Resistance of wire stretched n times becomes

(a) nR
(b) R/n
(c) n²R
(d) R/n²

Correct Answer: (c) n²R

Explanation

Length increases n times, area reduces n times → R ∝ n²

Notes for More Information

Concept

1. Resistance of a conductor is given by:

$$R=\rho \frac{l}{A}$$

2. Resistance is:

   • Directly proportional to length (l)

   • Inversely proportional to cross-sectional area (A)

3. When a wire is melted and stretched, its volume remains constant.

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Given

• Initial resistance = $R_1=R$

• Initial length = $l_1$

• Final length after stretching:

$$l_2=n{l}_1$$

• Initial area = $A_1$

• Final area = $A_2$

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Volume Condition

Since volume is constant:

$$A_1{l}_1=A_2{l}_2$$

Substitute $l_2=n{l}_1$:

$$A_1{l}_1=A_2(n{l}_1)$$$$A_2=\frac{A_1}{n}$$

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Calculation

Initial resistance:

$$R_1=\rho \frac{l_1}{A_1}$$

Final resistance:

$$R_2=\rho \frac{l_2}{A_2}$$

Substitute values:

$$R_2=\rho \frac{n{l}_1}{A_1\mathrm{/}n}$$$$R_2=\rho \frac{n^2{l}_1}{A_1}$$

Taking ratio:

$$\frac{R_2}{R_1}=n^2$$

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Final Answer

$$\boxed{{\displaystyle R_2=n^{2}R}}$$

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Correct Option

✅ (d) $n^{2}R$

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Exam Notes (Very Important)

• Stretching ⇒ Length increases, area decreases

• Volume constant problems always use:

  $$A_1{l}_1=A_2{l}_2$$

• If length becomes n times, resistance becomes n² times