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An inclined plane of height h and length l have the angle of inclination $θ$ . The time taken by a body to come from the top to the bottom of this inclined plane will be

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$\sin θ \sqrt{\frac{2 h}{g}}$

$\frac{1}{\sin θ} \sqrt{\frac{2 h}{g}}$

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

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

answer is B.

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Detailed Solution

Force down the plane $= m g \sin θ$

$∴$ Acceleration down the plane $= g \sin θ$

Using the relation, $h = u t + \frac{1}{2} g t^{2}$

$l = 0 + \frac{1}{2} (g \sin θ) t^{2}$

$\Rightarrow t^{2} = \frac{2 l}{g \sin θ} = \frac{2 h}{g \sin^{2} θ}$ $(∵ l = \frac{h}{\sin θ})$

$t = \frac{1}{\sin θ} \sqrt{\frac{2 h}{g}}$

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