A block of base $10 \mathrm{~cm} \times 10 \mathrm{~cm}$ and height $15 \mathrm{~cm}$ is kept on an inclined…

A block of base $10 \mathrm{~cm} \times 10 \mathrm{~cm}$ and height $15 \mathrm{~cm}$ 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,
  1. at $\theta=30^{\circ}$, the block will start sliding down the plane
  2. the block will remain at rest on the plane up to certain $\theta$ and then it will topple
  3. at $\theta=60^{\circ}$, the block will start sliding down the plane and continue to do so at higher angles
  4. at $\theta=60^{\circ}$, the block will start sliding down the plane and on further increasing $\theta$, it will topple at certain $\theta$

Solution

Condition of sliding is $m g \sin \theta>\mu m g \cos \theta$ or $\tan \theta>\mu$ or $\quad \tan \theta>\sqrt{3}$ Condition of toppling is
Torque of $m g \sin \theta$ about $0>$ torque of $m g \cos \theta$ about $\therefore \quad(m g \sin \theta)\left(\frac{15}{2}\right)>(m g \cos \theta)\left(\frac{10}{2}\right)$ or $\quad \tan \theta>\frac{2}{3}$ With increase in value of $\theta$, condition of sliding is satisfied first.

Asked in: JEE Advanced 2009 (Paper 1)

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