Question

The initial volume of water is $2000{\text{cm}}^3$. Determine the change in volume of the water when external pressure is changed from ${10}^5{\text{Nm}}^{-2}$ to ${10}^6{\text{Nm}}^{-2}$.
[Take compressibility of water as $60\times {10}^{-11}{\text{m}}^2/\text{N}$]

• A. $2.08\times {10}^{-5}{\text{m}}^3$
• B. $1.08\times {10}^{-6}{\text{m}}^3$
• C. $2.08\times {10}^{-6}{\text{m}}^3$
• D. $1.08\times {10}^{-5}{\text{m}}^3$

Answer 1

The correct option is B $1.08\times {10}^{-6}{\text{m}}^3$

From the data given in the question,
Change in pressure $\left(\Delta P\right)={10}^6{\text{Nm}}^{-2}-{10}^5{\text{Nm}}^{-2}=9\times {10}^5{\text{Nm}}^{-2}$
Initial volume $V=2\times {10}^{-3}{\text{m}}^3$
We know that,
Compressibility $=\frac{1}{\text{|Bulk modulus|}}$
From the definition of Bulk modulus of elasticity,
Bulk modulus $\left(B\right)=-\frac{V\times \Delta P}{\Delta V}$
$\Rightarrow$ Compressibility $=\frac{\left(\frac{\Delta V}{V}\right)}{\Delta P}$
$\Rightarrow 60\times {10}^{-11}=\frac{\Delta V}{2\times {10}^{-3}\times 9\times {10}^5}$
$\Rightarrow \Delta V=60\times {10}^{-11}\times 2\times {10}^{-3}\times 9\times {10}^5$
$\Rightarrow \Delta V=1.08\times {10}^{-6}{\text{m}}^3$
Hence, option (b) is the correct answer.