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Question

A block of mass m takes time t to slide down on a smooth inclined plane of angle of inclination θ and height h. If same block slides down on a rough inclined plane of same angle of inclination and same height and takes time n times of initial value, then coefficient friction between block and inclined plane is

A
[1+n2]tanθ
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B
[11n2]tanθ
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C
[1n2]tanθ
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D
[1+1n2]tanθ
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Solution

The correct option is B [11n2]tanθ
When an object is sliding down the incline plane of height h and inclination θ
then total length of the inclined plane is
L=hsinθ
Also the acceleration on the smooth inclined plane will be
a=gsinθ
now we have time to slide down
t=2×Lat=1sinθ2×hg
now when it slide down the rough plane the acceleration will be less due to friction force
now in that case acceleration is given by
a=sinθμgcosθ
again by above formula
t=1sinθ2×hg(1μtanθ)
given that the on rough plane is n time more than the smooth plane.
t=nt1sinθ2×hg(1μtanθ)=n×1sinθ2×hgg(1μtanθ)=gn2
so, friction coefficient is given by
μ=1tanθ(11n2)

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