Question

A pressure of 100 k.Pa. causes a decrease in volume of water by ${\text{5}} \times {\text{1}}{{\text{0}}^{ - 3}}$ percent. The speed of sound in water is
(A) $1414m{s^{ - 1}}$
(B) $1000m{s^{ - 1}}$
(C) $2000m{s^{ - 1}}$
(D) $3000m{s^{ - 1}}$

Answer 1

To solve this question, we need to use the formula for the speed of sound in a fluid. We have to determine the bulk modulus of water from the given information. Then on substituting the value of the bulk modulus in the formula for the speed of sound in fluid, we will get the final answer.

Formula used: The formulae used for solving this question are given by
1. $\beta = \dfrac{P}{{\Delta V/V}}$, here $\beta$ is the bulk modulus of a fluid, whose volume gets decrease by the amount of $\Delta V$ from the original amount of $V$ due to the pressure of $P$.
2. $v = \sqrt {\dfrac{\beta }{\rho }}$, here $v$ is the velocity of sound in a fluid having a bulk modulus of $\beta$ and a density of $\rho$.

Complete step-by-step solution:
We know that the bulk modulus of a fluid is defined as the ratio of the pressure acting on the fluid to the volumetric strain produced by the pressure in the fluid. So it can be written as
$\beta = \dfrac{P}{{\Delta V/V}}$...........(1)
According to the question we have
$P = 100{\text{kPa}}$
We know that $1{\text{kPa}} = 1000{\text{Pa}}$. So we have
$P = {10^5}{\text{Pa}}$.....(2)
Also, the volume is decreased by ${\text{5}} \times {\text{1}}{{\text{0}}^{ - 3}}$. This can be mathematically expressed as
$\Delta V = \dfrac{{{\text{5}} \times {\text{1}}{{\text{0}}^{ - 3}}}}{{100}} \times V$
$\Rightarrow \dfrac{{\Delta V}}{V} = {\text{5}} \times {\text{1}}{{\text{0}}^{ - 5}}$...........(3)
Substituting (2) and (3) in (1) we get
$\beta = \dfrac{{{{10}^5}}}{{5 \times {{10}^{ - 5}}}}$
$\Rightarrow \beta = 2 \times {10^9}{\text{Pa}}$...............(4)
Now, we know that the speed of sound in a fluid is given by
$v = \sqrt {\dfrac{\beta }{\rho }}$..............(5)
According to the question, the given fluid is water whose density is equal to $1000kg{m^{ - 3}}$. So we have
$\rho = 1000kg{m^{ - 3}}$.............(6)
Substituting (4) and (6) in (5) we get
$v = \sqrt {\dfrac{{2 \times {{10}^9}}}{{1000}}}$
On solving we get
$v = 1414m{s^{ - 1}}$
Thus, the speed of the sound in water is equal to $1414m{s^{ - 1}}$.

Hence the correct answer is option A.

Note: We must note that we are given the percentage decrease in the volume of water, and not the fractional decrease in the volume. But while calculating the bulk modulus, we have to substitute the value of the fractional change in the volume, not the percentage change.