Question

The compressibility of water is $6 \times {10^{ - 10}}{N^{ - 1}}m$. If one litre is subjected to a pressure of $4 \times {10^{ - 7}}N{m^{ - 2}}$, the decrease in its volume is then
(A) 10 cc
(B) 24 cc
(C) 15 cc
(D) 12 cc

Answer 1

When pressure of water increases then volume of water decreases and compressibility is a measure of the relative change of a fluid as a response to a pressure change.
So, for calculating the decrease in volume, we use following relation which is
$K = \dfrac{{ - \Delta V}}{{V\Delta P}}$
Here
$k =$ Compressibility
$\Delta V =$ Change in volume
$V =$ Initial volume
$\Delta P =$ Change in pressure

Complete step by step solution:
We know that the compressibility of any material is given by
$K = \dfrac{{ - \Delta V}}{{V\Delta P}}$
So, $\Delta V = - VK\Delta P$
Here, -ve sign represents the decrement in volume. i.e., if pressure increases then volume will be decreases given that V (initial volume) $= 1$ litre
$K = 6{\kern 1pt} \times {10^{ - 10}}{m^2}/N$
$\Delta P = 4 \times {10^7}N/{m^2}$
So, $\Delta V = - 1 \times 6 \times {10^{ - 10}} \times 4 \times {10^7}$
(decrease in volume) $\Delta V = 24 \times {10^{ - 3}}$ litre
$\Delta V = 24$m litre
$\Delta V = 24cc$

So, option (B) is correct.

Note: Compressibility is inversely proportional to bulk modulus, if bulk modulus is given then we can directly calculate the compressibility.