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 B35 Up the plane: s=ut−12at2 (v is the final velocity) and a=−g(sinθ+μcosθ) s=12g(sinθ+μcosθ)t21;
Down the plane: s=ut+12at2 ⇒s=12g(sinθ−μcosθ)t22
Given: t1=t22;⇒12g(sinθ+μcosθ)t21=12g(sinθ−μcosθ)t22 ⇒(sinθ+μcosθ)=(sinθ−μcosθ)4
Substituting θ=45∘ ⇒1+μ=(1−μ)4 ⇒μ=35