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Q.

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 $\times 10^{3} kg m^{- 3})$

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a

$4.9 \times 10^{10} {Nm}^{- 2}$

b

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

c

$4.9 \times 10^{9} {Nm}^{- 2}$

d

$9.8 \times 10^{9} {Nm}^{- 2}$

answer is B.

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Detailed Solution

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

$ρ = 1.0 \times 10^{3} {kgm}^{- 3}$

$P = hρg = 10^{3} \times 1 \times 10^{3} \times 9.8$

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

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

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

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