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A body is allowed to slide from the top along a smooth inclined plane of length 5 m at an angle of inclination
$30 °$ .If $g = 10 {ms}^{- 2}$ time taken by the body to reach the bottom of the plane is

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$\frac{\sqrt{3}}{2} s$

2 s

1.414 s

$\frac{1}{\sqrt{2}} s$

answer is B.

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Detailed Solution

$t = \sqrt{\frac{2 l}{g \sin θ}} = \sqrt{\frac{2 \times 5}{10 \times \frac{1}{2}}} = \sqrt{2} = 1.414$ $l = 5 m, θ = 30 °, g = 10 {m/s}^{2}$

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