Question

Two masses - ' $\sqrt{3}$m' and '$\sqrt{2}$m' are tied by a light string are placed on a wedge of mass '4m'. The wedge is placed on a smooth horizontal surface. Find out the value of $\theta$ such that the wedge does not move even after the system is set free from the state of rest.

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. none of these

Answer 1

The correct option is A

${30}^{\circ }$

Centre of mass will not shift in horizontal direction.
Let - $\sqrt{3}$ m moves a distance x on wedge in downward direction. - $\sqrt{2}$ m will also move on the other side in downward direction by a distance x .
Then $m_1$$x_1$ = $m_2$$x_2$
$\sqrt{3}$ mx cos${45}^{\circ }$=-$\sqrt{2}$mx cos $\theta$
$\Rightarrow Cos\theta =\frac{\sqrt{3}}{2}\Rightarrow \theta ={30}^{\circ }$