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 solve the problem of finding the normal stress on an air-filled balloon at a depth of 1 km below the water level in an ocean, we can follow these steps:

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

### Step 2: Calculate the pressure due to the water column
The pressure exerted by a column of fluid can be calculated using the formula:
\[ P_{\text{water}} = \rho g h \]
Substituting the values:
\[ P_{\text{water}} = (10^3 \text{ kg/m}^3)(9.8 \text{ m/s}^2)(1000 \text{ m}) \]
Calculating this:
\[ P_{\text{water}} = 10^3 \times 9.8 \times 1000 = 9.8 \times 10^6 \text{ Pa} \]

### Step 3: Calculate the total pressure at the depth
The total pressure at the depth of the balloon is the sum of the pressure due to the water column and the atmospheric pressure:
\[ P_{\text{total}} = P_{\text{water}} + P_{\text{atm}} \]
Substituting the values:
\[ P_{\text{total}} = (9.8 \times 10^6 \text{ Pa}) + (10^5 \text{ Pa}) \]
Calculating this:
\[ P_{\text{total}} = 9.8 \times 10^6 + 1 \times 10^5 = 9.9 \times 10^6 \text{ Pa} \]

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

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

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