A metal wire of resistance 3 ohm is elongated to make a uniform wire of double its previous length.

Q: Why does the area change when the wire is stretched?

The amount of metal in the wire (its volume) cannot change just by stretching it. Since Volume = Area $\times$ Length, if the length increases, the area must decrease proportionally to keep the volume constant.

Q: Is there a shortcut to find the new resistance when a wire is stretched?

Yes! If a wire is stretched to $n$ times its original length, its new resistance becomes $n^2$ times the original resistance. Here, $n=2$, so the new resistance is $2^2 \times 3 = 12 \Omega$.

Q: Why are the two arcs considered to be in parallel?

When you connect a battery across points A and B, the current has two distinct paths to travel from A to B: along the minor arc and along the major arc. Multiple paths between the same two nodes define a parallel connection.

Q: How do we know the resistance is proportional to the angle?

The resistance of a uniform wire is proportional to its length ($R \propto l$). For a circular arc, the arc length is $l = r\theta$. Since the radius $r$ is constant, the length (and thus the resistance) is directly proportional to the angle $\theta$.

Q: What if the points were diametrically opposite?

If the points were diametrically opposite, the angle would be $180^\circ$. Both arcs would have a resistance of $6 \Omega$. The equivalent parallel resistance would then be $\frac{6 \times 6}{6 + 6} = 3 \Omega$.

Question

A metal wire of resistance $3 \Omega$ is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle $60^\circ$ at the centre, the equivalent resistance between these two points will be

Options:

(A) $\frac{7}{2} \Omega$

(B) $\frac{5}{2} \Omega$

(C) $\frac{12}{5} \Omega$

(D) $\frac{5}{3} \Omega$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a metal wire with an initial resistance of $3 \Omega$. We know that the resistance of a wire is given by the formula $R = ho \frac{l}{A}$, where $ho$ is the resistivity, $l$ is the length, and $A$ is the cross-sectional area.

When a wire is stretched, its volume remains constant. The volume of a cylindrical wire is simply its cross-sectional area multiplied by its length, so $V = A \cdot l$.

### The Stretching Process
The problem states that the wire is elongated to double its previous length. This means our new length is $l' = 2l$.

Because the volume is conserved, the new area $A'$ must adjust accordingly. Equating the initial and final volumes:
$$A \cdot l = A' \cdot (2l)$$
$$A' = \frac{A}{2}$$

Now, let's calculate the new resistance $R'$ of this stretched wire:
$$R' = ho \frac{l'}{A'} = ho \frac{2l}{A/2} = 4 \left(ho \frac{l}{A}ight)$$

Since the original resistance was $3 \Omega$, the new resistance becomes:
$$R' = 4 	imes 3 = 12 \Omega$$

### Bending into a Circle
This $12 \Omega$ wire is now bent into a complete circle. We are asked to find the equivalent resistance between two points, $A$ and $B$, on this circle that subtend an angle of $60^\circ$ at the center.

The resistance of any arc of a uniform circular wire is directly proportional to the angle it subtends at the center. The entire circle corresponds to $360^\circ$.

Let's find the resistance of the smaller arc, $R_1$, which subtends $60^\circ$:
$$R_1 = \frac{60^\circ}{360^\circ} 	imes 12 = \frac{1}{6} 	imes 12 = 2 \Omega$$

The larger arc subtends the remaining angle, which is $360^\circ - 60^\circ = 300^\circ$. Its resistance, $R_2$, will be:
$$R_2 = \frac{300^\circ}{360^\circ} 	imes 12 = \frac{5}{6} 	imes 12 = 10 \Omega$$

### Final Calculation
When we measure the equivalent resistance between points $A$ and $B$, the current splits into two paths: the smaller arc and the larger arc. This means $R_1$ and $R_2$ are connected in parallel.

The formula for the equivalent resistance of two resistors in parallel is:
$$R_{	ext{eq}} = \frac{R_1 R_2}{R_1 + R_2}$$

Substituting our values:
$$R_{	ext{eq}} = \frac{2 	imes 10}{2 + 10} = \frac{20}{12}$$

Dividing the numerator and the denominator by $4$, we get our final answer:
**$$R_{	ext{eq}} = \frac{5}{3} \Omega$$**

Select Answer:

Visualized Solution

$R_{initial} = 3Ω$

$V = A l = constant$

$l^{'} = 2 l, A^{'} = \frac{A}{2}$

$R^{'} = 12Ω$

$θ = 6 0^{\circ}$

$R \propto θ$

$R_{1} = 2Ω$

$R_{2} = 10Ω$

$R_{eq} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}$

$R_{eq} = \frac{2 \times 10}{2 + 10}$

$R_{eq} = \frac{20}{12}$

$R_{eq} = \frac{5}{3} Ω$

\text{The Way Forward}

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

Analyzing the Setup Imagine a metal wire with an initial resistance of $3Ω$

The Stretching Process The problem states that the wire is elongated to double its previous length

Bending into a Circle This $12Ω$ wire is now bent into a complete circle

Final Calculation When we measure the equivalent resistance between points $A$ and $B$ , the current splits into two paths: the smaller arc and the larger arc

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rinitial​=3Ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V=Al=constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l′=2l,A′=2A​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R′=12Ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ=60∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R∝θ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​=2Ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​=10Ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Req​=R1​+R2​R1​R2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Req​=2+102×10​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Req​=1220​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Req​=35​Ω