Q. Compressibility of water is $5 \times 10^{10}\ m^{2}N^{- 1}.\ $ The change in volume of 100 mL water subjected to$15 \times 10^{6}\ Pa$ pressure will
A) No change
B) Increase by 0.75 mL
C) Decrease by 1.50 mL
D) Decrease by 0.75 mL
Explanation
(d) Compressibility, $K = \frac{1}{B} = \frac{\mathrm{\Delta}V}{V\mathrm{\Delta}P}$
$\therefore$ $5 \times 10^{- 10} = \frac{\mathrm{\Delta}V}{100 \times 10^{- 3} \times 15 \times 10^{6}}$
⟹ $\mathrm{\Delta}V = 5 \times 10^{- 10} \times 100 \times 10^{- 3} \times 15 \times 10^{6}$
=0.175 mL
Since, pressure increases, so volume will decrease.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.