Question

Find the change in the volume of 1.0-liter kerosene when it is subjected to an extra pressure of \[2.0 \times {10^5}N{m^{ - 2}}\]from the following data. Density of kerosene \[ = 800kg\,{m^{ - 3}}\] and the speed of sound in kerosene is \[1330m{s^{ - 1}}\].

Answer 1

Hint:First, we need to understand the concept of Bulk Modulus before proceeding with the question. When a uniform element is subjected to equal Stress in three mutually perpendicular directions then, the ratio of direct stress to Volumetric Strain is called Bulk Modulus. Basically denoted by K.

Formula used:
Using the formula for velocity in the form of Bulk Modulus and density:
\[v = \sqrt {\dfrac{k}{\rho }} \]
Where the value of Bulk Modulus is given as:
\[k = {v^2} \times \rho \]
With the help of the above two formulas, we can be able to get our desired answer.

Complete step by step solution:
Bulk modulus is defined as the ability of a material by which it can endure the change in its volume when the pressure is applied from all of its sides. Now, Using the formula for velocity in the form of Bulk Modulus and density:
\[v = \sqrt {\dfrac{k}{\rho }} \]
Where the value of the Bulk Modulus is given as:
\[k = {v^2} \times \rho \]
Substituting values of speed and density we get,
\[k = {(1300)^2} \times 800N{m^{ - 1}}\]

As we need to find the value for change in volume therefore the equation can be written in terms of volume and force by area as,
\[k = \dfrac{{F/A}}{{\Delta V/V}}\]
Since, \[\Delta V = \dfrac{{pressure}}{V} \times \dfrac{1}{k}\]
By solving and substituting values,
\[\Delta V = \dfrac{{2 \times 1 \times {{10}^5}}}{{1300 \times 1300 \times 800}}\]
\[\therefore \Delta V = \,0.14c{m^3}\]

Hence, the change in Volume for the given condition is found to be \[\Delta V = \,0.14c{m^3}\].

Note: By using the following relation between Bulk Modulus and change in Volume
\[\Delta V = \dfrac{{pressure}}{V} \times \dfrac{1}{k}\], we can get our required solution. But make sure you take values for k according to given material.