Question

# 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

A

$g,g$

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B

$g–a,g–a$

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C

$g–a,g$

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D

$a,g$

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Solution

## The correct option is C $g–a,g$Step 1: Given dataAcceleration of the lift w.r.t ground, ${a}_{lg}=a$Step 2: Find acceleration of ball w.r.t ground and the person who is standing in the liftAcceleration of ball w.r.t. ground, ${a}_{bg}=g$Acceleration of ball w.r.t. to the person who is standing in the lift, ${a}_{bl}={a}_{bg}-{a}_{lg}$ $=g-a$Hence, option (C) is correct.

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