A body is sliding down an inclined plane having coefficient of friction $0.5$. If the normal reaction is twice that of the resultant downward force along the incline, the angle between the inclined plane and the horizontal is ....... $^o$
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(c) Resultant downward force along the incline $ = mg(\sin \theta - \mu \cos \theta )$
Normal reaction $ = mg\cos \theta $
Given : $mg\cos \theta = 2mg(\sin \theta - \mu \cos \theta )$
By solving $\theta = {45^o}$.
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