How much pressure (in atm) is needed to compress a sample of water by 0.4 % ? (Assume Bulk modulus of water =2.0×10⁹Pa) 

Given,

Percentage change in volume = 0.4%

$\frac{\mathrm{\Delta V}}{\mathrm{V}}=-0.4\%=\frac{0.4}{100}$

Change in volume is,

$\mathrm{\Delta V}=-\frac{\mathrm{pV}}{\mathrm{B}}$

Pressure required ,

$\mathrm{p}=\mathrm{B}\left(-\frac{∆\mathrm{V}}{\mathrm{V}}\right)$

$\mathrm{p}=\frac{0.4}{100}\times 2\times {10}^9\mathrm{Pa}$

$\mathrm{p}=\frac{0.4}{100}\times \frac{2\times {10}^9}{101325}\mathrm{atm}$

$\mathrm{p}\approx 80\mathrm{atm}$

Hence the correct answer is 80atm.