Question

The bulk modulus of a sphere is ${}^{\prime }{K}^{\prime }$. If it is subjected to uniform pressure ${}^{\prime }{p}^{\prime }$ then fractional decrease in its radius is

• A. $\frac{K}{3p}$
• B. $\frac{3p}{K}$
• C. $\frac{p}{3K}$
• D. $\frac{p}{K}$

Answer 1

The correct option is C $\frac{p}{3K}$

$\text{Bulk modulus}\left(K\right)=\frac{-\text{Pressure}}{\text{Volumetric strain}}$
$\Rightarrow K=\frac{-p}{\left(\frac{\Delta V}{V}\right)}$
$\therefore K=\frac{-pV}{\Delta V}...\left(i\right)$
Now volume of sphere is given by:
$V=\frac{4}{3}\pi r^3$
$\Rightarrow \Delta V=\frac{4}{3}\pi \times 3r^2\times \Delta r...\left(ii\right)$
$\frac{-V}{\Delta V}=-\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi \times 3r^2\times \Delta r}$
$\frac{-V}{\Delta V}=\frac{-r}{3\Delta r}...\left(iii\right)$

From Eq. $\left(i\right)$ and $\left(iii\right)$:
$K=\frac{-pr}{3\Delta r}$
$\Rightarrow$ Fractional decrease in the radius is represented as:
$\frac{-\Delta r}{r}=\frac{p}{3K}$