Laws of Motion

  • Question 493
    CBSEENPH11020408

    A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45° with the initial vertical direction is

    • Mg left parenthesis square root of 2 minus 1 right parenthesis
    • Mg left parenthesis square root of 2 plus 1 right parenthesis
    • Mg square root of 2
    • fraction numerator Mg over denominator square root of 2 end fraction

    Solution

    A.

    Mg left parenthesis square root of 2 minus 1 right parenthesis straight F calligraphic l space sin space 45 space equals space M g left parenthesis calligraphic l minus calligraphic l space cos space 45 right parenthesis
F space equals space M g left parenthesis square root of 2 minus 1 right parenthesis
    Question 494
    CBSEENPH11020423

    A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. If the catching process is completed in 0.1 s, the force of the blow exerted by the ball on the hand of the player is equal to

    • 300 N

    • 150 N

    • 30

    • 3 N

    Solution

    C.

    30

    (mv-0)
    ⇒ 0.15 x 20
    F = 3/0.1 = 30 N

    Question 495
    CBSEENPH11020424

    A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m which applying the force and the ball goes upto 2 m height further, find the magnitude of the force. Consider g = 10 m/s2

    • 22 N

    • 4 N

    • 20 N

    • 30 N

    Solution

    C.

    20 N

    mgh = Fs
    F = 20 N

    Question 496
    CBSEENPH11020425

    Consider a two-particle system with particles having masses m1 and m2. If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved, so as to keep the centre of mass at the same position?

    • straight m subscript 1 over straight m subscript 2 straight d
    • straight m subscript 2 over straight m subscript 1 straight d
    • d

    • fraction numerator straight m subscript 1 over denominator straight m subscript 1 plus straight m subscript 2 end fraction space straight d

    Solution

    A.

    straight m subscript 1 over straight m subscript 2 straight d straight m subscript 1 straight d space plus straight m subscript 2 straight x space equals space 0
straight x space equals space fraction numerator straight m subscript 1 straight d over denominator straight m subscript 2 end fraction

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