A cubical vessel (open from top) of side L is filled with a liquid of density `rho`.
What is the ratio of magnitude of torque on one side wall to the torque on base about the same axis

Options

A. `1 : 1`
B. `1 : 2`
C. `1 : 3`
D. `1 : 4`

Answer

To solve the problem, we need to find the ratio of the torque on one side wall of a cubical vessel filled with liquid to the torque on the base about the same axis. Let's break down the solution step by step.

### Step 1: Define the Problem
We have a cubical vessel with side length \( L \) filled with a liquid of density \( \rho \). We need to calculate the torque on one side wall and the torque on the base about a vertical axis.

### Step 2: Torque on the Side Wall
1. **Consider an Element**: Take a horizontal strip of thickness \( dx \) at a distance \( x \) from the top of the liquid.
2. **Area of the Strip**: The area \( dA \) of this strip is given by:
\[ dA = L \cdot dx \]
3. **Pressure at Depth \( x \)**: The pressure \( P \) at this depth is:
\[ P = \rho g x \]
4. **Force on the Strip**: The differential force \( dF \) acting on this strip is:
\[ dF = P \cdot dA = \rho g x \cdot (L \cdot dx) = \rho g L x \, dx \]
5. **Distance to Axis**: The distance from the axis to the strip is \( L - x \).
6. **Torque Contribution**: The torque \( d\tau \) due to this strip is:
\[ d\tau = dF \cdot (L - x) = \rho g L x (L - x) \, dx \]
7. **Total Torque on Side Wall**: Integrate \( d\tau \) from \( 0 \) to \( L \):
\[ \tau_1 = \int_0^L \rho g L x (L - x) \, dx \]
\[ = \rho g L \int_0^L (Lx - x^2) \, dx \]
\[ = \rho g L \left[ \frac{Lx^2}{2} - \frac{x^3}{3} \right]_0^L \]
\[ = \rho g L \left( \frac{L^3}{2} - \frac{L^3}{3} \right) = \rho g L \left( \frac{3L^3 - 2L^3}{6} \right) = \frac{\rho g L^4}{6} \]

### Step 3: Torque on the Base
1. **Pressure at Base**: The pressure at the base (depth \( L \)) is:
\[ P = \rho g L \]
2. **Area of the Base**: The area \( A \) of the base is:
\[ A = L^2 \]
3. **Force on the Base**: The total force \( F \) on the base is:
\[ F = P \cdot A = \rho g L \cdot L^2 = \rho g L^3 \]
4. **Torque on the Base**: The torque \( \tau_2 \) about the axis is:
\[ \tau_2 = F \cdot \left(\frac{L}{2}\right) = \rho g L^3 \cdot \frac{L}{2} = \frac{\rho g L^4}{2} \]

### Step 4: Ratio of Torques
Now, we can find the ratio of the torque on the side wall to the torque on the base:
\[ \text{Ratio} = \frac{\tau_1}{\tau_2} = \frac{\frac{\rho g L^4}{6}}{\frac{\rho g L^4}{2}} = \frac{1/6}{1/2} = \frac{1}{3} \]

### Final Answer
Thus, the ratio of the magnitude of torque on one side wall to the torque on the base about the same axis is:
\[ \text{Ratio} = 1 : 3 \]