A copper wire has a resistance of 12 Ω. If the wire is stretch... | General Science MCQ | Examoogle

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General SciencePhysics: Electricity

A copper wire has a resistance of 12 Ω. If the wire is stretched to twice its original length, keeping the volume constant, what will be its new resistance?

24 $\Omega$

36 $\Omega$

48 $\Omega$

12 $\Omega$

Correct Answer

⚙️ TE • Technical Concept & PrincipleGeneral SciencePhysics: Electricity

Option C

48 $\Omega$

Quick Summary:

When a wire is stretched to $n$ times its original length while volume remains constant, its resistance increases by a factor of $n^2$ . Since the length is doubled ( $n = 2$ ), the new resistance becomes $2^2 = 4$ times the original resistance, i.e., $12 \Omega \times 4 = 48 \Omega$ .

🔬SCScience SolutionConcept & Principle

💡 Explanation

💡 Everyday Analogy (Real-World Intuition)

Think of a piece of playdough; if you roll it to be twice as long, it must become much thinner, making it harder for 'current' to flow through the now restricted, narrow path.

🧠 Memory Mnemonic / Shortcut Aid

R is proportional to L squared (when stretching): $R \propto l^2$ .

🔢 Key Formulas

$R = \rho \frac{l}{A}$ — formula for resistance

$R' = n^2 R$ — resistance scaling factor for uniform stretching

⚙️ Working Principle

The resistance of a wire is given by $R = \rho \frac{l}{A}$ . Since the volume $V = A \times l$ is constant, stretching the wire to $n$ times its length ( $l' = nl$ ) reduces the cross-sectional area to $A' = \frac{A}{n}$ . Substituting these into the formula, $R' = \rho \frac{nl}{A/n} = n^2 \rho \frac{l}{A} = n^2 R$ .

📌 Key Points

▸
Volume conservation is the key constraint for uniform stretching.
▸
Stretching a wire increases its resistance due to both increased length and decreased cross-sectional area.
▸
Resistivity ( $\rho$ ) remains constant as it is an intrinsic material property.

✅ Advantages

▸
Useful for calculating the effect of physical deformation on electrical components.

🛠️ Applications / Uses

▸
Strain gauges for measuring mechanical deformation.
▸
Design of variable resistors or potentiometers.

📄 Additional Information

▸
Constant: Volume ( $V$ ) is invariant during stretching.
▸
Option B (36 $\Omega$ ) is a common error assuming linear proportionality $R \propto l$ instead of $R \propto l^2$ .

📊 Diagram / Illustration

✅

C is correct — stretching the wire to twice its length increases the resistance by a factor of 4, resulting in 48 $\Omega$ .

Core Concepts Used

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Resistance Resistivity Volume conservation

💡 EXAM TIP

Always check if the problem specifies 'volume remains constant' before applying the $n^2$ rule; if a new wire is simply cut, only $l$ changes and $R$ changes linearly.

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