To What depth must a rubber ball be taken in deep sea so that its volume is decreased y 0.1 %. (The bulk modulus of rubber is `9.8 xx 10^8 Nm^(-2)` , and the density of seac water is `10^3 kg m^(-3).)`

Options

A. 100 m
B. 18 m
C. 90 m
D. 180 m

Explanation

To solve the problem, we need to find the depth \( h \) at which the volume of a rubber ball decreases by 0.1%. We will use the relationship between bulk modulus, pressure change, and volume change.

### Step-by-step Solution:

1. **Understand the Given Data:**
- Decrease in volume \( \Delta V = -0.1\% \) of original volume \( V \).
- This can be expressed as:
\[ \frac{\Delta V}{V} = -0.001 \]
- Bulk modulus of rubber \( B = 9.8 \times 10^8 \, \text{N/m}^2 \).
- Density of seawater \( \rho = 10^3 \, \text{kg/m}^3 \).
- Acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \).

2. **Use the Formula for Bulk Modulus:**
The bulk modulus \( B \) is defined as:
\[ B = -\frac{\Delta P}{\frac{\Delta V}{V}} \]
Rearranging this gives:
\[ \Delta P = -B \cdot \frac{\Delta V}{V} \]

3. **Calculate the Change in Pressure:**
Substitute the values into the equation:
\[ \Delta P = - (9.8 \times 10^8) \cdot (-0.001) = 9.8 \times 10^5 \, \text{N/m}^2 \]

4. **Relate Pressure Change to Depth:**
The change in pressure due to the depth \( h \) in a fluid is given by:
\[ \Delta P = h \cdot \rho \cdot g \]
Setting the two expressions for \( \Delta P \) equal gives:
\[ h \cdot \rho \cdot g = 9.8 \times 10^5 \]

5. **Substitute Known Values:**
Substitute \( \rho = 10^3 \, \text{kg/m}^3 \) and \( g = 9.8 \, \text{m/s}^2 \):
\[ h \cdot (10^3) \cdot (9.8) = 9.8 \times 10^5 \]

6. **Solve for Depth \( h \):**
\[ h \cdot 9800 = 9.8 \times 10^5 \]
\[ h = \frac{9.8 \times 10^5}{9800} = 100 \, \text{m} \]

### Final Answer:
The depth \( h \) at which the volume of the rubber ball decreases by 0.1% is **100 meters**.