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Question

# A block is placed on an inclined plane of inclination θ. The angle of inclination is such that the block slides down the plane at a constant speed. The coefficient of kinetic friction between the block and the inclined plane is equal to

A

sin θ

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B

cos θ

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C

tan θ

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D

cot θ

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Solution

## The correct option is C tan θ As the block moves down the inclined plane at a constant velocity, its acceleration will be zero. Hence, the net force acting on it will be zero. mg sinθ=μN⇒mg sinθ=μmg cos θ⇒μ=tan θ Hence, the correct choice is (c).

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