Question

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 the entire cross section of the 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, $(\dfrac{{dr}}{r})$ is?
A) $\dfrac{{mg}}{{3Ka}}$
B) $\dfrac{{mg}}{{Ka}}$
C) \[\dfrac{{Ka}}{{mg}}\]
D) $\dfrac{{Ka}}{{3mg}}$

Answer 1

Hint: Use the concepts of fluids, bulk modulus and differentiation to get the answer. At first find the value of a small change of volume. Then use the equations of bulk modulus and substitute the values of force, volume and small volume change in the equation. Rearrange the terms to get the final answer.


Formula Used:
Bulk Modulus, $K = \dfrac{{PV}}{{ - dv}}$
Pressure, $P = \dfrac{{force}}{{area}}$
Volume of sphere $V = \dfrac{4}{3}\pi {r^3}$

Complete step by step answer:
Given,
Radius of the solid sphere $ = r$
Bulk modulus of the material $ = K$
Area of piston $ = a$
Mass of the object placed on the surface of the piston to compress the liquid $ = m$
We know,
$V = \dfrac{4}{3}\pi {r^3}$
Differentiating both sides we ge3t,
\[ \Rightarrow - dV = 4\pi {r^2}dr\]
Here the negative sign is there because volume decreases on applying pressure.
Also pressure,
$P = \dfrac{{force}}{{area}}$
$ \Rightarrow P = \dfrac{{mg}}{a}$
Also Bulk Modulus,
 $K = \dfrac{{PV}}{{ - dv}}$
$ \Rightarrow K = \dfrac{{\dfrac{{mg}}{a} \times \dfrac{4}{3}\pi {r^3}}}{{ - dv}}$
We put \[ - dV = 4\pi {r^2}dr\] in the above equation and we get
$ \Rightarrow K = \dfrac{{\dfrac{{mg}}{a} \times \dfrac{4}{3}\pi {r^3}}}{{4\pi {r^2}dr}}$
$ \Rightarrow K = \dfrac{{mg{r^{}}}}{{3adr}}$
By rearranging the above terms, we get that,
$ \Rightarrow \dfrac{{dr}}{r} = \dfrac{{m{g^{}}}}{{3Ka}}$
Hence the correct option is (A).

Additional Information: Pressure is defined as the force per unit area. Its S.I Unit is Pascal or $Pa$. Although force is a vector quantity; pressure itself is a scalar quantity. Thus, it has no direction.
Modulus of elasticity is the measure of the stress–strain relationship on the object. There are 3 types of Modulus of elasticity: Bulk modulus, Young’s modulus and Shear modulus.

Note: Be careful while differentiating the terms. One should know how to use differentiation in such types of equations. Also the formulas involved in solving such tricky problems should be pretty clear. With practice you will be able to solve such types of questions.