Question

A uniform rod of density $\rho$ is placed in a wide tank containing a liquid of density ${\rho }_0({\rho }_0>\rho )$ . 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.

• A. $\sin \theta =\frac{1}{2}\sqrt{{\rho }_0/\rho }$
• B. $\sin \theta =\frac{1}{2}\frac{{\rho }_0}{\rho }$
• C. $\sin \theta =\sqrt{{\rho }_0/\rho }$
• D. $\sin \theta ={\rho }_0/\rho$

Answer 1

The correct option is A $\sin \theta =\frac{1}{2}\sqrt{{\rho }_0/\rho }$

Let the length of the rod is $L$.

The weight of the rod at point S is given as,

$W=Al\rho g$

The force of buoyancy is given as,

$F_B=Al{\rho }_{0}g$

For rotational equilibrium, it can be written as,

$Al{\rho }_{0}g\times \frac{l}{2}\cos \theta =AL\rho g\times \frac{L}{2}\cos \theta$

$Al{\rho }_{0}g\times \frac{l}{2}\cos \theta =AL\rho g\times \frac{L}{2}\cos \theta$

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

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

The angle is given as,

$\sin \theta =\frac{h}{l}$

$\sin \theta =\frac{L}{2l}$

$\sin \theta =\frac{1}{2}\sqrt{\frac{{\rho }_0}{\rho }}$