\(f =\left(\frac{330+15}{330-15}\right) \times f _0\)
\(=\frac{345}{315} f _0\)
\(\frac{345}{315} f _0- f _0=40\)
\(\frac{30}{315} f _0=40\)
\(f _0=\frac{4 \times 315}{3}=420\,Hz\)
(જયાં $ {I_0} $ થ્રેશોલ્ડ તીવ્રતા)
$\left(\mathrm{R}=8.3 \mathrm{JK}^{-1}, \gamma=1.4\right.$આપેલ છે.)