Question

At what temperature, the speed of sound in air will become double of its value at 27⁰C.
A.54⁰C
B.627⁰C
C.327⁰C
D.927⁰C

Answer 1

Hint: The speed of a sound is directly proportional to the square root of temperature of its medium i.e., with the increase in temperature, speed of sound increases and vice-versa.

Step by step answer: The speed of sound v in a gas (air) increases with the increase in temperature of the gas. The reason is that with the increase of temperature, there is a decrease in density as the volume of the gas increases. From the relation,
$v=\sqrt{{\displaystyle \frac{\gamma P}{\rho }}}$
where γ is the ratio of specific heat at constant pressure to the specific heat at constant volume, P is the pressure of the gas and ρ is the density of the gas. So, v is inversely proportional to the square root of density of the gas i.e. v is proportional to${\displaystyle \frac{1}{\sqrt{\rho }}}$ .
Initially let the speed of sound be v₁ , temperature be T₁ .
As v α√T. [v is speed of sound and T is temperature]
$v^2=kT$ [ where k is proportionality constant]
Now, we put the value of T₁ and v in the above equation,
${v_1}^2=k\times 27$
${v_1}^2=k\left(27+273\right)$ [ changing the temperature from ⁰C to K]
$$\begin{gathered} {v_1}^2=300k \\ k={\displaystyle \frac{{v_1}^2}{300}} \end{gathered}$$
Now, when the speed of sound is 2 v₁ and temperature be T₂. On substituting the value in the same equation,
${\left(2v_1\right)}^2=k{T}_2$
$4{v_1}^2=k{T}_2$
Substituting the value of k in the equation,
$4{v_1}^2=\left({\displaystyle \frac{{v_1}^2}{300}}\right)T_2$
$$\begin{gathered} T_2=4{v_1}^2\times {\displaystyle \frac{300}{{v_1}^2}} \\ \therefore T_2=1200K \end{gathered}$$
Temperature in ⁰C = $1200-273=927$⁰C
Hence Option (D) is correct.


Note: The temperature in degree Celsius should be converted to Kelvin. To convert temperature in degree Celsius to Kelvin, we have to add 273 ⁰C to the given temperature and to convert Kelvin to degree Celsius, subtract 273 ⁰C from the temperature in Kelvin.