In the P-V diagram shown, there are two adiabatic parts of the same gas intersecting two isothermals at \(T_1\) and \(T_2\). The ratio \(\left(\frac{V_b}{V_a}\right)\) is equal to

For adiabatic process
\(P V Y=C\) Ideal gas law states
\(P V=n R T\)
Dividing \(T V \gamma-1=\) constant above equations.
Consider the diagram as follows:

For path BC
\(\begin{aligned}
& T_1=V_b^{\gamma-1}=T_2=\underset{c}{V \gamma-1} \\
& \Rightarrow\left(\frac{V_b}{V_c}\right)^{Y-1}=\frac{T_2}{T_1}---(1)
\end{aligned}\)
Now, for path DA.
\(\begin{aligned}
& T_2=\underset{d}{V}-1=T_1=\underset{a}{V \gamma-1} \\
& \Rightarrow\left(\frac{V_a}{V_d}\right)^{\gamma-1}=\frac{T_2}{T_1} \quad--(2)
\end{aligned}\)
Using equation (1) & (2)
\(\begin{aligned}
& \frac{V_b}{V_c}=\frac{V_a}{V_d} \\
& \Rightarrow \frac{V_b}{V_a}=\frac{V_c}{V_d}
\end{aligned}\)
Option (B) is correct.