question_answer

A body takes time t to reach the bottom of an inclined plane of angle \[\theta \] with the horizontal. If the plane is made rough, time taken now is it. The coefficient of friction of the rough surface is:

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 :

 The free body diagram shows the various forces acting on the block. When plane is frictionless, then acceleration, \[a=g\,\sin \theta \] Let time taken is \[{{t}_{1}}\]. Since body starts from rest, initial velocity u = 0 \[\therefore \] \[s=ut+\frac{1}{2}at_{1}^{2}\] \[\Rightarrow \] \[s=\frac{1}{2}g\sin \theta t_{1}^{2}\] \[\Rightarrow \] \[{{t}_{1}}=\sqrt{\frac{2s}{g\,\sin \theta }}\] When plane is made rough, then acceleration\[a=g\,(\sin \theta -\mu \,cos\theta )\], where 4 is coefficient of friction. \[\therefore \] \[{{t}_{2}}=\sqrt{\frac{2s}{g\,(\sin \theta -\mu \cos \theta )}}\] Given, \[{{t}_{2}}=2{{t}_{1}}\] \[\therefore \] \[\sqrt{\frac{2s}{g\,(\sin \theta -\mu \cos \theta )}}=2\sqrt{\frac{2s}{g\sin \theta }}\] \[\therefore \] \[\frac{1}{\sin \theta -\mu \cos \theta }=\frac{4}{\sin \theta }\] \[\Rightarrow \] \[\sin \theta =4\sin \theta -4\mu \cos \theta \] \[\Rightarrow \] \[4\mu \cos \theta =3\sin \theta \] \[\Rightarrow \] \[\mu =\frac{3}{4}\,\tan \theta \]