JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Current Electricity

Stretching the wire to twice its length keeps its volume fixed, and since $R=\frac{\rho {\ell }^2}{\text{volume}}$, doubling $\ell$ multiplies the resistance by $2^2=4$, giving $3\times 4=12\Omega$ for the full loop. Bending it into a circle and tapping two points that subtend $60^{\circ }$ at the centre splits the ring into a minor arc carrying $\frac{60}{360}=\frac{1}{6}$ of the wire and a major arc carrying the other $\frac{5}{6}$. Their resistances are $12\times \frac{1}{6}=2\Omega$ and $12\times \frac{5}{6}=10\Omega$, and between the two tapping points these arcs sit in parallel: $R=\frac{2\times 10}{2+10}=\frac{20}{12}=\frac{5}{3}\Omega$.