A uniform rod of density ρ is placed in a wide tank containing a liquid of density ρ₀ (ρ₀ >ρ ). 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 θ with the horizontal

Let us analyze the equilibrium of the uniform rod step-by-step.

Let:
- $L$ be the total length of the uniform rod.
- $A$ be the uniform cross-sectional area of the rod.
- $\rho$ be the density of the rod.
- ${\rho }_0$ be the density of the liquid in the tank (${\rho }_0>\rho$).
- $h$ be the depth of the liquid in the tank, which is given as half the length of the rod, i.e., $h=\frac{L}{2}$.
- $\theta$ be the angle the rod makes with the horizontal.

Let the lower end of the rod resting on the bottom of the tank be point $O$. The rod is inclined at an angle $\theta$ to the horizontal, so its upper part is out of the liquid (since ${\rho }_0>\rho$ and the rod is in equilibrium, partly submerged). Let the submerged length of the rod be $x$.

From the geometry of the system, the vertical depth of the submerged part of the rod must equal the depth of the liquid $h$:
$x\sin \theta =h$
Since $h=\frac{L}{2}$, we have:
$x\sin \theta =\frac{L}{2}\Rightarrow x=\frac{L}{2\sin \theta }$

Now, let us identify the forces acting on the rod and their respective points of application to calculate the torque about the contact point $O$ at the bottom of the tank:

1. Weight of the rod ($W$):
The weight acts vertically downwards through the center of gravity of the rod, which is at its midpoint, i.e., at a distance of $\frac{L}{2}$ from the end $O$ along the rod.
$W=AL\rho g$
The perpendicular distance from the line of action of $W$ to the point $O$ is:
$d_W=\frac{L}{2}\cos \theta$
The clockwise torque about $O$ due to the weight is:
${\tau }_W=W·\left(\frac{L}{2}\cos \theta \right)=\left(AL\rho g\right)\frac{L}{2}\cos \theta =\frac{1}{2}A{L}^2\rho g\cos \theta$

2. Buoyant force ($F_B$):
The buoyant force acts vertically upwards through the center of buoyancy, which is the center of gravity of the submerged portion of the rod. Since the submerged length is $x$, this center is at a distance of $\frac{x}{2}$ from the end $O$ along the rod.
$F_B=Ax{\rho }_{0}g$
The perpendicular distance from the line of action of $F_B$ to the point $O$ is:
$d_B=\frac{x}{2}\cos \theta$
The counter-clockwise torque about $O$ due to the buoyant force is:
${\tau }_B=F_B·\left(\frac{x}{2}\cos \theta \right)=\left(Ax{\rho }_{0}g\right)\frac{x}{2}\cos \theta =\frac{1}{2}A{x}^2{\rho }_{0}g\cos \theta$

3. Normal force and friction at $O$:
Since these forces act directly at the point of rotation $O$, their torque about $O$ is zero.

For rotational equilibrium, the net torque about point $O$ must be zero:
${\tau }_W={\tau }_B$
$\frac{1}{2}A{L}^2\rho g\cos \theta =\frac{1}{2}A{x}^2{\rho }_{0}g\cos \theta$

Canceling common terms ($\frac{1}{2}$, $A$, $g$, and $\cos \theta$, assuming $\theta \ne \frac{\pi }{2}$), we get:
$L^2\rho =x^2{\rho }_0$

Substitute $x=\frac{L}{2\sin \theta }$ into the equation:
$L^2\rho ={\left(\frac{L}{2\sin \theta }\right)}^2{\rho }_0$
$L^2\rho =\frac{L^2}{4{\sin }^2\theta }{\rho }_0$

Divide both sides by $L^2$:
$\rho =\frac{{\rho }_0}{4{\sin }^2\theta }$

Rearrange the terms to solve for ${\sin }^2\theta$:
${\sin }^2\theta =\frac{{\rho }_0}{4\rho }$

Taking the square root on both sides:
$\sin \theta =\frac{1}{2}\sqrt{\frac{{\rho }_0}{\rho }}$

Thus, the correct option is indeed:
sin θ = (1/2)√(ρ₀/ρ)