A person standing on the floor of an elevator drops a coin. The coin reaches the floor in time t_1 if the elevator is moving uniformly and time t_2 if - PHYSICS MCQ NEET 2019 2026 | ExamTest.live

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A person standing on the floor of an elevator drops a coin. The coin reaches the floor in time t_{1} if the elevator is moving uniformly and time t_{2} if the elevator is stationary. Then:

A $t_{1} < t_{2} or t_{1} > t_{2} depending upon whether the lift is going up or down.$
B $t_{1} > t_{2}$
C $t_{1} < t_{2}$
D $t_{1} = t_{2}$

Solution & Step-by-step Explanation

In a frame moving uniformly (constant velocity), the acceleration is zero. The effective acceleration due to gravity g_{e ff} = g \pm a .Since a = 0 for uniform motion, g_{e ff} remains g .The time taken to fall a height h is t = \frac{2 h}{g _{e ff}} .Since g_{e ff} is the same in both cases (stationary and uniform motion), t_{1} = t_{2} .

Try it yourself before checking the explanation above.

A person standing on the floor of an elevator drops a coin. The coin reaches the floor in time t_{1} if the elevator is moving uniformly and time t_{2} if the elevator is stationary. Then:

t_{1} < t_{2} or t_{1} > t_{2} depending upon whether the lift is going up or down.

t_{1} > t_{2}

t_{1} < t_{2}

t_{1} = t_{2}

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