Question

$A$ sphere contracts in volume by $0.01$%, when taken to the bottom of sea $1\, km$ deep. The bulk modulus of the material of the sphere is (Given density of sea water may be taken as $1.0\times10^{3} kg \,m^{-3})$.

• A. $4.9\times10^{10} N$ $m^{-2}$
• B. $9.8\times10^{10} N$ $m^{-2}$
• C. $4.9\times10^{9} N$ $m^{-2}$
• D. $9.8\times10^{9} N$ $m^{-2}$

Answer 1

Answer: (B)

$9.8\times10^{10} N$ $m^{-2}$

Answer 2

Here, $\frac{\Delta V}{V}$ $=\frac{0.01}{100}$, $h=1 \,km$ $=10^{3}\,m$ $\rho$ $=1.0\times10^{3}kg$ $m^{-3}$ $p=h\rho g$ $=10^{3}\times1\times10^{3}\times9.8$ $=9.8\times10^{6} \,N$ $m^{-2}$ $\therefore\quad$ $B=\frac{p}{\Delta V/ V}$ $=\frac{9.8\times10^{6}\times100}{0.01}$ $=9.8\times10^{10}N$ $m^{-2}$