A wooden wedge of mass $M$ and inclination angle $\alpha$ rests on a smooth floor. A block of mass $m$ is kept on wedge. A force $\overrightarrow{\mathbf{P}}$ is applied on the wedge as shown in figure, such that a block remains stationary with respect to wedge. The magnitude of force $\overrightarrow{\mathbf{P}}$ is

• Correct answer
  A

  $\left(M+m\right)\mathrm{g}\mathrm{ }\mathrm{t}\mathrm{a}\mathrm{n} \alpha$
• B

  $\mathrm{g}\mathrm{ }\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{ }\mathrm{\alpha }$
• C

  $m\mathrm{g}\cos \alpha$
• D

  $(M+m)\mathrm{g}\mathrm{cosec}\mathrm{\alpha }$

Solution

Since, $P=(M+m)a$

Now as in free body diagram of block,



$ma\cos \alpha =m\mathrm{g}\sin \alpha$

$\therefore a=\mathrm{g}\frac{\sin \alpha }{\cos \alpha }=\mathrm{g}\tan \alpha$ or $P=\left(M+m\right)\mathrm{g}\tan \alpha$

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