A) \[\mu =\tan \theta \left( 1-\frac{1}{{{n}^{2}}} \right)\]
B) \[\mu =\cos \theta \left( 1-\frac{1}{{{n}^{2}}} \right)\]
C) \[\mu =\tan \theta \sqrt{1-\frac{1}{\sqrt{{{n}^{2}}}}}\]
D) \[\mu =\cot \theta \sqrt{1-\frac{1}{{{n}^{2}}}}\]
Correct Answer: A
Solution :
On smooth inclined plane: Acceleration of the body \[=g\sin \theta \] If s be the distance travelled, then \[s=\frac{1}{2}g\,\sin \theta \times t_{1}^{2}\] ... (i) On rough inclined plane: Acceleration, \[a=\frac{mg\sin \theta -\mu R}{m}\] or \[a=\frac{mg\sin \theta -\mu \,mg\cos \theta }{m}\] \[=g\,\sin \theta -\mu g\cos \theta \] \[\therefore \] \[s=\frac{1}{2}\,(g\,\sin \theta -\mu \,g\,\cos \theta )\,t_{2}^{2}\] ... (ii) From Eqs. (i) and (ii), we have \[\therefore \] \[\frac{t_{2}^{2}}{t_{1}^{2}}=\frac{\sin \theta }{\sin \theta -\mu cos\theta }\] But \[{{t}_{2}}=n{{t}_{1}}\] \[\therefore \] \[{{n}^{2}}=\frac{\sin \theta }{\sin \theta -\mu \,\cos \theta }\] or \[\mu =\frac{{{n}^{2}}-1}{{{n}^{2}}}\times \frac{\sin \theta }{\cos \theta }\] or \[\mu =\left( 1-\frac{1}{{{n}^{2}}} \right)\tan \theta \]
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