Q. The increase in pressure required to decrease the 200 L volume of a liquid by 0.004% (in kPa) is (Bulk modulus of the liquid ==2100 MPa)

Solution:

Bulk modulus $ \, \, B =\dfrac{normal \, stress}{volumetric \, strain }$
$\hspace40mm B = \dfrac{Δ p}{- Δ V/V}$
Here, negative sign shows that volume is decreased,
pressure is increased.
Here $\hspace30mm B = 2100 \times 10^6 \, Pa$
$\hspace40mm V = 200 \, L$
$\hspace40mm Δ V = 200 \times \dfrac{0.004}{100} = 0.008 \, L$
$\therefore \hspace20mm 2100 \times 10^6 = \dfrac{Δ p}{\big(\dfrac{0.008}{200}\big)}$
or $\hspace30mm Δ p = 84 \, kPa$