A block of mass m is placed on a smooth inclined wedge $A B C$ 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

• A

  $a=\frac{g}{\operatorname{cosec} \theta}$
• B

  $a=\frac{g}{\sin \theta}$
• Correct answer
  C

  $a=g \tan \theta$
• D

  $a=g \cos \theta$

Solution

Let the mass of block is m. It will remains stationary if forces acting on it are in equilibrium. i.e., $m a \cos \theta=m g \sin \theta$ $\Rightarrow a=g \tan \theta$
Here $\mathrm{ma}=$ Pseudo force on block, $\mathrm{mg}=$ weight

Hence, the correct answer is option: 'C'.