Q. A hollow cylinder of mass $m$ made heavy at its bottom is floating vertically in water. It is tilled from its vertical position through an angle $\theta$ and is left. The restoring force acting on it is
A) $mg\cos\theta$
B) $mg\sin\theta$
C) $mg\left\lbrack \frac{1}{\cos\theta} - 1 \right\rbrack$
D) $mg\left\lbrack \frac{1}{\cos\theta} + 1 \right\rbrack$
Explanation
(c) Let $l$ be the length of the cylinder in water it is in the vertical position and $A$ be the cross-sectional area of the cylinder. As cylinder is floating so
Weight of cylinder = upward thrust$mg = A\ l\ \rho\ g$ or $m = Al\rho$
When the cylinder is tilled through an angle $\theta$, length of cylinder in water $= \frac{l}{\cos\theta}$
Weight of water displaced $= \frac{l}{\cos\theta}A\ \rho\ g$
Restoring force $= \frac{l\ A\ \rho\ g}{\cos\theta} = l\ A\ \rho\ g$
$= l\ A\ \rho\ g\left\lbrack \frac{1}{\cos\theta} - 1 \right\rbrack = mg\left\lbrack \frac{1}{\cos\theta} - 1 \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.