Two bodies of different masses m_a and m_b ar | vedclass

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(C) For a body dropped from rest,the distance $h$ covered in time $t$ is given by the equation of motion: $h = \frac{1}{2}gt^2$.Rearranging for time $

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$\sqrt{a} : \sqrt{b}$

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(C) For a body dropped from rest,the distance $h$ covered in time $t$ is given by the equation of motion: $h = \frac{1}{2}gt^2$.Rearranging for time $t$,we get: $t = \sqrt{\frac{2h}{g}}$.For the first body dropped from height $a$,the time taken is $t_a = \sqrt{\frac{2a}{g}}$.For the second body dropped from height $b$,the time taken is $t_b = \sqrt{\frac{2b}{g}}$.The ratio of the time taken is $\frac{t_a}{t_b} = \frac{\sqrt{2a/g}}{\sqrt{2b/g}} = \sqrt{\frac{a}{b}}$.Thus,the ratio is $\sqrt{a} : \sqrt{b}$.

Two bodies of different masses $m_a$ and $m_b$ are dropped from two different heights $a$ and $b$. The ratio of the time taken by the two to cover these distances is

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