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$, δR/R,$ is ..........

Step 1: Calculate the increase in pressure.

The weight of the mass $M$ placed on the piston of area $A$ increases the pressure in the liquid.

$$\Delta P = \frac{Mg}{A}$$

Step 2: Relate bulk modulus to the fractional change in volume.

The bulk modulus $K$ of the material is defined by the ratio of pressure change to fractional volume change.

$$K = -\frac{\Delta P}{\frac{\Delta V}{V}}$$ $$\frac{\Delta V}{V} = -\frac{\Delta P}{K}$$

Step 3: Relate the fractional change in volume to the fractional change in radius.

The volume of a solid sphere of radius $R$ is given by its geometric formula.

$$V = \frac{4}{3}\pi R^3$$

Taking the differential of the volume yields the change in volume.

$$\Delta V = 4\pi R^2 \Delta R$$

Divide the change in volume by the original volume.

$$\frac{\Delta V}{V} = \frac{4\pi R^2 \Delta R}{\frac{4}{3}\pi R^3}$$ $$\frac{\Delta V}{V} = 3\frac{\Delta R}{R}$$

Step 4: Calculate the fractional change in the radius.

Equate the two expressions for the fractional change in volume.

$$3\frac{\delta R}{R} = -\frac{\Delta P}{K}$$

Substitute the expression for $\Delta P$ into the equation.

$$3\frac{\delta R}{R} = -\frac{Mg}{AK}$$ $$\frac{\delta R}{R} = -\frac{Mg}{3AK}$$

Taking the magnitude of the fractional change:

$$\frac{\delta R}{R} = \frac{Mg}{3AK}$$

Answer: $-\dfrac{Mg}{3AK}$ (or $\dfrac{Mg}{3AK}$ in magnitude)