Q. A block of base $10\ cm \times 10\ cm$ and height $15\ cm$ is kept on an inclined plane. The coefficient of friction between them is $\sqrt{3}$. The inclination $\theta$ of this inclined plane from the horizontal plane is gradually increased from $0{^\circ}$. Then
A) At $\theta = 30{^\circ}$, the block will start sliding down the plane
B) The block will remain at rest on the plane up to certain $\theta$ and then it will topple
C) At $\theta = 60{^\circ}$, the block will start sliding down the plane and continue to do so at higher angles
D) At $\theta = 60{^\circ}$, the block will start sliding down the plane and on further increasing $\theta$, it will topple at certain $\theta$
Explanation
(b) For rotational equilibrium about point “$P$”,
$mg\sin\theta\left( \frac{b}{2} \right) = mg\cos\theta\left( \frac{a}{2} \right)$
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$\Rightarrow \tan\theta = \frac{a}{b} = \frac{10}{15} = \frac{2}{3}$
$\Rightarrow \ \theta = 33.69{^\circ}$
$i.e$., toppling starts at $\theta = 33.69{^\circ}$
and angle of repose $= \tan^{- 1}{(\mu)} = \tan^{- 1}{(\sqrt{3}}) = 60{^\circ}$
It mean the block will remain at rest on the plane up to certain angle $\theta$ and then it will topple
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.