}\) = 0 ⇒ b = -1
Substituting the values of ‘a’ and ‘b’ in eq. (1), we get
∴ T = k r^3/2 g^-1/2 R^-1 = \(\frac{k}{R} \sqrt{\frac{r^3}{g}}\) This is the required relation.

Question 3.
State the number of significant figures in the following.
a) 6729
b) 0.024
c) 0.08240
d) 6.032
Solution:
a) In 6729 all are significant figures.
∴ Number of significant figures are Four.

b) In 0.024 the zeros to the left of 1st non-zero digit of a number less than one are not significant.
∴ Number of significant figures are Two.

c) 0.08240 – Significant figures are Four.

d) In 6.032 the zero between two non-zero digits is significant.
So, number of significant figures in 6.032 are 4.

e) 4.57 × 10⁸– Significant figures are Three.
[In the representation of powers of Ten our rule is only significant figures must be given].

Question 4.
A stick has a length of 12.132 cm and another has a length of 12.4 cm. (a) If the two sticks are placed end to end what is the total length ? (b)If the two sticks are placed side by side, what is the difference in their lengths?
Solution:
a) When placed end to end total length is l = l₁ + l₂

l₁ = 12.132 cm and l₂ = 12.4 cm.
∴ l₁ + l₂ = 12.132 + 12.4 = 24.532 cm.
In addition final answer must have least number of significant numbers in that addition, i.e., one after decimal point. So our answer is 24.5 cm.

b) For difference use, l₁ – l₂ ⇒ 12.4 – 12.132 = 0.268 cm.
In subtraction, final answer must be adjusted to least number of significant figures in that operation.
Here least number is one digit after decimal. By applying round off procedure our answer is 0.3 cm.

Question 5.
Each side of a cube is measured to be 7.203 m. What is
(i) the total surface area and
(ii) the volume of the cube, to appropriate significant figures ?
Solution:
Side of cube, a = 7.203 m.
So number of significant figures are Four.
i) Surface area of a cube = 6a² = 6 × 7.203 × 7.203 = 311.299
But our final answer must be rounded to least number of significant figures is four digits.
So surface area of a cube = 311.3 m²

ii) Volume of a cube, .
V = a³ = (7.203)³ = 373.147
But the answer must be limited to Four significant figures.
∴ Volume of a cube, V = 373.1 m³.

Units and Measurements Class 11 NCERT Solutions

Question 1.
Fill in the blanks.
(a) The volume of a cube of side 1 cm is equal to m³.
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ________ (mm)².
(c) A vehicle moving with a speed of 18 km h^-1 covers ______ m in 1 s.
(d) If the relative density of lead is 11.3. Its density is _________ g cm^-3 or _______ kg m^-3.
Answer:
(a) Volume of a cube = l³
1 cm = 10^-2 m
Side of a cube, l = 1 cm = 10^-2 m
Volume of a cube of side ‘l’ is V = l³ = (10^-2)³ = 10^-6 m³.

(b) Surface area of a solid cylinder of radius r and length l is A = 2πrl + 2πr²
Here, r = 2.0 cm, l = 10 cm ; As 1 cm = 10 mm; r = 2.0 × 10 mm = 20 mm
l = 10.0 × 10 mm = 100 mm
A = 2πrl + 2πr² = 2πr (l + r)
= 2 × 3.14 × 20(100 + 20) = 125.6 (120) = 15072 mm² = 1.5 x 10⁴ mm² .

(c) Distance = Velocity × Time ⇒ 1 km/h = \(\frac{1000}{3600}\) m/s = \(\frac{5}{18}\)m/s.
Given velocity of vehicle, v = 18 km/h = 18 × \(\frac{5}{18}\) m/s = 5 m/s.
Distance traversed = velocity × time = 5 × 1 = 5 m.

(d)
Density, (p) = Relative Density × Density of water = 11.3 × 1 g cm^-3 = 11.3 g cm^-3
But 1 g cm^-3 = 1000 kg m^-3; p = 11.3 × 1000 kg/m³ = 1.13 × 10⁴ kgm^-3.

Question 2.
Fill in the blanks by suitable conversion of units.
(a) 1 kg m² s^-2 = g cm² s^-2.
(b) 1 m = _______ light years.
(c) 3 ms^-2 = _________ km h^-2.
(d) G = 6.67 × 10^-11 Nm²kg^-2 = g^-1 cm³ s^-2.
Ans .
(a) 1 kg = 1000 g, 1 m = 100 cm
1 kg m² s^-2 = 1000 g (100 cm)² s^-2 = 10⁷ g cm²s^-2
1 kg m² s^-2 = 10⁷ g cm² s^-2

(b) 1 light year = 9.46 × 10¹⁵ m
1m = \(\frac{1}{9.46 \times 10^{15}}\) light year = 1.057 × 10^-16 light year.

(c) 1 km = 1000m ⇒ 1 m = \(\frac{1}{1000}\)km
1 h = 3600s ⇒ 1s = \(\frac{1}{3600}\) h
3 ms^-2 = 3 × \(\left(\frac{1}{1000} \mathrm{~km}\right) \times\left(\frac{1}{3600} \mathrm{~h}\right)^{-2}=3 \times \frac{1}{1000} \times(3600)^2 \mathrm{~km} \mathrm{~h}^{-2}\)
= 3.888 × 10⁴ km h^-2.

(d) 1 N = 1 kg ms^-2
G = 6.67 × 10^-11 Nm² kg^-2
= 6.67 × 10^-11(kg ms^-2) m²kg^-2
= 6.67 × 10^-11kg^-1 m³ s^-2
= 6.67 × 10^-11 × \(\frac{1}{1000}\) × (10)⁶ g^-1 cm³s^-2 = 6.67 × 10^-8 g^-1 cm³ s^-2

Question 3.
A calorie is a unit of heat (energy in transit) and it equals about 4.2 J, where 1 J = 1 kg m² s^-2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m and the unit of time is γ s. Show that a calorie has a magnitude 4.2 α^-1 β^-2 γ² in terms of new units.
Answer:
Dimension of energy are [ML²T^-2]
n₁[u₁] = m₂[u₂]
n₁[M₁L₁²T₁^-2] = n₂[M₂L₂²T₂^-2] ………(i)
In SI system, n₁ = 4.2 J, M₁ = 1 kg, L₁ = 1 m, T₁ = 1 s
In new system, M₂ = α kg, L₂ = β m, T₂ = γ s
So equation (i), gives
n₂ = n₁ \(\left(\frac{\mathrm{M}_1}{\mathrm{M}_2}\right) \cdot\left(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right)^2 \cdot\left(\frac{\mathrm{~T}_1}{\mathrm{~T}_2}\right)^{-2}\) = 4.2 \(\left(\frac{1 \mathrm{~kg}}{\alpha \mathrm{~kg}}\right) \cdot\left(\frac{1 \mathrm{~m}}{\beta \mathrm{~m}}\right)^2 \cdot\left(\frac{1 \mathrm{~s}}{\gamma \mathrm{~s}}\right)^{-2}\) = 4.2 α^-1 β^-2 γ²
i.e., 1 calorie = 4.23 J = 4.2 α^-1 β^-2 γ².

Question 4.
Explain this statement clearly :
“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary.
(a) atoms are very small objects.
(b) a jet plane moves with great speed.
(c) the mass of Jupiter is very large.
(d) the air inside this room contains a large number of molecules.
(e) a proton is much more massive than an electron.
(f) the speed of sound is much smaller than the speed of light.
Answer:
The statement is true; it is meaningless to say that Meerut is a big city because it is bigger as compared to Bijnor and smaller as compared to Calcutta. Hence, to call a dimensionless quantity larger or small it is essential to specify a standard for comparison.
(a) Atoms are very small objects as compared to edge of a sharp blade.
(b) The jet plane moves with greater speed than that of a superfast train.
(c) The mass of Jupiter is very large as compared to that of the Earth.
(d) The air inside a room contains more number of molecules than the Avogadro number.

(e) Mass of proton = 1.67 × 10^-27 kg; Mass of electron = 9.1 × 10^-31 kg
\(\frac{\text { Mass of proton }}{\text { Mass of electron }}=\frac{1.67 \times 10^{-27} \mathrm{~kg}}{9.1 \times 10^{-31} \mathrm{~kg}} \approx 1836 .\)
Hence, the given statement is correct.

(f) Speed of sound in air, v = 332 m/s; Speed of light, c = 3 × 10⁸ m/s
\(\frac{v}{c}=\frac{332}{3 \times 10^8} \approx 10^{-6}\)
Hence, the statement is correct.

Question 5.
A new unit of length is chosen such that the speed of light in vacuum is unity. What
is the distance between the Sun and the Earth in terms of the new unit of light takes 8 min and 20 s to cover this distance ?
Answer:
Given, c = 1 u_n s^-1 where u_n is new unit
Time, t = 8 min 20 s = 8 × 60 + 20 = 500 s
Distance between the Sun and the Earth
x = ct = 1 u_n s^-1 × 500 u_n = 500 new unit of length.

Question 6.
Which of the following is the most precise device for measuring length ?
(a) a vernier callipers with 20 divisions on the sliding scale.
(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale.
(c) an optical instrument that can measure length to within a wavelength of light ?
Answer:
The most precise measurement is that in which least count of the measuring instrument is minimum.
(a) Least count of Vernier callipers = \(\frac{s}{n}=\frac{\text { Value of one part of main scale }(s)}{\text { Number of divisions on Vernier scale }}\)
= \(\frac{0.1 \mathrm{~cm}}{20}=\frac{1}{200} \mathrm{~cm}\) = 0.005 cm

(b) Least count of screw guage \(=\frac{\text { Pitch }}{\text { Number of divisions on circular scale }}\)
= \(\frac{1 \mathrm{~mm}}{100}\) = 0.01 mm = 0.001 cm

(c) Least count of optical instrument = Wavelength of visible light = 5 × 10^-7 m
Clearly, the least count of optical instrument is minimum, hence optical instrument is more precise device for measuring length.