When a sphere is taken to bottom of sea 1 km deep, it contracts by 0.01%. The bulk modulus of elasticity of the material of sphere is
(Given density of water = 1 g `cm^(-3)`)

Options

A. `9.8 xx 10^(10) Nm^(-2)`
B. `10.2 xx 10^(10) Nm^(-2)`
C. `0.98 xx 10^(10) Nm^(-2)`
D. `8.4 xx 10^(10) Nm^(-2)`

Explanation

To find the bulk modulus of elasticity of the material of the sphere when it is taken to the bottom of the sea 1 km deep and contracts by 0.01%, we can follow these steps:

### Step 1: Understand the given data
- Depth of the sea, \( h = 1 \text{ km} = 1000 \text{ m} \)
- Volume contraction, \( \frac{\Delta V}{V} = -0.01\% = -0.0001 \)
- Density of water, \( \rho = 1 \text{ g/cm}^3 = 1000 \text{ kg/m}^3 \)

### Step 2: Calculate the change in pressure (\( \Delta P \))
The change in pressure when going to a depth \( h \) in a fluid is given by:
\[ \Delta P = \rho g h \]
Where:
- \( g = 9.8 \text{ m/s}^2 \) (acceleration due to gravity)

Substituting the values:
\[ \Delta P = 1000 \text{ kg/m}^3 \times 9.8 \text{ m/s}^2 \times 1000 \text{ m} \]
\[ \Delta P = 9.8 \times 10^6 \text{ Pa} \]

### Step 3: Use the formula for bulk modulus
The bulk modulus \( B \) is defined as:
\[ B = -\frac{\Delta P}{\frac{\Delta V}{V}} \]
Substituting the values we have:
\[ B = -\frac{9.8 \times 10^6 \text{ Pa}}{-0.0001} \]
\[ B = \frac{9.8 \times 10^6}{0.0001} \]
\[ B = 9.8 \times 10^{10} \text{ Pa} \]

### Conclusion
The bulk modulus of elasticity of the material of the sphere is:
\[ B = 9.8 \times 10^{10} \text{ N/m}^2 \]