Starting from rest, a body slides down a 45° inclined plane in twice the time it takes to slide the same distance in the absence of friction. What is the coefficient of friction between the body and the inclined plane ?
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a
1/4
b
1/2
c
3/4
d
√3/2
answer is C.
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Detailed Solution
For a frictionless surface, o =g sin 45° = g/√2∴ l=12×g2×t22 …(1)In presence of friction.a=g2−μg2=g2(1−μ)∴ l=12×g2(1−μ)×t12 …(2)Dividing eq. (2) by eq. (1), we get1=(1−μ)t12t22=(1−μ)2t22t22=(1−μ)×4 or (1−μ)=1/4 or μ=3/4.