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Step 1: Understand the Problem
We start with a metal wire that has an initial resistance of 3 Ω. This wire is elongated to double its original length. Because the wire’s volume remains constant when it is elongated, its cross-sectional area decreases accordingly. We then bend this elongated wire into a circle. Two points on the circle subtend an angle of 60° at the center. We want to find the equivalent resistance between these two points.
Step 2: Determine the New Resistance After Elongation
The resistance of a wire is given by the formula
$R = \rho \frac{l}{A}$,
where $l$ is the length of the wire, $A$ is its cross-sectional area, and $\rho$ is the resistivity (a constant for the same material).
When the length is doubled ($l \to 2l$) and the volume is unchanged, the area becomes $\frac{A}{4}$. Hence, new resistance becomes:
$R_{\text{new}} = R_{\text{old}} \times \left(\frac{\text{length factor}}{\text{area factor}}\right) = 3 \times \left(\frac{2}{\frac{1}{4}}\right) = 3 \times 4 = 12 \, \Omega$
Step 3: Form the Circular Wire and Identify the Two Sections
The wire of total resistance 12 Ω is then bent into a circle. The 60° angle at the center means that one arc corresponds to $\frac{60^\circ}{360^\circ} = \frac{1}{6}$ of the circle, and the other arc is $\frac{5}{6}$ of the circle.
Since the resistance is uniform along the length of the circle,
Resistance of 60° arc = $\frac{1}{6} \times 12 = 2 \, \Omega$,
Resistance of remaining 300° arc = $\frac{5}{6} \times 12 = 10 \, \Omega$.
Step 4: Combine the Two Arcs in Parallel
The two arcs are connected across the same two points, effectively forming two parallel resistors of 2 Ω and 10 Ω. The equivalent resistance $R_{\text{eq}}$ for two resistors $R_1$ and $R_2$ in parallel is:
$R_{\text{eq}} = \frac{R_1 \, R_2}{R_1 + R_2}$
Substituting $R_1 = 2\,\Omega$ and $R_2 = 10\,\Omega$:
$R_{\text{eq}} = \frac{2 \times 10}{2 + 10} = \frac{20}{12} = \frac{5}{3} \, \Omega$
Final Answer
The equivalent resistance between the two points is
$ \frac{5}{3}\,\Omega$.
