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 determine the value of Young's modulus for the wire, we will use the formula:

\(Y = \frac{F \cdot L}{A \cdot \Delta L}\)

where:

• \(Y\) is Young's modulus,
• \(F\) is the force applied,
• \(L\) is the original length of the wire,
• \(A\) is the cross-sectional area of the wire,
• \(\Delta L\) is the extension in the length.

Given values:

• Mass, \(m = 50 \, \text{kg}\)
• Acceleration due to gravity, \(g = 3\pi \, \text{m/s}^2\)
• Original length, \(L = 3 \, \text{m}\)
• Extension, \(\Delta L = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m}\)
• Radius, \(r = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m}\)

First, calculate the force applied, \(F\):

\(F = m \cdot g = 50 \times 3\pi \, \text{N} = 150\pi \, \text{N}\)

Next, calculate the cross-sectional area, \(A\):

\(A = \pi r^2 = \pi \times (3 \times 10^{-3})^2 = 9\pi \times 10^{-6} \, \text{m}^2\)

Substitute these values into the formula for Young's modulus:

\(Y = \frac{150\pi \times 3}{9\pi \times 10^{-6} \cdot 0.1 \times 10^{-3}}\)

Simplifying gives:

\(Y = \frac{450\pi}{9\pi \times 10^{-9}}\)

\(Y = \frac{450}{9} \times 10^{9}\)

\(Y = 50 \times 10^{9} \, \text{N/m}^2 = 5 \times 10^{10} \, \text{N/m}^2\)

Therefore, when expressed as \(P \times 10^{11} \, \text{N/m}^2\), the value of \(P\) is 5.

Thus, the correct answer is:

• 5

Answer 2

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).