1. Initial resistance and stretching of the wire

Given original resistance: $R_0 = 3\,\Omega$

The wire is elongated to double its original length.

For stretching of the same material and same volume:

• $L' = 2L$
• Area $A' = A/2$

Since $R = \rho \frac{L}{A},$ new resistance becomes $R' = \rho \frac{2L}{A/2} = 4\rho\frac{L}{A} = 4R_0 = 4\times 3 = 12\,\Omega.$

So the total resistance of the new wire is: $R_{\text{total}} = 12\,\Omega$

2. Wire bent into a circle

This entire wire is bent into a circle. Two points on the circle subtend an angle $60^\circ$ at the centre.

Hence the circle is divided into two arcs:

• smaller arc: $60^\circ$
• larger arc: $300^\circ$

Since resistance is proportional to length, the resistances of the two arcs are in the same ratio: $60:300 = 1:5$

Let smaller arc resistance be $R_1$ and larger arc resistance be $R_2$. Then $R_1:R_2 = 1:5$ and $R_1 + R_2 = 12\,\Omega$

So, $R_1 = 12\times \frac{1}{6} = 2\,\Omega$ $R_2 = 12\times \frac{5}{6} = 10\,\Omega$

3. Equivalent resistance between the two points

The two arcs connect the same pair of points, so they are in parallel.

Therefore, $R_{\text{eq}} = \frac{R_1R_2}{R_1+R_2}$

Substitute values: $R_{\text{eq}} = \frac{2\times 10}{2+10} = \frac{20}{12} = \frac{5}{3}\,\Omega$

4. Option check

• A: $\frac{5}{2}\,\Omega$ ❌
• B: $\frac{12}{5}\,\Omega$ ❌
• C: $\frac{7}{2}\,\Omega$ ❌
• D: $\frac{5}{3}\,\Omega$ ✅

Final Answer: $\boxed{\frac{5}{3}\,\Omega}$