question_answer

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 :                                [JEE Online 15-04-2018]

A) \[\sqrt{1-\frac{1}{{{n}^{2}}}}\]                

B) \[1-\frac{1}{{{n}^{2}}}\]   

C) \[\frac{1}{2-{{n}^{2}}}\]               

D) \[\sqrt{\frac{1}{1-{{n}^{2}}}}\]    

Correct Answer: B

Solution :

For a body moving with constant acceleration, the kinematics equation is \[s=ut+\frac{1}{2}a{{t}^{2}}\] If the initial speed is zero, then the time taken to reach a distance \[s\] is \[t=\sqrt{\frac{2s}{a}}\] i.e.,\[t\propto {{a}^{-0.5}}\] In the case of a smooth inclined plane,\[{{a}_{1}}=g\sin \theta \] In the case of rough inclined plane,\[{{a}_{2}}=g(\sin \theta -\mu \cos \theta )\] Time taken to travel down the smooth inclined plane is \[{{t}_{1}}=t\] Time taken to travel down the smooth inclined plane is \[{{t}_{2}}=nt\] \[\frac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\frac{{{a}_{2}}}{{{a}_{1}}}}\Rightarrow \frac{t_{1}^{2}}{t_{2}^{2}}=\frac{{{a}_{2}}}{{{a}_{1}}}\Rightarrow \frac{1}{{{n}^{2}}}=\frac{\sin \theta -\mu cos\theta }{\sin \theta }\Rightarrow \mu =\tan \theta (1-\frac{1}{{{n}^{2}}})\]