Question

Bulk modulus of water is $ 2\times 10^{9}N/m^{2}$ . The change in pressure required to increase the density of water by $0.1\%$ is

• A. $ 2\times 10^{9}N/m^{2}$
• B. $2 \times 10^{8} N / m ^{2}$
• C. $2 \times 10^{6} N / m ^{2}$
• D. $2 \times 10^{4} N / m ^{2}$

Answer 1

Answer: (C)

Mass $=$ Volume $\times$ density $=V d=$ constant
$V \Delta d +d \Delta V=0$
$\frac{\Delta V}{V} =-\frac{\Delta d}{d}$
$=-\frac{0.1}{100}=10^{-3}$
Bulk modulus $K=-\frac{\Delta p}{\Delta V / V}$
$\Delta p =-K \frac{\Delta V}{V}$
$\Delta p =2 \times 10^{9} \times 10^{-3}$
$=2 \times 10^{6} N / m ^{2}$