Question

To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by $0.4 \%$ [density of sea water $=1000 kg m ^{-3},$ Bulk modulus of rubber $\left.=9 \times 10^{8} Nm ^{-2}, g =10 m / s ^{2}\right]$

• A. $5 \times 10^{2} m$
• B. $30 \times 10^{6} m$
• C. $40 \times 10^{6} m$
• D. $36 \times 10^{1} m$

Answer 1

Answer: (D)

$B=\frac{P}{\frac{-\Delta V}{V}}$
$\Rightarrow B=\frac{h \rho g}{0.4 \times 10^{-2}}$
$h=\frac{9 \times 10^{8} \times 0.4 \times 10^{-2}}{10^{3} \times 10}$
$=36 \times 10^{1} m$