On taking a solid ball of rubber from the surface to the bottom of a lake of 200m depth, the reduction in the volume of ball is 0.1%. The density of water of the lakes is `10^(3) kgm^(-3)`. Calculate the value off bulk modulus elasticity of rubber, `g = 10ms^(-2)`.

Explanation

To calculate the bulk modulus of elasticity of rubber when a solid ball is taken from the surface to the bottom of a lake, we will follow these steps:

### Step 1: Understand the given data
- Depth of the lake (H) = 200 m
- Reduction in volume (ΔV/V) = 0.1% = 0.1/100 = 0.001
- Density of water (ρ) = \(10^3 \, \text{kg/m}^3\)
- Acceleration due to gravity (g) = \(10 \, \text{m/s}^2\)

### Step 2: Calculate the pressure at the depth of 200 m
The pressure (P) at a depth in a fluid is given by the formula:
\[ P = H \cdot \rho \cdot g \]
Substituting the values:
\[ P = 200 \, \text{m} \cdot 10^3 \, \text{kg/m}^3 \cdot 10 \, \text{m/s}^2 \]
\[ P = 200 \cdot 10^3 \cdot 10 \]
\[ P = 2 \times 10^6 \, \text{Pa} \]

### Step 3: Use the formula for bulk modulus
The bulk modulus (K) is defined as:
\[ K = \frac{P}{\frac{\Delta V}{V}} \]
Where:
- \( P \) is the pressure calculated in Step 2.
- \( \frac{\Delta V}{V} \) is the fractional change in volume.

Substituting the values:
\[ K = \frac{2 \times 10^6 \, \text{Pa}}{0.001} \]

### Step 4: Calculate the bulk modulus
\[ K = 2 \times 10^6 \, \text{Pa} \times 1000 \]
\[ K = 2 \times 10^9 \, \text{Pa} \]

### Final Result
Thus, the bulk modulus of elasticity of rubber is:
\[ K = 2 \times 10^9 \, \text{N/m}^2 \]

---