Two identical flutes produce fundamental notes of frequency 300Hz at $27°C$. If the temperature in the air in one of the flutes is increased to $31°C$, the number of beats heard per second will be:
A). 3
B). 2
C). 1
D). 4

Hint: Use the formula for the velocity of sound in air. Deduce the equation to show proportionality between temperature and velocity. Now, use the relation between frequency and velocity. Substitute and get the value of constant. Now subtract the frequencies to obtain the number of beats.

Complete step-by-step solution:
Given: ${ T }_{ 1 }$= $27°C$= 300K
            ${ T }_{ 2 }$= $31°C$= 304K
Velocity of sound in air is given by,
$v=\sqrt { \dfrac { \gamma RT }{ M } }$.............(1)
Where T is the absolute temperature in K
             M is the molecular mass
             $\gamma$ is the adiabatic index
From the equation. (1), it can be inferred
$ v=k\sqrt { T }$ …………….…(2)
Where k is a constant
We know, velocity is related to frequency and wavelength by,
$v=f\lambda$
$\therefore v\propto f$ ……………….…(3)
From equation.(2) and (3), we get,
$ f=k\sqrt { T }$
Substituting the values in above equation to get the value of k,
$\therefore 300=k\sqrt { 300 }$
$\Rightarrow k=\sqrt { 300 }$
Number of beats at $31°C$ will be:
$n={ f }_{ 2 }-{ f }_{ 1 }$
$\Rightarrow n=k\sqrt { { T }_{ 2 } } -k\sqrt { { T }_{ 1 } }$
$\Rightarrow n=k(\sqrt { { T }_{ 2 } } -\sqrt { { T }_{ 1 } } )$
Substituting the values in above equation we get.,
$\Rightarrow n=\sqrt { 300 } (\sqrt { 304 } -\sqrt { 300 } )$
$\Rightarrow n=17.32(17.44-17.32)$
$\Rightarrow n=17.32(0.12)$
$\Rightarrow n=2.08\simeq 2$
Therefore, the number of beats heard per second will be 2.
Hence, the correct answer is option B i.e.2.

Note: From the equation of velocity of sound in the air given above, it can be inferred velocity is directly proportional to the temperature, If the temperature increases, the velocity of sound also increases. Frequency is directly proportional to velocity.
There’s an alternate method to this problem which is given below.
At temperature $27°C$,
${ v }_{ 1 }=f\lambda$
At temperature $31°C$,
$ { v }_{ 1 }=\left( f+x \right) \lambda$
We know, $ v\propto f\propto \sqrt { T }$
$ \Rightarrow \dfrac { { v }_{ 2 } }{ { v }_{ 1 } } =\sqrt { \dfrac { { T }_{ 2 } }{ { T }_{ 1 } } } =\dfrac { f+x }{ f }$
Now, substituting the values we get,
$ \Rightarrow \sqrt { \cfrac { 304 }{ 300 } } =\cfrac { 300+x }{ 300 }$
$ \Rightarrow \sqrt { 304 } \times \sqrt { 300 } =300+x$
$ \Rightarrow 17.44\times 17.32=300+x$
$ \Rightarrow 302.06=300+x$
$ \Rightarrow x=2.06\simeq 2$
Thus, the number of beats heard per second will be 2.