To what depth below the surface of sea should a rubber ball be taken as to decreases its volume by 0.1% (Given denisty of sea water = 1000 kg `m^(-3)`, Bulk modulus of rubber `= 9 xx 10^(8) Nm^(-2)`, acceleration due to gravity `= 10 ms^(-2)`)

Options

A. 9m
B. 18 m
C. 180 m
D. 90 m

Explanation

To find the depth below the surface of the sea at which a rubber ball's volume decreases by 0.1%, we can use the relationship between bulk modulus, pressure, and volume strain. Here’s a step-by-step solution:

### Step 1: Understand the relationship between bulk modulus, pressure, and volume strain
The bulk modulus \( B \) is defined as:
\[ B = \frac{\text{Normal Stress}}{\text{Volumetric Strain}} = \frac{P}{\frac{\Delta V}{V}} \]
Where:
- \( P \) is the pressure applied,
- \( \Delta V \) is the change in volume,
- \( V \) is the original volume.

### Step 2: Identify the given values
- Bulk modulus of rubber, \( B = 9 \times 10^8 \, \text{N/m}^2 \)
- Decrease in volume, \( \Delta V/V = 0.1\% = \frac{0.1}{100} = 0.001 \)
- Density of seawater, \( \rho = 1000 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 3: Express pressure in terms of depth
The pressure \( P \) at a depth \( d \) in a fluid is given by:
\[ P = \rho g d \]

### Step 4: Set up the equation using bulk modulus
From the definition of bulk modulus, we can express pressure as:
\[ P = B \cdot \frac{\Delta V}{V} \]
Substituting the values we have:
\[ \rho g d = B \cdot \frac{\Delta V}{V} \]
This can be rearranged to find \( d \):
\[ d = \frac{B \cdot \frac{\Delta V}{V}}{\rho g} \]

### Step 5: Substitute the known values into the equation
Substituting the values into the equation:
\[ d = \frac{9 \times 10^8 \, \text{N/m}^2 \cdot 0.001}{1000 \, \text{kg/m}^3 \cdot 10 \, \text{m/s}^2} \]

### Step 6: Calculate the depth
Calculating the numerator:
\[ 9 \times 10^8 \cdot 0.001 = 9 \times 10^5 \, \text{N/m}^2 \]
Calculating the denominator:
\[ 1000 \cdot 10 = 10000 \, \text{kg/(m}^2\text{s}^2) = 10^4 \, \text{N/m}^2 \]
Now substituting back:
\[ d = \frac{9 \times 10^5}{10^4} = 90 \, \text{m} \]

### Final Answer
Thus, the depth below the surface of the sea to which the rubber ball should be taken to decrease its volume by 0.1% is:
\[ \boxed{90 \, \text{m}} \]