A ball falling in a lake of depth 200m shown 0.1% decrease in its volume at the bottom .What is the bulk modulus of the materialof the ball

Options

A. `19.6xx10^(8) Nm^(-2)`
B. `19.6xx 10^(9) Nm^(-2)`
C. `19.6xx10^(10) Nm^(-2)`
D. `19.6 xx 10^(12) Nm^(-2)`

Explanation

To find the bulk modulus of the material of the ball, we can follow these steps:

### Step 1: Understand the Problem
We know that a ball is falling in a lake of depth 200 m and experiences a 0.1% decrease in its volume at the bottom. We need to find the bulk modulus of the material of the ball.

### Step 2: Write the Formula for Bulk Modulus
The bulk modulus (B) is defined as the ratio of stress to volumetric strain:
\[ B = \frac{\text{Stress}}{\text{Volumetric Strain}} \]
Where:
- Stress can be expressed as the change in pressure (ΔP).
- Volumetric strain is given by the change in volume (ΔV) over the original volume (V).

### Step 3: Calculate the Volumetric Strain
Given that there is a 0.1% decrease in volume:
\[ \text{Volumetric Strain} = \frac{\Delta V}{V} = \frac{-0.1}{100} = -0.001 \]
(Note: The negative sign indicates a decrease in volume, but we will use the absolute value for the calculation of bulk modulus.)

### Step 4: Calculate the Pressure at the Depth of 200 m
The pressure at a depth (h) in a fluid is given by:
\[ P = \rho g h \]
Where:
- \(\rho\) (density of water) = \(10^3 \, \text{kg/m}^3\)
- \(g\) (acceleration due to gravity) = \(9.8 \, \text{m/s}^2\)
- \(h\) (depth) = \(200 \, \text{m}\)

Substituting the values:
\[ P = 10^3 \times 9.8 \times 200 \]
\[ P = 1960000 \, \text{Pa} = 1.96 \times 10^6 \, \text{Pa} \]

### Step 5: Substitute Values into the Bulk Modulus Formula
Now we can substitute the values into the bulk modulus formula:
\[ B = \frac{P}{\text{Volumetric Strain}} = \frac{1.96 \times 10^6}{0.001} \]
\[ B = 1.96 \times 10^9 \, \text{Pa} = 19.6 \times 10^8 \, \text{Pa} \]

### Step 6: Final Result
Thus, the bulk modulus of the material of the ball is:
\[ B = 19.6 \times 10^8 \, \text{N/m}^2 \]