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Q.

A smooth plane inclined at an angle $θ$ with the horizontal as shown in the given figure. A body starts from rest and slides down on the inclined surface. The time taken by the body to reach the bottom is

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a

$\sqrt{\frac{2 h}{g}}$

b

$\sqrt{\frac{2 l}{g}}$

c

$\frac{1}{sinθ} \sqrt{\frac{2 h}{g}}$

d

$sinθ \sqrt{\frac{2 h}{g}}$

answer is A.

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Detailed Solution

${\begin{array}{l} u = 0 \\ a = gsin θ \\ s = 1 \\ s = ut + \frac{1}{2} {at}^{2} \\ I = \frac{1}{2} gsin {θt}^{2} \\ (∵ \sin θ = \frac{h}{I}) \\ t^{2} = \frac{2 I}{gsin θ} \\ t = \sqrt{\frac{2 I}{gsin θ}} \\ t = \sqrt{\frac{2 h}{g}} \end{array}}^{}$

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