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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 tanθ
The 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=mgsinθf=mgsinθμmgcosθ
acceleration aR=mg(sinθμcosθ)m=g(sinθμcosθ)

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
v2u2=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
v21v2=2aRS
02v2=2g(sinθμcosθ)S
v2=2g(sinθμcosθ)S
2gSsinθ=2gS(sinθμcosθ)
sinθ=sinθμcosθ
2sinθ=μcosθ
μ=2sinθcosθ=2tanθ

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