Class 11

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.5

Question 1.
Show that C₀ + C₁ + C₂ + ….. + C₈ = 256
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 8, we get
C₀ + C₁ + C₂ + ….. + C₈ = 2⁸
∴ C₀ + C₁ + C₂ + ….. + C₈ = 256

Question 2.
Show that C₀ + C₁ + C₂ + …… + C₉ = 512
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 9, we get
C₀ + C₁ + C₂ + ….. + C₉ = 2⁹
∴ C₀ + C₁ + C₂ + …… + C₉ = 512

Question 3.
Show that C₁ + C₂ + C₃ + ….. + C₇ = 127
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 7, we get
C₀ + C₁ + C₂ + ….. + C₇ = 2⁷
∴ C₀ + C₁ + C₂ +….. + C₇ = 128
But, C₀ = 1
∴ 1 + C₁ + C₂ + ….. + C₇ = 128
∴ C₁ + C₂ + ….. + C₇ = 128 – 1 = 127

Question 4.
Show that C₁ + C₂ + C₃ + ….. + C₆ = 63
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
Putting n = 6, we get
C₀ + C₁ + C₂ + ….. + C₆ = 2⁶
∴ C₀ + C₁ + C₂ + …… + C₆ = 64
But, C₀ = 1
∴ 1 + C₁ + C₂ + ….. + C₆ = 64
∴ C₁ + C₂ + ….. + C₆ = 64 – 1 = 63

Question 5.
Show that C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇ = 128
Solution:
Since C₀ + C₁ + C₂ + C₃ + …… + C_n = 2ⁿ
Putting n = 8, we get
C₀ + C₁ + C₂ + C₃ + …… + C₈ = 2⁸
But, sum of even coefficients = sum of odd coefficients
∴ C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇
Let C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇ = k
Now, C₀ + C₁ + C₂ + C₃ + C₄ + C₅ + C₆ + C₇ + C₈ = 256
∴ (C₀ + C₂ + C₄ + C₆ + C₈) + (C₁ + C₃ + C₅ + C₇) = 256
∴ k + k = 256
∴ 2k = 256
∴ k = 128
∴ C₀ + C₂ + C₄ + C₆ + C₈ = C₁ + C₃ + C₅ + C₇ = 128

Question 6.
Show that C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ – 1
Solution:
Since C₀ + C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ
But, C₀ = 1
∴ 1 + C₁ + C₂ + C₃ + …… + C_n = 2ⁿ
∴ C₁ + C₂ + C₃ + ….. + C_n = 2ⁿ – 1

Question 7.
Show that C₀ + 2C₁ + 3C₂ + 4C₃ + ….. + (n + 1)C_n = (n + 2) 2^n-1
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.4

Question 1.
State, by writing the first four terms, the expansion of the following, where |x| < 1.
(i) (1 + x)^-4
Solution:

(ii) (1 – x)^1/3
Solution:

(iii) (1 – x²)^-3
Solution:

(iv) (1 + x)^-1/5
Solution:

(v) (1 + x²)^-1
Solution:

Question 2.
State by writing first four terms, the expansion of the following, where |b| < |a|.
(i) (a – b)^-3
Solution:
(a – b)^-3 = \(\left[a\left(1-\frac{b}{a}\right)\right]^{-3}\)

(ii) (a + b)^-4
Solution:

(iii) (a + b)^1/4
Solution:

(iv) (a – b)^-1/4
Solution:
(a – b)^-1/4 = \(\left[a\left(1-\frac{b}{a}\right)\right]^{\frac{-1}{4}}\)

(v) (a + b)^-1/3
Solution:

Question 3.
Simplify the first three terms in the expansion of the following:
(i) (1 + 2x)^-4
Solution:

(ii) (1 + 3x)^-1/2
Solution:

(iii) (2 – 3x)^1/3
Solution:

(iv) (5 + 4x)^-1/2
Solution:

(v) (5 – 3x)^-1/3
Solution:

Question 4.
Use the binomial theorem to evaluate the following upto four places of decimals.
(i) √99
Solution:

= 10 [1 – 0.005 – 0.0000125 – ……]
= 10(0.9949875)
= 9.94987 5
= 9.9499

(ii) \(\sqrt[3]{126}\)
Solution:

(iii) \(\sqrt[4]{16.08}\)
Solution:

(iv) (1.02)^-5
Solution:

(v) (0.98)^-3
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.3

Question 1.
In the following expansions, find the indicated term.
(i) \(\left(2 x^{2}+\frac{3}{2 x}\right)^{8}\), 3rd term
Solution:

(ii) \(\left(x^{2}-\frac{4}{x^{3}}\right)^{11}\), 5th term
Solution:

(iii) \(\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{9}\), 7th term
Solution:

(iv) In \(\left(\frac{1}{3}+a^{2}\right)^{12}\), 9th term
Solution:

(v) In \(\left(3 a+\frac{4}{a}\right)^{13}\), 10th term
Solution:

Question 2.
In the following expansions, find the indicated coefficients.
(i) x³ in \(\left(x^{2}+\frac{3 \sqrt{2}}{x}\right)^{9}\)
Solution:

(ii) x⁸ in \(\left(2 x^{5}-\frac{5}{x^{3}}\right)^{8}\)
Solution:

(iii) x⁹ in \(\left(\frac{1}{x}+x^{2}\right)^{18}\)
Solution:

(iv) x^-3 in \(\left(x-\frac{1}{2 x}\right)^{5}\)
Solution:

(v) x^-20 in \(\left(x^{3}-\frac{1}{2 x^{2}}\right)^{15}\)
Solution:

Question 3.
Find the constant term (term independent of x) in the expansion of
(i) \(\left(2 x+\frac{1}{3 x^{2}}\right)^{9}\)
Solution:

(ii) \(\left(x-\frac{2}{x^{2}}\right)^{15}\)
Solution:

(iii) \(\left(\sqrt{x}-\frac{3}{x^{2}}\right)^{10}\)
Solution:

(iv) \(\left(x^{2}-\frac{1}{x}\right)^{9}\)
Solution:

(v) \(\left(2 x^{2}-\frac{5}{x}\right)^{9}\)
Solution:

Question 4.
Find the middle terms in the expansion of
(i) \(\left(\frac{x}{y}+\frac{y}{x}\right)^{12}\)
Solution:

(ii) \(\left(x^{2}+\frac{1}{x}\right)^{7}\)
Solution:

(iii) \(\left(x^{2}-\frac{2}{x}\right)^{8}\)
Solution:

(iv) \(\left(\frac{x}{a}-\frac{a}{x}\right)^{10}\)
Solution:

(v) \(\left(x^{4}-\frac{1}{x^{3}}\right)^{11}\)
Solution:

Question 5.
In the expansion of (k + x)⁸, the coefficient of x⁵ is 10 times the coefficient of x⁶. Find the value of k.
Solution:

Question 6.
Find the term containing x⁶ in the expansion of (2 – x) (3x + 1)⁹.
Solution:

Question 7.
The coefficient of x² in the expansion of (1 + 2x)^m is 112. Find m.
Solution:
The coefficient of x² in (1 + 2x)^m = ^mC₂ (2²)
Given that the coefficient of x² = 112
∴ ^mC₂ (4) = 112
∴ ^mC₂ = 28
∴ \(\frac{\mathrm{m} !}{2 !(\mathrm{m}-2) !}=28\)
∴ \(\frac{m(m-1)(m-2) !}{2 \times(m-2) !}=28\)
∴ m(m – 1) = 56
∴ m² – m – 56 = 0
∴ (m – 8) (m + 7) = 0
As m cannot be negative.
∴ m = 8

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.2

Question 1.
Expand:
(i) (√3 + √2)⁴
Solution:
Here, a = √3, b = √2 and n = 4.
Using binomial theorem,

∴ (√3 + √2)⁴ = 1(9) (1) + 4(3√3) (√2) + 6(3)(2) + 4(√3) (2√2) + 1(1)(4)
= 9 + 12√6 + 36 + 8√6 + 4
= 49 + 20√6

(ii) (√5 – √2)⁵
Solution:
Here, a = √5, b = √2 and n = 5.
Using binomial theorem,

Question 2.
Expand:
(i) (2x² + 3)⁴
Solution:
Here, a = 2x², b = 3 and n = 4.
Using binomial theorem,

(ii) \(\left(2 x-\frac{1}{x}\right)^{6}\)
Solution:
Here, a = 2x, b = \(\frac{1}{x}\) and n = 6.
Using binomial theorem,

Question 3.
Find the value of
(i) (√3 + 1)⁴ – (√3 – 1)⁴
Solution:

(ii) (2 + √5)⁵ + (2 – √5)⁵
Solution:


Adding (i) and (ii), we get
∴ (2 + √5 )⁵ + (2 – √5)⁵ = (32 + 80√5 + 400 + 200√5 + 250 + 25√5) + (32 – 80√5 + 400 – 200√5+ 250 – 25√5 )
= 64 + 800 + 500
= 1364

Question 4.
Prove that:
(i) (√3 + √2)⁶ + (√3 – √2)⁶ = 970
Solution:

(ii) (√5 + 1)⁵ – (√5 – 1)⁵ = 352
Solution:

Question 5.
Using binomial theorem, find the value of
(i) (102)⁴
Solution:

(ii) (1.1)⁵
Solution:

Question 6.
Using binomial theorem, find the value of
(i) (9.9)³
Solution:

(ii) (0.9)⁴
Solution:

Question 7.
Without expanding, find the value of
(i) (x + 1)⁴ – 4(x + 1)³ (x – 1) + 6(x + 1)² (x – 1)² – 4(x + 1) (x – 1)³ + (x – 1)⁴
Solution:

(ii) (2x – 1)⁴ + 4(2x – 1)³ (3 – 2x) + 6(2x – 1)² (3 – 2x)² + 4(2x – 1)¹ (3 – 2x)³ + (3 – 2x)⁴
Solution:

Question 8.
Find the value of (1.02)⁶, correct upto four places of decimals.
Solution:

Question 9.
Find the value of (1.01)⁵, correct upto three places of decimals.
Solution:

Question 10.
Find the value of (0.9)⁶, correct upto four places of decimals.
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 4 Methods of Induction and Binomial Theorem Ex 4.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Methods of Induction and Binomial Theorem Ex 4.1

Prove by the method of induction, for all n ∈ N.

Question 1.
2 + 4 + 6 + …… + 2n = n(n + 1)
Solution:
Let P(n) = 2 + 4 + 6 + …… + 2n = n(n + 1), for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 2
R.H.S. = 1(1 + 1) = 2
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 2 + 4 + 6 + ….. + 2k = k(k + 1) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
2 + 4 + 6 + …… + 2(k + 1) = (k + 1) (k + 2)
L.H.S. = 2 + 4 + 6 + …+ 2(k + 1)
= 2 + 4 + 6+ ….. + 2k + 2(k + 1)
= k(k + 1) + 2(k + 1) …..[From (i)]
= (k + 1).(k + 2)
= R.H.S.
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 2 + 4 + 6 + …… + 2n = n(n + 1) for all n ∈ N.

Question 2.
3 + 7 + 11 + ……… to n terms = n(2n + 1)
Solution:
Let P(n) = 3 + 7 + 11 + ……… to n terms = n(2n +1), for all n ∈ N.
But 3, 7, 11, …. are in A.P.
∴ a = 3 and d = 4
Let t_n be the nth term.
∴ t_n = a + (n – 1)d = 3 + (n – 1)4 = 4n – 1
∴ P(n) = 3 + 7 + 11 + ……. + (4n – 1) = n(2n + 1)

Step I:
Put n = 1
L.H.S. = 3
R.H.S. = 1[2(1)+ 1] = 3
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 3 + 7 + 11 + ….. + (4k – 1) = k(2k + 1) …..(i)

Sept III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
3 + 7 + 11 + …+ [4(k + 1) – 1] = (k + 1)(2k + 3)
L.H.S. = 3 + 7 + 11 + …… + [4(k + 1) – 1]
= 3 + 7 + 11 + ….. + (4k – 1) + [4(k+ 1) – 1]
= k(2k + 1) + (4k + 4 – 1) …..[From (i)]
= 2k² + k + 4k + 3
= 2k² + 2k + 3k + 3
= 2k(k + 1) + 3(k + 1)
= (k + 1) (2k + 3)
= R.H.S.
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 3 + 7 + 11 + ….. to n terms = n(2n + 1) for all n ∈ N.

Question 3.
1² + 2² + 3² +…..+ n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Solution:
Let P(n) = 1² + 2² + 3² +…..+ n² = \(\frac{n(n+1)(2 n+1)}{6}\) for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 1² = 1
RHS = \(\frac{1(1+1)[2(1)+1]}{6}=\frac{6}{6}\) = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1² + 2² + 3² +…+ k² = \(\frac{k(k+1)(2 k+1)}{6}\) …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1² + 2² + 3² + …+ n² = \(\frac{n(n+1)(2 n+1)}{6}\) for all n ∈ N.

Question 4.
1² + 3² + 5² + ….. + (2n – 1)² = \(\frac{n}{3}\) (2n – 1)(2n + 1)
Solution:
Let P(n) = 1² + 3² + 5²+…..+ (2n – 1)² = \(\frac{n}{3}\) (2n – 1)(2n + 1), for all n ∈ N.
Step I:
Put n = 1
L.H.S. = 1² = 1
R.H.S. = \(\frac{1}{3}\) [2(1) – 1][2(1) + 1] = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1² + 3² + 5² +….+(2k – 1)² = \(\frac{k}{3}\) (2k – 1)(2k + 1) …….(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1² + 3² + 5² + …+ (2n – 1)² = \(\frac{n}{3}\) (2n – 1)(2n + 1) for all n ∈ N.

Question 5.
1³ + 3³ + 5³ + ….. to n terms = n² (2n² – 1)
Solution:
Let P(n) = 1³ + 3³ + 5³ + …. to n terms = n² (2n² – 1), for all n ∈ N.
But 1, 3, 5, are in A.P.
∴ a = 1, d = 2
Let tn be the nth term.
t_n = a + (n – 1) d = 1 + (n – 1) 2 = 2n – 1
∴ P(n) = 1³ + 3³ + 5³ +…..+ (2n – 1)³ = n² (2n² – 1)

Step I:
Put n = 1
L.H.S. = 1³ = 1
R.H.S. = 1² [2(1)² – 1] = 1
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1³ + 3³ + 5³ +…+ (2k – 1)³ = k² (2k² – 1) …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1³ + 3³ + 5³ + … to n terms = n² (2n² – 1) for all n ∈ N.

Question 6.
1.2 + 2.3 + 3.4 +… + n(n + 1) = \(\frac{n}{3}\) (n + 1)(n + 2)
Solution:
Let P(n) = 1.2 + 2.3 + 3.4 +….+n(n + 1) = \(\frac{n(n+1)(n+2)}{3}\), for all n ∈ N.

Step I:
Put n = 1
L.H.S. = 1.2 = 2
R.H.S. = \(\frac{1}{3}\) (1 + 1)(1 + 2) = 2
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1.2 + 2.3 + 3.4 + ….. + k(k + 1) = \(\frac{k}{3}\) (k + 1)(k + 2) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1.2 + 2.3 + 3.4 + … + n(n + 1) = \(\frac{n}{3}\) (n + 1)(n + 2), for all n ∈ N.

Question 7.
1.3 + 3.5 + 5.7 +… to n terms = \(\frac{n}{3}\) (4n² + 6n – 1)
Solution:
Let P(n) = 1.3 + 3.5 + 5.7 +… to n terms = \(\frac{n}{3}\) (4n² + 6n -1), for all n ∈ N.
But first factor in each term, i.e., 1, 3, 5,… are in A.P. with a = 1 and d = 2.
∴ nth term = a + (n – 1)d = 1 + (n – 1) 2 = (2n – 1)
Also, second factor in each term,
i.e., 3, 5, 7, … are in A.P. with a = 3 and d = 2.
∴ nth term = a + (n – 1) d = 3 + (n – 1) 2 = (2n + 1)
∴ nth term, t_n = (2n – 1) (2n + 1)
∴ P(n) ≡ 1.3 + 3.5 + 5.7 + …. + (2n – 1) (2n + 1) = \(\frac{n}{3}\) (4n² + 6n – 1)

Step I:
Put n = 1
L.H.S. = 1.3 = 3
R.H.S. = \(\frac{1}{3}\) [4(1)² + 6(1) – 1] = 3
∴ L.H.S. = R.H.S.
∴ P(n) is trae for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 1.3 + 3.5 + 5.7 +….+ (2k – 1)(2k + 1) = \(\frac{k}{3}\) (4k² + 6k – 1) ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 1.3 + 3.5 + 5.7 +… to n terms = \(\frac{n}{3}\) (4n² + 6n – 1) for all n ∈ N.

Question 8.
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\)
Solution:
Let P(n) ≡ \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\), for all n ∈ N.

Step I:
Put n = 1
L.H.S. = \(\frac{1}{1.3}=\frac{1}{3}\)
R.H.S. = \(\frac{1}{2(1)+1}=\frac{1}{3}\)
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 k-1)(2 k+1)}=\frac{k}{2 k+1}\) …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\), for all n ∈ N.

Question 9.
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots \text { to } n \text { terms }=\frac{n}{3(2 n+3)}\)
Solution:
Let P(n) ≡ \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots \text { to } n \text { terms }=\frac{n}{3(2 n+3)}\), for all n ∈ N.
But first factor in each term of the denominator,
i.e., 3, 5, 7, ….. are in A.P. with a = 3 and d = 2.
∴ nth term = a + (n – 1)d = 3 + (n – 1) 2 = (2n + 1)
Also, second factor in each term of the denominator,
i.e., 5, 7, 9, … are in A.P. with a = 5 and d = 2.
∴ nth term = a + (n – 1) d = 5 + (n – 1) 2 = (2n + 3)
∴ nth term, t_n = \(\frac{1}{(2 n+1)(2 n+3)}\)
P(n) ≡ \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}\) = \(\frac{n}{3(2 n+3)}\)

Step I:
Put n = 1
L.H.S. = \(\frac{1}{3.5}=\frac{1}{15}\)
R.H.S. = \(\frac{1}{3[2(1)+3]}=\frac{1}{3(2+3)}=\frac{1}{15}\)
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 k+1)(2 k+3)}\) = \(\frac{k}{3(2 k+3)}\) ….(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that


∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots \text { to } n \text { terms }=\frac{n}{3(2 n+3)}\), for all n ∈ N.

Question 10.
(2³ⁿ – 1) is divisible by 7.
Solution:
(2³ⁿ – 1) is divisible by 7 if and only if (2³ⁿ – 1) is a multiple of 7.
Let P(n) ≡ (2³ⁿ – 1) = 7m, where m ∈ N.

Step I:
Put n = 1
∴ 2³ⁿ – 1 = 2³⁽¹⁾ – 1 = 2³ – 1 = 8 – 1 = 7
∴ (2³ⁿ – 1) is a multiple of 7.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
i.e., 2^3k – 1 is a multiple of 7.
∴ 2^3k – 1 = 7a, where a ∈ N
∴ 2^3k = 7a + 1 ……(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
2^3(k+1) – 1 = 7b, where b ∈ N.
∴ P(k + 1) = 2^3(k+1) – 1
= 2^3k+3 – 1
= 2^3k . (2³) – 1
= (7a + 1)8 – 1 …..[From (i)]
= 56a + 8 – 1
= 56a + 7
= 7(8a + 1)
7b, where b = (8a + 1) ∈ N
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ (2⁴ⁿ – 1) is divisible by 7, for all n ∈ N.

Question 11.
(2⁴ⁿ – 1) is divisible by 15.
Solution:
(2⁴ⁿ – 1) is divisible by 15 if and only if (2⁴ⁿ – 1) is a multiple of 15.
Let P(n) ≡ (2⁴ⁿ – 1) = 15m, where m ∈ N.

Step I:
Put n = 1
∴ 2⁴⁽¹⁾ – 1 = 16 – 1 = 15
∴ (2⁴ⁿ – 1) is a multiple of 15.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 2^4k – 1 = 15a, where a ∈ N
∴ 2^4k = 15a + 1 …..(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
∴ 2^4(k+1) – 1 = 15b, where b ∈ N
∴ P(k + 1) = 2^4(k+1) – 1 = 2^4k+4 – 1
= 2^4k . 2⁴ – 1
= 16 . (2^4k) – 1
= 16(15a + 1) – 1 …..[From (i)]
= 240a + 16 – 1
= 240a + 15
= 15(16a + 1)
= 15b, where b = (16a + 1) ∈ N
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ (2⁴ⁿ – 1) is divisible by 15, for all n ∈ N.

Question 12.
3ⁿ – 2n – 1 is divisible by 4.
Solution:
(3ⁿ – 2n – 1) is divisible by 4 if and only if (3ⁿ – 2n – 1) is a multiple of 4.
Let P(n) ≡ (3ⁿ – 2n – 1) = 4m, where m ∈ N.

Step I:
Put n = 1
∴ (3ⁿ – 2n – 1) = 3⁽¹⁾ – 2(1) – 1 = 0 = 4(0)
∴ (3ⁿ – 2n – 1) is a multiple of 4.
∴ P(n) is tme for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 3^k – 2k – 1 = 4a, where a ∈ N
∴ 3^k = 4a + 2k + 1 ….(i)

Step III:
We have to prove that P(n) is tme for n = k + 1,
i.e., to prove that
3^(k+1) – 2(k + 1) – 1 = 4b, where b ∈ N
P(k + 1) = 3^k+1 – 2(k + 1) – 1
= 3^k . 3 – 2k – 2 – 1
= (4a + 2k + 1) . 3 – 2k – 3 …….[From (i)]
= 12a + 6k + 3 – 2k – 3
= 12a + 4k
= 4(3a + k)
= 4b, where b = (3a + k) ∈ N
∴ P(n) is tme for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is tme for all n ∈ N.
∴ 3ⁿ – 2n – 1 is divisible by 4, for all n ∈ N.

Question 13.
5 + 5² + 5³ + ….. + 5ⁿ = \(\frac{5}{4}\) (5ⁿ – 1)
Solution:
Let P(n) ≡ 5 + 52 + 53 +…..+ 5n = \(\frac{5}{4}\) (5n – 1), for all n ∈ N.

Step I:
Put n = 1
L.H.S. = 5
R.H.S. = \(\frac{5}{4}\) (5¹ – 1) = 5
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ 5 + 5² + 5³ + ….. + 5^k = \(\frac{5}{4}\) (5^k – 1) …….(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ 5 + 5² + 5³ + … + 5ⁿ = \(\frac{5}{4}\) (5ⁿ – 1), for all n ∈ N.

Question 14.
(cos θ + i sin θ)ⁿ = cos (nθ) + i sin (nθ)
Solution:
Let P(n) ≡ (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ, for all n ∈ N.
Step I:
Put n = 1
L.H.S. = (cos θ + i sin θ)¹ = cos θ + i sin θ
R.H.S. = cos[(1)θ] + i sin[(1)θ] = cos θ + i sin θ
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.

Step II:
Let us assume that P(n) is true for n = k.
∴ (cos θ + i sin θ)^k = cos kθ + i sin kθ …….(i)

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that
(cos θ + i sin θ)^k+1 = cos (k + 1)θ + i sin (k + 1)θ
L.H.S. = (cos θ + i sin θ)^k+1
= (cos θ + i sin θ)^k . (cos θ + i sin θ)
= (cos kθ + i sin kθ) . (cos θ + i sin θ) ……[From (i)]
= cos kθ cos θ + i sin θ cos kθ + i sin kθ cosθ – sin kθ sin θ ……[∵ i² = -1]
= (cos kθ cos θ – sin k θ sin θ) + i(sin kθ cos θ + cos kθ sin θ)
= cos(kθ + θ) + i sin(kθ + θ)
= cos(k + 1) θ + i sin (k + 1) θ
= R.H.S.
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ (cos θ + i sin θ)ⁿ = cos (nθ) + i sin (nθ), for all n ∈ N.

Question 15.
Given that t_n+1 = 5 t_n+4, t₁ = 4, prove by method of induction that t_n = 5ⁿ – 1.
Solution:
Let the statement P(n) has L.H.S. a recurrence relation t_n+1 = 5 t_n+4, t₁ = 4 and R.H.S. a general statement t_n = 5ⁿ – 1.
Step I:
Put n = 1
L.H.S. = 4
R.H.S. = 5¹ – 1 = 4
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 1.
Put n = 2
L.H.S. = t₂ = 5t₁ + 4 = 24
R.H.S. = t₂ = 5² – 1 = 24
∴ L.H.S. = R.H.S.
∴ P(n) is true for n = 2.

Step II:
Let us assume that P(n) is true for n = k.
∴ t_k+1 = 5 t_k+4 and t_k = 5^k – 1

Step III:
We have to prove that P(n) is true for n = k + 1,
i.e., to prove that t_k+1 = 5^k+1 – 1
Since t_k+1 = 5 t_k+4 and t_k = 5^k – 1 …..[From Step II]
t_k+1 = 5 (5^k – 1) + 4 = 5^k+1 – 1
∴ P(n) is true for n = k + 1.

Step IV:
From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.
∴ t_n = 5ⁿ – 1, for all n ∈ N.

Question 16.
Prove by method of induction
\(\left(\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right)^{n}=\left(\begin{array}{cc}
1 & 2 n \\
0 & 1
\end{array}\right) \forall n \in N\)
Solution:

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Miscellaneous Exercise 3

(I) Select the correct answer from the given alternatives.

Question 1.
A college offers 5 courses in the morning and 3 in the evening. The number of ways a student can select exactly one course, either in the morning or in the evening is
(A) 5
(B) 3
(C) 8
(D) 15
Answer:
(C) 8
Hint:
Number of ways to select one course from available 8 courses
(i.e., 5 courses in the morning and 3 in the evening) = 5 + 3 = 8

Question 2.
A college has 7 courses in the morning and 3 in the evening. The possible number of choices with the student if he wants to study one course in the morning and one in the evening is
(A) 21
(B) 4
(C) 42
(D) 10
Answer:
(A) 21
Hint:
Number of ways to select one morning and one evening course = ⁷C₁ × ³C₁ = 21

Question 3.
In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all persons of the same nationality sit together?
(A) 3! 8!
(B) 3! 4! 8! 4!
(C) 4! 4!
(D) 8! 4! 4!
Answer:
(B) 3! 4! 8! 4!
Hint:
8 Indians take their seats in 8! ways, 4 Americans take their seats in 4! ways, 4 Englishmen take their seats in 4! ways.
Three groups of Indians, Americans and Englishmen can be permuted in 3! ways.
Required number = 3! × 8! × 4! × 4!

Question 4.
In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together?
(A) 9 × 8!
(B) 8 × 8!
(C) 9 × 9!
(D) 8 × 9!
Answer:
(D) 8 × 9!
Hint:
Arrange 8 papers in 8! ways and two papers in 9 gaps are arranged in ⁹P₂ ways.
Required number = 8! ⁹P₂
= 8! × 9 × 8
= 9! × 8

Question 5.
In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate.
(A) 12
(B) 288
(C) 144
(D) 256
Answer:
(C) 144
Hint:
B G B G B G B
4 boys take their seats in 4! ways.
3 girls take their seats in 3! ways.
Required number = 4! × 3!
= 24 × 6
= 144

Question 6.
Find the number of triangles which can be formed by joining the angular points of a polygon of 8 sides as vertices.
(A) 16
(B) 56
(C) 24
(D) 8
Answer:
(B) 56
Hint:
A triangle is obtained by joining three vertices.
Number of ways of selecting 3 vertices out of 8 vertices = ⁸C₃
= \(\frac{8 \times 7 \times 6}{1 \times 2 \times 3}\)
= 56

Question 7.
A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 8 from part A and 5 from part B, in how many ways can he choose the questions?
(A) 320
(B) 750
(C) 40
(D) 11340
Answer:
(D) 11340
Hint:
Number of ways to choose 8 questions from Part A and 5 from Part B = ¹⁰C₈ × ¹⁰C₅
= ¹⁰C₂ × ¹⁰C₅
= 45 × 252
= 11340

Question 8.
There are 10 persons among whom two are brothers. The total number of ways in which these persons can be seated around a round table so that exactly one person sits between the brothers is equal to:
(A) 2! × 7!
(B) 2! × 8!
(C) 3! × 7!
(D) 3! × 8!
Answer:
(B) 2! × 8!
Hint:
Select a person from 8 people (i.e., the people excluding two brothers).
This is done in 8 ways.
2 brothers sit adjacent to the selected person on two sides, they may interchange their seats.
Remaining 7 people sit in 7! ways
Required number = 8 × 2 × 7! = 2! × 8!

Question 9.
The number of arrangements of the letters of the word BANANA in which two N’s do not appear adjacently is
(A) 80
(B) 60
(C) 40
(D) 100
Answer:
(C) 40
Hint:
Arrange B, A, A, A in \(\frac{4 !}{3 !}\) ways.
These four letters create 5 gaps in which 2 N are to be filled, this can be done in 5C2 ways, we do not permute those 2N as they are identical.
∴ Required number = \(\frac{4 !}{3 !}\) × ⁵C₂ = 40

Question 10.
The number of ways in which 5 male and 2 female members of a committee can be seated around a round table so that the two females are not seated together is
(A) 840
(B) 600
(C) 720
(D) 480
Answer:
(D) 480
Hint:
5 males take their seats in 4! ways, creating 5 gaps.
In these 5 gaps, 2 females are to be seated.
∴ The number of ways to do this = ⁵C₂ × 2!
Required number = 4! × ⁵C₂ × 2! = 480

(II) Answer the following.

Question 1.
Find the value of r if ⁵⁶P_r+2 : ⁵⁴P_r-1 = 30800 : 1.
Solution:

Question 2.
How many words can be formed by writing letters in the word CROWN in a different order?
Solution:
Five Letters of the word CROWN are to be permuted.
∴ Number of different words = 5! = 120

Question 3.
Find the number of words that can be formed by using all the letters in the word REMAIN. If these words are written in dictionary order, what will be the 40th word?
Solution:
There are 6 letters A, E, I, M, N, R.
Number of words that can be formed by using all these letters = 6! = 720
When a word starts with ‘A’,
‘A’ can be arranged in 1 way and the remaining 5 letters can be arranged among themselves in 5! ways.
The number of words starting with A = 5!
∴ Similarly,
The number of words starting with E = 5!
The number of words starting with I = 5!
The number of words starting with M = 5!
The number of words starting with N = 5!
The number of words starting with R = 5!
Total number of words = 6 × 5! = 720
Number of words starting with AE = 4! = 24
Number of words starting with AIE = 3! = 6
Number of words starting with AIM = 3! = 6
Number of words starting with AINE = 2!
Total words = 24 + 6 + 6 + 2 = 38
39th word is AINMER
40th word is AINMRE

Question 4.
The Capital English alphabet has 11 symmetric letters that appear the same when looked at in a mirror. These letters are A, H, I, M, O, T, U, V, W, X, and Y. How many symmetric three letters passwords can be formed using these letters?
Solution:
There are 11 symmetric letters.
∴ Number of 3 Letter passwords = ¹¹P₃
= 11 × 10 × 9
= 990

Question 5.
How many numbers formed using the digits 3, 2, 0, 4, 3, 2, 3 exceed one million?
Solution:
A number that exceeds one million is to be formed from the digits 3, 2, 0, 4, 3, 2, 3.
Then the numbers should be any number of 7 digits which can be formed from these digits.
Also, among the given numbers 2 is repeated twice and 3 is repeated thrice.
∴ Required number of numbers = Total number of arrangements possible among these digits – number of arrangements of 7 digits which begin with 0.
= \(\frac{7 !}{2 ! 3 !}-\frac{6 !}{2 ! 3 !}\)
= \(\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 3 !}-\frac{6 \times 5 \times 4 \times 3 !}{2 \times 3 !}\)
= 7 × 6 × 5 × 2 – 6 × 5 × 2
= 6 × 5 × 2(7 – 1)
= 60 × 6
= 360

Question 6.
Ten students are to be selected for a project from a class of 30 students. There are 4 students who want to be together either in the project or not in the project. Find the number of possible selections.
Solution:
Ten students are to be selected for a project from a class of 30 students.
Case I:
If 4 students join the project, then from remaining 26 students, rest of the 6 students are to be selected.
Which can be done in ²⁶C₆
= \(\frac{26 !}{6 !(26-6) !}\)
= \(\frac{26 \times 25 \times 24 \times 23 \times 22 \times 21 \times 20 !}{6 ! \times 20 !}\)
= 230230 ways.

Case II:
If 4 students does not join the project, then from remaining 26 students, all the 10 students are to be selected.
Which can be done in ²⁶C₁₀
= \(\frac{26 !}{10 !(26-10) !}\)
= \(\frac{26 \times 25 \times 24 \times 23 \times 22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 !}{10 ! \times 16 !}\)
= 5311735 ways.
∴ Required number of selections = ²⁶C₆ + ²⁶C₁₀
= 230230 + 5311735
= 5541965

Question 7.
A student finds 7 books of his interest but can borrow only three books. He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed. Find the number of ways he can choose three books that he wants to borrow.
Solution:
There are 7 books of student’s interest, but he can borrow only three books.
He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed.
Consider the following table:

Required number of selections = 5 + 10 = 15

Question 8.
30 objects are to be divided into three groups containing 7, 10, 13 objects. Find the number of distinct ways of doing so.
Solution:
First we can select 7 objects out of 30 for the first group in ³⁰C₇ ways.
Now there are 23 objects left out of which we can select 10 objects for the second group in ²³C₁₀ ways.
Remaining 13 objects can be selected for the third group in ⁵C₅ ways.
∴ Required number of ways = ³⁰C₇ × ²³C₁₀ × ¹³C₁₃
= \(\frac{30 !}{23 ! 7 !} \times \frac{23 !}{10 ! 13 !} \times 1\)
= \(\frac{30 !}{7 ! 10 ! 13 !}\)

Question 9.
A student passes an examination if he secures a minimum in each of the 7 subjects. Find the number of ways a student can fail.
Solution:
Every subject a student may pass or fail.
∴ Total number of outcomes = 2⁷ = 128
This number includes one case when the student passes in all subjects.
Required number of ways = 128 – 1 = 127

Question 10.
Nine friends decide to go for a picnic in two groups. One group decides to go by car and the other group decides to go by train. Find the number of different ways of doing so if there must be at least 3 friends in each group.
Solution:
Nine friends decide to go for a picnic in two groups and there must be at least 3 friends in each group.
Consider the following table:

Question 11.
A hall has 12 lamps and every lamp can be switched on independently. Find the number of ways of illuminating the hall.
Solution:
Every lamp is either ON or OFF.
There are 12 lamps
Number of instances = 2¹²
This number includes one case in when all 12 lamps are OFF.
∴ Required Number of ways = 2¹² – 1 = 4095

Question 12.
How many quadratic equations can be formed using numbers from 0, 2, 4, 5 as coefficients if a coefficient can be repeated in an equation?
Solution:
A quadratic equation is to be formed using numbers 0, 2, 4, 5 as coefficients and a coefficient can be repeated.
Let the quadratic equation be ax² + bx + c = 0, a ≠ 0
Consider the following table:

Number of quadratic equations can be formed = 3 × 4 × 4 = 48

Question 13.
How many six-digit telephone numbers can be formed if the first two digits are 45 and no digit can appear more than once?
Solution:
There are total of 10 digits.
Let the telephone number be 45abcd.
There are 8 digits left for the choice of a, b, c, d as repetition is not allowed.
Consider the following table:

∴ Required number of numbers formed = 8 × 7 × 6 × 5 = 1680

Question 14.
A question paper has 6 questions. How many ways does a student have to answer if he wants to solve at least one question?
Solution:
Every question is ‘SOLVED’ or ‘NOT SOLVED’.
There are 6 questions.
Number of outcomes = 2⁶
This number includes one case when the student solves NONE of the questions.
∴ Required number of ways = 2⁶ – 1 = 64 – 1 = 63

Question 15.
Find the number of ways of dividing 20 objects into three groups of sizes 8, 7, and 5.
Solution:
First we can select 8 objects our of 20 for the first group in ²⁰C₈ ways.
Now there are 12 objects left out of which we can select 7 objects for the second group in ¹²C₇ ways.
Remaining 5 objects can be selected for the third group in ⁵C₅ ways.
∴ Required number of ways = ²⁰C₈ × ¹²C₇ × ⁵C₅
= \(\frac{20 !}{8 ! 12 !} \times \frac{12 !}{7 ! 5 !} \times 1\)
= \(\frac{20 !}{8 ! 7 ! 5 !}\)

Question 16.
There are 4 doctors and 8 lawyers in a panel. Find the number of ways for selecting a team of 6 if at least one doctor must be in the team.
Solution:
There are 4 doctors and 8 lawyers in a panel.
A team of 6 with at least one doctor is to be formed.
We count the number by the INDIRECT method of counting.
Number of ways to select a team of 6 people = ¹²C₆
Number of teams with No doctor in any team = ⁸C₆
∴ Required number of ways = ¹²C₆ – ⁸C₆
= 924 – 28
= 896

Question 17.
Four parallel lines intersect another set of five parallel lines. Find the number of distinct parallelograms formed.
Solution:
The first set has 4 parallel lines and another set has 5 parallel lines.
To form a parallelogram, we need 2 lines from each set.
∴ Required number of distinct parallelograms formed = ⁴C₂ × ⁵C₂
= 6 × 10
= 60

Question 18.
There are 12 distinct points A, B, C, …, L, in order, on a circle. Lines are drawn passing through each pair of points.
(i) How many lines are there in total?
(ii) How many lines pass through D?
(iii) How many triangles are determined by lines?
(iv) How many triangles have on vertex C?
Solution:
(i) We need two points to draw a line.
∴ Total number of lines = ¹²C₂ = 66

(ii) Lines are drawn passing through each pair of points.
∴ Lines from point D will pass through all the remaining 11 points.
∴ 11 lines pass through D.

(iii) We need three points to draw a triangle.
∴ Number of triangles = ¹²C₃ = 220

(iv) To get the triangles with one vertex as C,
we need two vertices from the remaining 11 vertices.
∴ Number of triangles with vertex at C = ¹¹C₂
= \(\frac{11 \times 10}{2}\)
= 55

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.6 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.6

Question 1.
Find the value of
(a) ¹⁵C₄
Solution:

(b) ⁸⁰C₂
Solution:

(c) ¹⁵C₄ + ¹⁵C₅
Solution:

(d) ²⁰C₁₆ – ¹⁹C₁₆
Solution:

Question 2.
Find n if
(a) ⁶P₂ = n(⁶C₂)
Solution:

(b) ²ⁿC₃ : ⁿC₂ = 52 : 3
Solution:

(c) ⁿC_n-3 = 84
Solution:

Question 3.
Find r if ¹⁴C_2r : ¹⁰C_2r-4 = 143 : 10.
Solution:

∴ 2r(2r – 1) (2r – 2) (2r – 3) = 14 × 12 × 10
∴ 2r(2r – 1) (2r – 2) (2r – 3) = 8 × 7 × 6 × 5
Comparing on both sides, we get
∴ r = 4

Question 4.
Find n and r if,
(a) ⁿP_r = 720 and ⁿC_n-r = 120
Solution:

(b) ⁿC_r-1 : ⁿC_r : ⁿC_r+1 = 20 : 35 : 42
Solution:

Question 5.
If ⁿP_r = 1814400 and ⁿC_r = 45, find ⁿ⁺⁴C_r+3.
Solution:

Question 6.
If ⁿC_r-1 = 6435, ⁿC_r = 5005, ⁿC_r+1 = 3003, find ^rC₅.
Solution:

Question 7.
Find the number of ways of drawing 9 balls from a bag that has 6 red balls, 8 green balls, and 7 blue balls so that 3 balls of every colour are drawn.
Solution:
9 balls are to be selected from 6 red, 8 green, 7 blue balls such that the selection consists of 3 balls of each colour.
∴ 3 red balls can be selected from 6 red balls in ⁶C₃ ways.
3 reen balls can be selected from 8 green balls in ⁸C₃ ways.
3 blue balls can be selected from 7 blue balls in ⁷C₃ ways.
∴ Number of ways selection can be done if the selection consists of 3 balls of each colour

Question 8.
Find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls.
Solution:
There are 6 boys and 4 girls.
A team of 3 boys and 2 girls is to be selected.
∴ 3 boys can be selected from 6 boys in ⁶C₃ ways.
2 girls can be selected from 4 girls in ⁴C₂ ways.
∴ Number of ways the team can be selected

Question 9.
After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting.
Solution:
Let there be n participants present in the meeting.
A handshake occurs between 2 persons.
∴ Number of handshakes = ⁿC₂
Given 66 handshakes were exchanged.
66 = ⁿC₂
66 = \(\frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}\)
66 × 2 = \(\frac{n(n-1)(n-2) !}{(n-2) !}\)
132 = n (n – 1)
n(n – 1) = 12 × 11
Comparing on both sides, we get n = 12
∴ 12 participants were present at the meeting.

Question 10.
If 20 points are marked on a circle, how many chords can be drawn?
Solution:
To draw a chord we need to join two points on the circle.
There are 20 points on a circle.
∴ Total number of chords possible from these points

Question 11.
Find the number of diagonals of an n-sided polygon. In particular, find the number of diagonals when
(i) n = 10
(ii) n = 15
(iii) n = 12
(iv) n = 8
Solution:
In n-sided polygon, there are ‘n’ points and ‘n’ sides.
∴ Through ‘n’ points we can draw ⁿC₂ lines including sides.
∴ Number of diagonals in n sided polygon = ⁿC₂ – n (n = number of sides)

Question 12.
There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.
Solution:
There are 20 lines such that no two of them are parallel and no three of them are concurrent.
Since no two lines are parallel, they intersect at a point.
∴ Number of points of intersection if no two lines are parallel and no three lines are concurrent = ²⁰C₂
= \(\frac{20 !}{2 ! 18 !}\)
= \(\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}\)
= 190

Question 13.
Ten points are plotted on a plane. Find the number of straight lines obtained by joining these points if (a) no three points are collinear (b) four points are collinear
Solution:
There are 10 points on a plane.
(a) When no three of them are collinear.
A line is obtained by joining 2 points.
∴ Number of lines passing through these points = ¹⁰C₂
= \(\frac{10 !}{2 ! 8 !}\)
= \(\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}\)
= 5 × 9
= 45

(b) When 4 of them are collinear.
If no three points are collinear, we get a total of ¹⁰C₂ = 45 lines by joining them. …..[From (i)]
Since 4 points are collinear, only one line passes through these points instead of ⁴C₂ lines.
∴ ⁴C₂ – 1 extra lines are included in 45 lines.
Number of lines passing through these points
= 45 – (⁴C₂ – 1)
= 45 – \(\frac{4 !}{2 ! 2 !}\) + 1
= 45 – \(\frac{4 \times 3 \times 2 !}{2 \times 2 !}\) + 1
= 45 – 6 + 1
= 40

Question 14.
Find the number of triangles formed by joining 12 points if
(a) no three points are collinear
(b) four points are collinear
Solution:
There are 12 points on the plane.
(a) When no three of them are collinear.
A triangle can be drawn by joining any three non-collinear points.
∴ Number of triangles that can be obtained from these points = ¹²C₃
= \(\frac{12 !}{3 ! 9 !}\)
= \(\frac{12 \times 11 \times 10 \times 9 !}{3 \times 2 \times 1 \times 9 !}\)
= 220

(b) When 4 of these points are collinear.
If no three points are collinear, total we get ¹²C₃ = 220 triangles by joining them. ……[From (i)]
Since 4 points are collinear, no triangle can be formed by joining these four points.
∴ ⁴C₃ extra triangles are included in 220 triangles.
∴ Number of triangles that can be obtained from these points = ¹²C₃ – ⁴C₃
= 220 – \(\frac{4 !}{3 ! \times 1 !}\)
= 220 – \(\frac{4 \times 3 !}{3 !}\)
= 220 – 4
= 216

Question 15.
A word has 8 consonants and 3 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are chosen?
Solution:
There are 8 consonants and 3 vowels.
From 8 consonants, 4 can be selected in ⁸C₄
= \(\frac{8 !}{4 ! 4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 \times 3 \times 2 \times 1 \times 4 !}\)
= 70 ways.
From 3 vowels, 2 can be selected in ³C₂
= \(\frac{3 !}{2 ! 1 !}\)
= \(\frac{3 \times 2 !}{2 !}\)
= 3 ways.
Now, to form a word, these 6 ietters (i.e., 4 consonants and 2 vowels) can be arranged in ⁶P₆ = 6! ways.
∴ Total number of words that can be formed = 70 × 3 × 6!
= 70 × 3 × 720
= 151200
∴ 151200 words of 4 consonants and 2 vowels can be formed.

Question 16.
Find n if,
(i) ⁿC₈ = ⁿC₁₂
Solution:
ⁿC₈ = ⁿC₁₂
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ 8 = 12 or 8 = n – 12
But 8 = 12 is not possible
∴ 8 = n – 12
∴ n = 20

(ii) ²³C_3n = ²³C_2n+3
Solution:
²³C_3n = ²³C_2n+3
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ 3n = 2n + 3 or 3n = 23 – 2n – 3
∴ n = 3 or n = 4

(iii) ²¹C_6n = \({ }^{21} \mathrm{C}_{\left(\mathrm{n}^{2}+5\right)}\)
Solution:
²¹C_6n = \({ }^{21} \mathrm{C}_{\left(\mathrm{n}^{2}+5\right)}\)
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ 6n = n² + 5 or 6n = 21 – (n² + 5)
∴ n² – 6n + 5 = 0 or 6n = 21 – n² – 5
∴ n² – 6n + 5 = 0 or n² + 6n – 16 = 0
If n² – 6n + 5 = 0, then (n – 1)(n – 5) = 0
∴ n = 1 or n = 5
If n = 5 then
n² + 5 = 30 > 21
∴ n ≠ 5
∴ n = 1
If n² + 6n – 16 = 0, then (n + 8)(n – 2) = 0
n = -8 or n = 2
n ≠ -8
∴ n = 2
∴ n = 1 or n = 2

Check:
n = 2
∴ n² + 5 = 2² + 5 = 9
²¹C_6n = ²¹C₁₂
and \({ }^{21} \mathrm{C}_{\left(\mathrm{n}^{2}+5\right)}\) = ²¹C₉
∴ ⁿC_r = ⁿC_n-r
∴ ²¹C₁₂ = ²¹C₉
∴ n = 2 is a right answer.

(iv) ²ⁿC_r-1 = ²ⁿC_r+1
Solution:
²ⁿC_r-1 = ²ⁿC_r+1
If ⁿC_x = ⁿC_y, then either x = y or x = n – y
∴ r – 1 = r + 1 or r – 1 = 2n – (r + 1)
But r – 1 = r + 1 is not possible
∴ r – 1 = 2n – (r + 1)
∴ r + r = 2n
∴ r = n

Check:
²ⁿC_r-1 = ²ⁿC_n-1
and ²ⁿC_r+1 = ²ⁿC_n+1
using ⁿC_r = ⁿC_n-r, we have
²ⁿC_n+1 = ²ⁿC_2n-(n+1) = ²ⁿC_n-1
∴ ²ⁿC_r-1 = ²ⁿC_r+1

(v) ⁿC_n-2 = 15
Solution:
ⁿC_n-2 = 15
∴ ⁿC₂ = 15 …..[∵ ⁿC_r = ⁿC_n-r]
∴ \(\frac{n !}{(n-2) ! 2 !}=15\)
∴ \(\frac{n(n-1)(n-2) !}{(n-2) ! \times 2 \times 1}=15\)
∴ n(n – 1) = 30
∴ n(n – 1) = 6 × 5
Equating both sides, we get
∴ n = 6

Question 17.
Find x if ⁿP_r = x ⁿC_r.
Solution:

Question 18.
Find r if ¹¹C₄ + ¹¹C₅ + ¹²C₆ + ¹³C₇ = ¹⁴C_r.
Solution:

Question 19.
Find the value of \(\sum_{r=1}^{4}{ }^{(21-r)} \mathrm{C}_{4}\).
Solution:

Question 20.
Find the differences between the greatest values in the following:
(a) ¹⁴C_r and ¹²C_r
Solution:
Greatest value of ¹⁴C_r.
Here, n = 14, which is even.
Greatest value of nCr occurs at r = \(\frac{n}{2}\) if n is even.
∴ r = \(\frac{n}{2}\)
∴ r = \(\frac{14}{2}\) = 7

∴ Difference between the greatest values of ¹⁴C_r and ¹²C_r = ¹⁴C_r – ¹²C_r
= 3432 – 924
= 2508

(b) ¹³C_r and ⁸C_r
Solution:
Greatest value of ¹³C_r.
Here n = 13, which is odd.
Greatest value of nCr occurs at r = \(\frac{n-1}{2}\) if n is odd.
∴ r = \(\frac{\mathrm{n}-1}{2}\)
∴ r = \(\frac{13-1}{2}\) = 6

∴ Difference between the greatest values of ¹³C_r and ⁸C_r = ¹³C_r – ⁸C_r
= 1716 – 70
= 1646

(c) ¹⁵C_r and ¹¹C_r
Solution:
Greatest value of ¹⁵C_r.
Here n = 15, which is odd.
Greatest value of nCr occurs at r = \(\frac{n-1}{2}\) if n is odd.
∴ r = \(\frac{n-1}{2}\)
∴ r = \(\frac{15-1}{2}\) = 7

∴ Difference between the greatest values of ¹⁵C_r and ¹¹C_r = ¹⁵C_r – ¹¹C_r
= 6435 – 462
= 5973

Question 21.
In how many ways can a boy invite his 5 friends to a party so that at least three join the party?
Solution:
Boy can invite = (3 or 4 or 5 friends)
Consider the following table:

∴ Number of ways a boy can invite his friends to a party so that three or more of join the party = 10 + 5 + 1 = 16

Question 22.
A group consists of 9 men and 6 women. A team of 6 is to be selected. How many of possible selections will have at least 3 women?
Solution:
There are 9 men and 6 women.
A team of 6 persons is to be formed such that it consist of at least 3 women.
Consider the following table:

∴ No. of ways this can be done = 1680 + 540 + 54 + 1 = 2275
∴ 2275 teams can be formed if team consists of at least 3 women.

Question 23.
A committee of 10 persons is to be formed from a group of 10 women and 8 men. How many possible committees will have at least 5 women? How many possible committees will have men in majority?
Solution:
(i) A committee of 10 persons is to be formed from 10 women and 8 men such that the committee contains at least 5 women.
Consider the following table:

∴ Number of committees with at least 5 women
= 14112 + 14700 + 6720 + 1260 + 81
= 36873

(ii) Number of committees with men in majority = Total number of committees – (Number of committees with women in majority + women and men equal in number)
= ¹⁸C₁₀ – 36873
= ¹⁸C₈ – 36873
= 43758 – 36873
= 6885

Question 24.
A question paper has two sections. Section I has 5 questions and section II has 6 questions. A student must answer at least two questions from each section among 6 questions he answers. How many different choices does the student have in choosing questions?
Solution:
There are 11 questions, out of which 5 questions are from section I and 6 questions are from section II.
The student has to select 6 questions taking at least 2 questions from each section.
Consider the following table:

∴ Number of choices = 150 + 200 + 75 = 425
∴ In 425 ways students can select 6 questions, taking at least 2 questions from each section.

Question 25.
There are 3 wicketkeepers and 5 bowlers among 22 cricket players. A team of 11 player is to be selected so that there is exactly one wicketkeeper and at least 4 bowlers in the team. How many different teams can be formed?
Solution:
There are 22 cricket players, of which 3 are wicketkeepers and 5 are bowlers.
A team of 11 players is to be chosen such that exactly one wicket keeper and at least 4 bowlers are to be included in the team.
Consider the following table:

∴ Number of ways a team of 11 players can be selected
= 45045 + 6006
= 51051

Question 26.
Five students are selected from 11. How many ways can these students be selected if
(a) two specified students are selected?
(b) two specified students are not selected?
Solution:
5 students are to be selected from 11 students.
(a) When 2 specified students are included,
then remaining 3 students can be selected from (11 – 2) = 9 students.
∴ Number of ways of selecting 3 students from 9 students = ⁹C₃
= \(\frac{9 !}{3 ! \times 6 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 !}{3 \times 2 \times 1 \times 6 !}\)
= 84
∴ Selection of students is done in 84 ways when 2 specified students are included.

(b) When 2 specified students are not included, then 5 students can be selected from the remaining (11 – 2) = 9 students.
∴ Number of ways of selecting 5 students from 9 students = ⁹C₅
= \(\frac{9 !}{5 ! 4 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 \times 3 \times 2 \times 1}\)
= 126
∴ Selection of students is done in 126 ways when 2 specified students are not included.

Balbharti Maharashtra State Board 11th Maths Book Solutions Pdf Chapter 3 Permutations and Combination Ex 3.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 3 Permutations and Combination Ex 3.5

Question 1.
In how many different ways can 8 friends sit around a table?
Solution:
We know that ‘n’ persons can sit around a table in (n – 1)! ways.
∴ 8 friends can sit around a table in 7!
= 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040 ways.

Question 2.
A party has 20 participants. Find the number of distinct ways for the host to sit with them around a circular table. How many of these ways have two specified persons on either side of the host?
Solution:
A party has 20 participants.
All of them and the host (i.e., 21 persons) can be seated at a circular table in (21 – 1)! = 20! ways.
When two particular participants are seated on either side of the host.
The host takes the chair in 1 way.
These 2 persons can sit on either side of the host in 2! ways.
Once the host occupies his chair, it is not circular permutation more.
The remaining 18 people occupy their chairs in 18! ways.
∴ A total number of arrangements possible if two particular participants are seated on either side of the host = 2! × 18! = 2 × 18!

Question 3.
Delegates from 24 countries participate in a round table discussion. Find the number of seating arrangements where two specified delegates are (a) always together. (b) never together.
Solution:
(a) Delegates of 24 countries are to participate in a round table discussion such that two specified delegates are always together.
Let us consider these 2 delegates as one unit. They can be arranged among themselves in 2! ways.
Also, these two delegates are to be seated with 22 other delegates (i.e. total of 23) which can be done in (23 – 1)! = 22! ways.
∴ Required number of arrangements = 2! × 22!

(b) When 2 specified delegates are never together then, other 22 delegates can be participate in a round table discussion in (22 – 1)! = 21! ways.
∴ There are 22 places of which any 2 places can be filled by those 2 delegates so that they are never together.
∴ Two specified delegates can be arranged in ²²P₂ ways.
∴ Required number of arrangements = ²²P₂ × 21!
= \(\frac{22 !}{(22-2) !} \times 21 !\)
= \(\frac{22 !}{20 !}\) × 21!
= 22 × 21 × 21!
= 21 × 22 × 21!
= 21 × 22!

Question 4.
Find the number of ways for 15 people to sit around the table so that no two arrangements have the same neighbours.
Solution:
There are 15 people to sit around a table.
∴ They can be arranged in(15 – 1)! = 14! ways.
But, they should not have the same neighbour in any two arrangements.
Around the table, arrangements (i.e., clockwise and anticlockwise) coincide.

∴ Required number of arrangements = \(\frac{14 !}{2}\)

Question 5.
A committee of 10 members sits around a table. Find the number of arrangements that have the President and the Vice-president together.
Solution:
A committee of 10 members sits around a table.
But, President and Vice-president sit together.
Let us consider President and Vice-president as one unit.
They can be arranged among themselves in 2! ways.
Now, this unit with the other 8 members of the committee is to be arranged around a table, which can be done in (9 – 1)! = 8! ways.
∴ Required number of arrangements = 8! × 2! = 2 × 8!

Question 6.
Five men, two women, and a child sit around a table. Find the number of arrangements where the child is seated (a) between the two women. (b) between two men.
Solution:
5 men, 2 women, and a child sit around a table.
(a) When the child is seated between two women.
5 men, 2 women, and a child are to be seated around a round table such that the child is seated between two women.
∴ the two women can be seated on either side of the child in 2! ways.
Let us consider these 3 (two women and a child) as one unit.
Now, this one unit is to be arranged with the remaining 5 men,
i.e., a total of 6 units are to be arranged around a round table, which can be done in (6 – 1)! = 5! ways.
∴ Required number of arrangements = 5! × 2!
= 120 × 2
= 240

(b) Two men can be selected from 5 men in
⁵C₂ = \(\frac{5 !}{2 !(5-2) !}=\frac{5 \times 4 \times 3 !}{2 \times 3 !}\) = 10 ways.
Also, these two men can sit on either side of the child in 2! ways.
Let us take two men and a child as one unit.
Now, this one unit is to be arranged with the remaining 3 men and 2 women,
i.e., a total of 6 units (3 + 2 + 1) are to be arranged around a round table, which can be done in (6 – 1)! = 5! ways.
∴ Required number of arrangements = 10 × 2! × 5!
= 10 × 2 × 120
= 2400

Question 7.
Eight men and six women sit around a table. How many sitting arrangements will have no two women together?
Solution:
8 men can be seated around a table in (8 – 1)! = 7! ways.
No two women should sit together.
There are 8 gaps created by 8 men’s seats.
∴ Women can be seated in 8 gaps in ⁸P₆ ways.
∴ Required number of arrangements = 7! × ⁸P₆

Question 8.
Find the number of seating arrangements for 3 men and 3 women to sit around a table so that exactly two women are together.
Solution:
2 women (who wish to sit together) can be selected from 3 in
³C₂ = \(\frac{3 !}{2 !(3-2) !}=\frac{3 \times 2 !}{2 ! \times 1 !}\) = 3 ways.
Also, these two women can sit together in 2! ways.
Let us take two women as one unit.
Now, this one unit is to be arranged with the remaining 3 men and 1 woman,
i.e., a total of 5 units are to be arranged around a round table, which can be done in (5 – 1)! = 4! ways.
∴ Required number of arrangements = 3 × 2! × 4!
= 3 × 2 × 24
= 144

Question 9.
Four objects in a set of ten objects are alike. Find the number of ways of arranging them in a circular order.
Solution:
Ten things can be arranged in a circular order of which 4 are alike in \(\frac{9 !}{4 !}\) ways.
∴ Required number of arrangements = \(\frac{9 !}{4 !}\)

Question 10.
Fifteen persons sit around a table. Find the number of arrangements that have two specified persons not sitting side by side.
Solution:
Since 2 particular persons can’t be sitting side by side,
the other 13 persons can be arranged around the table in (13 – 1)! = 12! ways.
The two persons who are not sitting side by side may take 13 positions created by 3 persons in ¹³P₂ ways.
∴ Required number of arrangements = 12! × ¹³P₂
= 12! × 13 × 12
= 13 × 12! × 12
= 12 × 13!

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 15 Hydrocarbons Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 15 Hydrocarbons

1. Choose correct options

Question A.
Which of the following compound has the highest boiling point?
a. n-pentane
b. iso-butane
c. butane
d. neopentane
Answer:
a. n-pentane

Question B.
Acidic hydrogen is present in :
a. acetylene
b. ethane
c. ethylene
d. dimethyl acetylene
Answer:
a. acetylene

Question C.
Identify ‘A’ in the following reaction:

a. KMnO₄/H⁺
b. alkaline KMnO₄
c. dil. H₂SO₄/1% HgSO₄
d. NaOH/H₂O₂
Answer:
a. KMnO₄/H⁺

Question D.
Major product of chlorination of ethyl benzene is :
a. m-chlorethyl benzene
b. p-chloroethyl benzene
c. chlorobenzene
d. o-chloroethylbenzene
Answer:
b. p-chloroethyl benzene

Question E.
1 – chloropropane on treatment with alc. KOH produces :
a. propane
b. propene
c. propyne
d. propyl alcohol
Answer:
b. propene

2. Name the following :

Question A.
The type of hydrocarbon that is used as lubricant.
Answer:
Waxes

Question B.
Alkene used in the manufacture of polythene bags.
Answer:
Ethene

Question C.
The hydrocarbon said to possess carcinogenic property.
Answer:
Benzene

Question D.
What are the main natural sources of alkane?
Answer:
Crude petroleum and natural gas.

Question E.
Arrange the three isomers of alkane with malecular formula C₅H₁₂ in increasing order of boiling points and write their IUPAC names.
Answer:
The three isomers of alkane with molecular formula C₅H₁₂ are as follows:

The increasing order of their boiling point is I > II > III.

Question F.
Write IUPAC names of the products obtained by the reaction of cold concentrated sulphuric acid followed by water with the following compounds.
a. propene
b. but-1-ene
Answer:
a. propene:

b. but-1-ene:

Question G.
Write the balanced chemical reaction for preparation of ethane from
a. Ethyl bromide
b. Ethyl magnesium iodide
Answer:
a. Preparation of ethane from ethyl bromide:

b. Preparation of ethane from ethyl magnesium iodide:

Question H.
How many monochlorination products are possible for
a. 2-methylpropane ?
b. 2-methylbutane ?
Draw their structures and write their IUPAC names.
Answer:
a. Possible monochlorination products for 2-methylpropane:

b. Possible monochlorination products for 2-methylbutane:

Question I.
Write all the possible products for pyrolysis of butane.
Answer:
Possible products for pyrolysis of butane are:

Question J.
Which of the following will exhibit geometical isomerism ?

Answer:
Compound (c) will exhibit geometrical isomerism.

Question K.
What is the action of following on ethyl iodide ?
a. alc. KOH
b. Zn, HCl
Answer:
a. Action of alc. KOH on ethyl iodide:

b. Action of Zn/HCl on ethyl iodide:

Question L.
An alkene ‘A’ an ozonolysis gives 2 moles of ethanal. Write the structure and IUPAC name of ‘A’.
Answer:
Structure of A: CH₃ – CH = CH – CH₃
IUPAC name of A: But-2-ene

Question M.
Acetone and acetaldehyde are the ozonolysis products of an alkene. Write the structural formula of an alkene and give IUPAC name of it.
Answer:
The structural formula of alkene:

IUPAC name is 2-methylbut-2-ene.

Question N.
Write the reaction to convert
a. propene to n-propyl alcohol.
b. propene to isoproyl alcohol.
Answer:
a.

b.

Question O.
What is the action of following on but-2-ene ?
a. dil alkaline KMnO₄
b. acidic KMnO₄
Answer:
a. Action of dil. alkaline KMnO₄ on but-2-ene:

b. Action of acidic KMnO₄ on but-2-ene:

Question P.
Complete the following reaction sequence :

Answer:

Question Q.
Write the balanced chemical reactions to get benzene from
a. Sodium benzoate.
b. Phenol.
Answer:
a. Sodium benzoate:
When anhydrous sodium benzoate is heated with soda lime, it undergoes decarboxylation and gives benzene.

b. Phenol:
When vapours of phenol are passed over heated zinc dust, it undergoes reduction and gives benzene.

Question R.
Predict the possible products of the following reaction.
a. chlorination of nitrobenzene,
b. sulfonation of chlorobenzene,
c. bromination of phenol,
d. nitration of toluene.
Answer:
a. Nitro group is meta directing group. So, chlorination of nitrobenzene gives m-chloronitrobenzene.

b. Chloro group is ortho and para directing group. So, sulphonation of chlorobenzene gives p-chlorobenzene sulphonic acid and o- chlorobenzene sulphonic acid.

c. Phenolic -OH group is ortho and para directing group. So, bromination of phenol gives p-bromophenol and o-bromophenol.

d. Methyl group is ortho and para directing group. So, nitration of toluene gives p-nitrotoluene and o-nitrotoluene.

3. Identify the main product of the reaction

Answer:
a.

b.

c.

d.

4. Read the following reaction and answer the questions given below.

a. Write IUPAC name of the product.
b. State the rule that governs formation of this product.
Answer:
a. IUPAC name of the product: 1 -Bromo-2-methylpropane
b. Anti-Markownikov’s rule/Kharasch effect/peroxide effect: It states that, the addition of HBr to unsymmetrical alkene in the presence of organic peroxide (R-O-O-R) takes place in the opposite orientation to that suggested by Markovnikov’s rule.

5. Identify A, B, C in the following reaction sequence :

Answer:

6. Identify giving reason whether the following compounds are aromatic or not.

Answer:
A.

Compound is non-aromatic since it has 4π electrons and hence, does not obey Huckel rule of aromaticity.

B.

Compound is non-aromatic since it has 4π electrons and hence, does not obey Huckel rule of aromaticity.

C.

Compound is aromatic since it has 6π electrons and hence, obeys Huckel rule of aromaticity.

D.

Compound is aromatic since it has 6n electrons and hence, obeys Huckel rule of aromaticity.

7. Name two reagents used for acylation of benzene.
Answer:
The two reagents used for acylation of benzene are:
i. CH₃COCl (acetyl chloride) and anhydrous AlCl₃
ii. (CH₃CO)₂O (acetic anhydride) and anhydrous AlCl₃

8. Read the following reaction and answer the questions given below.

A. Write the name of the reaction.
B. Identify the electrophile in it.
C. How is this electrophile generated?
Answer:
A. The name of the reaction is Friedel-Craft’s alkylation reaction.
B. The electrophile in the reaction is ⁺CH₃.
C. The electrophile ⁺CH₃ is generated as follows:

Activity:

Prepare chart of hydrocarbons and note down the characteristics.
Answer:

Characteristics of hydrocarbons:

• They are chemical compounds that are formed from only hydrogen and carbon atoms.
• Both ‘C’ and ‘H’ share an electron pair forming covalent bonds.
• One of the special properties of carbon is its ability to form double and triple bonds (unsaturation). Saturated hydrocarbons are alkanes and cycloalkanes while the unsaturated hydrocarbons are the aromatics, alkenes and alkynes.
• All hydrocarbons are insoluble in water, their boiling point increases as the size of alkane increases.
• All hydrocarbons can reach complete oxidation.
• Hydrocarbons are mainly used as fuel for transport and industry.

[Note: Students are expected to collect additional information on hydrocarbons on their own.]

11th Chemistry Digest Chapter 15 Hydrocarbons Intext Questions and Answers

Can you recall? (Textbook Page No. 233)

Question i.
What are hydrocarbons?
Answer:
The compounds which contain carbon and hydrogen as the only elements are called hydrocarbons.

Question ii.
Write structural formulae of the following compounds: propane, ethyne, cyclobutane, ethene, benzene.
Answer:

Do you know? (Textbook Page No. 233)

Question 1.
Why are alkanes called paraffins?
Answer:
i. Alkanes contain only carbon-carbon and carbon-hydrogen single covalent bonds.
ii. They are chemically less reactive and do not have much affinity for other chemicals.
Hence, they are called paraffins.

Internet my friend. (Textbook Page No. 233)

Question 1.
Collect information about hydrocarbon.
Answer:

• In organic chemistry, a hydrocarbon is an organic compound consisting of carbon and hydrogen as the only elements.
• They are examples of group 14 hydrides.
• Alkanes, cycloalkanes, aromatic hydrocarbons are different types of hydrocarbons.
• Most of the hydrocarbons found on earth occur naturally in crude oil.
• They mainly undergo substitution, addition or combustion reactions.
• Most hydrocarbons are flammable and toxic.
• They are the primary energy source in the form of combustible fuel source.

[Note: Students are expected to collect additional information on their own]

Use your brain power! (Textbook Page No. 234)

Question 1.
i. Write the structures of all the chain isomers of the saturated hydrocarbon containing six carbon atoms.
ii. Write IUPAC names of all the above structures.
Answer:
The structural formulae and names of all possible isomers having molecular formula C₆H₁₄ are as follows:

Note:
Alkanes and isomer number

Number of Carbon	Alkane	Number of isomers
1	Methane	No structural isomer
2	Ethane	No structural isomer
3	Propane	No structural isomer
4	Butane	Two
5	Pentane	Three
6	Hexane	Five

Can you recall? (Textbook Page No. 235)

Question i.
What is a catalyst?
Answer:
A catalyst is a substance that can be added to a reaction to increase the reaction rate without getting consumed in the process.
e.g. Ni is used as a catalyst in the catalytic hydrogenation of alkenes or alkynes.

Question ii.
What is addition reaction?
Answer:
When a compound combines with another compound to form a product that contain all the atoms in both the reactants, it is called an addition reaction.

Try this (Textbook Page No. 235)

Question 1.
Transform the following word equation into balanced chemical equation and write at least 3 changes that occur at molecular level during this chemical change.
\(\text { 2-Methylpropene + Hydrogen } \stackrel{\text { catalyst }}{\longrightarrow} \text { Isobutane }\)
Answer:

Three changes which occur at molecular level include:
Step 1: Adsorption of reactants: Reactants (alkene and hydrogen) get adsorbed on the catalytic surface.
Step 2: Formation of a product: Hydrogen atoms are added across the double bond of 2-methylpropene which results in the formation of product isobutane.
Step 3: Desorption: Product formed on the catalytic surface is readily desorbed making catalytic surface available for other molecules.

Use your brain power! (Textbook Page No. 236)

Question 1.
Why are alkanes insoluble in water and readily soluble in organic solvents like chloroform or ether?
Answer:

• The solubility of any substance is governed by the principle of like dissolves like. This means polar compounds are soluble in polar solvents while nonpolar compounds are soluble in nonpolar solvents.
• Alkanes consist of C – C and C – H nonpolar covalent bonds and thus, they are nonpolar in nature, whereas water is a polar solvent.
• The dipole-dipole forces that exist between water molecules is much stronger than the forces of attraction between alkane and water molecules.

Hence, alkanes are insoluble in water and readily soluble in organic solvents like chloroform or ether.

Can you recall? (Textbook Page No. 238)

Question 1.
What is the product which is poisonous and causes air pollution formed by incomplete combustion of alkane?
Answer:
When alkanes are subjected to incomplete combustion, it forms carbon monoxide and carbon (soot).
i. 2CH_4(g) + 3O_2(g) → 2CO_(g) + 4H₂O_(g)
ii. CH_4(g) + O_2(g) → C_(s) + 2H₂O_(l)

Can you recall? (Textbook Page No. 238)

Question i.
What are alkenes?
Answer:
Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond.

Question ii.
Calculate the number of sigma (σ) and pi (π) bonds in 2-methylpropene.
Answer:

Question iii.
Write the structural formula of pent-2-ene.
Answer:

Can you tell? (Textbook Page No. 241)

Question i.
Explain by writing a reaction, the main product formed on heating 2-methylbutan-2-ol with concentrated sulphuric acid.
Answer:

Question ii.
Will the main product in the above reaction show geometrical isomerism?
Answer:
No, the major product, i.e., 2-methylbut-2-ene does not show geometrical (or cis-trans) isomerism.

Can you tell? (Textbook Page No. 244)

Question 1.
Propan-1-ol and 2-methypropan-1-ol are not prepared by hydration method. Why?
Answer:
Propan-1-ol and 2-methylpropan-1-ol cannot be prepared by hydration of propene and 2-methylprop-1-ene because the addition reaction follows Markovnikov’s rule.

Use your brainpower. (Textbook Page No. 244)

Question 1.
On ozonolysis, an alkene forms the following carbonyl compounds. Draw the structure of unknown alkene from which these compounds are formed: HCHO and CH₃COCH₂CH₃
Answer:
The structure of alkene which produces a mixture of HCHO and CH₃COCH₂CH₃ on ozonolysis is

Use your brain power! (Textbook Page No. 245)

Question 1.
Write the structure of monomer from which each of the following polymers are obtained.

Answer:

	Polymer	Monomeric unit
i.	Teflon	CF2 – CF2 Tetrafluoroethene
ii.	Polypropene	H3C – CH = CH2 Propene
iii.	Polyvinyl chloride	H2C = CHCl Vinyl chloride

Can you tell? (Textbook Page No. 246)

Question i.
What are aliphatic hydrocarbons?
Answer:
Aliphatic hydrocarbons are hydrocarbons containing carbon and hydrogen joined together in straight chain or branched chain. They may be saturated (alkanes) or unsaturated (alkenes or alkynes).

Question ii.
Compare the proportion of carbon and hydrogen atoms in ethane, ethene and ethyne. Which compound is most unsaturated with hydrogen?
Answer:
Ethane
C : H = 2 : 6 = 1 : 3
Ethene
C : H = 2 : 4 = 1 : 2
Ethyne
C : H = 2 : 2 = 1 : 1
From the above proportion it is clear that ethyne with 1 : 1 ratio of C : H is most unsaturated with hydrogen (50%) as compared to ethane (25%) and ethene (33.33%).

Can you tell? (Textbook Page No. 247)

Question 1.
Why is sodamide used in dehydrohalogenation of vicinal dihalides to remove HX from alkenyl halide in place of alcoholic KOH?
Answer:

• Sodamide (NaNH₂) is a strong base and hence, helps in complete conversion of alkenyl halide formed in the first step to form alkynes.
• The base (KOH or NaOH) used in first step gives alkynes in poor yield and hence, stronger bases such as NaNH₂ on KNH₂ are used in second step.

Use your brainpower! (Textbook Page No. 247)

Question 1.
Convert: 1-Bromobutane to hex-1-yne
Answer:

Can you tell? (Textbook Page No. 248)

Question 1.
Alkanes and alkenes do not react with lithium amide. Give reason.
Answer:
i. The sp hybrid carbon atom in terminal alkynes is more electronegative than the sp² carbon in ethene or the sp³ carbon in ethane.
ii. Due to high electronegative character of carbon in terminal alkynes, hydrogen atom can be given away as proton (H⁺) to very strong base as shown in the reactions below.

iii. Further, since s-character decreases from sp to sp² to sp³ carbon atom, the relative acidity of alkanes, alkenes and alkynes is in the following order: H – C = C – H > H₂C = CH₂ > H₃C – CH₃
Hence, alkenes and alkanes do not react with lithium amide.

Use your brain power! (Textbook Page No. 248)

Question 1.
Arrange following hydrocarbons in the increasing order of acidic character: propane, propyne, propene.
Answer:
Propyne > propene > propane

Use your brain power! (Textbook Page No. 249)

Question 1.
Convert: 3-Methylbut-l-yne into 3-methylbutan-2-one
Answer:

Can you recall? (Textbook Page No. 249)

Question i.
What are aromatic hydrocarbons?
Answer:
Benzene and all compounds that have structures and chemical properties resembling benzene are called as aromatic hydrocarbons.

Question ii.
What are benzenoid and non-benzenoid aromatics?
Answer:
Benzenoid aromatics are compounds having at least one benzene ring in the structure.
e.g. Benzene, naphthalene, anthracene, phenol, etc.,
Non-benzenoid aromatics are compounds that contain an aromatic ring, other than benzene. e.g. Tropone, etc.

Can you recall? (Textbook Page No. 254)

Question 1.
What is decarboxylation?
Answer:
The reaction which involves removal of a carboxyl group (-COOH) in the form of carbon dioxide (CO₂) is known as decarboxylation reaction.
R – COOH → R – H + CO₂

Balbharti Maharashtra State Board 11th Physics Textbook Solutions Chapter 13 Electromagnetic Waves and Communication System Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Physics Solutions Chapter 13 Electromagnetic Waves and Communication System

1. Choose the correct option.

Question 1.
The EM wave emitted by the Sun and responsible for heating the Earth’s atmosphere due to green house effect is
(A) Infra-red radiation
(B) X ray
(C) Microwave
(D) Visible light
Answer:
(A) Infra-red radiation

Question 2.
Earth’s atmosphere is richest in
(A) UV
(B) IR
(C) X-ray
(D) Microwaves
Answer:
(B) IR

Question 3.
How does the frequency of a beam of ultraviolet light change when it travels from air into glass?
(A) depends on the values of p and e
(B) increases
(C) decreases
(D) remains same
Answer:
(D) remains same

Question 4.
The direction of EM wave is given by
(A) \(\bar{E}\) × \(\bar{B}\)
(B) \(\bar{E}\).\(\bar{B}\)
(C) along \(\bar{E}\)
(D) along \(\bar{B}\)
Answer:
(A) \(\bar{E}\) × \(\bar{B}\)

Question 5.
The maximum distance upto which TV transmission from a TV tower of height h can be received is proportional to
(A) h½
(B) h
(C) h^3/2
(D) h²
Answer:
(A) h½

Question 6.
The waves used by artificial satellites for communication purposes are
(A) Microwave
(B) AM radio waves
(C) FM radio waves
(D) X-rays
Answer:
(A) Microwave

Question 7.
If a TV telecast is to cover a radius of 640 km, what should be the height of transmitting antenna?
(A) 32000 m
(B) 53000 m
(C) 42000 m
(D) 55000 m
Answer:
(A) 32000 m

2. Answer briefly.

Question 1.
State two characteristics of an EM wave.
Answer:
i. The electric and magnetic fields, \(\vec{E}\) and \(\vec{B}\) are always perpendicular to each other and also to the direction of propagation of the EM wave. Thus, the EM waves are transverse waves.

ii. The cross product (\(\vec{E}\) × \(\vec{B}\)) gives the direction in which the EM wave travels. (\(\vec{E}\) × \(\vec{B}\)) also gives the energy carried by EM wave.

Question 2.
Why are microwaves used in radar?
Answer:
Microwaves are used in radar systems for identifying the location of distant objects like ships, aeroplanes etc.

Question 3.
What are EM waves?
Answer:
Waves that are caused by the acceleration of charged particles and consist of electric and magnetic fields vibrating sinusoidally at right angles to each other and to the direction of propagation are called EM waves or EM radiation.

Question 4.
How are EM waves produced?
Answer:

1. According to quantum theory, an electron, while orbiting around the nucleus in a stable orbit does not emit EM radiation even though it undergoes acceleration.
2. It will emit an EM radiation only when it falls from an orbit of higher energy to one of lower energy.
3. EM waves (such as X-rays) are produced when fast moving electrons hit a target of high atomic number (such as molybdenum, copper, etc.).
4. An electric charge at rest has an electric field in the region around it but has no magnetic field.
5. When the charge moves, it produces both electric and magnetic fields.
6. If the charge moves with a constant velocity, the magnetic field will not change with time and hence, it cannot produce an EM wave.
7. But if the charge is accelerated, both the magnetic and electric fields change with space and time and an EM wave is produced.
8. Thus, an oscillating charge emits an EM wave which has the same frequency as that of the oscillation of the charge.

Question 5.
Can we produce a pure electric or magnetic wave in space? Why?
Answer:
No.
In vacuum, an electric field cannot directly induce another electric field so a “pure” electric field wave cannot exist and same can be said for a “pure” magnetic wave.

Question 6.
Does an ordinary electric lamp emit EM waves?
Answer:
Yes, ordinary electric lamp emits EM waves.

Question 7.
Why light waves travel in vacuum whereas sound wave cannot?
Answer:
Light waves are electromagnetic waves which can travel in vacuum whereas sound waves travel due to the vibration of particles of medium. Without any particles present (like in a vacuum) no vibrations can be produced. Hence, the sound wave cannot travel through the vacuum.

Question 8.
What are ultraviolet rays? Give two uses.
Answer:
Production:

1. Ultraviolet rays can be produced by the mercury vapour lamp, electric spark and carbon arc lamp.
2. They can also be obtained by striking electrical discharge in hydrogen and xenon gas tubes.
3. The Sun is the most important natural source of ultraviolet rays, most of which are absorbed by the ozone layer in the Earth’s atmosphere.

Uses:

1. Ultraviolet rays destroy germs and bacteria and hence they are used for sterilizing surgical instruments and for purification of water.
2. Used in burglar alarms and security systems.
3. Used to distinguish real and fake gems.

Question 9.
What are radio waves? Give its two uses.
Answer:

1. Radio waves are produced by accelerated motion of charges in a conducting wire. The frequency of waves produced by the circuit depends upon the magnitudes of the inductance and the capacitance.
2. Thus, by choosing suitable values of the inductance and the capacitance, radio waves of desired frequency can be produced.

Uses:

1. Radio waves are used for wireless communication purpose.
2. They are used for radio broadcasting and transmission of TV signals.
3. Cellular phones use radio waves to transmit voice communication in the ultra high frequency (UHF) band.

Question 10.
Name the most harmful radiation entering the Earth’s atmosphere from the outer space.
Answer:
Ultraviolet radiation.

Question 11.
Give reasons for the following:
i. Long distance radio broadcast uses short wave bands.
ii. Satellites are used for long distance TV transmission.
Answer:
i. Long distance radio broadcast uses short wave bands because electromagnetic waves only in the frequency range of short wave bands only are reflected by the ionosphere.

ii. a. It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere.
b. Hence, satellites are helpful in long distance TV transmission.

Question 12.
Name the three basic units of any communication system.
Answer:
Three basic (essential) elements of every communication system are transmitter, communication channel and receiver.

Question 13.
What is a carrier wave?
Answer:
The high frequency waves on which the signals to be transmitted are superimposed are called carrier waves.

Question 14.
Why high frequency carrier waves are used for transmission of audio signals?
Answer:
An audio signal has low frequency (<20 kHz) and low frequency signals cannot be transmitted over large distances. Because of this, a high frequency carrier waves are used for transmission.

Question 15.
What is modulation?
Answer:
The signals in communication system (e.g. music, speech etc.) are low frequency signals and cannot be transmitted over large distances. In order to transmit the signal to large distances, it is superimposed on a high frequency wave (called carrier wave). This process is called modulation.

Question 16.
What is meant by amplitude modulation?
Answer:
When the amplitude of carrier wave is varied in accordance with the modulating signal, the process is called amplitude modulation.

Question 17.
What is meant by noise?
Answer:

1. A random unwanted signal is called noise.
2. The source generating the noise may be located inside or outside the system.
3. Efforts should be made to minimize the noise level in a communication system.

Question 18.
What is meant by bandwidth?
Answer:
The bandwidth of an electronic circuit is the range of frequencies over which it operates efficiently.

Question 19.
What is demodulation?
Answer:
The process of regaining signal from a modulated wave is called demodulation. This is the reverse process of modulation.

Question 20.
What type of modulation is required for television broadcast?
Answer:
Amplitude modulation is required for television broadcast.

Question 21.
How does the effective power radiated by an antenna vary with wavelength?
Answer:

1. To transmit a signal, an antenna or an aerial is needed.
2. Power radiated from a linear antenna of length l is, P ∝ (\(\frac {l}{λ}\))²
   where, λ is the wavelength of the signal.

Question 22.
Why should broadcasting programs use different frequencies?
Answer:
If broadcasting programs run on same frequency, then the information carried by these waves will get mixed up with each other. Hence, different broadcasting programs should run on different frequencies.

Question 23.
Explain the necessity of a carrier wave in communication.
Answer:

1. Without a carrier wave, the input signals could be carried by very low frequency electromagnetic waves but it will need quite a bit of amplification in order to transmit those very low frequencies.
2. The input signals themselves do not have much power and need a fairly large antenna in order to transmit the information.
3. Hence, it is necessary to impose the input signal on carrier wave as it requires less power in order to transmit the information.

Question 24.
Why does amplitude modulation give noisy reception?
Answer:
i. In amplitude modulation, carrier is varied in accordance with the message signal.

ii. The higher the amplitude, the greater is magnitude of the signal. So even if due to any reason, the magnitude of the signal changes, it will lead to variation in the amplitude of the signal. So its easy for noise to disturb the amplitude modulated signal.

Question 25.
Explain why is modulation needed.
Answer:
Modulation helps in avoiding mixing up of signals from different transmitters as different carrier wave frequencies can be allotted to different transmitters. Without the use of these waves, the audio signals, if transmitted directly by different transmitters, would get mixed up.

3. Solve the numerical problem.

Question 1.
Calculate the frequency in MHz of a radio wave of wavelength 250 m. Remember that the speed of all EM waves in vacuum is 3.0 × 10⁸ m/s.
Answer:
Given: λ = 250 m, c = 3 × 10⁸ m/s
To find: Frequency (v)
Formula: c = v⁸
Calculation: From formula,
v = \(\frac {c}{λ}\) = \(\frac {3×10^8}{250}\) = 1.2 × 10⁶ Hz
= 1.2 MHz

Question 2.
Calculate the wavelength in nm of an X-ray wave of frequency 2.0 × 10¹⁸ Hz.
Solution:
Given: c = 3 × 10⁸, v = 2 × 10¹⁸ Hz
To find: Wavelength (λ)
Formula: c = vλ
Calculation. From formula,
λ = \(\frac {c}{v}\) = \(\frac {3×10^8}{2×10^{18}}\) = 1.5 × 10^-10
= 0.15 nm

Question 3.
The speed of light is 3 × 10⁸ m/s. Calculate the frequency of red light of wavelength of 6.5 × 10^-7 m.
Answer:
Given: c = 3 × 10⁸ m/s, λ = 6.5 × 10^-7 m
To find: Frequency (v)
Formula: c = vλ
Calculation: From formula,
v = \(\frac {c}{λ}\) = \(\frac {3×10^8}{6.5×10^{-7}}\) = 4.6 × 10¹⁴ Hz

Question 4.
Calculate the wavelength of a microwave of frequency 8.0 GHz.
Answer:
Given: v = 8 GHz = 8 × 10⁹ Hz,
c = 3 × 10⁸ m/s
To find: Wavelength (λ)
Formula: c = vλ
Calculation: From formula,
λ = \(\frac {c}{λ}\) = \(\frac {3×10^8}{8×10^9}\) = 3.75 × 10^-2
= 3.75 cm

Question 5.
In a EM wave the electric field oscillates sinusoidally at a frequency of 2 × 10¹⁰ What is the wavelength of the wave?
Answer:
Given: v = 2 × 10¹⁰ Hz, c = 3 × 10⁸ m
To find: Wavelength (λ)
Formula: c = vλ
Calculation: From formula,
λ = \(\frac {c}{λ}\) = \(\frac {3×10^8}{2×10^{10}}\) = 1.5 × 10^-2

Question 6.
The amplitude of the magnetic field part of a harmonic EM wave in vacuum is B0 = 5 X 10-7 T. What is the amplitude of the electric field part of the wave?
Answer:
Given: B₀ = 5 × 10^-7 T, c = 3 × 10⁸
To find: Amplitude of electric field (E₀)
Formula: c = \(\frac {E_0}{B_0}\)
Calculation /From formula,
E₀ = c × B₀
= 3 × 10⁸ × 5 × 10^-7
= 150 V/m

Question 7.
A TV tower has a height of 200 m. How much population is covered by TV transmission if the average population density around the tower is 1000/km²? (Radius of the Earth = 6.4 × 10⁶ m)
Answer:
Given: h = 200 m,
Population density (n)
= 1000/km² = 1000 × 10^-6/m² = 10^-3/m²
R = 6.4 ×10⁶ m
To find: Population covered
Formulae: i. A = πd² = π(\(\sqrt{2Rh}\))² = 2πRh
ii. Population covered = nA
Calculation /From formula (i),
A = 2πRh
= 2 × 3.142 × 6.4 × 10⁶ × 200
≈ 8 × 10⁹ m²
From formula (ii),
Population covered = nA
= 10^-3 × 8 × 10⁹
= 8 × 10⁶

Question 8.
Height of a TV tower is 600 m at a given place. Calculate its coverage range if the radius of the Earth is 6400 km. What should be the height to get the double coverage area?
Answer:
Given: h = 600 m, R = 6.4 × 10⁶ m
To find: Range (d)
Height to get the double coverage (h’)
Formula: d = \(\sqrt{2hR}\)
Calculation: From formula,
d = \(\sqrt{2×600×6.4×10^6}\) = 87.6 × 10³ = 87.6 km
Now, for A’ = 2A
π(d’)² = 2 (πd²)
∴ (d’)² = 2d²
From formula,
h’ = \(\frac{(d’)^2}{2R}\)
= \(\frac{2d^2}{2R}\)
= 2 × h ……….. (∵ h = \(\frac{d^2}{2R}\))
= 2 × 600
=1200 m

Question 9.
A transmitting antenna at the top of a tower has a height 32 m and that of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in line of sight mode? Given radius of Earth is 6.4 × 10⁶ m.
Answer:
Given: h_t = 32 m, h_r = 50 m, R = 6.4 × 10⁶ m
To find: Maximum distance or range (d)
Formula: d = \(\sqrt{2Rh}\)
Calculation: From formula,
d_t = \(\sqrt{2Rh_t}\) = \(\sqrt{2×6.4×10^6×32}\)
= 20.238 × 10³ m
= 20.238 km
d_r = \(\sqrt{2Rh_t}\)
= \(\sqrt{2×6.4×10^6×50}\)
= 25.298 × 10³ m
= 25.298 km
Now, d = d_t + d_r
= 20.238 + 25.298
= 45.536 km

11th Physics Digest Chapter 13 Electromagnetic Waves and Communication System Intext Questions and Answers

Can you recall? (Textbookpage no. 229)

Question 1.
i. What is a wave?
Answer:
Wave is an oscillatory disturbance which travels through a medium without change in its form.

ii. What is the difference between longitudinal and transverse waves?
Answer:
a. Transverse wave: A wave in which particles of the medium vibrate in a direction perpendicular to the direction of propagation of wave is called transverse wave.
b. Longitudinal wave: A wave in which particles of the medium vibrate in a direction parallel to the direction of propagation of wave is called longitudinal wave.

iii. What are electric and magnetic fields and what are their sources?
Answer:
a. Electric field is the force experienced by a test charge in presence of the given charge at the given distance from it.
b. A magnetic field is produced around a magnet or around a current carrying conductor.

iv. By which mechanism heat is lost by hot bodies?
Answer:
Hot bodies lose the heat in the form of radiation.

Question 2.
What are Lenz’s law, Ampere’s law and Faraday’s law?
Answer:
Lenz’s law:
Whereas, Lenz’s law states that, the direction of the induced emf is such that the change is opposed.

Ampere’s law:
Ampere’s law describes the relation between the induced magnetic field associated with a loop and the current flowing through the loop.

Faraday’s law:
Faraday’s law states that, time varying magnetic field induces an electromotive force (emf) and an electric field.

Internet my friend. (Tpxtboakpage no. 240)

https//www.iiap.res.in/centers/iao
[Students are expected to visit the above mentioned website and collect more information about different EM wave propagations used by astronomical observatories.]