A solid sphere of radius R made of a material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless pistion of area A floats on the surface of the liquid. When a mass M is placed on the piston to compress the liquid the fractional change in the radius of the sphere, `deltaR//R`, is .............

Options

A. `(mg)/(Ka)`
B. `(Ka)/(mg)`
C. `(Ka)/(3mg)`
D. `(mg)/(3Ka)`

Answer

To solve the problem, we need to find the fractional change in the radius of a solid sphere when a mass \( M \) is placed on a piston that compresses the surrounding liquid. We will use the concepts of bulk modulus and the relationship between volume and radius.

### Step-by-Step Solution

1. **Understanding the Bulk Modulus**:
The bulk modulus \( K \) is defined as the ratio of the change in pressure to the fractional change in volume:
\[ K = -\frac{P}{\frac{\Delta V}{V}} \]
where \( P \) is the change in pressure, \( \Delta V \) is the change in volume, and \( V \) is the original volume.

2. **Calculating the Change in Pressure**:
When a mass \( M \) is placed on the piston, the force due to the mass is \( F = Mg \), where \( g \) is the acceleration due to gravity. The pressure increase \( P \) due to this force is given by:
\[ P = \frac{F}{A} = \frac{Mg}{A} \]

3. **Relating Pressure to Volume Change**:
From the bulk modulus definition, we can express the fractional change in volume \( \frac{\Delta V}{V} \) as:
\[ \frac{\Delta V}{V} = -\frac{P}{K} \]
Substituting the expression for \( P \):
\[ \frac{\Delta V}{V} = -\frac{Mg/A}{K} \]

4. **Volume of the Sphere**:
The volume \( V \) of a solid sphere of radius \( R \) is given by:
\[ V = \frac{4}{3} \pi R^3 \]

5. **Relating Volume Change to Radius Change**:
The change in volume \( \Delta V \) can also be related to the change in radius \( \Delta R \) using the formula for the volume of a sphere:
\[ \Delta V = V \frac{\Delta R}{R} \]
Thus, we can write:
\[ \frac{\Delta V}{V} = \frac{\Delta R}{R} \]

6. **Combining the Equations**:
Setting the two expressions for \( \frac{\Delta V}{V} \) equal to each other:
\[ \frac{\Delta R}{R} = -\frac{Mg}{AK} \]

7. **Final Expression**:
To find the fractional change in radius, we can express it as:
\[ \frac{\Delta R}{R} = -\frac{Mg}{3K} \]
Here, we have used the factor of \( 3 \) from the differentiation of the volume with respect to the radius.

### Conclusion
The fractional change in the radius of the sphere when a mass \( M \) is placed on the piston is:
\[ \frac{\Delta R}{R} = -\frac{Mg}{3K} \]