Q.

The upper half of an inclined plane of inclination θ is perfectly smooth while the lower half rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and the lower half of the plane is given by

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a

μ = 2 tanθ

b

μ = tanθ

c

μ = 2 / tanθ

d

μ = 1 / tanθ

answer is A.

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Detailed Solution

Suppose length of plane is L. When block descends the plane, rise in kinetic energy = fall in potential energy= mg LsinθWork done by friction= μ x (Reaction) x Distance=0+μ(mgcosθ)×(L/2)=μ(mgcosθ)×(L/2)Now, work done = change in KEmgLsinθ=μ(mgcosθ)×(L/2)⇒tanθ=μ/2∴  μ = 2 tanθ
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The upper half of an inclined plane of inclination θ is perfectly smooth while the lower half rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and the lower half of the plane is given by