61. ​Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously the number of beats is

1. 1. 7

   
   
   

   2. 8

   
   
   

   3. 3

   
   
   

   4. 5

   
   
   

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1. View Hint View Answer Discuss in Forum

   The frequency of vibration of a string is given by,
   

   f =	1	√	T
   	2I		m

   
   where m is mass per unit length.
   

   f1 =	1	√	T	f2 =	1	√	T
   	2I1		m		2I2		m

   
   

   f2 - f1 =	1	√	T		(l1 - l2)
   	2		m		l1l2

   
   

   √	T	= √	20	= √2 10 × 10² = 1.414 × 100
   	m		10-3

   
   = 141.4
   

   	l1 + l2	=	(51.6 - 49.1) × 10²
   	l1l2		51.6 × 49.1

   
   

   =	2.5 × 10²	=	1
   	50 × 50		10

   
   

   ∴ f2 - f1 =	1	× 141.4 ×	1	= 7 beats
   	2		10

   Correct Option: A

   The frequency of vibration of a string is given by,
   

   f =	1	√	T
   	2I		m

   
   where m is mass per unit length.
   

   f1 =	1	√	T	f2 =	1	√	T
   	2I1		m		2I2		m

   
   

   f2 - f1 =	1	√	T		(l1 - l2)
   	2		m		l1l2

   
   

   √	T	= √	20	= √2 10 × 10² = 1.414 × 100
   	m		10-3

   
   = 141.4
   

   	l1 + l2	=	(51.6 - 49.1) × 10²
   	l1l2		51.6 × 49.1

   
   

   =	2.5 × 10²	=	1
   	50 × 50		10

   
   

   ∴ f2 - f1 =	1	× 141.4 ×	1	= 7 beats
   	2		10

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62. ​A string of 7 m length has a mass of 0.035 kg. If tension in the string is 60.5 N, then speed of a wave on the string is

1. 1. 77 m/s

   
   
   

   2. 102 m/s

   
   
   

   3. )110 m/s

   
   
   

   4. 165 m/s

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given : Length (l) = 7 m
   Mass (M) = 0.035 kg and tension (T) = 60.5 N.
   Therefore, mass of string per unit length (m) =
   

   	0.035	= 0.005 kg/m
   	7

   
   speed of wave
   

   = √	T	√	60.5	= 110 m/s
   	m		0.005

   Correct Option: C

   Given : Length (l) = 7 m
   Mass (M) = 0.035 kg and tension (T) = 60.5 N.
   Therefore, mass of string per unit length (m) =
   

   	0.035	= 0.005 kg/m
   	7

   
   speed of wave
   

   = √	T	√	60.5	= 110 m/s
   	m		0.005

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63. An organ pipe P₁ closed at one end vibrating in its first overtone and another pipe P₂, open at both ends vibrating in its third overtone are in resonance with a given tuning fork. The ratio of lengths of P₁ and P₂ respectively are given by

1. 1. 1 : 2

   
   
   

   2. 1 : 3

   
   
   

   3. 3 : 8

   
   
   

   4. 3 : 4

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that the length of pipe closed at one end for first overtone (l₁) = 3λ/4 and length of the open pipe for third overtone (l₂) = 4λ/2 = 2λ.
   

   Therefore, the ratio of lengths	l1	=	3λ/4	=	3	or l1 : l2 = 3 : 8 .
   	l2		2λ		8

   Correct Option: C

   We know that the length of pipe closed at one end for first overtone (l₁) = 3λ/4 and length of the open pipe for third overtone (l₂) = 4λ/2 = 2λ.
   

   Therefore, the ratio of lengths	l1	=	3λ/4	=	3	or l1 : l2 = 3 : 8 .
   	l2		2λ		8

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64. Two trains move towards each other with the same speed. The speed of sound is 340 m/s. If the height of the tone of the whistle of one of them heard on the other changes 9/8 times, then the speed of each train should be

1. 1. 20 m/s

   
   
   

   2. 2 m/s

   
   
   

   3. 200 m/s

   
   
   

   4. 2000 m/s

   
   
   

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1. View Hint View Answer Discuss in Forum

   Here, v' =	9	v
   	8

   
   Source and observer are moving in opposite direction, therefore, apparent frequency
   

   v' = v &time;	(v + u)
   	(v - u)

   
   

   9	v = v &time;	340 + u
   8		340 - u

   
   ⇒ 9 &time; 340 - 9u = 8 &time; 340 + 8u
   

   ⇒ 17u = 340&time; 1 ⇒ u =	340	= 20 m/sec.
   	17

   
   

   Correct Option: A

   Here, v' =	9	v
   	8

   
   Source and observer are moving in opposite direction, therefore, apparent frequency
   

   v' = v &time;	(v + u)
   	(v - u)

   
   

   9	v = v &time;	340 + u
   8		340 - u

   
   ⇒ 9 &time; 340 - 9u = 8 &time; 340 + 8u
   

   ⇒ 17u = 340&time; 1 ⇒ u =	340	= 20 m/sec.
   	17

   
   

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65. A stretched string resonates with tuning fork frequency 512 Hz when length of the string is 0.5 m. The length of the string required to vibrate resonantly with a tuning fork of frequency 256 Hz would be

1. 1. 0.25 m

   
   
   

   2. 0.5 m

   
   
   

   3. 1 m

   
   
   

   4. 2 m

   
   
   

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1. View Hint View Answer Discuss in Forum

   ƒ =	1		T		1/2
   	2l		μ

   
   When f is halved, the length is doubled.

   Correct Option: D

   ƒ =	1		T		1/2
   	2l		μ

   
   When f is halved, the length is doubled.

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