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Question

A body of mass m is launched up a rough inclined plane of angle 45 with the horizontal. If the time of ascent is half the time of descent in covering the same distance, the coefficient of friction is

A
25
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B
35
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C
34
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D
58
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Solution

The correct option is B 35
Up the plane:

F=ma(mgsinθ+f)=ma1
mgsinθ+μmgcosθ=ma1
a1=g(sinθ+μcosθ)
s1=ut+12at2 (u is the initial velocity)
s1=12g(sinθ+μcosθ)t21;
Down the plane:

F=ma(mgsinθf)=ma2
mgsinθμmgcosθ=ma2
a2=g(sinθμcosθ)
s2=ut+12at2
s2=12g(sinθμcosθ)t22
Given: t1=t22;12g(sinθ+μcosθ)t21 =12g(sinθμcosθ)t22
(sinθ+μcosθ)=(sinθμcosθ)4μ=35tanθ
Since θ=45
μ=35tan45=35

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