Question

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 $\theta$

Answer 1

Answer: (B)

For sliding, $\tan \theta \geq \sqrt{3} \quad(=1.732)$
For toppling, $\tan \theta \geq \frac{2}{3} \quad(=0.67)$