A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 ~m / s^2 . Find the displacement of the block during the first 0.2 s after the start. (Take, g=10 ~m / s^2 )

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Correct answer: C. 20 cm

The acceleration (downward) of the elevator is 12 ~m / s^2 which is greater than gravitational acceleration g . So, the block will act as a freely falling body. From the motion of the block aligned s & =u t+(1)/(2) g t^2 & =0 × t+(1)/(2) × 10 ×(0.2)^2 & =(1)/(2) × 10 × 0.2 × 0.2 & =20 ~cm aligned

A block is kept on the floor of an elevator at rest. The elevator...

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