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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, 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:
s=ut12at2 (v is the final velocity) and
a=g(sinθ+μcosθ)
s=12g(sinθ+μcosθ)t21;

Down the plane: s=ut+12at2
s=12g(sinθμcosθ)t22
Given: t1=t22;12g(sinθ+μcosθ)t21=12g(sinθμcosθ)t22
(sinθ+μcosθ)=(sinθμcosθ)4
Substituting θ=45
1+μ=(1μ)4
μ=35

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