Question

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

• A. 8.4
• B. 84
• C. 92.4
• D. 168

Answer 1

Answer: (B)

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 $B=2100\times 10^{6}Pa$
$V=200L$
$\Delta V=200\times \frac{0.004}{100}=0.008L$
$\therefore 2100\times 10^6=\frac{\Delta p}{\left(\frac{0.008}{200}\right)}$
or $\Delta p=84kPa$