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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