Question

A pressure of \[100\,kPa\] causes a decrease in volume of water by \[5 \times {10^{ - 3}}\% \] . The speed of sound in water is:
A. \[1414\,m{s^{ - 1}}\]
B. \[1000\,m{s^{ - 1}}\]
C. \[2000\,m{s^{ - 1}}\]
D. \[3000\,m{s^{ - 1}}\]

Answer 1

Hint: The speed of sound in liquid medium depends upon the bulk modulus of the liquid. Therefore, we should calculate the bulk modulus using given quantities. Use the formula for velocity of sound as a liquid medium to calculate the speed of sound.

Formula used:
\[v = \sqrt {\dfrac{B}{\rho }} \]
Here, \[\rho \] is the density of fluid and B is the bulk modulus.
\[B = - \dfrac{{\Delta p}}{{\left( {\dfrac{{\Delta V}}{V}} \right)}}\]
Here, \[\Delta p\] is the applied pressure, \[\dfrac{{\Delta V}}{V}\] relative change in the volume.

Complete step by step answer:
We know that the speed of sound varies medium to medium. It is \[330\,m{s^{ - 1}}\] in air medium.
We have the speed of sound in fluid of density \[\rho \] and bulk modulus B is,
\[v = \sqrt {\dfrac{B}{\rho }} \] …… (1)
We know that the bulk modulus is the ratio of applied pressure on a body to the relative change in the volume of the body due to tensile stress applied in the all direction.
We have the expression for bulk modulus,
\[B = - \dfrac{{\Delta p}}{{\left( {\dfrac{{\Delta V}}{V}} \right)}}\]
Here, \[\Delta p\] is the applied pressure, \[\dfrac{{\Delta V}}{V}\] relative change in the volume.
We have given that the percentage decrease in the volume of water is \[ - 5 \times {10^{ - 3}}\% \].
Therefore, the relative change in the volume is,
\[\dfrac{{ - 5 \times {{10}^{ - 3}}\% }}{{100\% }} = - 5 \times {10^{ - 5}}\]
Now, we can substitute \[100\,kPa\] for \[\Delta p\] and \[ - 5 \times {10^{ - 5}}\] for \[\dfrac{{\Delta V}}{V}\] in the above equation.
\[B = - \dfrac{{100 \times {{10}^3}\,Pa}}{{ - 5 \times {{10}^{ - 5}}}}\]
\[ \Rightarrow B = 2 \times {10^9}\,Pa\]
We know that the density of water is \[1000\,kg/{m^3}\]. Therefore, we can substitute \[2 \times {10^9}\,Pa\] for B and \[1000\,kg/{m^3}\] for \[\rho \] in equation (1).
\[v = \sqrt {\dfrac{{2 \times {{10}^9}}}{{1000}}} \]
\[ \Rightarrow v = \sqrt {2 \times {{10}^6}} \]
\[ \therefore v = 1414\,m/s\]

So, the correct answer is “Option A”.

Additional Information:
The velocity of sound in solid if given as,
\[v = \sqrt {\dfrac{Y}{\rho }} \]
Here, Y is the Young’s modulus of the solid and \[\rho \] is the density of solid.
The Young’s modulus of solid is the ratio of change in length of the material to the original uncompressed length.

Note:
 The relative decrease in the volume of water is given in percentage. Therefore, you should divide it by 100 to get the corresponding decrease in the relative volume. Since we have given that the volume of the water is decreased, so the sign of relative change in the volume should be negative. While solving these types of questions, make sure that the units of given quantities are in S.I. units.