Ratio of Elongation in Two Wires

❓ Question

Two wires A and B are made of the same material.
The ratio of their lengths is

$$\frac{L_A}{L_B}=\frac{1}{3}$$

and the ratio of their diameters is

$$\frac{d_A}{d_B}=2.$$

If both wires are stretched using the same force, what is the ratio of their elongations?

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🖼️ Question Image

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✍️ Short Solution

This is a direct Young’s modulus application.
No heavy maths — bas elongation dependence samajh lo, answer khud nikal jaata hai 😎

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🔹 Step 1 — Elongation Formula (FOUNDATION 💯)**

For a wire under tension:

$$\Delta L = \frac{FL}{YA}$$

Where:

• $F$ = applied force

• $L$ = original length

• $Y$ = Young’s modulus

• $A$ = cross-sectional area

📌 Same material ⇒ same $Y$
📌 Same force ⇒ same $F$

So,

$$\Delta L \propto \frac{L}{A}$$

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🔹 Step 2 — Area in terms of Diameter

Cross-sectional area:

$$A=\frac{\pi d^2}{4}$$

So:

$$A \propto d^2$$

Hence:

$$\Delta L \propto \frac{L}{d^2}$$

🧠 Golden relation to remember:

Elongation ∝ Length / (Diameter)²

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🔹 Step 3 — Write Ratio of Elongations

$$\frac{\Delta L_A}{\Delta L_B} = \frac{L_A}{L_B} \times \frac{d_B^2}{d_A^2}$$

Now substitute given ratios.

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🔹 Step 4 — Substitute Given Data

Given:

$$\frac{L_A}{L_B}=\frac{1}{3}$$
$$\frac{d_A}{d_B}=2 \Rightarrow \frac{d_B}{d_A}=\frac{1}{2}$$

So:

$$\frac{d_B^2}{d_A^2}=\left(\frac{1}{2}\right)^2=\frac{1}{4}$$

Putting everything together:

$$\frac{\mathrm{\Delta }{L}_A}{\mathrm{\Delta }{L}_B}=\frac{1}{3}\times \frac{1}{4}=\frac{1}{12}$$

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✅ Final Answer

$$\boxed{{\displaystyle \frac{\mathrm{\Delta }{L}_A}{\mathrm{\Delta }{L}_B}=\frac{1}{12}}}$$

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⭐ Golden JEE Insight

• Longer wire ⇒ more elongation

• Thicker wire ⇒ less elongation

• Diameter affects elongation much more strongly (square effect!)

🧠 One-line memory:

Elongation ∝ L / d²