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, we need to find the value of α in the fractional change in radius of a solid sphere when a mass is placed on a piston floating on the surface of a liquid. The steps to derive the solution are as follows:

### Step 1: Understand the setup
We have a solid sphere of radius \( R \) 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, it compresses the liquid, leading to a change in pressure.

### Step 2: Calculate the change in pressure
The pressure change \( \Delta P \) due to the mass \( M \) placed on the piston can be calculated using the formula:
\[ \Delta P = \frac{Mg}{A} \]
where \( g \) is the acceleration due to gravity.

### Step 3: Relate pressure change to volume change
Using the definition of bulk modulus \( B \):
\[ B = -\frac{\Delta P}{\frac{\Delta V}{V}} \]
we can rearrange this to find the change in volume:
\[ \frac{\Delta V}{V} = -\frac{\Delta P}{B} \]
Substituting for \( \Delta P \):
\[ \frac{\Delta V}{V} = -\frac{Mg/A}{B} \]

### Step 4: Express the volume of the sphere
The volume \( V \) of the sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the change in volume \( \Delta V \) can also 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:
\[ \Delta V \approx 4 \pi R^2 \Delta R \]

### Step 5: Substitute into the bulk modulus equation
Now we can substitute \( \Delta V \) into the bulk modulus equation:
\[ \frac{4 \pi R^2 \Delta R}{\frac{4}{3} \pi R^3} = -\frac{Mg/A}{B} \]
Simplifying gives:
\[ \frac{3 \Delta R}{R} = -\frac{3Mg}{4AB} \]

### Step 6: Solve for the fractional change in radius
From the equation above, we can express the fractional change in radius:
\[ \frac{\Delta R}{R} = -\frac{Mg}{4AB} \]
This indicates that the fractional change in radius is:
\[ \frac{\Delta R}{R} = -\frac{Mg}{4AB} \]

### Step 7: Identify the value of α
Comparing this with the given expression for fractional change in radius:
\[ \frac{\Delta R}{R} = \frac{Mg}{\alpha AB} \]
we can see that:
\[ \alpha = 4 \]

### Final Answer
Thus, the value of \( \alpha \) is:
\[ \alpha = 4 \]