
A wire of resistance $R$ is bent into a triangular pyramid as shown in figure with each segment having same…

- 16
- 14
- 10
- 12
Solution

(As balanced wheat stone bridge is formed) Now, Equivalent resistance between A and B can be written as
$\frac{1}{\mathrm{R}_{\mathrm{AB}}}=\frac{1}{2 \mathrm{r}}+\frac{1}{2 \mathrm{r}}+\frac{1}{\mathrm{r}}=\frac{2}{\mathrm{r}}$
$\mathrm{R}_{\mathrm{AB}}=\frac{\mathrm{R}}{12}$
Asked in: JEE Main 2025 (07 Apr Shift 1)