A submarine experiences a pressure of `5.05xx10^(6)Pa` at a depth of `d_(1)` in a sea When it goes futher to a depth of `d_(2)`. It experiences a pressure of `8.08xx10^(6)Pa`. The `d_(2)-d_(1)` is approximately (density of water `=10^(3)kg//m^(3)` and acceleration due to gravity `=10ms^(-2))`

Options

A. 300m
B. 400m
C. 600m
D. 500m

Answer

To solve the problem, we need to determine the difference in depth \( d_2 - d_1 \) experienced by the submarine based on the pressures at those depths.

### Step-by-Step Solution:

1. **Identify the given values:**
- Pressure at depth \( d_1 \) (P1) = \( 5.05 \times 10^6 \) Pa
- Pressure at depth \( d_2 \) (P2) = \( 8.08 \times 10^6 \) Pa
- Density of water (\( \rho \)) = \( 10^3 \) kg/m³
- Acceleration due to gravity (\( g \)) = \( 10 \) m/s²

2. **Use the hydrostatic pressure formula:**
The pressure at a depth in a fluid is given by:
\[ P = P_0 + \rho g h \]
where \( P_0 \) is the atmospheric pressure (which we can ignore since it cancels out), \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the depth.

3. **Set up the equations for both depths:**
For depth \( d_1 \):
\[ P_1 = \rho g d_1 \]
For depth \( d_2 \):
\[ P_2 = \rho g d_2 \]

4. **Subtract the two equations:**
To find \( d_2 - d_1 \), we can subtract the first equation from the second:
\[ P_2 - P_1 = \rho g (d_2 - d_1) \]

5. **Rearranging the equation:**
\[ d_2 - d_1 = \frac{P_2 - P_1}{\rho g} \]

6. **Substitute the known values:**
- Calculate \( P_2 - P_1 \):
\[ P_2 - P_1 = 8.08 \times 10^6 \, \text{Pa} - 5.05 \times 10^6 \, \text{Pa} = 3.03 \times 10^6 \, \text{Pa} \]
- Now substitute into the equation:
\[ d_2 - d_1 = \frac{3.03 \times 10^6 \, \text{Pa}}{(10^3 \, \text{kg/m}^3)(10 \, \text{m/s}^2)} \]

7. **Calculate the depth difference:**
\[ d_2 - d_1 = \frac{3.03 \times 10^6}{10^4} = 303 \, \text{m} \]

8. **Conclusion:**
The difference in depth \( d_2 - d_1 \) is approximately \( 303 \, \text{m} \).

### Final Answer:
The approximate difference in depth \( d_2 - d_1 \) is **300 m** (rounding to the nearest option).

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