A solid sphere of radius `R` made of a material of bulk modulus `B` is surrounded by a liquid in a cylindrical container. `A` massless piston of area `A` (the area of container is also `A`) floats on the surface of the liquid. When a mass `M` is placed on the piston to compress the liquid , fractional change in radius of the sphere is `(Mg)/(alpha AB)`. Find the value of `alpha`.

Explanation

To solve the problem step by step, we need to analyze the situation involving a solid sphere submerged in a liquid, surrounded by a piston. Here’s the solution:

### Step 1: Understand the System
We have a solid sphere of radius \( R \) made of a material with bulk modulus \( B \). The sphere is surrounded by a liquid in a cylindrical container, and a mass \( M \) is placed on a massless piston that floats on the liquid.

### Step 2: Determine the Change in Pressure
When the mass \( M \) is placed on the piston, it exerts a force on the liquid, creating a change in pressure (\( \Delta P \)). The change in pressure can be calculated as:
\[ \Delta P = \frac{F}{A} = \frac{Mg}{A} \]
where \( g \) is the acceleration due to gravity and \( A \) is the area of the piston.

### Step 3: Relate Bulk Modulus to Pressure and Volume Change
The bulk modulus \( B \) is defined as:
\[ B = -\frac{\Delta P}{\frac{\Delta V}{V}} \]
Rearranging gives:
\[ \frac{\Delta V}{V} = -\frac{\Delta P}{B} \]

### Step 4: Substitute the Change in Pressure
Substituting the expression for \( \Delta P \) into the bulk modulus equation:
\[ \frac{\Delta V}{V} = -\frac{\frac{Mg}{A}}{B} = -\frac{Mg}{AB} \]
Taking the absolute value, we have:
\[ \left| \frac{\Delta V}{V} \right| = \frac{Mg}{AB} \]

### Step 5: Relate Volume Change to Radius Change
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
The change in volume \( \Delta V \) can be expressed in terms of the change in radius \( \Delta R \):
\[ \Delta V = V' - V = \frac{4}{3} \pi (R + \Delta R)^3 - \frac{4}{3} \pi R^3 \]
Using the binomial expansion for small changes, we can approximate:
\[ \Delta V \approx 4 \pi R^2 \Delta R \]
Thus:
\[ \frac{\Delta V}{V} \approx \frac{4 \pi R^2 \Delta R}{\frac{4}{3} \pi R^3} = \frac{3 \Delta R}{R} \]

### Step 6: Set the Two Expressions for Volume Change Equal
From the previous steps, we have:
\[ \frac{Mg}{AB} = \frac{3 \Delta R}{R} \]
Rearranging gives:
\[ \Delta R = \frac{MgR}{3AB} \]

### Step 7: Find the Fractional Change in Radius
The fractional change in radius is given by:
\[ \frac{\Delta R}{R} = \frac{Mg}{3AB} \]

### Step 8: Compare with Given Expression
The problem states that the fractional change in radius is also given by:
\[ \frac{Mg}{\alpha AB} \]
By comparing the two expressions:
\[ \frac{Mg}{3AB} = \frac{Mg}{\alpha AB} \]
This implies:
\[ \alpha = 3 \]

### Final Answer
Thus, the value of \( \alpha \) is:
\[ \alpha = 3 \]