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 $0 = 30{^\circ}$, the block will start sliding down the plane
B) The block will remain at the 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
(d) Condition of sliding is
$mg\sin{\theta > \mu\ mg\cos\theta}$
$or\ \tan{\theta > \mu}$
$or\tan{\theta > \sqrt{3}}$ …(i)
condition of toppling is
[image:5_image183.jpeg]
Torque of $mg\sin\theta\ $about $O >$ torque of $mg\cos\theta\ $about.
$\therefore\ \ \ \ (mg\sin{\theta)\left( \frac{15}{2} \right) > (mg\cos{\theta)\left( \frac{10}{2} \right)}}$
$or\ \ \tan{\theta > \frac{2}{3}\ \ \ldots.(ii)}$
With increase in value of $\theta,$ condition of sliding is satisfied first.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.