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

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(B) The time taken by the coin to reach the floor is given by the kinematic equation: $h = \frac{1}{2} g_{eff} t^2$ (where $u = 0$).Thus,$t = \sqrt{\f

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$t_2 > t_1$

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(B) The time taken by the coin to reach the floor is given by the kinematic equation: $h = \frac{1}{2} g_{eff} t^2$ (where $u = 0$).Thus,$t = \sqrt{\frac{2h}{g_{eff}}}$.In a stationary lift,the effective acceleration is $g_{eff} = g$. Therefore,$t_1 = \sqrt{\frac{2h}{g}}$.In a lift moving downward with constant acceleration $a$,the effective acceleration experienced by the coin relative to the lift is $g_{eff} = g - a$.Therefore,$t_2 = \sqrt{\frac{2h}{g - a}}$.Since $g - a Consequently,$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 down with constant acceleration $a$. Then:

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