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Q.

The time taken by a particle to slide down a smooth inclined plane is double the time it would take in falling down through a height equal to the vertical height of the plane. The inclination of the plane with horizontal is:

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a

45°

b

90°

c

30°

d

60°

answer is A.

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Detailed Solution

$time taken along the inclined plane = \sqrt{\frac{2 L}{gsinθ}} I f it falls f r o m t h e s a m e h e i g h t t' = \sqrt{\frac{2 h}{g}} given t = 2 t' but h = Lsinθ \sqrt{\frac{2 L}{gsinθ}} = 2 \sqrt{\frac{2 h}{g}} \frac{2 L}{sinθg} = 4 \frac{2 Lsinθ}{g} \sin^{2} θ = \frac{1}{4}; θ = 30^{0}$

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