Explanation:
When qbserver is at rest and source of sound id moving towards observer then observed frequency n'.
$\text{n}'=\Big(\frac{\text{v}}{\text{V}-\text{v}_\text{s}}\Big)\text{n}_0$
Where n0 original frequency of source of sound
v = speed of sound in medium
$\therefore\text{n}'>\text{n}_0\ \ \text{v}_\text{s}=$ speed of source
When source is moving away from observer
$\text{n}'=\frac{\text{v}}{(\text{v}+\text{v}_\text{s})}\text{n}_0\ \text{n}''<\text{n}_0$
Hence, the frequencies in both cases are same and $\text{n}'>\text{n}''.$ so graph (c) verifies the answer.
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Match the following
$\begin{array}{|l|l|} \hline Column\,\,-\,\,1 & Column\,\,-\,\,2 \\ \hline P\,:\,Process\,\,-\,\,I & \,\,A\,\,:\,\,Adiabatic \\ \hline Q\,:\,Process\,\,-\,\,II & \,\,B\,\,:\,\,Isobaric \\ \hline R\,:\,Process\,\,-\,\,III & \,\,C\,\,:\,\,Isochoric \\ \hline S\,:\,Process\,\,-\,\,IV & \,\,D\,\,:\,\,Isothermal \\ \hline \end{array}$
$Reason$ : $\frac{{\Delta E}}{E} = \frac{{\Delta m}}{m} + \frac{{2\Delta v}}{v}$
