At what temperature, the velocity of sound in air is increased by 50% to that at 27°C?
1 Answer
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.
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