Question

An uniform rod of density $\rho$ is placed in a wide tank containing a liquid of density ${\rho }_0({\rho }_0>\rho )$. The depth of the 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, identify the correct relation :

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


Let $L=PQ=$ length of rod
$\therefore SP=SQ=\frac{L}{2}$
Weight of rod, $W=V\times \rho \times g=AL\rho g...\left(i\right)$, acting at the mid point $S$
where $A$ is cross-sectional area of the rod.

Buoyant force acting on the rod,,
$F_B=V_f\times {\rho }_0\times g=Al{\rho }_{0}g...\left(ii\right)$
where $l=PR\&V_f=$ volume of fluid displaced
$F_B$ acts at the mid-point of $PR$ (length of rod inside liquid).
For rotational equilibrium of rod, balancing the torques of weight and $F_B$ about point $P$,
${\tau }_{F_B}={\tau }_W$
where, $\tau =F\times r_{\perp }$
$\Rightarrow Al{\rho }_{0}g\times \left(\frac{l}{2}\cos \theta \right)=AL\rho g\times \frac{L}{2}\cos \theta$
$\Rightarrow \frac{l^2}{L^2}=\frac{\rho }{{\rho }_0}$
$\Rightarrow \frac{l}{L}=\sqrt{\frac{\rho }{{\rho }_0}}...\left(iii\right)$
From figure, $\sin \theta =\frac{h}{PR}$
Or, $\sin \theta =\frac{h}{l}=\frac{L/2}{l}=\frac{L}{2l}$
Substituting from Eq.$\left(iii\right)$,
$\Rightarrow \sin \theta =\frac{1}{2}\sqrt{\frac{{\rho }_0}{\rho }}$