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 find the difference in depth \( d_2 - d_1 \) based on the pressures experienced by the submarine at two different depths.

### Step-by-Step Solution:

1. **Understand the Pressure Formula**:
The pressure at a depth in a fluid is given by the formula:
\[ P = P_0 + \rho g h \]
where:
- \( P \) is the pressure at depth,
- \( P_0 \) is the atmospheric pressure (which we can ignore here since we are dealing with changes in pressure),
- \( \rho \) is the density of the fluid (water in this case),
- \( g \) is the acceleration due to gravity,
- \( h \) is the depth.

2. **Identify Given Values**:
- Pressure at depth \( d_1 \):
\[ P_1 = 5.05 \times 10^6 \, \text{Pa} \]
- Pressure at depth \( d_2 \):
\[ P_2 = 8.08 \times 10^6 \, \text{Pa} \]
- Density of water:
\[ \rho = 10^3 \, \text{kg/m}^3 \]
- Acceleration due to gravity:
\[ g = 10 \, \text{m/s}^2 \]

3. **Calculate the Change in Pressure**:
\[ \Delta P = P_2 - P_1 = (8.08 \times 10^6) - (5.05 \times 10^6) = 3.03 \times 10^6 \, \text{Pa} \]

4. **Relate Change in Pressure to Change in Depth**:
Using the formula for pressure change:
\[ \Delta P = \rho g (d_2 - d_1) \]
We can rearrange this to find \( d_2 - d_1 \):
\[ d_2 - d_1 = \frac{\Delta P}{\rho g} \]

5. **Substitute the Values**:
\[ d_2 - d_1 = \frac{3.03 \times 10^6}{(10^3)(10)} = \frac{3.03 \times 10^6}{10^4} \]

6. **Calculate the Result**:
\[ d_2 - d_1 = 303 \, \text{m} \]

### Final Answer:
The difference in depth \( d_2 - d_1 \) is approximately \( 303 \, \text{m} \).

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