A smooth inclined plane is inclined at an angle q with the horizontal. A body starts from rest and slides down the inclined surface. Then the time taken by it to reach the bottom is
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a
2hg
b
2θg
c
1sinθ2hg
d
sinθ{(2h)/g}
answer is C.
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Detailed Solution
See fig.Here F=mgsinθ or ma=mgsinθa=gsinθ …(1)Now, v = u + at = 0 + at or t = v/a … (2)But v=(2gh) …(3)t=(2gh)gsinθ=1sinθ2hg