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

The correct option is C $2\times {10}^{6}N/m^2$

$B=-V\frac{dP}{dV}$

Thus, $\Delta V=-\frac{\Delta pV}{B}$, where p is change in pressure.

Thus, $V^{\prime }-V=-\frac{\Delta pV}{B}$ or, $V^{\prime }=V(1-\frac{\Delta p}{B})$

Now, ${\rho }^{\prime }=\frac{m}{V^{\prime }}=\frac{m}{V(1-\frac{\Delta p}{B})}=\frac{\rho }{(1-\frac{\Delta p}{B})}$

Given ${\rho }^{\prime }=1.001\rho$, or $\frac{1}{1.001}=1-\frac{p}{2\times {10}^9}$

or, $\Delta p=\frac{1}{1001}\times 2\times {10}^9$, which is nearly $2\times {10}^6$