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Question

# Starting from rest, a body slides down a 45 degree inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is:

A
0.33
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B
0.25
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C
0.75
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D
0.80
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Solution

## The correct option is D 0.75When body slides down without friction , mgsinθ=ma1or gsinθ=a1 .............................eq1When body slides down with friction , mgsinθ−fk=ma2but, frictional force fk=μmgcosθ Hence mgsinθ−μmgcosθ=ma2 or gsinθ−μgcosθ=a2.............................eq2Dividing eq1 by eq2 , a1a2=gsinθgsinθ−μgcosθor a1a2=sinθsinθ−μcosθ ..........eq3If time taken by body without friction is t1 and with friction is t2 , then distance covered by body is given by , s=1/2a1t21=1/2a2t22or a1a2=t22t21=22=4 (because t2=2t1)putting this value in eq3 , 4=sinθsinθ−μcosθ or 4=1/√21/√2−μ/√2 (as given θ=45o)or 4(1−μ)=1or μ=3/4=0.75

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