Solution:
Bulk modulus \(B = -\frac{\Delta P}{\Delta V/V}\). Ignoring the negative sign for magnitude, \(Delta P = B \frac{\Delta V}{V} = (2.1 \times 10^9) \times \left(\frac{0.001}{100}\right) = 2.1 \times 10^4 \text{ Pa} = 21 \text{ kPa}\).
The increase in pressure required to decrease the 400 litre volume of a liquid by 0.001% is (Bulk modulus of the liquid is \(2.1 \times 10^9 \text{ N/m}^2\))
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