The force acting on a window of are 50`xx`50 cm 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) kg m^(-3)` ,g =`10 m s^(-2)` )

Options

A. `5 xx10^(5)N`
B. `25 xx10^(5)` N
C. `5 xx 10^(6)`N
D. `25 xx 10^(6)`N

Explanation

To solve the problem, we need to calculate the force acting on a window of a submarine at a depth of 2000 m in the ocean. The steps to find the solution are as follows:

### Step 1: Determine the Pressure at Depth
The pressure at a depth in a fluid can be calculated using the formula:
\[ P = P_a + \rho g h \]
where:
- \( P_a \) = atmospheric pressure at sea level (approximately \( 1.01 \times 10^5 \, \text{Pa} \))
- \( \rho \) = density of seawater (\( 10^3 \, \text{kg/m}^3 \))
- \( g \) = acceleration due to gravity (\( 10 \, \text{m/s}^2 \))
- \( h \) = depth (2000 m)

Substituting the values:
\[ P = 1.01 \times 10^5 + (10^3)(10)(2000) \]

Calculating the hydrostatic pressure:
\[ P = 1.01 \times 10^5 + 2 \times 10^7 \]
\[ P = 2.0101 \times 10^7 \, \text{Pa} \]

### Step 2: Calculate the Area of the Window
The area \( A \) of the window is given by:
\[ A = \text{length} \times \text{width} \]
Given that the dimensions of the window are \( 50 \, \text{cm} \times 50 \, \text{cm} \):
\[ A = 0.5 \, \text{m} \times 0.5 \, \text{m} = 0.25 \, \text{m}^2 \]

### Step 3: Calculate the Force Acting on the Window
The force \( F \) acting on the window can be calculated using the formula:
\[ F = P \times A \]
Substituting the values:
\[ F = (2.0101 \times 10^7) \times (0.25) \]
\[ F = 5.02775 \times 10^6 \, \text{N} \]

### Step 4: Round the Result
For simplicity, we can round this to:
\[ F \approx 5 \times 10^6 \, \text{N} \]

### Final Answer
The force acting on the window of the submarine is approximately \( 5 \times 10^6 \, \text{N} \).

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