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 piston 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//2AK`
B. `Mg//3AK`
C. `Mg//AK`
D. `2Mg//3AK`

Explanation

To find the fractional change in the radius of a solid sphere when a mass is placed on a piston floating on a liquid, we can follow these steps:

### Step 1: Understand the Bulk Modulus
The bulk modulus \( K \) is defined as:
\[ K = -\frac{\Delta P}{\frac{\Delta V}{V}} \]
where \( \Delta P \) is the change in pressure, \( \Delta V \) is the change in volume, and \( V \) is the original volume.

### Step 2: Relate Change in Pressure to the Mass on the Piston
When a mass \( M \) is placed on the piston, the change in pressure \( \Delta P \) can be expressed as:
\[ \Delta P = \frac{Mg}{A} \]
where \( g \) is the acceleration due to gravity and \( A \) is the area of the piston.

### Step 3: Substitute Change in Pressure into the Bulk Modulus Equation
Substituting \( \Delta P \) into the bulk modulus equation gives:
\[ K = -\frac{Mg/A}{\frac{\Delta V}{V}} \]
Rearranging this, we can express the change in volume as:
\[ \frac{\Delta V}{V} = -\frac{Mg}{AK} \]

### Step 4: Determine the Volume Change of the Sphere
The volume \( V \) of a sphere with radius \( R \) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
The change in volume \( \Delta V \) when the radius changes by \( \Delta R \) can be approximated using the formula for the volume of a sphere:
\[ \Delta V = V_f - V_i = \frac{4}{3} \pi (R - \Delta R)^3 - \frac{4}{3} \pi R^3 \]
Using the binomial approximation for small changes, we can simplify this to:
\[ \Delta V \approx -4 \pi R^2 \Delta R \]

### Step 5: Substitute Volume Change into the Volume Fraction Equation
Now, substituting \( \Delta V \) into the equation we derived from the bulk modulus:
\[ \frac{-4 \pi R^2 \Delta R}{\frac{4}{3} \pi R^3} = -\frac{Mg}{AK} \]
This simplifies to:
\[ \frac{-3 \Delta R}{R} = -\frac{Mg}{AK} \]

### Step 6: Solve for the Fractional Change in Radius
Rearranging gives us:
\[ \frac{\Delta R}{R} = \frac{Mg}{3AK} \]

### Final Answer
Thus, the fractional change in the radius of the sphere is:
\[ \frac{\Delta R}{R} = \frac{Mg}{3AK} \]