Question:

A given object takes \(n\) times more time to slide down a \(45^\circ\) 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 :

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For the same distance from rest, \(t\propto\frac{1}{\sqrt a}\). Compare the accelerations \(g\sin45^\circ\) and \(g(\sin45^\circ-\mu\cos45^\circ)\).
Updated On: Nov 11, 2026
  • \(\frac{1}{2-n^2}\)
  • \(1-\frac{1}{n^2}\)
  • \(\sqrt{1-\frac{1}{n^2}}\)
  • \(\sqrt{\frac{1}{1-n^2}}\)
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The Correct Option is B

Solution and Explanation

Smooth incline: \(a_1=g\sin45^\circ\). Rough incline: \(a_2=g(\sin45^\circ-\mu\cos45^\circ)=g\sin45^\circ(1-\mu)\) (since \(\sin45^\circ=\cos45^\circ\)).
For the same distance \(s\) from rest, \(s=\frac12at^2\), so \(t=\sqrt{\frac{2s}{a}}\).
\(\frac{t_2}{t_1}=\sqrt{\frac{a_1}{a_2}}=\frac{1}{\sqrt{1-\mu}}=n\).
\(1-\mu=\frac{1}{n^2}\), so \(\mu=1-\frac{1}{n^2}\).
Answer: (2) \(1-\frac{1}{n^2}\)
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