question_answer

A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively                [AIEEE 2002]

A) g, g

B) g-a, g-a

C) g-a, g

D) a, g

Correct Answer: C

Solution :

[c] The net force in any direction equal the product of mass and acceleration in the direction of net force. Also net force means the vector sum of forces acting on a body.

Apparent weight of ball

\[w'=w-R\]

\[R=ma\]                       (acts upward)

\[mg-R=ma\]

\[w'=mg-ma=m(g-a)\]

Hence, apparent acceleration in the lift is g – a. Now, if the man is standing stationary on the ground, then the apparent acceleration of the falling ball is g.