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)Nm^-2`
B. `9.8xx10^(10)Nm^-2`
C. `4.9xx10^(9)Nm^-2`
D. `9.8xx10^(9)Nm^-2`

Explanation

To find the bulk modulus of the material of the sphere, we will follow these steps:

### Step 1: Understand the relationship for bulk modulus
The bulk modulus \( K \) is defined as:
\[ K = -\frac{\Delta P}{\frac{\Delta V}{V}} \]
where:
- \( \Delta P \) is the change in pressure,
- \( \Delta V \) is the change in volume,
- \( V \) is the original volume.

### Step 2: Calculate the change in volume
The sphere contracts in volume by \( 0.01\% \). This can be expressed as:
\[ \frac{\Delta V}{V} = \frac{0.01}{100} = 0.0001 \]

### Step 3: Calculate the change in pressure
To find the change in pressure \( \Delta P \) when the sphere is submerged 1 km deep in seawater, we use the formula:
\[ \Delta P = \rho g h \]
where:
- \( \rho \) is the density of seawater, which is given as \( 1.0 \times 10^3 \, \text{kg/m}^3 \),
- \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \),
- \( h \) is the depth, which is \( 1 \, \text{km} = 1000 \, \text{m} \).

Substituting the values:
\[ \Delta P = (1.0 \times 10^3 \, \text{kg/m}^3)(9.8 \, \text{m/s}^2)(1000 \, \text{m}) \]
\[ \Delta P = 9.8 \times 10^6 \, \text{Pa} \]

### Step 4: Substitute values into the bulk modulus formula
Now, substituting \( \Delta P \) and \( \frac{\Delta V}{V} \) into the bulk modulus formula:
\[ K = -\frac{9.8 \times 10^6 \, \text{Pa}}{0.0001} \]
\[ K = -\frac{9.8 \times 10^6}{10^{-4}} = -9.8 \times 10^{10} \, \text{Pa} \]

### Step 5: Final result
Since bulk modulus is a positive quantity, we take the absolute value:
\[ K = 9.8 \times 10^{10} \, \text{Pa} \]

### Conclusion
The bulk modulus of the material of the sphere is \( 9.8 \times 10^{10} \, \text{Pa} \).

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