Question

When $$200ml$$of water is subjected to a pressure of $$2\times {10}^{6}Pa$$.The decrease in its volume is $$0.2ml$$. The compressibility of water is --
A. $$5\times {10}^{-8}{m}^2{N}^{-1}$$
B. $$5\times {10}^{-10}{m}^2{N}^{-1}$$
C. $$5\times {10}^{-12}{m}^2{N}^{-1}$$
D. None

Answer 1

Hint:Stress is the force on an object which causes a change in the object while a strain is the change in the shape of an object when stress is applied, it means that Stress can occur independent of strain, bur strain depends on stress.Compressibility is the ability of compression of any fluid; it is the reciprocal of Bulk Modulus (K).

Formula used:
Bulk modulus = K = (Direct Stress)/(Volumetric Strain) = $$(dP)/(-dV/V)$$.
Coefficient of Compressibility = $$\beta$$= $$1/K$$

Complete step by step answer:
The relation between stress and strain is given by Hooke’s Law, which states that the strain of the material is proportional to the applied stress within the elastic limit of that material. We know that,
$$\rho =m/V$$,
where$$\rho$$ is the density of a substance, m is its mass and V is the volume.
For constant mass, $$\rho V=C$$ , where C is any constant.
Upon differentiating both sides, $$\Rightarrow \rho dV+Vd\rho =0$$
$$dV/V=-d\rho /\rho$$
$$\Rightarrow -dV/V=d\rho /\rho$$
$$K=\rho dP/d\rho$$.
In the above question, $$dV=0.2ml$$,$$V=200ml$$,$$dP=2\times {10}^{6}Pa$$.
$$\begin{gathered} K=(dP)/(-dV/V) \\ \Rightarrow K={\displaystyle \frac{2\times {10}^6}{0.2/200}} \\ \Rightarrow K=2\times {10}^{-9} \end{gathered}$$
$$\therefore \beta =1/K=5\times {10}^{-10}$$

Hence the option B is the correct answer.

Note:If $$\beta =0$$, then compressibility of the fluid is zero, hence the fluid is incompressible. If $$\beta \ne 0$$, then the fluid is compressible, gases are compressible while liquids are considered as incompressible fluids. Stress is a measurable quantity and has a unit Newton per square meter. Strain on the other hand is a dimensionless quantity with no unit because it is the ratio of change in length to the original length.