Question

A ball falling in a lake of depth 200 m shows 0.1% decrease in its volume at the bottom. The bulk modulus of the material of the ball is

• A. $1.96 \times 10^9 \; N/m^2$
• B. $1.96 \times 10^{11} \; N/m^2$
• C. $1.96 \times 10^{-9} \; N/m^2$
• D. $1.96 \times 10^{-7} \; N/m^2$

Answer 1

Answer: (A)

$1.96 \times 10^9 \; N/m^2$

Answer 2

A ball at a depth of $200 \,m$ in a tank is shown in figure below

Here, depth $h = 200 \,m$ and percentage decrease in volume $= 0.1\%$
As, pressure at depth $h$ is given by
$p =\rho gh$
where, $g = 9.8 ms^{-2}$
and $\rho = 10^{3} kg \,m^{-3}$
So, $p = 10^{3}\times 9.8 \times 200$
$\Rightarrow p = 19.6 \times 10^{5} N/m^{2}$
Now, the bulk modulus
$k =\frac{p}{\frac{\Delta V}{V}}$
$\Rightarrow k =\frac{ 19.6 \times 10^{5}}{\frac{0.1}{100}}$
$= 1.96 \times 10^{9} Nm^{-2}$
Hence, the bulk modulus of the material of ball is
$1.96 \times 10^{9} Nm^{-2}$