Q. The compressibility of water is $4 \times 10^{5}\ $per unit atmospheric pressure. The decrease in volume of 100 $cm^{3}$ of water under a pressure of 100 atmosphere will be
A) $0.4\ cm^{3}$
B) $0.025\ m^{3}$
C) $4\ \times 10^{5}\ cm^{3}$
D) $0.04\ cm^{3}$
Explanation
(a) $K = \ \frac{\mathrm{\Delta}p}{\mathrm{\Delta}V/V}$ or $\frac{1}{K} = \frac{\mathrm{\Delta}p}{V\mathrm{\Delta}p}\ $
Or $\mathrm{\Delta}V = \frac{1}{K}\ V\mathrm{\Delta}p\ $
= $4\ \times 10^{- 5} \times 100\ \times 100\ cm^{3}$
= 4 $\times 10^{- 1}cm^{3} = 0.4\ \ cm^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.