A body is experiencing an atmospheric pressure of `1.01xx10^(5)N//m^(2)` on the surface of a pond. The body will experience double of the previous pressure if it is brought to a depth of (given density of pond water `=1.03xx10^(3)kg//m^(2)`, `g=10m//s^(2)`)

Options

A. `8m`
B. `8.6m`
C. `9.7m`
D. `10.3m`

Answer

To solve the problem, we need to find the depth at which the pressure on the body will be double the atmospheric pressure it experiences at the surface of the pond.

### Step-by-Step Solution:

1. **Identify the Initial Pressure**:
The body is experiencing an atmospheric pressure of \( P_0 = 1.01 \times 10^5 \, \text{N/m}^2 \).

2. **Calculate the Target Pressure**:
If the body is to experience double the initial pressure, the target pressure \( P \) will be:
\[ P = 2 \times P_0 = 2 \times 1.01 \times 10^5 \, \text{N/m}^2 = 2.02 \times 10^5 \, \text{N/m}^2 \]

3. **Use the Hydrostatic Pressure Formula**:
The pressure at a depth \( h \) in a fluid is given by the formula:
\[ P = P_0 + \rho g h \]
where \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the depth.

4. **Substitute Known Values**:
We know:
- \( P_0 = 1.01 \times 10^5 \, \text{N/m}^2 \)
- \( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \)
- \( g = 10 \, \text{m/s}^2 \)
We can substitute these values into the hydrostatic pressure formula:
\[ 2.02 \times 10^5 = 1.01 \times 10^5 + (1.03 \times 10^3)(10)h \]

5. **Rearranging the Equation**:
To isolate \( h \), we first subtract \( P_0 \) from both sides:
\[ 2.02 \times 10^5 - 1.01 \times 10^5 = (1.03 \times 10^3)(10)h \]
Simplifying the left side:
\[ 1.01 \times 10^5 = (1.03 \times 10^3)(10)h \]

6. **Calculate \( h \)**:
Now, we can solve for \( h \):
\[ h = \frac{1.01 \times 10^5}{(1.03 \times 10^3)(10)} \]
\[ h = \frac{1.01 \times 10^5}{1.03 \times 10^4} \]
\[ h = \frac{1.01}{1.03} \times 10^{5-4} = \frac{1.01}{1.03} \times 10^1 \]
\[ h \approx 0.9806 \times 10 = 9.806 \, \text{m} \]

### Final Answer:
The body will experience double the previous pressure at a depth of approximately \( 9.8 \, \text{m} \).