Q.

A block released on a rough inclined plane of inclination θ=30 slides down the plane with an acceleration g4, where g is the acceleration due to gravity. What is the coefficient of friction between the block and the inclined plane ?

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a

23

b

13

c

123

d

32

answer is C.

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Detailed Solution

The acceleration of block moving down the inclined plane is 

a = g sin θ - µg cos θ 

Putting a = g4 and θ = 30°, we get μ=123

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