Question

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, $\dfrac{{\delta R}}{R}$ is ______________.

Answer 1

Hint:When the mass is placed on the piston, the pressure in the container increases causing the volume of the sphere to change as well. The excess pressure in the container depends on the weight of the mass acting on the area of the piston. The hydraulic strain of the sphere will give us the fractional change in its radius.

Formulas used:
-The increase or decrease in pressure is given by, $\Delta P = \dfrac{F}{A}$ where $F$ is the applied force and $A$ is the area on which the force acts.
-The bulk modulus of a material is given by, $K = \dfrac{P}{{\left( {\dfrac{{ - \delta V}}{V}} \right)}}$ where $P$ is the increase in pressure and $\dfrac{{\delta V}}{V}$ is the hydraulic strain of the body.

Complete step by step answer.
Step 1: Sketch a figure depicting the arrangement mentioned in the question.
As the mass $M$ placed on the piston causes the piston to push down, the volume of the sphere decreases.
The radius of the sphere is $R$ and let $V = \dfrac{4}{3}\pi {R^3}$ be its original volume. The bulk modulus of the material of the sphere is given to be $K$ .
The area of the piston is $A$ .
Here the force applied to increase the pressure is given by the weight of the mass.
i.e., $F = W = Mg$
Step 2: Express the excess pressure in the container and the bulk modulus of the material.
The excess pressure can be expressed as $P = \dfrac{W}{A} = \dfrac{{Mg}}{A}$ --------- (1)
The bulk modulus of the material of the sphere is given by, $K = \dfrac{P}{{\left( {\dfrac{{ - \delta V}}{V}} \right)}}$ ------- (2) where $P$ is the excess pressure and $\dfrac{{\delta V}}{V}$ is the hydraulic strain of the solid sphere.
Substituting equation (1) in (2) we get, $K = \dfrac{{Mg}}{{A\left( {\dfrac{{ - \delta V}}{V}} \right)}} = \dfrac{{ - MgV}}{{A\delta V}}$
We can neglect the negative sign for the time being.
$ \Rightarrow \dfrac{{\delta V}}{V} = \dfrac{{ - Mg}}{{AK}}$
Then the hydraulic strain of the sphere will be $\dfrac{{\delta V}}{V} = \dfrac{{Mg}}{{AK}}$ ---------- (3)
Step 3: Differentiate the volume of the sphere to obtain the fractional change in its radius.
The volume of the sphere is given by, $V = \dfrac{4}{3}\pi {R^3}$ ------- (4)
Differentiating the above relation with respect to the radius $R$ we get,
$dV = \dfrac{4}{3}\pi 3{R^2}dR = 4\pi {R^2}dR$ -------- (5)
Then on dividing equation (5) by (4) we have $\dfrac{{dV}}{V} = \dfrac{{4\pi {R^2}dR}}{{\dfrac{4}{3}\pi {R^3}}}$
$ \Rightarrow \dfrac{{dV}}{V} = \dfrac{{3dR}}{R}$ or we have $\dfrac{{\delta V}}{V} = \dfrac{{3\delta R}}{R}$ ---------- (6)
Substituting equation (6) in (3) we get, $\dfrac{{3\delta R}}{R} = \dfrac{{Mg}}{{AK}}$
$ \Rightarrow \dfrac{{\delta R}}{R} = \dfrac{{Mg}}{{3AK}}$
So the fractional change in the radius of the sphere will be $\dfrac{{\delta R}}{R} = \dfrac{{Mg}}{{3AK}}$

Note:The negative sign in equation (2) is a mere reminder that when there is an increase in pressure, there will be a corresponding decrease in the volume of the sphere. When the system was in equilibrium, the bulk modulus was positive. It is only because the piston is massless that it can float on the surface of the liquid and that the force applied to increase the pressure becomes solely the weight of the mass placed on the piston. The derivative of ${R^3}$ is written as $3{R^2}$ in equation (5) based on the formula $\dfrac{d}{{dx}}\left( {{x^n}} \right) = \left( {n - 1} \right){x^{n - 1}}$ .