Question

To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by $0.1 \%$ ? (Take density of sea water $=1000\, kg m^{-3}$, Bulk modulus of rubber $=9 \times 10^{8} Nm ^{-2}$, acceleration due to gravity $=10\, m\,s^{-2}$ )

• A. 9 m
• B. 18 m
• C. 180 m
• D. 90 m
• E. 900 m

Answer 1

Answer: (D)

Bulk modulus $B = \frac{\text{normal stress}}{\text{volume strain}}$
$\therefore B = \frac{p}{\Delta V/V}$
Given $\frac{\Delta V}{V} = \frac{0.1}{100} , B = 9 \times 10^8 \, Nm^{-2}$
$p = hdg = h \times 1000 \times 10$
$\therefore \frac{BV}{V} = p = hdg$
$\Rightarrow h = - \frac{B \Delta V}{Vdg}$
$h = - \frac{9 \times 10^8 \times 0.1}{1000 \times 100 \times 10}$
$h = - 90 \, m$