A cubical vessel (open from top) of side L is filled with a liquid of density `rho`.
What is the torque of hydrostatic force on a side wall about an axis passing through one of the bottom edges?

Options

A. `(rho gL^(4))/(4)`
B. `(rho gL^(4))/(6)`
C. `(2rho gL^(4))/(3)`
D. `(rho gL^(4))/(3)`

Answer

To find the torque of the hydrostatic force on a side wall of a cubical vessel about an axis passing through one of the bottom edges, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Problem**:
We have a cubical vessel of side length \( L \) filled with a liquid of density \( \rho \). We need to calculate the torque about an axis at one of the bottom edges of the side wall.

2. **Consider an Element of the Wall**:
At a depth \( x \) from the top of the liquid, consider a thin horizontal strip of thickness \( dx \) on the side wall. The area of this strip is:
\[ dA = L \cdot dx \]

3. **Calculate the Pressure at Depth \( x \)**:
The pressure at this depth due to the liquid is given by:
\[ P = \rho g x \]
where \( g \) is the acceleration due to gravity.

4. **Determine the Hydrostatic Force on the Strip**:
The hydrostatic force \( dF \) acting on the strip is given by:
\[ dF = P \cdot dA = \rho g x \cdot (L \cdot dx) = \rho g L x \, dx \]

5. **Calculate the Distance from the Axis of Rotation**:
The distance from the axis (bottom edge) to the strip at depth \( x \) is:
\[ r = L - x \]

6. **Find the Torque Contribution from the Strip**:
The torque \( d\tau \) due to the force \( dF \) about the axis is:
\[ d\tau = dF \cdot r = (\rho g L x \, dx) \cdot (L - x) = \rho g L x (L - x) \, dx \]

7. **Integrate to Find Total Torque**:
To find the total torque \( \tau \), integrate \( d\tau \) from \( x = 0 \) to \( x = L \):
\[ \tau = \int_0^L \rho g L x (L - x) \, dx \]

8. **Simplify the Integral**:
Expanding the integrand:
\[ \tau = \rho g L \int_0^L (Lx - x^2) \, dx \]
Now, calculate the integral:
\[ \int_0^L Lx \, dx = L \cdot \frac{x^2}{2} \bigg|_0^L = L \cdot \frac{L^2}{2} = \frac{L^3}{2} \]
\[ \int_0^L x^2 \, dx = \frac{x^3}{3} \bigg|_0^L = \frac{L^3}{3} \]
Thus,
\[ \tau = \rho g L \left( \frac{L^3}{2} - \frac{L^3}{3} \right) \]

9. **Combine the Results**:
Finding a common denominator (6):
\[ \tau = \rho g L \left( \frac{3L^3}{6} - \frac{2L^3}{6} \right) = \rho g L \cdot \frac{L^3}{6} = \frac{\rho g L^4}{6} \]

### Final Result:
The torque of the hydrostatic force on a side wall about an axis passing through one of the bottom edges is:
\[ \tau = \frac{\rho g L^4}{6} \, \text{Newton meter} \]