Question

When a rubber ball is taken to the bottom of a sea of depth $2000\text{m}$, the volume of the rubber ball decreases by $1.5\%$. Find the bulk modulus of the rubber ball.
Take the density of water $\rho =1000{\text{kg/m}}^3$ and $g=10{\text{m/s}}^2$.

• A. $2.66\times {10}^9{\text{N/m}}^2$
• B. $3.3\times {10}^9{\text{N/m}}^2$
• C. $6.6\times {10}^9{\text{N/m}}^2$
• D. $1.33\times {10}^9{\text{N/m}}^2$

Answer 1

The correct option is D $1.33\times {10}^9{\text{N/m}}^2$

We known that,
Bulk modulus , $K=\frac{P}{\frac{-\Delta V}{V}}...\left(1\right)$
where, $P=$ Change in pressure
$V=$ Original volume
$\Delta V=$ Change in the volume
given, $\frac{\Delta V}{V}=-\frac{1.5}{100}$
(Negative sign indicate the decrease in the volume of the rubber ball.)
Pressure at the bottom of the sea at the depth of $h=2000\text{m}$ is given by
$P=\rho gh$
$P=1000\times 10\times 2000$
$P=2\times {10}^7{\text{N/m}}^2$
Substituting the values in equation $\left(1\right)$, we get
$K=\frac{2\times {10}^7}{\left(\frac{1.5}{100}\right)}=1.33\times {10}^9{\text{N/m}}^2$
$K=1.33\times {10}^9{\text{N/m}}^2$