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Question

A block placed on a rough inclined plane of inclination (θ=30) can just be pushed upwards by applying a force F as shown. If the angle of inclination of the inclined plane is increased to (θ=60), the same block can just be prevented from sliding down by the application of a force of the same magnitude. The coefficient of friction between the block and the inclined plane is

A
3+131
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B
2313+1
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C
313+1
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D
None of these
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Solution

The correct option is C 313+1

The friction force in the first case is in the downward direction and in the second case it is in the upward direction.
From free body diagram.
When
θ=30,Fmgsin30μmgcos30=0
F=mgsin30+μmgcos30 ... (1)
When
θ=60,F+μmgcos60mgsin60=0
F=mgsin60μmgcos60 ... (2)
Now from equation (1) and (2)
mgsin30+μmgcos30=mgsin60μmgcos60
mg(12)+μmg(32)=mg(32)μmg(12)
μ[32+12]=[3212]
μ(3+1)=(31)
μ=(313+1)

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