Question

The bulk modulus of a spherical object is $B$. If it is subjected to uniform pressure $p$,the fractional decrease in radius is?

Answer 1

Hint:When pressure is applied on a body or an object, it is subjected to compression. The parameter related to compression is $B$, it shows the resistance offered by the object on being subjected to a pressure. Calculate the fractional decrease in radius using the given bulk modulus and applied pressure.

Formula to be used: $B = - \dfrac{{\Delta P}}{{\dfrac{{\Delta V}}{V}}}$

Complete step by step solution:
Consider a sphere of radius $r$ having bulk modulus as $B$.
The volume of the sphere will be $V = \dfrac{4}{3}\pi {r^3}$
Change in volume will be given as
$
\Delta V = \dfrac{4}{3}\pi \left( {{{\left( {r + \Delta r} \right)}^3} - {r^3}} \right) \\
\Delta V = \dfrac{4}{3}\pi \left( {{r^3} + {{\left( {\Delta r} \right)}^3} + 3{r^2}\Delta r + 3r{{\left(
{\Delta r} \right)}^2} - {r^3}} \right) \\
\Delta V = \dfrac{4}{3}\pi \left( {{{\left( {\Delta r} \right)}^3} + 3{r^2}\Delta r + 3r{{\left( {\Delta
r} \right)}^2}} \right) \\
$
It is very hard to achieve compression of solids and liquids. So, whenever compression of solids and liquid are involved, the change in any quantity is very small. Hence, here $\Delta r$ is very small and the higher powers can be neglected.
$
\therefore \Delta V = \dfrac{4}{3}\pi \left( {3{r^2}\Delta r} \right) \\
\therefore \Delta V = 4\pi {r^2}\Delta r \\
$
Now, the fractional change in volume will be given by change in volume divide by initial volume.
Mathematically,
$
\dfrac{{\Delta V}}{V} = \dfrac{{4\pi {r^2}\Delta r}}{{\dfrac{4}{3}\pi {r^3}}} \\
\dfrac{{\Delta V}}{V} = \dfrac{{3\Delta r}}{r} \\
$
Now, the bulk modulus is defined as the ratio of change in pressure and relative change in volume.

$B = - \dfrac{{\Delta P}}{{\dfrac{{\Delta V}}{V}}}$
The pressure to which the object is subjected is $p$ and hence $\Delta P = p$
Substituting value of $\dfrac{{\Delta V}}{V}$ in $B$,
$
\therefore B = - \dfrac{p}{{\dfrac{{3\Delta r}}{r}}} \\
\therefore \dfrac{{\Delta r}}{r} = \dfrac{{ - p}}{{3B}} \\
$

Hence, if a spherical object having bulk modulus $B$ is subjected to uniform pressure $p$, the fractional decrease in radius is given as $\dfrac{{ - p}}{{3B}}$.

Note: Actually, the bulk change in pressure involved in bulk modulus is infinitesimally small and so is the change in volume. Therefore, $B = -\dfrac{{dP}}{{\dfrac{{dV}}{V}}}$. So, another way to solve this question was to differentiate to find the change in volume and radius.