A smooth inclined plane is inclined at an angle $θ$ with horizontal. A body starts from rest and slides down the inclined surface. Then the time taken by it to reach the bottom is:

\sqrt{(\frac{2 h}{g})}

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\sqrt{(\frac{2 l}{g})}

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\frac{1}{sin θ} \sqrt{\frac{2 h}{g}}

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sin θ \frac{\sqrt{(2 h)}}{g}

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Solution

The correct option is C $\frac{1}{sin θ} \sqrt{\frac{2 h}{g}}$
So by second equation of motion, we get $S = u t + \frac{1}{2} a t^{2}$
here $S = l, u = 0, a = g sin θ$
$t = \sqrt{\frac{2 l}{a}} = \sqrt{\frac{2 h}{g {sin}^{2} θ}} = \frac{1}{sin θ} \sqrt{\frac{2 h}{g}} (∵ sin θ = \frac{h}{l})$

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A smooth inclined plane is inclined at an angle θ with horizontal. A body starts from rest and slides down the inclined surface. Then the time taken by it to reach the bottom is:

The correct option is C 1sinθ√2hgSo by second equation of motion, we get S=ut+12at2here S=l,u=0,a=gsinθt=√2la=√2hgsin2θ=1sinθ√2hg(∵sinθ=hl)

A smooth inclined plane is inclined at an angle $θ$ with horizontal. A body starts from rest and slides down the inclined surface. Then the time taken by it to reach the bottom is:

The correct option is C $\frac{1}{sin θ} \sqrt{\frac{2 h}{g}}$
So by second equation of motion, we get $S = u t + \frac{1}{2} a t^{2}$
here $S = l, u = 0, a = g sin θ$
$t = \sqrt{\frac{2 l}{a}} = \sqrt{\frac{2 h}{g {sin}^{2} θ}} = \frac{1}{sin θ} \sqrt{\frac{2 h}{g}} (∵ sin θ = \frac{h}{l})$