Question

# A given object takes $$n$$ times more time to slide down $$45^\circ$$ rough inclined plane as it takes to slide down a perfectly smooth $$45^\circ$$ incline. The coefficient of kinetic friction between the object and the incline is:

A
12n2
B
11n2
C
11n2
D
11n2

Solution

## The correct option is B $$1 - \dfrac{1}{n^2}$$Let the length of inclination $$=d$$Acceleration at the smooth surface$$=g\sin45^{\circ}$$Time taken by smooth surface $$=t=\sqrt{\cfrac{2d}{a}}=\sqrt{\cfrac{2d}{\cfrac{g}{\sqrt 2}}}$$Now, for rough surface, let the co-efficient of Kinetic friction , then acceleration will be $$a^{'}=g\sin45^{\circ}-\mu g\cos 45^{\circ}=\cfrac{g(1-\mu)}{\sqrt 2}$$Time taken$$t^{'}=\sqrt{\cfrac{2d}{\cfrac{g(1-\mu)}{\sqrt 2}}}$$$$\therefore$$ As we know, $$\cfrac{t^{'}}{t}=n$$$$\sqrt{\cfrac{2d}{g\sqrt2}}\div\sqrt{\cfrac{2d}{\cfrac{g(1-\mu)}{\sqrt 2}}}=n$$$$\Rightarrow \mu=[1-\cfrac{1}{n^{2}}]$$Physics

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