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 inclination $\theta$ of the rod with the horizontal in equilibrium position is

• A. $si{n}^{-1}\left[\sqrt{\frac{{\rho }_0}{\rho }}\right]$
• B. $si{n}^{-1}\left[n\sqrt{\frac{{\rho }_0}{\rho }}\right]$
• C. $si{n}^{-1}\left[\frac{1}{n}\sqrt{\frac{{\rho }_0}{\rho }}\right]$
• D. $si{n}^{-1}\left[\frac{1}{n}\sqrt{\frac{\rho }{{\rho }_0}}\right]$

Answer 1

The correct option is C $si{n}^{-1}\left[\frac{1}{n}\sqrt{\frac{{\rho }_0}{\rho }}\right]$

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

The weight of rod $=mg=Al\rho g$ acting vertically downwards off the centre of gravity $c$ of the rod.

The upward thrust on rod $F_s=Al\rho g$ acting vertically upwards at the centre of buoyancy $D$;

Which is the midpoint of the length of the rod inside the liquid

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

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

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

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

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

or $\theta =si{n}^{-1}\left[\frac{1}{n}\sqrt{\frac{{\rho }_0}{\rho }}\right]$