Question

# The upper half of an inclined plane of inclination θ is perfectly smooth while lower half is 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 lower half of plane is given by

A
2 tanθ
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B
tanθ
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C
2 cotθ
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D
2 cosθ
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Solution

## The correct option is A 2 tanθThe given condition is shown in the figure as First of all we have to calculate the acceleration of the block in both the regions i.e. smooth (aS) and rough (aR) Force acting down the plane in smooth region is F=mgsinθ ∴ acceleration aS=mgsinθm=gsinθ Similarly, force acting down the plane in rough region is F=mgsinθ−f=mgsinθ−μmgcosθ ∴ acceleration aR=mg(sinθ−μcosθ)m=g(sinθ−μcosθ) Now, applying equation of motion in both the regions: 1. For smooth plane Let the velocity gained by block at the end of smooth surface be v So, from equation we have v2−u2=2aSS v2−(0)2=2(gsinθ)S v2=2gSsinθ 2. For rough plane The initial velocity of the block at the rough surface will be the velocity gained by block at the end of smooth surface. Let the final velocity of the block at the end of the rough surface be v1 then, from the equation we have v21−v2=2aRS ⇒ 02−v2=2g(sinθ−μcosθ)S ⇒ −v2=2g(sinθ−μcosθ)S ⇒ −2gSsinθ=2gS(sinθ−μcosθ) ⇒ −sinθ=sinθ−μcosθ ⇒ −2sinθ=−μcosθ ⇒ μ=2sinθcosθ=2tanθ

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