A triangular lamina of area `A` and height `h` is immersed in a liquid of density `rho` in a vertical plane with its base on the surface of the liquid. The thrust on the lamina is.

Options

A. `(1)/(2) A rho gh`
B. `(1)/(3) A rho gh`
C. `(1)/(6) A rho gh`
D. `(2)/(3) A rho gh`

Explanation

To find the thrust on a triangular lamina immersed in a liquid, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a triangular lamina with area \( A \) and height \( h \) immersed in a liquid of density \( \rho \).
- The base of the triangle is on the surface of the liquid.

2. **Identify the Pressure at Depth**:
- The pressure \( P \) at a depth \( y \) in the liquid is given by:
\[ P = \rho g y \]
where \( g \) is the acceleration due to gravity.

3. **Consider a Differential Strip**:
- Consider a small horizontal strip of the triangle at depth \( y \) with thickness \( dy \) and width \( b \).
- The area of this differential strip \( dA \) is:
\[ dA = b \, dy \]

4. **Calculate the Thrust on the Differential Strip**:
- The thrust \( dF \) on this small strip due to pressure is:
\[ dF = P \cdot dA = \rho g y \cdot b \, dy \]

5. **Relate Width \( b \) to Height \( y \)**:
- By similarity of triangles, we can express \( b \) in terms of \( y \):
\[ \frac{y}{b} = \frac{h}{B} \implies b = \frac{y B}{h} \]
where \( B \) is the base length of the triangle.

6. **Substitute \( b \) into the Thrust Equation**:
- Substitute \( b \) into the thrust equation:
\[ dF = \rho g y \cdot \left(\frac{y B}{h}\right) dy = \frac{\rho g B}{h} y^2 \, dy \]

7. **Integrate to Find Total Thrust**:
- To find the total thrust \( F \), integrate \( dF \) from \( 0 \) to \( h \):
\[ F = \int_0^h \frac{\rho g B}{h} y^2 \, dy \]
- The integral of \( y^2 \) is:
\[ \int y^2 \, dy = \frac{y^3}{3} \]
- Evaluating the integral from \( 0 \) to \( h \):
\[ F = \frac{\rho g B}{h} \cdot \left[\frac{h^3}{3} - 0\right] = \frac{\rho g B h^2}{3} \]

8. **Express in Terms of Area**:
- The area \( A \) of the triangular lamina is:
\[ A = \frac{1}{2} B h \]
- Rearranging gives \( B = \frac{2A}{h} \).
- Substitute \( B \) back into the thrust equation:
\[ F = \frac{\rho g}{3} \cdot \left(\frac{2A}{h}\right) h^2 = \frac{2}{3} \rho g A h \]

### Final Answer:
The thrust on the triangular lamina is:
\[ F = \frac{2}{3} \rho g A h \]