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(B) Let $t_1$ be the time taken to fall the first $1 \ m$ and $t$ be the time taken to fall the first $2 \ m$.Using the equation of motion $s = ut + \

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$\sqrt{2}+1$

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(B) Let $t_1$ be the time taken to fall the first $1 \ m$ and $t$ be the time taken to fall the first $2 \ m$.Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ and $a = g$:For the first $1 \ m$: $1 = \frac{1}{2} g t_1^2 \implies t_1 = \sqrt{\frac{2}{g}}$.For the first $2 \ m$: $2 = \frac{1}{2} g t^2 \implies t = \sqrt{\frac{4}{g}}$.The time taken to fall through the second metre is $t_2 = t - t_1 = \sqrt{\frac{4}{g}} - \sqrt{\frac{2}{g}}$.The ratio of the time taken for the first metre to the second metre is $\frac{t_1}{t_2} = \frac{\sqrt{2/g}}{\sqrt{4/g} - \sqrt{2/g}}$.Simplifying this: $\frac{t_1}{t_2} = \frac{\sqrt{2}}{\sqrt{4} - \sqrt{2}} = \frac{\sqrt{2}}{2 - \sqrt{2}}$.Rationalizing the denominator: $\frac{\sqrt{2}(2 + \sqrt{2})}{(2 - \sqrt{2})(2 + \sqrt{2})} = \frac{2\sqrt{2} + 2}{4 - 2} = \frac{2(\sqrt{2} + 1)}{2} = \sqrt{2} + 1$.

$A$ stone is allowed to fall freely from rest. The ratio of the times taken to fall through the first metre and the second metre distance is

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