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A wedge of height 'h is released from rest with a light particle P placed on it as shown. The wedge slides down an incline which makes an angle $θ$ with the horizontal. All the surfaces are smooth, P will reach the surface of the incline in time:

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$\sqrt{\frac{2 h}{g \tan θ}}$

$\sqrt{\frac{2 h}{g \sin θ \cos θ}}$

$\sqrt{\frac{2 h}{g \sin^{2} θ}}$

$\sqrt{\frac{2 h}{g \cos^{2} θ}}$

answer is A.

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Detailed Solution

Acceleration of wedge = g sin $θ$
FBD of particle w.r.t. wedge:
Displacement of Particle on wedge

$x = (\frac{h}{\tan θ})$

Acceleration of particle along displacement = g sin $θ$ cos $θ$

$t = \sqrt{\frac{2 s}{a}} = \sqrt{\frac{\frac{2 h}{\tan θ}}{g \sin θ \cos θ}} = \sqrt{\frac{2 h}{g \sin^{2} θ}}$

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