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, then

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

Answer 1

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

$R=$ Point on rod at level of water
$PQ=LSQ=SP=\frac{L}{2}$
w = $AL\rho g$
$PR=l$
$F_B=Al{\rho }_{0}g$ .....(1)
For rotational equilibrium,
$Al{\rho }_{0}g\left(\frac{l}{2}\right)cos\theta =AL\rho g\times \frac{L}{2}cos\theta \frac{l}{L}=\sqrt{\frac{\rho }{{\rho }_0}}sin\theta =\frac{L}{2l}=\frac{1}{2}\sqrt{\frac{{\rho }_0}{\rho }}$