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, ( dr \over r ) is:

Properties of Matter PYQ — JEE Main 2018 · Shift 0 · Q62 with Solution

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Correct answer: D

When a mass $m$ is placed on the massless piston of area $a$ , the अतिरिक्त pressure transmitted to the liquid is

P = \frac{mg}{a}.

By Pascal's law, this pressure is transmitted equally throughout the liquid, so the sphere experiences a uniform external pressure

\Delta P = \frac{mg}{a}.

Bulk modulus is defined as

K = -\frac{\Delta P}{\Delta V/V}.

So, in magnitude,

\frac{\Delta V}{V} = \frac{\Delta P}{K} = \frac{mg}{Ka}.

Since the sphere is compressed, volume decreases:

\frac{dV}{V} = -\frac{mg}{Ka}.

For a sphere,

V = \frac{4}{3}\pi r^3.

Differentiating,

dV = 4\pi r^2 \, dr.

Hence,

\frac{dV}{V} = \frac{4\pi r^2 dr}{\frac{4}{3}\pi r^3} = 3\frac{dr}{r}.

Thus,

3\frac{dr}{r} = -\frac{mg}{Ka}.

So the fractional decrement in radius (magnitude) is

\left|\frac{dr}{r}\right| = \frac{mg}{3Ka}.

Therefore,

\boxed{\frac{dr}{r} = \frac{mg}{3Ka}}

in magnitude of decrement.

So the correct option is D.