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Find the pressure to be applied to increase the density of water by 3% if the bulk modulus of water is $2.2 \times 10^{9} N m^{- 2}$ .

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$2.2 \times 10^{7} N / m^{2}$

$1.1 \times 10^{7} N / m^{2}$

$6.6 \times 10^{7} N / m^{2}$

$3.3 \times 10^{7} N / m^{2}$

answer is C.

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Detailed Solution

$k = \frac{- Δ P}{\frac{Δ V}{V}} = \frac{Δ P}{\frac{Δ d}{d}}$

$\frac{Δ d}{d} = \frac{Δ P}{k}$

$\frac{3}{100} = \frac{Δ P}{2.2 \times 10^{9}} \Rightarrow Δ P = \frac{6.6 \times 10^{9}}{100}$

$= 6.6 \times 1 0^{7}$

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