question_answer

A block of mass m is placed on a smooth inclined wedge ABC of inclination \[\theta \] as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and \[\theta \] for the block to remain stationary on the wedge is [NEET - 2018]

A)  \[a=g\,\,\cos \theta \]              

B)  \[a=\frac{g}{\sin \theta }\]

C)  \[a=\frac{g}{\text{cosec}\theta }\]               

D)  \[a=g\tan \theta \]

Correct Answer: D

Solution :

In non-inertial frame, \[\text{N }sin\text{ }\theta =ma...(i)\] \[\text{N cos }\theta =mg...(ii)\] \[\tan \theta =\frac{a}{g}\] \[a=g\tan \theta \]