In taking a solid ball of rubber from the surface to the bottom of a lake of `180 m` depth, reduction in the volume of the ball is `0.01 %`. The density of water of the lake is `1xx10^(3) kg//m^(3)`. Determine the value of the bulk modulus of elasticity of rubber . `(g = 9.8 m//s^(2))`

Explanation

To determine the bulk modulus of elasticity of rubber based on the given information, we can follow these steps:

### Step 1: Understand the given data
- Depth of the lake, \( h = 180 \, \text{m} \)
- Reduction in volume, \( \Delta V = 0.01\% \) of the original volume \( V \)
- Density of water, \( \rho = 1 \times 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Calculate the change in volume
The change in volume can be expressed as:
\[ \frac{\Delta V}{V} \times 100 = 0.01 \]
From this, we can find \( \Delta V \):
\[ \Delta V = \frac{0.01}{100} \times V = 0.0001 V \]

### Step 3: Calculate the change in pressure
The change in pressure \( \Delta P \) at a depth \( h \) in a fluid is given by:
\[ \Delta P = \rho g h \]
Substituting the values:
\[ \Delta P = (1 \times 10^3 \, \text{kg/m}^3)(9.8 \, \text{m/s}^2)(180 \, \text{m}) \]
Calculating this gives:
\[ \Delta P = 1 \times 10^3 \times 9.8 \times 180 = 1764000 \, \text{Pa} = 1.764 \times 10^6 \, \text{Pa} \]

### Step 4: 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:
\[ B = -\frac{1.764 \times 10^6 \, \text{Pa}}{0.0001} \]
Calculating this gives:
\[ B = -1.764 \times 10^6 \times 10^4 = 1.764 \times 10^{10} \, \text{Pa} \]

### Step 5: Final result
The bulk modulus of elasticity of rubber is:
\[ B \approx 1.76 \times 10^9 \, \text{N/m}^2 \]