A ball falling in a lake of depth `200 m` shows a decrease of `0.1% ` in its volume at the bottom. The bulk modulus of the elasticity of the material of the ball is (take `g = 10 ms^(-12)`)

Options

A. `10^(9)N//m^(2)`
B. `2xx10^(9)N//m^(2)`
C. `3xx10^(9)N//m^(2)`
D. `4xx10^(9)N//m^(2)`

Explanation

To find the bulk modulus of elasticity of the material of the ball, we can follow these steps:

### Step 1: Understand the given data
- Depth of the lake, \( h = 200 \, m \)
- Fractional change in volume, \( \frac{\Delta V}{V} = -0.1\% = -0.001 \)
- Acceleration due to gravity, \( g = 10 \, m/s^2 \)

### Step 2: Calculate the change in pressure (\( \Delta P \))
The change in pressure at a depth \( h \) in a fluid is given by:
\[ \Delta P = \rho g h \]
where \( \rho \) is the density of the fluid (water in this case). For water, we can use \( \rho \approx 1000 \, kg/m^3 \).

Now, substituting the values:
\[ \Delta P = 1000 \, kg/m^3 \times 10 \, m/s^2 \times 200 \, m \]
\[ \Delta P = 1000 \times 10 \times 200 = 2,000,000 \, Pa \]

### Step 3: Relate bulk modulus to change in pressure and volume
The bulk modulus \( K \) is defined as:
\[ K = -\frac{\Delta P}{\frac{\Delta V}{V}} \]
Substituting the values we have:
\[ K = -\frac{2,000,000 \, Pa}{-0.001} \]
\[ K = \frac{2,000,000}{0.001} = 2,000,000,000 \, Pa \]

### Step 4: Final answer
Thus, the bulk modulus of elasticity of the material of the ball is:
\[ K = 2 \times 10^9 \, Pa \]

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