A 3 m long wire of radius 3 mm shows an extension of 0.1 mm when loaded vertically by a mass of 50 kg in an experiment to determine Young's modulus. The value of Young’s modulus of the wire as per this experiment is P × 10¹¹ N/m², where the value of P is: (Take g = 3π m/s²)

 

🧪 Young’s Modulus from Wire Extension | JEE Physics | Doubtify

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📌 Question:

A 3 m long wire of radius 3 mm shows an extension of 0.1 mm when loaded vertically by a mass of 50 kg in an experiment to determine Young's modulus.
The value of Young’s modulus of the wire as per this experiment is P × 10¹¹ N/m², where the value of P is:
(Take g = 3π m/s²)

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🖼️ Question Image:

🧠 Solution Image:

📖 Concept:

Young’s modulus (Y) is given by:

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

Where:

• $F=m\mathrm{g\; =\; force\; applied\; (N)}$

• $L$ = original length of wire

• $\Delta L$ = extension in wire

• $A$ = cross-sectional area = $\pi r^2$
  

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🧮 Given Data:

• $L = 3 \, m$
  

• $\Delta L = 0.1 \, mm = 0.1 \times 10^{-3} \, m$
  

• $r=3mm=3\times {10}^{-3}m$

• $m = 50 \, kg$
  

• $g = 3\pi \, m/s^2$
  

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📐 Step-by-Step Calculation:

1. Force $F = mg = 50 \times 3\pi = 150\pi \, N$
   

2. Area $A = \pi r^2 = \pi (3 \times 10^{-3})^2 = \pi \times 9 \times 10^{-6}$
   

3. Now, plug into the formula:

$$Y = \frac{150\pi \times 3}{\pi \times 9 \times 10^{-6} \times 0.1 \times 10^{-3}} = \frac{450}{9 \times 10^{-10}} = \frac{50}{10^{-10}} = 5 \times 10^{11} \, N/m^2$$

✅ Final Answer:

P = 5

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🔍 Why This Question Is Important:

• Classic formula-based question from Elasticity chapter

• Appears often in JEE Mains & Advanced

• Teaches how to apply formulae involving unit conversions and SI consistency

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🧠 Pro Tip:

Always convert mm to meters and use SI units throughout the calculation to avoid silly mistakes!

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