Question

A sphere contracts in volume by $0.01$% when taken to the bottom of sea $1km$ deep. The bulk modulus of the material of the sphere is then(Given density of sea water may be taken as $1.0\times {10}^{3}kg{m}^{-3}$)

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

Answer 1

The correct option is B $9.8\times {10}^{10}N{m}^{-2}$

Here, $\frac{\Delta V}{V}=\frac{0.01}{100},h=1km={10}^{3}m\rho =1.0\times {10}^{3}kg{m}^{-3}$
$P=h\rho g={10}^3\times 1\times {10}^3\times 9.8=9.8\times {10}^{6}N{m}^{-2}$

$\therefore B=\frac{P}{\Delta V/V}=\frac{9.8\times {10}^6\times 100}{0.01}=9.8\times {10}^{10}N{m}^{-2}$