# 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)

MHT CET MHT CET 2006Mechanical Properties Of Solids

Solution:

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

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