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A sphere contracts in volume by $0.01 %$ when taken to the bottom of sea 1km deep. If the density of water is $1 g m / c c$ .Then the bulk modulus of water is

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$9.8 \times 10^{6} N m^{- 2}$

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

$9.8 \times 10^{8} N m^{- 2}$

$9.8 \times 10^{5} N m^{- 2}$

answer is C.

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Detailed Solution

$k = \frac{h ρ g}{∆ v / v} = \frac{10^{3} \times 10^{3} \times 9.8}{10^{- 4}} = 9.8 \times 10^{10} N / m^{2}$

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