A) \[\frac{3}{4}\tan \theta \]
B) \[\frac{1}{2}\tan \theta \]
C) \[\frac{2}{3}\tan \theta \]
D) \[\frac{1}{4}\tan \theta \]
Correct Answer: A
Solution :
When the plane is smooth, \[a=g\sin \theta ,\,{{t}_{1}}=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2s}{g\,\sin \theta }},\] Where s is the length of the inclined plane. When plane is rough, \[a=g(sin\theta -\mu cos\theta )\] \[\therefore \]\[{{t}_{2}}=\sqrt{\frac{2\,s}{g(sin\theta -\mu cos\theta )}}\] As \[{{t}_{1}}=2{{t}_{1}}\] \[\therefore \] \[\sqrt{\frac{2\,s}{g(sin\theta -\mu \cos \theta )}}=2\sqrt{\frac{2s}{g\sin \theta }}\] Squaring, we get \[\frac{1}{\sin \theta -\mu \cos \theta }=\frac{4}{\sin \theta }\] \[\sin \theta -4\sin \theta -4\mu \cos \theta \] \[4\mu \cos \theta =3\sin \theta \] or \[\mu =\frac{3}{4}\tan \theta \] Hence, the correction option is [a].
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