A sphere contracts in volume by `0.01%` when taken to the bottom of sea `1km` keep. The bulk modulus of the material of the sphere is (Given density of sea water may be taken as `1.0xx10^3kgm^-3`).

Options

A. `4.9xx10^(10) N m^(-2)`
B. `9.8 xx 10^(10)N m^(-2)`
C. `4.9xx10^(9) N m^(-2)`
D. `9.8 xx 10^(9)N m^(-2)`

Explanation

To solve the problem, we need to find the bulk modulus of the material of the sphere given that it contracts in volume by 0.01% when taken to a depth of 1 km in the sea.

### Step-by-step Solution:

1. **Understanding the Volume Change**:
- The sphere contracts in volume by 0.01%.
- This can be expressed mathematically as:
\[ \frac{dV}{V} = -0.01\% = -\frac{0.01}{100} = -0.0001 \]
- Here, \(dV\) is the change in volume and \(V\) is the original volume of the sphere.

2. **Calculating the Change in Pressure**:
- The change in pressure (\(dp\)) at a depth \(h\) in a fluid can be calculated using the formula:
\[ dp = \rho g h \]
- Given:
- Density of sea water, \(\rho = 1.0 \times 10^3 \, \text{kg/m}^3\)
- Acceleration due to gravity, \(g = 9.8 \, \text{m/s}^2\)
- Depth, \(h = 1 \, \text{km} = 1000 \, \text{m}\)
- Substituting the values:
\[ dp = (1.0 \times 10^3) \times (9.8) \times (1000) = 9.8 \times 10^6 \, \text{N/m}^2 \]

3. **Calculating the Bulk Modulus**:
- The bulk modulus (\(K\)) is defined as:
\[ K = -\frac{dp}{\frac{dV}{V}} \]
- Substituting the values we have:
\[ K = -\frac{9.8 \times 10^6}{-0.0001} \]
- Simplifying this:
\[ K = \frac{9.8 \times 10^6}{0.0001} = 9.8 \times 10^{10} \, \text{N/m}^2 \]

4. **Final Result**:
- The bulk modulus of the material of the sphere is:
\[ K = 9.8 \times 10^{10} \, \text{N/m}^2 \]