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 .............

Explanation

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

### Step-by-Step Solution:

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

2. **Change in Pressure**:
When a mass \( M \) is placed on the piston, it exerts a force \( F = Mg \) on the liquid. The pressure change \( \Delta P \) caused by this force over the area \( A \) of the piston is given by:
\[ \Delta P = \frac{F}{A} = \frac{Mg}{A} \]

3. **Volume Change of the Sphere**:
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
The change in volume \( \Delta V \) due to a change in radius \( \Delta R \) can be approximated using the derivative:
\[ \Delta V \approx 4 \pi R^2 \Delta R \]

4. **Substituting into the Bulk Modulus Equation**:
We can substitute \( \Delta P \) and \( \Delta V \) into the bulk modulus equation:
\[ K = -\frac{\Delta P}{\frac{\Delta V}{V}} = -\frac{\frac{Mg}{A}}{\frac{4 \pi R^2 \Delta R}{\frac{4}{3} \pi R^3}} = -\frac{Mg}{A} \cdot \frac{3R}{4 \Delta R} \]

5. **Rearranging for \( \Delta R \)**:
Rearranging the above equation gives:
\[ \Delta R = -\frac{3R \cdot \Delta P \cdot A}{4K} \]
Since we are interested in the fractional change \( \frac{\Delta R}{R} \), we can express it as:
\[ \frac{\Delta R}{R} = -\frac{3 \Delta P}{4K} \]

6. **Substituting \( \Delta P \)**:
Now substituting \( \Delta P = \frac{Mg}{A} \):
\[ \frac{\Delta R}{R} = -\frac{3 \cdot \frac{Mg}{A}}{4K} = -\frac{3Mg}{4AK} \]

7. **Final Result**:
The negative sign indicates a decrease in radius, but we are interested in the magnitude:
\[ \frac{\Delta R}{R} = \frac{3Mg}{4AK} \]

### Conclusion:
The fractional change in the radius of the sphere is:
\[ \frac{\Delta R}{R} = \frac{3Mg}{4AK} \]