The bulk modulus of rubber is `9.1xx10^(8)N//m^(2)`. To what depth a rubber ball be taken in a lake so that its volume is decreased by `0.1%`?

Options

A. 25m
B. 91m
C. 200m
D. 500m

Answer

To solve the problem, we need to determine the depth a rubber ball must be submerged in a lake for its volume to decrease by 0.1%. We will use the bulk modulus of rubber and the relationship between pressure, volume change, and depth.

### Step-by-step Solution:

1. **Understanding the Bulk Modulus**: The bulk modulus (K) is defined as the ratio of the change in pressure (ΔP) to the relative change in volume (ΔV/V). Mathematically, it is expressed as:
\[ K = -\frac{\Delta P}{\frac{\Delta V}{V}} \]

2. **Given Values**:
- Bulk modulus of rubber, \( K = 9.1 \times 10^8 \, \text{N/m}^2 \)
- Volume decrease, \( \Delta V/V = -0.1\% = -0.001 \)

3. **Calculating Change in Pressure**: Rearranging the bulk modulus formula to find ΔP:
\[ \Delta P = -K \cdot \frac{\Delta V}{V} \]
Substituting the known values:
\[ \Delta P = - (9.1 \times 10^8) \cdot (-0.001) = 9.1 \times 10^5 \, \text{N/m}^2 \]

4. **Relating Pressure Change to Depth**: The change in pressure (ΔP) when a rubber ball is submerged to a depth \( h \) in a lake can be expressed as:
\[ \Delta P = \rho g h \]
where:
- \( \rho \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \))
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))

5. **Setting Up the Equation**: Equating the two expressions for ΔP:
\[ 9.1 \times 10^5 = 1000 \cdot 10 \cdot h \]

6. **Solving for Depth (h)**:
\[ 9.1 \times 10^5 = 10000h \]
\[ h = \frac{9.1 \times 10^5}{10000} = 91 \, \text{m} \]

7. **Conclusion**: The depth to which the rubber ball must be submerged in the lake for its volume to decrease by 0.1% is:
\[ h = 91 \, \text{m} \]

### Final Answer:
The depth is \( 91 \, \text{m} \).