A) \[\frac{3}{4}\,\tan \,\theta \]
B) \[\frac{2}{3}\,\tan \,\theta \]
C) \[\frac{1}{4}\,\tan \,\theta \]
D) \[\frac{1}{2}\,\tan \,\theta \]
Correct Answer: A
Solution :
When the plane is smooth, then acceleration of the body \[a=g\,\sin \theta \] \[\therefore \]Distance covered \[s=\frac{1}{2}at_{1}^{2}\] \[{{t}_{1}}=\sqrt{\frac{2s}{a}}\] \[{{t}_{1}}=\sqrt{\left( \frac{2s}{g\sin \theta } \right)}\] When plane is rough, then acceleration\[a=g(\sin \theta -\mu \cos \theta )\] where\[\mu =\]coefficient of friction \[\therefore \]New time \[{{t}_{2}}=\sqrt{\left[ \frac{2s}{g(\sin \theta -\mu \cos \theta )} \right]}\] But\[{{t}_{2}}=2{{t}_{1}}\] \[\therefore \]\[\sqrt{\left[ \frac{2s}{g(sin\theta -\mu cos\theta )} \right]}=2\sqrt{\left( \frac{2s}{g\sin \theta } \right)}\] \[\frac{1}{\sin \theta -\mu \cos \theta }=\frac{4}{\sin \theta }\] \[\sin \theta =4\sin \theta -4\mu \cos \theta \] \[4\mu \cos \theta =3\sin \theta \] \[\mu =\frac{3}{4}\tan \theta \]
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