WBJEE 2019 Physics Solved Paper

The circuit diagram is as shown below.According to question, when given pair of point in options are connected through a resistance $R$, then equivalent resistance between point A and B $(R_{AB})$ remains unchanged. It is only possible when current does not flows through resistance $R$ and circuit become Wheatstone bridge as shown in the figure below,At the balanced condition, $\;\frac{P}{Q}=\;\frac{R}{S}$ The current flow in $C D$ branch will be zero. Now by checking each option, from option $(a)$, circuit is,Now, $\because \frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{2r}{4r}=\frac{r}{2r} \Rightarrow \frac{1}{2}=\frac{1}{2}$ Hence, option (a) is correct. From option (b), circuit isNow, $\;\frac{P}{Q}=\;\frac{R}{S} \Rightarrow \;\frac{3r}{3r} \neq \;\frac{r}{2r}$ So, option (b) is also incorrect. From option (c) circuit is,Now. $\;\frac{P}{Q}=\;\frac{R}{S} \Rightarrow \;\frac{4r}{2}=\;\frac{2}{r} \Rightarrow \;\frac{2}{1}=\;\frac{2}{1}$ So, option $(c)$ is also correct. From option (d), circuit is