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  4. The increase in pressure required to decrease the 200 L volume of a liquid by 0.008 % (in kPa) is (Bulk modulus of the liquid = 2100 MPa)

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Q. The increase in pressure required to decrease the $200\, L$ volume of a liquid by $0.008\%$ (in kPa) is (Bulk modulus of the liquid $= 2100\, MPa)$

MHT CETMHT CET 2006Mechanical Properties of Solids

Solution:

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