A coin is dropped in a lift. It takes time t_ | vedclass

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(C) The time taken by a coin to reach the floor is given by the kinematic equation $h = \frac{1}{2} gt^2$ (since initial velocity $u = 0$).Thus,$t = \

Vedclass · Accepted Answer

$t_2 < t_1$

Vedclass · Answer

(C) The time taken by a coin to reach the floor is given by the kinematic equation $h = \frac{1}{2} gt^2$ (since initial velocity $u = 0$).Thus,$t = \sqrt{\frac{2h}{g}}$.In a stationary lift,the effective acceleration is $g$,so $t_1 = \sqrt{\frac{2h}{g}}$.In a lift moving upward with constant acceleration $a$,the effective acceleration experienced by the coin is $g' = g + a$.Therefore,$t_2 = \sqrt{\frac{2h}{g+a}}$.Since $g' = g + a > g$,it follows that the denominator in the expression for $t_2$ is larger than that for $t_1$.Thus,$t_2 < t_1$.

$A$ coin is dropped in a lift. It takes time $t_1$ to reach the floor when the lift is stationary. It takes time $t_2$ when the lift is moving up with constant acceleration $a$. Then:

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