Consider a mass-pulley system as shown in the figure. There is a wedge of mass $M$ and equal wedge angles $\theta$ lying on a rigid horizontal table. The coefficient of friction between the wedge and the table is $\mu$. There are two blocks of mass $m_1$ and $m_2$ lying on the incline of the wedge. The coefficients of friction between the blocks and wedge are $\mu_1$ and $\mu_2$ as shown in the figure. Consider $m_1 \gt m_2$ and the coefficients of friction $\left(\mu, \mu_1\right.$ and $\left.\mu_2\right)$ to be less than $\tan \theta$. Gravity is acting downwards with acceleration due to gravity $g$. What should be the value of $\frac{m_1}{m_2}$ so that the system is in equilibrium?

• A

  $\frac{\mu_1 \mu_2 \cos \theta+\sin \theta}{\sin \theta-\mu \cos \theta}$
• B

  $\frac{\mu \cos \theta+\sin \theta}{\sin \theta-\mu_1 \cos \theta}$
• C

  $\frac{\mu_1 \cos \theta+\sin \theta}{\sin \theta-\mu_2 \cos \theta}$
• Correct answer
  D

  $\frac{\mu_2 \cos \theta+\sin \theta}{\sin \theta-\mu_1 \cos \theta}$

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