A ball of mass $m$ and another ball of mass $2 m$ are dropped from equal height. If times taken by the balls to reach the ground are $t_{1} a n d t_{2}$ respectively then

t_{1} = \frac{t_{2}}{2}

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t_{1} = t_{2}

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t_{1} = 4 t_{2}

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t_{1} = \frac{t_{2}}{4}

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Solution

The correct option is B $t_{1} = t_{2}$
Time taken by any body of mass $m$ to reach the ground when dropped from a height $h$ is $t = \sqrt{\frac{2 h}{g}}$
From this equation time is independent of mass so $t_{1} = t_{2}$

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