Question

Find the increase in pressure required to decrease the volume of a water sample by 0.01%. Bulk modulus of water = 2.1 × 109 N m−2.

Answer 1

The increase in pressure is 210000($\frac{N}{m^{2}}$).

Explanation:

1. Given data

  Bulk modulus of water (B) = $2.1\times 10^{9}(\frac{N}{m^{2}})$

  Decrease in volume of water sample $(\frac{\Delta V}{V})= \frac{0.01}{100}$

 $(\frac{\Delta V}{V})$ is also known as volume metric strain($\varepsilon _{v}$).

 So $\varepsilon _{v}$ = $\frac{0.01}{100}$

2. Now from relation

    Bulk modulus of water (B) = $\frac{\Delta P}{\varepsilon _{v}}$

    Where $\Delta P$ is increase in pressure.

   $2.1\times 10^{9}= \frac{\Delta P}{\frac{0.01}{100}}$  

3. After calculating we get

  $\Delta P$ = 210000 = 2.1× $10^{5}(\frac{N}{m^{2}})$

   

Answer 2

Increase in Pressure is 2.1 x $10^5$

Explanation:

Given,  

Bulk Modulus of Water (B) = $2.1 \times 10^9 Nm^-^2$

Let P be the increase in Pressure, so as to decrease the volume of the water sample by 0.01%  

$\frac {v \times 0.01}{100} = \Delta V$

$\frac {\Delta V}{V} = 10^-^4$

From $B = \frac {PV}{\Delta V}$, we get-

$P = B(\frac {\Delta V}{V})$

   = $2.1 \times 10^9 \times 10^-^4$

   = $2.1 \times 10^5 N/m^2$

So, the required increase in pressure is 2.1 $\times 10^5 Nm^-^2$.