An air filled balloon is at a depth of `1 km` below the water level in an ocean. The normal stress of the balloon (in Pa) is
(Given,` rho_("water") = 10^(3) kgm^(-3), g = 9.8 ms^(-2)` and `P_(atm) = 10^(5) Pa)`

Options

A. `10^(6)`
B. `9.9xx10^(5)`
C. `9.9xx10^(7)`
D. `9.9xx10^(6)`

Explanation

To find the normal stress on an air-filled balloon at a depth of 1 km below the water level in the ocean, we need to calculate the total pressure acting on the balloon. The total pressure at a depth in a fluid is given by the formula:

\[ P_{\text{total}} = P_{\text{atm}} + \rho g h \]

Where:
- \( P_{\text{atm}} \) is the atmospheric pressure,
- \( \rho \) is the density of the fluid (water in this case),
- \( g \) is the acceleration due to gravity,
- \( h \) is the depth in the fluid.

### Step-by-Step Solution:

1. **Identify the given values:**
- Density of water, \( \rho_{\text{water}} = 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
- Depth, \( h = 1 \, \text{km} = 1000 \, \text{m} \)
- Atmospheric pressure, \( P_{\text{atm}} = 10^5 \, \text{Pa} \)

2. **Calculate the hydrostatic pressure due to the water column:**
\[ P_{\text{water}} = \rho_{\text{water}} \cdot g \cdot h \]
Substituting the values:
\[ P_{\text{water}} = 10^3 \, \text{kg/m}^3 \cdot 9.8 \, \text{m/s}^2 \cdot 1000 \, \text{m} \]
\[ P_{\text{water}} = 10^3 \cdot 9.8 \cdot 1000 = 9.8 \times 10^6 \, \text{Pa} \]

3. **Calculate the total pressure at the depth:**
\[ P_{\text{total}} = P_{\text{atm}} + P_{\text{water}} \]
Substituting the values:
\[ P_{\text{total}} = 10^5 \, \text{Pa} + 9.8 \times 10^6 \, \text{Pa} \]
\[ P_{\text{total}} = 10^5 + 9.8 \times 10^6 = 9.9 \times 10^6 \, \text{Pa} \]

4. **Conclusion:**
The normal stress on the balloon at a depth of 1 km below the water level is:
\[ \text{Normal Stress} = P_{\text{total}} = 9.9 \times 10^6 \, \text{Pa} \]

### Final Answer:
The normal stress of the balloon is \( 9.9 \times 10^6 \, \text{Pa} \).

---