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 \times 10^{11} \, \text{N/m}^2 $, where the value of $ P $ is: (Take $ g = 3\pi \, \text{m/s}^2 $)

• A. 25
• B. 10
• C. 2.5
• D. 5

Hint:

When calculating Young's modulus for a wire, use the formula \( Y = \frac{F L}{A \Delta L} \), where \( F \) is the force, \( L \) is the length, \( A \) is the cross-sectional area, and \( \Delta L \) is the extension.

Answer 1

The Correct Option is D

To calculate the Young's modulus \( Y \), we use the formula for Young's modulus for a wire under tension: \[ Y = \frac{F L}{A \Delta L} \] Where: - \( F \) is the force applied on the wire (equal to the weight of the mass),
- \( L \) is the original length of the wire,
- \( A \) is the cross-sectional area of the wire,
- \( \Delta L \) is the extension of the wire.

Step 1: Calculate the force \( F \) The force applied on the wire is the weight of the mass: \[ F = m \cdot g = 50 \, \text{kg} \times 3\pi \, \text{m/s}^2 = 150\pi \, \text{N} \]
Step 2: Calculate the cross-sectional area \( A \) The wire is circular, so the area is given by: \[ A = \pi r^2 = \pi \times (3 \, \text{mm})^2 = \pi \times (3 \times 10^{-3} \, \text{m})^2 = 9\pi \times 10^{-6} \, \text{m}^2 \]
Step 3: Use the values in the formula for Young's modulus Substitute the values into the formula for Young's modulus: \[ Y = \frac{150\pi \times 3 \, \text{m}}{9\pi \times 10^{-6} \, \text{m}^2 \times 0.1 \times 10^{-3} \, \text{m}} \] Simplifying: \[ Y = \frac{450\pi}{9\pi \times 10^{-7}} = \frac{450}{9 \times 10^{-7}} = 5 \times 10^{10} \, \text{N/m}^2 \] Therefore, the value of \( P \) is 5. Thus, the correct answer is Option (4).