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

Let the rod have length $L$ and cross-sectional area $A$. Since the liquid depth is $L/2$, the immersed length of the rod is found from geometry: $$L\sin\theta=\frac{L}{2}$$ only if the whole rod just reaches the surface. But here the rod is longer than the depth, so the immersed length $l$ satisfies $$l\sin\theta=\frac{L}{2}$$ Hence $$l=\frac{L}{2\sin\theta}$$ Weight of the rod is $$W=\rho A L g$$ Buoyant force is $$B=\rho_0 A l g$$ For rotational equilibrium about the lower end, the weight acts at the midpoint of the rod and the buoyant force acts at the midpoint of the immersed part. Both forces are vertical, so their moment arms are the corresponding horizontal distances: $$B\left(\frac{l}{2}\cos\theta\right)=W\left(\frac{L}{2}\cos\theta\right)$$ This gives $$Bl=WL$$ Substituting $B$ and $W$, $$\rho_0 A l g\cdot l=\rho A L g\cdot L$$ $$\rho_0 l^2=\rho L^2$$ Using $l=\dfrac{L}{2\sin\theta}$, $$\rho_0\left(\frac{L}{2\sin\theta}\right)^2=\rho L^2$$ $$\frac{\rho_0}{4\sin^2\theta}=\rho$$ $$\sin^2\theta=\frac{\rho_0}{4\rho}$$ $$\sin\theta=\frac{1}{2}\sqrt{\frac{\rho_0}{\rho}}$$ Now checking the options, this matches option $\mathrm{(A)}$ exactly.