1. A system consists of three masses m₁, m₂ and m₃ connected by a string passing over a pulley P. The mass m1 hangs freely and m₂ and m₃ are on a rough horizontal table (the coefficient of friction = µ). The pulley is frictionless and of negligible mass. The downward acceleration of mass m1 is : (Assume m₁ = m₂ = m₃ = m)
   
   

1. 1. g(1 - gμ)

      g

   
   
   

   2. 2gμ

      3

   
   
   

   3. g(1 - 2μ)

      3

   
   
   

   4. g(1 - 2μ)

      2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Acceleration =	Net force in the direction of motion
   	Totla mass of system

   
   

   =	m1g - μ(m2 + m3)g	=	g	(1 - 2μ)
   	m1 + m2 + m3		3

   
   (∵ m₁ = m₂ = m₃ = m given)

   Correct Option: C

   Acceleration =	Net force in the direction of motion
   	Totla mass of system

   
   

   =	m1g - μ(m2 + m3)g	=	g	(1 - 2μ)
   	m1 + m2 + m3		3

   
   (∵ m₁ = m₂ = m₃ = m given)

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2. Three blocks A, B and C of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block then the contact force between A and B is​​[2015]
   

1. 1. 6 N

   
   
   

   2. 8 N

   
   
   

   3. ​18 N

   
   
   

   4. ​2 N

   
   
   

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1. View Hint View Answer Discuss in Forum

   Acceleration of system a =	Fnet
   	MTotal

   
   

   =	14	=	14	= 2m/s²
   	4 + 2 + 1		7

   
   
   The contact force between A and B
   = (m_B + m_C) × a = (2 + 1) × 2 = 6N

   Correct Option: A

   Acceleration of system a =	Fnet
   	MTotal

   
   

   =	14	=	14	= 2m/s²
   	4 + 2 + 1		7

   
   
   The contact force between A and B
   = (m_B + m_C) × a = (2 + 1) × 2 = 6N

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3. One end of string of length l is connected to a particle of mass 'm' and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed 'v' the net force on the particle (directed towards centre) will be (T represents the tension in the string) :-

1. 1. T -	mv²
      	1

   
   
   

   2. T -	mv²
      	1

   
   
   

   3. Zero

   
   
   

   4. T

   
   
   

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1. View Hint View Answer Discuss in Forum

   Net force on particle in uniform circular motion is centripetal force
   

   		mv²
   		l

   
   which is  provided by tension in string so the net force will be equal to tension i.e., T.

   Correct Option: D

   Net force on particle in uniform circular motion is centripetal force
   

   		mv²
   		l

   
   which is  provided by tension in string so the net force will be equal to tension i.e., T.

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4. Physical independence of force is a consequence of

1. 1. third law of motion

   
   
   

   2. second law of motion

   
   
   

   3. first law of motion

   
   
   

   4. all of these laws

   
   
   

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1. View Hint View Answer Discuss in Forum

   Newton’s first law of motion is related to physical independence of force.
   

   Correct Option: C

   Newton’s first law of motion is related to physical independence of force.
   

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5. If the force on a rocket moving with a velocity of 300 m/sec is 345 N, then the rate of combustion of the fuel, is

1. 1. ​0.55 kg/sec

   
   
   

   2. 0.75 kg/sec

   
   
   

   3. ​1.15 kg/sec

   
   
   

   4. 2.25 kg/sec

   
   
   

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1. View Hint View Answer Discuss in Forum

   Velocity of the rocket (u) = 300 m/s and force (F) = 345N. Rate of combustion of fuel
   

   		dm		=	F	= 1.15 kg/sec
   		dt			u

   Correct Option: C

   Velocity of the rocket (u) = 300 m/s and force (F) = 345N. Rate of combustion of fuel
   

   		dm		=	F	= 1.15 kg/sec
   		dt			u

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