Young Modulus Calculation from Load and Extension

❓ Question

A $3\text{ m}$ long wire of radius $3\text{ mm}$ shows an extension of $0.1\text{ mm}$ when loaded vertically by a mass of $50\text{ kg}$ in an experiment to determine Young’s modulus.

The value of Young’s modulus of the wire for this experiment is:

(Take $g=3\pi \text{ m/s}^2$)

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Young’s modulus
👉 Stress and strain
👉 Elasticity of solids.

Main idea:

$$Y=\frac{\text{Stress}}{\text{Strain}}$$

For a wire:

$$Y=\frac{FL}{A\Delta L}$$

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🔷 Step 1 — Write Given Values 💯

Length:

$$L=3\text{ m}$$

Radius:

$$r=3\text{ mm}=3\times10^{-3}\text{ m}$$

Extension:

$$\Delta L=0.1\text{ mm}=10^{-4}\text{ m}$$

Mass:

$$m=50\text{ kg}$$

Acceleration due to gravity:

$$g=3\pi$$

Force applied:

$$F=mg$$
$$F=50(3\pi)$$
$$F=150\pi \text{ N}$$

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🔷 Step 2 — Find Cross-Sectional Area

$$A=\pi r^2$$
$$=\pi(3\times10^{-3})^2$$
$$=9\pi\times10^{-6}\text{ m}^2$$

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🔷 Step 3 — Apply Young’s Modulus Formula

$$Y=\frac{FL}{A\Delta L}$$

Substitute values:

$$Y= \frac{(150\pi)(3)} {(9\pi\times10^{-6})(10^{-4})}$$
$$= \frac{450\pi}{9\pi\times10^{-10}}$$
$$= 50\times10^{10}$$
$$=5\times10^{11}\text{ N/m}^2$$

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🔷 Step 4 — Express in Required Form

Question asks value in:

$$\times10^{11}\text{ N/m}^2$$

Hence:

$$\boxed{5}$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ mm ko meter me convert na karna

❌ Area me $r^2$ bhool jaana

❌ Force ko directly mass le lena

Remember:

$$\boxed{ Y=\frac{FL}{A\Delta L} }$$

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✅ Final Answer

$$\boxed{5}$$

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📚 Related Topics

• Extension of Wire: Length vs Thickness
• Ratio of Elongation in Two Wires
• Bulk Modulus and Volume Change Calculation