Question

A liquid of density ${\rho }_0$ is filled in a wide tank to a height $h$. A solid rod of length $L$, cross-section $A$ and density $\rho$ is suspended freely in the tank. The lower end of the rod touches the base of the tank and $h=L/n$ (where $n>1$). Then the angle of indication $\theta$ of the rod with the horizontal in equilibrium position is:

• A. ${\sin }^{-1}\left(\sqrt{\frac{{\rho }_0}{\rho }}\right)$
• B. ${\sin }^{-1}\left(n\sqrt{\frac{{\rho }_0}{\rho }}\right)$
• C. ${\sin }^{-1}\left(\frac{1}{n}\sqrt{\frac{{\rho }_0}{\rho }}\right)$
• D. ${\sin }^{-1}\left(\frac{1}{n}\sqrt{\frac{\rho }{{\rho }_0}}\right)$

Answer 1

The correct option is C ${\sin }^{-1}\left(\frac{1}{n}\sqrt{\frac{{\rho }_0}{\rho }}\right)$

Let $l$ be the length of rod immersed in liquid. $\theta$ be the angle of inclination of rod with horizontal in equilibrium position

Here $L$ is the length of the rod.

Weight of rod ,$mg=AL\rho g$ acting vertically downwards at the center of gravity $C$ of the rod.

Upward thrust on rod, $F_B=Al{\rho }_{0}g$ acting vertically upwards at the center of buoyancy $D$, which is the mid point of length of rod inside the liquid.

As the rod is in equilibrium position, then net torque on the rod about point $A$ is zero, i.e,

$\left(AL\rho g\right)\frac{L}{2}cos\theta -\left(Al{\rho }_{0}g\right)\frac{L}{2}cos\theta =0$

$⟹\frac{L^2}{l^2}=\frac{{\rho }_0}{\rho }$

$⟹\frac{L}{l}=\sqrt{\frac{{\rho }_0}{\rho }}$

Now, $sin\theta =\frac{h}{l}=\frac{1}{n}\frac{L}{l}=\frac{1}{n}\sqrt{\frac{{\rho }_0}{\rho }}$

$⟹\theta =si{n}^{-1}(\frac{1}{n}\sqrt{\frac{{\rho }_0}{\rho }})$