Question

On a litre of water to compress it by $0.10\%$ how much should the pressure be changed? (Bulk Modulus of water, ( $B = 2.2 \times {10^9}N{m^{ - 2}}$ )

Answer 1

Here Basically we are going to the formula of Bulk modulus. Bulk modulus is defined as, the measure of the ability of a substance to hold the capability of changes in volume when under pressure on all sides. It is equal to the applied pressure divided by the relative deformation. Sometimes referred to as incompressibility also.

Formula Used:
$B = \left( {\dfrac{P}{{(\dfrac{{\Delta V}}{V})}}} \right)$ Where P is the pressure and V is the volume.

Complete step by step answer:
According to the question given, we have to find the pressure on the given volume with the use of the concept of Bulk Modulus.
Volume of water given is $1L$
Let the pressure be P
It is mention that with the application of pressure P, the volume changes by $0.10\%$ which is equal to ${10^{ - 3}}$
As $\Delta V$ and $V$ has same units so $\dfrac{{\Delta V}}{V}$ is dimensionless
Coming back to the calculations,
Formula of Bulk Modulus is $B = \left( {\dfrac{P}{{(\dfrac{{\Delta V}}{V})}}} \right)$
With small modifications,
$P = B \times \dfrac{{\Delta V}}{V}$
Bulk Modulus of Water is fixed equals to $2.2 \times {10^9}N{m^{ - 2}}$
So, putting the value of Bulk Modulus and fractional change in volume
$P = 2.2 \times {10^9} \times {10^{ - 3}} = 2.2 \times {10^6}$
Hence, the correct solution is $2.2 \times {10^6}N{m^{ - 2}}$

Note: Higher is the bulk modulus of a substance smaller is the volumetric strain it goes for the same pressure, which means higher is the bulk modulus the more difficult it is to change the shape of the material/component.