A uniform thin hemispherical shall is kept at rest and in equilibrium on an inclined plane of angle of inclination $\mathrm{\theta }={30}^{\mathrm{o}}$ as shown in figure. If the surface of the inclined plane is sufficiently rough to prevent sliding then the angle $\mathrm{\alpha }$ made by the plane of hemisphere with inclined plane is:

• A

  Value of $\mathrm{\mu }$ is needed
• B

  ${30}^{\mathrm{o}}$
• C

  ${45}^{\mathrm{o}}$
• Correct answer
  D

  ${60}^{\mathrm{o}}$

Solution

O is the centre of mass of the hollow hemisphere and is $\frac{\mathrm{R}}{2}$ from C.





$\mathrm{f}=\mathrm{m}\mathrm{g}\sin \mathrm{\theta }$ ...(i)

$\mathrm{N}-\mathrm{m}\mathrm{g}\cos \mathrm{\theta }$ ...(ii)

$\mathrm{N}\times \frac{\mathrm{R}}{2}\sin \mathrm{\alpha }=\left[\mathrm{R}\frac{\mathrm{R}}{2}\cos \mathrm{\alpha }\right]\mathrm{f}$ ...(iii)

$\therefore \tan \mathrm{\theta }=\frac{\sin \mathrm{\alpha }}{2-\cos \mathrm{\alpha }}\Rightarrow \mathrm{\alpha }={60}^{\mathrm{o}}$

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