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$1.0 \times \text{1}\text{0}^{3}\text{ kg }\text{m}^{- 3}).$

• A. $4.9 \times \text{1}\text{0}^{\text{10}}\text{ N }\text{m}^{- 2}$
• B. $9.8 \times \text{1}\text{0}^{\text{10}}\text{ N }\text{m}^{- 2}$
• C. $4.9 \times \text{1}\text{0}^{9}\text{ N }\text{m}^{- 2}$
• D. $9.8 \times \text{1}\text{0}^{9}\text{ N }\text{m}^{- 2}$

Answer 1

Answer: (B)

$9.8 \times \text{1}\text{0}^{\text{10}}\text{ N }\text{m}^{- 2}$

Answer 2

: Here, $\frac{\Delta V}{V} = \frac{0.01}{100},h = 1km = 10^{3}m$

$\rho = 1.0 \times 10^{3}kgm^{- 3}$

$p = h\rho g = 10^{3} \times 1 \times 10^{3} \times 9.8$

$= 9.8 \times 10^{6}Nm^{- 2}$

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

$= 9.8 \times 10^{10}Nm^{- 2}$