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In the circult shown in the figure all the resistance are identical and each has the value $r \Omega$. The equivalent resistance of the combination between the points $A$ and $B$ will remain unchanged even when the followring pairs of points marked in the figure are connected through a resistance $\mathbb{R}$. 
Options:
Solution:
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Verified Answer
The correct answers are:
2 and 6, 4 and 7
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 $B\left(R_{A}\right)$ 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, $\because \frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{2 r}{4 r}=\frac{r}{2 r} \Rightarrow \frac{1}{2}=\frac{1}{2}$
Hence, option is correct. From option (b), circuit is
Now, $\frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{3 r}{3 r} \neq \frac{r}{2 r}$
So, option (b) is also incorrect. From option (c) circuit is,
Now, $\frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{4 r}{2 r}=\frac{2 r}{r} \Rightarrow \frac{2}{1}=\frac{2}{1}$
So, option (c) is also correct. From option (d), circuit is
Now, $\frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{4 r}{2 r} \neq \frac{r}{2 r}$
So, option (d) is also incorrect. Hence, option (a) and (c) are correct.
According to question, when given pair of point in options are connected through a resistance $R$, then equivalent resistance between point $A$ and $B\left(R_{A}\right)$ 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, $\because \frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{2 r}{4 r}=\frac{r}{2 r} \Rightarrow \frac{1}{2}=\frac{1}{2}$
Hence, option is correct. From option (b), circuit is
Now, $\frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{3 r}{3 r} \neq \frac{r}{2 r}$
So, option (b) is also incorrect. From option (c) circuit is,
Now, $\frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{4 r}{2 r}=\frac{2 r}{r} \Rightarrow \frac{2}{1}=\frac{2}{1}$
So, option (c) is also correct. From option (d), circuit is
Now, $\frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{4 r}{2 r} \neq \frac{r}{2 r}$
So, option (d) is also incorrect. Hence, option (a) and (c) are correct.
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