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 finding the total force against the side of a glass aquarium filled with water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Dimensions**:
- The height (h) of the aquarium is 1 meter.
- The length (L) of the aquarium is 2 meters.

2. **Understand 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. **Calculate the Force on a Differential Strip**:
- Consider a small horizontal strip of height \( dy \) at depth \( y \).
- The area (dA) of this strip is:
\[ dA = L \cdot dy = 2 \, \text{m} \cdot dy \]
- The differential force (dF) acting on this strip is:
\[ dF = P \cdot dA = (\rho g y) \cdot (2 \, dy) = 2 \rho g y \, dy \]

4. **Integrate to Find Total Force**:
- To find the total force (F) against the side of the aquarium, integrate \( dF \) from \( y = 0 \) to \( y = 1 \):
\[ F = \int_0^1 dF = \int_0^1 2 \rho g y \, dy \]
- This integral can be computed as follows:
\[ F = 2 \rho g \int_0^1 y \, dy = 2 \rho g \left[ \frac{y^2}{2} \right]_0^1 = 2 \rho g \cdot \frac{1^2}{2} = \rho g \]

5. **Substitute Values**:
- Now substitute the values of \( \rho \) and \( g \):
\[ F = 1000 \, \text{kg/m}^3 \cdot 9.8 \, \text{m/s}^2 = 9800 \, \text{N} \]

6. **Final Result**:
- The total force against the side of the aquarium is:
\[ F = 9800 \, \text{N} \]