Q. A uniform rod of density $\rho$ is placed in a wide tank containing a liquid $\sigma(\sigma > \rho)$. The depth of liquid in the tank is half the length of the rod. The rod is in equilibrium, with its lower end resting on the bottom of the tank. In this position, the rod makes an angle $\theta$ with the horizontal. Then $\sin\theta$ is equal to
A) $\frac{1}{2}\sqrt{\frac{\sigma}{\rho}}$
B) $\frac{1}{2}\frac{\sigma}{\rho}$
C) $\sqrt{\frac{\rho}{\sigma}}$
D) $\sqrt{\frac{\sigma}{\rho}}$
Explanation
(a) $AB = L,\ AC = \frac{L}{2};AD = l\ (say)$
Let $A$ = area of cross-section of the rod
Weight of the rod $= AL\ \rho\ g$ acting vertically downwards at $C$
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Upthrust of liquid on rod $= Al\ \sigma\ g$, acting upwards through the mid-point of $AD$
For rotational equilibrium of rod net torque about point $A$ should be zero. So
$(LA\rho\ g)\frac{L}{2}\cos\theta = (l\ A\ \sigma g)\frac{l}{2}\cos\theta\ or\frac{l^{2}}{L^{2}} = \frac{\rho}{\sigma}\ $
Or $\sin{\theta = \frac{1}{2}\sqrt{\frac{\sigma}{\rho}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.