A liquid of density `rho_0` is filled in a wide tank to a height h. A solid rod of length L, cross-section A and density `rho` is suspended freely in the tank. The lower end of the rod touches the base of the tank and h=L/n (where n gt 1). Then the angle of inclination `theta` of the rod with the horizontal in equilibrium position is

Options

A. `sin^(-1)(sqrt((rho_0)/rho))`
B. `sin^(-1)(n sqrt((rho_0)/rho))`
C. `sin^(-1)(1/n sqrt((rho_0)/rho))`
D. `sin^(-1)(1/n sqrt((rho)/rho_0))`

Answer

To solve the problem step by step, we will analyze the forces acting on the rod and use the equilibrium conditions to find the angle of inclination \( \theta \).

### Step 1: Understand the setup
We have a tank filled with a liquid of density \( \rho_0 \) to a height \( h \). A solid rod of length \( L \), cross-sectional area \( A \), and density \( \rho \) is suspended in the tank such that its lower end touches the base of the tank, and the height of the liquid is given by \( h = \frac{L}{n} \) where \( n > 1 \).

### Step 2: Identify the forces acting on the rod
The forces acting on the rod are:
1. The weight of the rod \( W \) acting downwards at its center of gravity.
2. The buoyant force \( F_B \) acting upwards at the center of the submerged part of the rod.

### Step 3: Calculate the weight of the rod
The weight \( W \) of the rod can be expressed as:
\[ W = \rho \cdot A \cdot L \cdot g \]
where \( g \) is the acceleration due to gravity.

### Step 4: Calculate the buoyant force
The buoyant force \( F_B \) is given by Archimedes' principle:
\[ F_B = \rho_0 \cdot A \cdot V_{immersed} \cdot g \]
The volume of the rod that is immersed in the liquid can be calculated as:
\[ V_{immersed} = A \cdot h \]
Given \( h = \frac{L}{n} \), we have:
\[ V_{immersed} = A \cdot \frac{L}{n} \]
Thus, the buoyant force becomes:
\[ F_B = \rho_0 \cdot A \cdot \left( A \cdot \frac{L}{n} \right) \cdot g = \rho_0 \cdot A^2 \cdot \frac{L}{n} \cdot g \]

### Step 5: Set up the equilibrium condition
For the rod to be in equilibrium, the sum of torques about any point must be zero. We can choose the lower end of the rod as the pivot point. The torques due to the weight and buoyant force must balance each other.

Let’s denote the length of the rod submerged in the liquid as \( L_{immersed} \). The center of gravity of the rod is located at \( \frac{L}{2} \) from the bottom, and the center of buoyancy is at \( \frac{L_{immersed}}{2} \).

The torque due to the weight \( W \) is:
\[ \tau_W = W \cdot \left(\frac{L}{2} \cos \theta\right) \]
The torque due to the buoyant force \( F_B \) is:
\[ \tau_B = F_B \cdot \left(\frac{L_{immersed}}{2} \cos \theta\right) \]

Setting the torques equal for equilibrium:
\[ W \cdot \left(\frac{L}{2} \cos \theta\right) = F_B \cdot \left(\frac{L_{immersed}}{2} \cos \theta\right) \]

### Step 6: Substitute the expressions for weight and buoyant force
Substituting the expressions for \( W \) and \( F_B \):
\[ \rho \cdot A \cdot L \cdot g \cdot \left(\frac{L}{2} \cos \theta\right) = \rho_0 \cdot A^2 \cdot \frac{L}{n} \cdot g \cdot \left(\frac{L_{immersed}}{2} \cos \theta\right) \]

### Step 7: Cancel common terms and simplify
Canceling \( g \), \( A \), and \( \frac{1}{2} \):
\[ \rho \cdot L = \rho_0 \cdot \frac{L_{immersed}}{n} \]

### Step 8: Relate \( L_{immersed} \) to \( L \)
Since \( L_{immersed} = \frac{L}{n} \):
\[ \rho \cdot L = \rho_0 \cdot \frac{L}{n^2} \]

### Step 9: Solve for \( \sin \theta \)
From the geometry of the setup, we can relate \( \sin \theta \) to the height of the liquid:
\[ \sin \theta = \frac{h}{L} = \frac{L/n}{L} = \frac{1}{n} \]

### Step 10: Final expression for the angle of inclination
Using the relationship derived earlier:
\[ \sin \theta = \frac{1}{n} \sqrt{\frac{\rho_0}{\rho}} \]
Thus, the angle of inclination \( \theta \) is given by:
\[ \theta = \sin^{-1}\left(\frac{1}{n} \sqrt{\frac{\rho_0}{\rho}}\right) \]