A solid sphere of radius r made of a soft 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, covering entire cross-section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere $$\left(\frac{dr}{r}\right)$$, is:

First, let us find the extra pressure applied on the liquid when the mass $$m$$ is gently placed on the mass-less piston of area $$a$$. The weight of the mass is $$mg$$, so the force transmitted to the liquid is also $$mg$$. Since pressure is force per unit area, the increase in pressure is given by

$$P=\dfrac{\text{Force}}{\text{Area}}=\dfrac{mg}{a}\;.$$

This pressure is conveyed undiminished throughout the liquid (Pascal’s law), and hence the soft solid sphere experiences the same additional pressure $$P$$ on all its surfaces.

Now we recall the definition of bulk modulus. For any isotropic solid,

$$K=-\dfrac{P}{\dfrac{\Delta V}{V}},$$

where $$K$$ is the bulk modulus, $$P$$ is the applied pressure, $$V$$ is the original volume and $$\Delta V$$ is the change in volume (a decrease in volume makes $$\Delta V$$ negative, hence the minus sign in the definition).

The sphere has initial radius $$r$$, so its initial volume is

$$V=\dfrac{4}{3}\pi r^{3}\;.$$

If, because of the pressure, the radius decreases by a small amount $$dr$$ (so $$dr<0$$), then the new radius becomes $$r+dr$$ and the new volume is

$$V+\Delta V=\dfrac{4}{3}\pi(r+dr)^{3}\;.$$

Expanding the cube and keeping only the first-order term in the small quantity $$dr$$, we have

$$\begin{aligned} (r+dr)^{3}&=r^{3}+3r^{2}dr+\text{(higher-order terms in }dr) \\ &\approx r^{3}+3r^{2}dr. \end{aligned}$$

Thus,

$$\Delta V=\dfrac{4}{3}\pi\bigl[r^{3}+3r^{2}dr-r^{3}\bigr]=4\pi r^{2}dr.$$

Dividing by the original volume to obtain the fractional change, we get

$$\dfrac{\Delta V}{V}=\dfrac{4\pi r^{2}dr}{\dfrac{4}{3}\pi r^{3}}=3\,\dfrac{dr}{r}.$$

Putting this result into the bulk modulus relation, we write

$$K=-\dfrac{P}{\dfrac{\Delta V}{V}}=-\dfrac{P}{3\,\dfrac{dr}{r}}.$$

Rearranging to isolate the fractional change in radius,

$$\dfrac{dr}{r}=-\dfrac{P}{3K}.$$

Here $$dr<0$$ (a decrease), so the magnitude of the fractional decrement in radius is

$$\left|\dfrac{dr}{r}\right|=\dfrac{P}{3K}.$$

Substituting the earlier expression for $$P$$, namely $$P=\dfrac{mg}{a}$$, we finally obtain

$$\left|\dfrac{dr}{r}\right|=\dfrac{1}{3K}\,\dfrac{mg}{a} =\dfrac{mg}{3Ka}.$$

Therefore, the fractional decrease in the radius of the sphere equals $$\dfrac{mg}{3Ka}$$.

Hence, the correct answer is Option D.