Solution:

Let, v be the velocity of sound at 27°C. Then, the velocity of sound increased by 50% will be

\(v_2=\left(v+\frac{50v}{100}\right)\)

Here,

T₁ = 27°C = 273 + 27 K = 300 K, v₁ = v
T₂ = ( 273 + T ) K = 300 K and \(v_2=v+\frac{50v}{100}=\frac{3v}2\)

Now, we have,

\(\frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}\)
Or, \(\frac v{3v/2}=\sqrt{\frac{300}{T+273}}\)
Or, \(\frac23=\sqrt{\frac{300}{T+273}}\)
Or, \(\frac49=\frac{300}{T+273}\)
Or, \(T+273=675\)
Or, \(T=675-273=402^\circ C\)

Thus, the required temperature is 42°C.