The side of glass aquarium is 1 m high and 2m long, when the aquarium is filled to the top with water what is the total force against the side-

Options

A. `980xx10^(3)N`
B. `9.8xx10^(3)N`
C. `0.98xx10^(3)N`
D. `0.098xx10^(3)N`

Answer

To solve the problem of calculating the total force against the side of a glass aquarium filled with water, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
- The aquarium has a height (h) of 1 meter and a length (L) of 2 meters.
- We need to find the total force exerted by the water on the side wall of the aquarium.

2. **Identify the Pressure at a Depth**:
- The pressure (P) at a depth (y) in a fluid is given by the formula:
\[ P = \rho g y \]
- Where:
- \(\rho\) = density of water (approximately \(1000 \, \text{kg/m}^3\))
- \(g\) = acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\))
- \(y\) = depth from the surface of the water.

3. **Consider a Differential Strip**:
- Consider a small horizontal strip of the aquarium wall at a depth \(y\) with a small height \(dy\).
- The area (A) of this strip is:
\[ A = L \cdot dy = 2 \cdot dy \]

4. **Calculate the Differential Force**:
- The differential force (dF) acting on this strip due to the pressure is:
\[ dF = P \cdot A = (\rho g y) \cdot (2 \cdot dy) \]
- Thus, substituting the expression for pressure:
\[ dF = 2 \rho g y \, dy \]

5. **Integrate to Find Total Force**:
- To find the total force (F) on the wall, integrate \(dF\) from \(y = 0\) to \(y = 1\) (the height of the aquarium):
\[ F = \int_0^1 2 \rho g y \, dy \]

6. **Perform the Integration**:
- The integral can be calculated as follows:
\[ F = 2 \rho g \int_0^1 y \, dy = 2 \rho g \left[\frac{y^2}{2}\right]_0^1 = 2 \rho g \cdot \frac{1^2}{2} = \rho g \]

7. **Substitute Values**:
- Now substitute the values of \(\rho\) and \(g\):
\[ F = 1000 \cdot 9.8 = 9800 \, \text{N} \]

8. **Final Answer**:
- The total force against the side of the aquarium is:
\[ F = 9800 \, \text{N} \quad \text{(or } 9.8 \times 10^3 \text{ N)} \]