A smooth inclined plane is inclined at an angle θ with the horizontal. A body starts from rest and slides down the inclined surface, then the time taken by the body to reach the bottom is:
A
√2hg
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B
√2lg
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C
1sinθ√2hg
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D
sinθ√2hg
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Solution
The correct option is C1sinθ√2hg ma=mgsinθa=gsinθS=ut+12at2u=0(startsfromrest)l=0+12gsinθt2t=√2lgsinθ[hl=sinθ]t=√2gsinθ−hsinθt=1sinθ√2hg