JEE Main 2025 Apr 7 Shift 1, Physics Q3: Wheatstone Bridge, Ammeter, Voltmeter And Potentiometer

The frame has 6 segments of equal length; each carries resistance $r=R/6$.

Symmetry observation. With current entering at A and leaving at B, the two remaining vertices are at the same potential due to the tetrahedral symmetry, producing a balanced Wheatstone bridge. Any conductor joining those two equipotential nodes carries no current and can be removed.

Removing this bridging edge leaves three parallel branches between A and B:

• Direct edge AB: resistance $r$.
• Path through one intermediate vertex (two edges): $2r$.
• Path through the other intermediate vertex (two edges): $2r$.

Combine in parallel:

$\frac{1}{R_{AB}}=\frac{1}{r}+\frac{1}{2r}+\frac{1}{2r}=\frac{2}{r}$

$R_{AB}=\frac{r}{2}=\frac{R/6}{2}=\frac{R}{12}$

Matching with $R_{AB}=R/n$: $n=12$.

$\therefore$ Option (4).