Solution:

Given,

Difference between the velocity of sound at 27°C and -13°C = ?
i.e., v₂₇ - v_-13 = ?
The velocity of sound at 0°C (v₀) = ?

Now, the velocity of sound in air is given by

\(v0=\sqrt{\frac{\gamma P}\rho}=\sqrt{\frac{1.4\times1.01\times10^5}{1.29}}=331.1\) m/s

Since, \(v\propto\sqrt T\), then 

\(\frac{v_{27}}{v_0}=\sqrt{\frac{T_{27}}{T_0}}\)
Or, \(v_{27}=v_{0\times}\sqrt{\frac{T_{27}}{T_0}}=331.1\times\sqrt{\frac{273+27}{273}}=347.087\) m/s

Also, \(\frac{v_{-13}}{v_0}=\sqrt{\frac{T_{-13}}{T_0}}\)

Or, \(v_{-13}=v_0\times\sqrt{\frac{T_{-13}}{T_0}}=331.1\times\sqrt{\frac{273-13}{273}}=323.12\) m/s

Now, the difference in velocity at these temperatures is

= \(v_{27}-v_{-13}=347.087-323.12=23.96\) m/s