If a rubber ball is taken to 100 m deep in a lake and its volume changes by `0.1%` then the bulk modulus of rubber will be `(g=10m//s^(2))`

Options

A. `1xx10^(6)N//m^(2)`
B. `1xx10^(8)N//m^(2)`
C. `1xx10^(7)N//m^(2)`
D. `1xx10^(9)N//m^(2)`

Explanation

To find the bulk modulus of rubber when a rubber ball is taken to a depth of 100 m in a lake and its volume changes by 0.1%, we can follow these steps:

### Step 1: Understand the given information
- Depth (h) = 100 m
- Volume change (ΔV/V) = -0.1% = -0.001 (as a fraction)
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the change in pressure (ΔP)
The change in pressure when the ball is submerged in water can be calculated using the formula:
\[ \Delta P = \rho g h \]
where:
- ρ (density of water) = \(10^3 \, \text{kg/m}^3\) (approximately)
- g = 10 m/s²
- h = 100 m

Substituting the values:
\[ \Delta P = (10^3 \, \text{kg/m}^3)(10 \, \text{m/s}^2)(100 \, \text{m}) = 10^6 \, \text{Pa} \]

### Step 3: Relate bulk modulus (B) to volume change
The bulk modulus is defined as:
\[ B = -\frac{\Delta P}{\frac{\Delta V}{V}} \]
Substituting the values we have:
\[ B = -\frac{10^6 \, \text{Pa}}{-0.001} \]

### Step 4: Calculate the bulk modulus
Now, calculating the bulk modulus:
\[ B = \frac{10^6}{0.001} = 10^9 \, \text{Pa} \]

### Step 5: Convert to appropriate units
Since \(1 \, \text{Pa} = 1 \, \text{N/m}^2\), we can express the bulk modulus as:
\[ B = 10^9 \, \text{N/m}^2 \]

### Conclusion
Thus, the bulk modulus of rubber is \(10^9 \, \text{N/m}^2\).

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