A uniform rod of density `rho` is placed in a wide tank containing a liquid of density `rho_0 (rho_0 gt 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.

Options

A. `sin theta = (1)/(2) sqrt(rho_0//rho)`
B. `sin theta = (1)/(2).(rho_0)/(rho)`
C. `sin theta = sqrt(rho_0//rho)`
D. `sin theta = rho//rho_0`

Explanation

To solve the problem step by step, we will analyze the forces acting on the rod and apply the principles of equilibrium and torque.

### Step 1: Understand the Setup
We have a uniform rod of density \( \rho \) placed in a liquid of density \( \rho_0 \) (where \( \rho_0 > \rho \)). The depth of the liquid is half the length of the rod, which we denote as \( L/2 \) (where \( L \) is the length of the rod). The rod is in equilibrium, with its lower end resting on the bottom of the tank and making an angle \( \theta \) with the horizontal.

### Step 2: Identify Forces Acting on the Rod
1. **Weight of the Rod (W)**: The weight of the rod can be calculated as:
\[ W = \rho \cdot V \cdot g = \rho \cdot A \cdot L \cdot g \]
where \( A \) is the cross-sectional area of the rod.

2. **Buoyant Force (F_B)**: The buoyant force acting on the rod can be calculated using Archimedes' principle:
\[ F_B = \rho_0 \cdot V_{displaced} \cdot g = \rho_0 \cdot A \cdot (L/2) \cdot g \]

### Step 3: Set Up the Torque Equation
For the rod to be in equilibrium, the torque about any point must be zero. We can take torques about the point where the rod touches the bottom of the tank.

- The torque due to the weight of the rod (acting at its center of mass, which is at \( L/2 \)):
\[ \tau_W = W \cdot \left(\frac{L}{2} \cos \theta\right) \]

- The torque due to the buoyant force (acting at the midpoint of the submerged part of the rod, which is at \( L/4 \)):
\[ \tau_B = F_B \cdot \left(\frac{L}{4} \cos \theta\right) \]

### Step 4: Equate the Torques
Setting the torques equal for equilibrium:
\[ W \cdot \left(\frac{L}{2} \cos \theta\right) = F_B \cdot \left(\frac{L}{4} \cos \theta\right) \]

Substituting the expressions for \( W \) and \( F_B \):
\[ (\rho A L g) \cdot \left(\frac{L}{2} \cos \theta\right) = \left(\rho_0 A \cdot \frac{L}{2} g\right) \cdot \left(\frac{L}{4} \cos \theta\right) \]

### Step 5: Simplify the Equation
Canceling common terms:
\[ \rho L \cdot \frac{L}{2} = \rho_0 \cdot \frac{L}{4} \]
This simplifies to:
\[ \rho L = \frac{\rho_0}{2} \]

### Step 6: Solve for \( \sin \theta \)
From the equilibrium condition, we find:
\[ \sin \theta = \frac{L/2}{L} \cdot \sqrt{\frac{\rho}{\rho_0}} = \frac{1}{2} \sqrt{\frac{\rho}{\rho_0}} \]

### Final Answer
Thus, the angle \( \theta \) can be expressed in terms of the densities:
\[ \sin \theta = \frac{1}{2} \sqrt{\frac{\rho}{\rho_0}} \]