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

Young's modulus (Y) for the wire is calculated using the formula:

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

Where:

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

Provided values:

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

The applied force, \(F\), is calculated as:

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

The cross-sectional area, \(A\), is calculated as:

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

Substituting these values into the Young's modulus formula:

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

Simplification yields:

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

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

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

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

The final answer is:

• 5