Question

Ablockofbase $l0 cm\times10 cm$ and height 15 cm is kept on an inclined plane. The coefficient of friction between them is $\sqrt3$. 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 0, it will topple at certain $\theta$

Answer 1

The Correct Option is B

Condition of sliding is
$mg \sin\theta > \mu mg \cos \theta$
or $\tan\theta > \mu$
or $\tan\theta > \sqrt3$ ...(i)
Condition of toppling is
Torque of $mg \sin \theta about 0 > torque of mg \cos\theta about$
$\therefore (mg \sin\theta)\bigg(\frac{15}{2}\bigg) > (mg \cos \theta)\bigg(\frac{10}{2}\bigg)$
or $\tan \theta > \frac{2}{3}$ .....(ii)
With increase in value of $\theta$, condition of sliding is satisfied first.