Question

To what depth must a rubber ball be taken in deep sea so that its volume is decreased by $0.1 \%$. (Take, density of sea water =$10^{3} \, kg$ $m^{-3}$, bulk modulus of rubber =$9\times10^{8}N$ $m^{-2}$, $g=10 \,m$ $s^{-2})$

• A. $9 \, m$
• B. $18\, m$
• C. $90\, m$
• D. $180\, m$

Answer 1

Answer: (C)

$90\, m$

Answer 2

Let $h$ be the depth at which the rubber ball is taken. Then $p=h\rho g \quad\ldots\left(i\right)$ By definition of bulk modulus, $B=-\frac{P}{\Delta V /V}$ The negative sign shows that with increase in pressure, a decrease in volume occurs. $\therefore\quad$ $P=B \frac{\Delta V}{V}$ Using $\left(i\right)$, $h\rho g$=$B \frac{\Delta V}{V}$ $\quad$ or $\quad$ $h=\frac{B}{\rho g}$ $\frac{\Delta V}{V}$ Substituting the given values, we get $h=\frac{9\times10^{8} N\,m^{-2}}{10^{3}kg \,m^{-3}\times10 \,ms^{-2}}$ $\left(\frac{0.1}{100}\right)$ $=90\, m\quad$