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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
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B
11n2
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C
11n2
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D
11n2
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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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