Laws of Motion

  • Question 561

    A cone filled with water is revolved in a vertical circle of radius 4 m and the water does not fall down. What must be the maximum period of revolution?

    • 2 s

    • 4 s

    • 1 s

    • 6 s



    4 s

    When a cone filled with water is revolved in a vertical circle, then the velocity at the highest point is given by

    ν g ralso ν = rω =r2πTHence, 2πrTr gT2πrr g=2πrg=2π49.8=2×3.14×0.6389=4.0 secso Tmax =4 sec

    Question 562

    Which is true for rolling frictrion   μr , static friction μs  and  kinetic friction  μk :

    • μs > μk >  μr

    • μs <  μk <  μr

    • μs <  μk >  μr

    • μs > μr  > μk



    μs > μk >  μr

    Static friction is a force that keeps an object at rest. It must be overcome to start moving the object. When external force exceeds the maximum limit of static friction body begins to move.
    Once the body is in motion it is subjected to sliding or kinetic friction which opposes relative motion between the two surfaces in contact. At every instant, there is just one point of contact
    between the body and the plane and this point has no motion relative to the plane. In this ideal situation, kinetic or static friction is zero and the body should continue to roll with constant velocity. 
    In practice, this will not happen and some resistance to motion (rolling friction) does occur, i.e. to keep the body rolling. Static friction is always greater than kinetic friction and rolling friction is less than kinetic friction.
                            μs >  μk > μr
    Question 563

    Two weights w1 and w2 are suspended to the two strings on a frictionless pulley. When the pulley is pulled up with an acceleration g, then the tension in the string is :

    • 4 w1w2w1 + w2

    • w1w2w1 + w2

    • 2 w1 w2w1 w2

    • w1 + w22



    4 w1w2w1 + w2

    Equation of motion for first weight

    This is a frictionless and inextensible pulley.

    for first weight

    T - m1g = m1 ( a - g)

    ⇒ T - 2m1 g = m1a

    For second weight 

    m2g -T = m2 ( a - g)

    2 mg - T = m2 a

    on solving equations (1) & (2)

    T =4 m1 m2m1 + m2T = 4 m1 m2w1 + w2

    Question 564

    A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is 6 ms'.The kinetic energy of the other mass is

    • 256 J

    • 486 J

    • 524 J

    • 324 J



    486 J

    The linear momentum of the exploding part will remain conserved.

    Applying coservation of linear momentum,

    m1 u1 =m2 u2

    Here, m1 = 18kg, m2 = 12kg

    u1 = 6ms-1, u2 = ?

    u2 = 18×612 9 ms-1

    Thus kinetic energy of 12 kg mass

    K2 =12m1u22       = 12×12×92       = 6× 81K2   = 486 J

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