question_answer

A smooth inclined plane of length L having inclination 9 with the horizontal is inside a lift which is moving down with a retardation a. The time taken by a body to slide down the inclined plane from rest will be : 

A) \[\sqrt{\frac{2L}{(g+a)\sin \theta }}\]                    

B) \[\sqrt{\frac{2L}{(g-a)\sin \theta }}\]

C) \[\sqrt{\frac{2L}{a\sin \theta }}\]                                             

D) \[\sqrt{\frac{2L}{g\sin \theta }}\]

Correct Answer: B

Solution :

                 Key Idea: Net acceleration in downward motion decreases. The free body diagram of the set up is shown. Net acceleration is \[(g-a)\sin \theta \] Using equation of motion \[s=ut+\frac{1}{2}a'{{t}^{2}}\] where, s is displacement, u is initial velocity, a is acceleration and t is time. We have, \[u=0,\text{ }a'=(g-a)\text{ }sin\,\theta ,s=L\] \[\therefore \]                \[L=0+\frac{1}{2}(g-a)sin\theta .{{t}^{2}}\] \[\Rightarrow \]    \[t=\sqrt{\frac{2L}{(g-a)sin\theta }}\]