Solution

Given,

Beats per second = 10,
So, the frequency of the pipe = 440 ± 10 Hz = 430 Hz or 450 Hz,

When the fork is loaded with a small piece of wax, the beat frequency decreases, so the air column must have a lower frequency. i.e., f = 430 Hz

Let t°C be the temperature at which the air column will be in unison.

At this temperature, the frequency of air column = 440 Hz

∴ v₁ = 440 x λ --------- (i) 

And at 20°C the frequency of the air column is given by

∴ v₂ = 430 x λ --------- (ii)

Dividing (i) by (ii), we get

\(\frac{v_t}{v_{20}}=\frac{440\times\lambda}{430\times\lambda}=\frac{44}{43}\) --------- (iii)

Since, \(v\propto\sqrt T\), so the equation (iii) can be written as

\(\frac{\sqrt{273+t}}{\sqrt{273+20}}=\frac{44}{43}\Rightarrow\frac{273+t}{293}=\left(\frac{44}{43}\right)^2\)

∴ \(t=\left(\frac{44}{33}\right)^2\times293-273=33.73\) °C = 306.33 K

Hence, the required temperature is 306.33 K