Question

A ball falling in a lake of depth $200 \,m$ shows $0.1 \%$ decrease in its volume at the bottom. What is the bulk modulus of the ball material ? (Take density of water $=1000 \,kg / m ^{3}$ )

• A. $ 19.6\times 10^{8}N/m^{2} $
• B. $ 19.6\times 10^{-10}N/m^{2} $
• C. $ 19.6\times 10^{10}N/m^{2} $
• D. $ 19.6\times 10^{-8}N/m^{2} $

Answer 1

Answer: (A)

When strain is small, the ratio of normal stress to the volume-strain is called the bulk modulus of the material of the body.
$K=\frac{p}{\Delta V / V}$
where $p$ is pressure $\left(=\frac{F}{A}\right)$, and $\frac{\Delta V}{V}$ the change in volume
Given, $\Delta V=0.1 \%, V=0.001$ volt,
$p=h \rho g$
$p=h \rho g \,\, p =200 \times 1000 \times 9.8$
$ =1.96 \times 10^{6} N / m ^{2} $
$\therefore K= \frac{1.96 \times 10^{6}}{0.001}$
$ =19.6 \times 10^{8} N / m ^{2}$