Question

The speed of sound in air under ordinary conditions is around\[330\dfrac{m}{s}\]. The speed of sound in hydrogen under similar conditions will be nearest to
 (a) \[330\dfrac{m}{s}\]
 (b) \[1200\dfrac{m}{s}\]
 (c) \[600\dfrac{m}{s}\]
 (d) \[900\dfrac{m}{s}\]

Answer 1

Hint: The given question asks us what will be tha speed of sound in hydrogen if in the question we have been given that speed of sound. We know that all the gases have different densities so, to solve this we will find the relationship between the densities of different gases with respect to speed of sound.

Complete step-by-step solution:
In the above question we have to find the speed of sound in hydrogen if the speed of sound in air is\[330\dfrac{m}{s}\]. So, we first need to find what is the relation between the speed of sound of different gases.
The speed of sound in a medium depends on the fact that how quickly the vibrational energy can be transferred through different mediums. For this reason, the derivation of the speed of sound in a medium depends on the medium and on the state of the medium. In general, the equation for the speed of a mechanical wave in a medium depends on the square root of the restoring force, or the elastic property, divided by the inertial property,
Which can be written as-
\[v=\sqrt{\dfrac{\text{elastic property}}{\text{inertial property}}}\]
Also, if we consider What is Speed of Sound?
The speed of sound will be defined as the dynamic propagation of sound waves, which depends on the characteristics of the medium through which the propagation takes place. Speed of sound is used for describing the speed of sound waves in an elastic medium.
Now, let us consider the speed of Sound Formula
The formula for speed of sound is given with respect to different gases. It is the square root of the product of the coefficient of adiabatic expansion and pressure of the gas divided by the density of the medium. Its mathematical expression is given by:
\[v=\sqrt{\dfrac{\gamma P}{\rho }}\]
where, \[v\] is the speed of sound
\[\gamma \] is the coefficient of adiabatic expansion
\[P\] is the pressure of the gas
\[\rho \] is the density of the medium
If we apply this relationship,
\[v\prec \dfrac{1}{\sqrt{\rho }}\]
\[\dfrac{{{v}_{{{H}_{2}}}}}{{{v}_{air}}}=\sqrt{\dfrac{{{\rho }_{air}}}{{{\rho }_{{{H}_{2}}}}}}\]
We know that
\[{{\rho }_{air}}\] \[=1.2\]\[\dfrac{gm}{L}\]and \[{{\rho }_{{{H}_{2}}}}\] \[=0.089\]\[\dfrac{gm}{L}\]
\[{{v}_{{{H}_{2}}}}\] \[=\]\[330\times \sqrt{\dfrac{1.2}{0.089}}\]
= \[1205\dfrac{m}{s}\]
So, option (b) \[1200\dfrac{m}{s}\] is the correct answer.

Note:In questions like these always remember that the speed of sound is an essential parameter that is used in a variety of fields in Physics. The speed of sound refers to the distance travelled per unit time by any sound wave propagating through any of the given medium.