Laws of Motion

  • Question 513
    CBSEENPH11020463

    A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds?

    • h/9 metres from the ground

    • 7h/9 metres from the ground

    • 8h/9 metres from the ground

    • 17h/18 metres from the ground.

    Solution

    C.

    8h/9 metres from the ground


    second law of motion
    straight s space equals ut space plus space 1 half space plus gT squared
or space straight h equals 0 plus 1 half gT squared space left parenthesis because space straight u equals 0 right parenthesis
therefore space straight T equals square root of open parentheses fraction numerator 2 straight h over denominator straight g end fraction close parentheses end root
At space straight t space equals space straight T over 3 straight s comma
straight s space equals space 0 plus space 1 half straight g space open parentheses straight T over 3 close parentheses squared
rightwards double arrow space straight s space equals space 1 half space straight g. straight T squared over 9
rightwards double arrow space straight s space equals space straight g over 18 space straight x space fraction numerator 2 straight h over denominator straight g end fraction space space space space open parentheses therefore space equals space square root of fraction numerator 2 straight h over denominator straight g end fraction end root close parentheses
    therefore s = h/9 m
    Hence, the position of ball from the ground= h- h/9 = 8h/9 m
    Question 514
    CBSEENPH11020467

    An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 cm. If the car is going twice as fast, i.e 120 km/h, the stopping distance will be

    • 20 m

    • 40 m

    • 60 m

    • 80 m

    Solution

    D.

    80 m

    Third equation of motion gives
    v2 = u2 + 2as ⇒ 2
    s ∝ u (since v = 0)
    where a = retardation of body in both the cases
    therefore, straight s subscript 1 over straight s subscript 2 space equals space fraction numerator straight u subscript 1 superscript 2 over denominator straight u subscript 2 superscript 2 end fraction .... (i)
    Here, s1 = 20 m, u1 = 60 km/h, u2 = 120 km/h. Putting the given values in eq. (i), we get
    20 over straight s subscript 2 space equals space open parentheses 60 over 120 close parentheses squared
space straight s subscript 2 space equals space 20 space straight x space open parentheses 120 over 60 close parentheses squared
space equals space 20 space straight x space 4 space
space equals space 80 space straight m

    Question 515
    CBSEENPH11020468

    A machine gun fires a bullet of mass 40 g with a velocity 1200 ms−1. The man holding it can exert a maximum force of 144 N on the gun. How many bullets can he fire per second at the most?

    • one 

    • Four

    • Two

    • Three

    Solution

    D.

    Three

    The force exerted by machine gun on man's hand in firing a bullet = change in momentum per second on a bullet or rate of change of momentum
    space equals open parentheses 40 over 1000 close parentheses space straight x space 1200 space equals space 48 space straight N
    The force exerted by man on machine gun = 144 N Hence, number of bullets fired =144/48 = 3

    Question 516
    CBSEENPH11020470

    A block rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (take g = 10 m/s2 )
    • 2.0

    • 4.0

    • 1.6 

    • 2.4

    Solution

    A.

    2.0

    Let the mass of block be m.
    Frictional force in rest position
    F = mg sin 30

    10 space equals space straight m space straight x space 10 space straight x space 1 half
space straight m space equals space fraction numerator 2 space straight x space 10 over denominator 10 end fraction space equals space 2 space kg

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