The force acting on a window of area `50cm xx 50cm` of a submarine at a depth of `2000 m` in an ocean, the interior of which is maintained at sea level atmospheric pressure is (Density of sea water = `10^(3)kgm^(-3)`, `g = 10ms^(-2)`)

Options

A. `5xx10^(5)N`
B. `25xx10^(5)N`
C. `5xx10^(6)N`
D. `25xx10^(6)N`

Explanation

To find the force acting on the window of the submarine, we can follow these steps:

### Step 1: Determine the area of the window
The area of the window is given as \(50 \, \text{cm} \times 50 \, \text{cm}\). We need to convert this to square meters for consistency with SI units.

\[ \text{Area} = 50 \, \text{cm} \times 50 \, \text{cm} = 0.5 \, \text{m} \times 0.5 \, \text{m} = 0.25 \, \text{m}^2 \]

### Step 2: Calculate the pressure at a depth of 2000 m
The pressure at a depth \(h\) in a fluid is given by the formula:

\[ P = P_0 + \rho g h \]

Where:
- \(P_0\) is the atmospheric pressure at sea level, approximately \(10^5 \, \text{Pa}\) (or \(1 \, \text{atm}\)).
- \(\rho\) is the density of the fluid (sea water), given as \(10^3 \, \text{kg/m}^3\).
- \(g\) is the acceleration due to gravity, given as \(10 \, \text{m/s}^2\).
- \(h\) is the depth, given as \(2000 \, \text{m}\).

Substituting the values:

\[ P = 10^5 \, \text{Pa} + (10^3 \, \text{kg/m}^3)(10 \, \text{m/s}^2)(2000 \, \text{m}) \]

Calculating the second term:

\[ P = 10^5 \, \text{Pa} + 2 \times 10^7 \, \text{Pa} = 2.1 \times 10^7 \, \text{Pa} \]

### Step 3: Calculate the net pressure acting on the window
The net pressure acting on the window is the pressure from the outside minus the pressure from the inside. The inside pressure is maintained at atmospheric pressure \(P_0\):

\[ \text{Net Pressure} = P - P_0 = (2.1 \times 10^7 \, \text{Pa}) - (10^5 \, \text{Pa}) = 2.09 \times 10^7 \, \text{Pa} \]

### Step 4: Calculate the force acting on the window
The force \(F\) acting on the window can be calculated using the formula:

\[ F = \text{Net Pressure} \times \text{Area} \]

Substituting the values:

\[ F = (2.09 \times 10^7 \, \text{Pa}) \times (0.25 \, \text{m}^2) \]

Calculating the force:

\[ F = 5.225 \times 10^6 \, \text{N} \]

### Conclusion
The force acting on the window of the submarine at a depth of 2000 m is approximately \(5.225 \times 10^6 \, \text{N}\).