41. If the length of a simple pendulum is increased by 2%, then the time period​
    

1. 1. increases by 2%

   
   
   

   2. decreases by 2%

   
   
   

   3. ​increases by 1%

   
   
   

   4. decreases by 1%

   
   
   

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1. View Hint View Answer Discuss in Forum

   We know that , T = 2π
   		l
   		g

   
   

   	∆T	× 100 =	1	+	∆ℓ	× 100
   	T		2		ℓ

   
   If length is increased by 2%., time period increases by 1%.

   Correct Option: C

   We know that , T = 2π
   		l
   		g

   
   

   	∆T	× 100 =	1	+	∆ℓ	× 100
   	T		2		ℓ

   
   If length is increased by 2%., time period increases by 1%.

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42. A mass m is vertically suspended from a spring of negligible mass; the system oscillates with a frequency n. What will be the frequency of the system, if a mass 4 m is suspended from the same spring?​​

1. 1. 	n
      	4

   
   
   

   2. 4n

   
   
   

   3. 	n
      	2

   
   
   

   4. 2n

   
   
   

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1. View Hint View Answer Discuss in Forum

   n =	1
   				k
   	2π			m

   
   

   n' =	1				=	1	×	1
   				k						k
   	2π			4m		2		2π		m

   
   On putting the value of n  we get
   

   n' =	n
   	2

   
   

   Correct Option: C

   n =	1
   				k
   	2π			m

   
   

   n' =	1				=	1	×	1
   				k						k
   	2π			4m		2		2π		m

   
   On putting the value of n  we get
   

   n' =	n
   	2

   
   

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43. Two simple pendulums of length 5m and 20m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed ....... oscillations
    

1. 1. 5

   
   
   

   2. 1

   
   
   

   3. 2

   
   
   

   4. 3

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the pendulums be in phase after t sec of start. Within this time, if the bigger pendulum executes n oscillations, the smaller one will have executed (n + 1) oscillations.
   

   Now, the time of n oscillation = 2π
   		20	× n
   		g

   
   

   & the time of (n + 1) oscillation = 2π
   		5	× (n + 1)
   		g

   
   To be in phase
   

   2π			× n = 2π
   		20			5	× (n + 1)
   		g			g

   
   or,  2n = n + 1 ​
   or,  n = 1 ​
   Hence, the no. of oscillations executed by shorter pendulum = n + 1 = 1 + 1 = 2
   

   Correct Option: C

   Let the pendulums be in phase after t sec of start. Within this time, if the bigger pendulum executes n oscillations, the smaller one will have executed (n + 1) oscillations.
   

   Now, the time of n oscillation = 2π
   		20	× n
   		g

   
   

   & the time of (n + 1) oscillation = 2π
   		5	× (n + 1)
   		g

   
   To be in phase
   

   2π			× n = 2π
   		20			5	× (n + 1)
   		g			g

   
   or,  2n = n + 1 ​
   or,  n = 1 ​
   Hence, the no. of oscillations executed by shorter pendulum = n + 1 = 1 + 1 = 2
   

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44. The time period of a simple pendulum is 2 seconds. If its length is increased by 4-times, then its period becomes​​
    

1. 1. 16 s

   
   
   

   2. 12 s

   
   
   

   3. 8 s

   
   
   

   4. 4 s
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   T = 2π
   		ℓ	T ∝ √ℓ
   		g

   
   If is increased by 4 times, time period will  increase by two times.
   

   Correct Option: D

   T = 2π
   		ℓ	T ∝ √ℓ
   		g

   
   If is increased by 4 times, time period will  increase by two times.
   

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45. Masses M_A and M_B hanging from the ends of strings of lengths L_A and L_B are executing simple harmonic motions. If their frequencies are f_A = 2f_B, then​​
    

1. 1. L_A = 2L_B and M_A = M_B/2 ​

   
   
   

   2. L_A = 4L_B regardless of masses

   
   
   

   3. L_A = L_B/4 regardless of masses ​

   
   
   

   4. L_A = 2L_B and M_A = 2M_B

   
   
   

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1. View Hint View Answer Discuss in Forum

   fA =	1
   				g
   	2π			LA

   
   

   and fB =	fA	=	1
   						g
   	2		2π			LB

   
   

   ∴	fA	=	1				× 2π
   						g			g
   	fA / 2		2π			LA			LB

   
   

   ⇒ 2 =			⇒ 4 =
   		LA		LB	, regardless of mass.
   		LB		LA

   
   

   Correct Option: C

   fA =	1
   				g
   	2π			LA

   
   

   and fB =	fA	=	1
   						g
   	2		2π			LB

   
   

   ∴	fA	=	1				× 2π
   						g			g
   	fA / 2		2π			LA			LB

   
   

   ⇒ 2 =			⇒ 4 =
   		LA		LB	, regardless of mass.
   		LB		LA

   
   

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