Question

A 4 kg stone is attached to a steel wire being whirled at a constant speed of \( 12 \) m/s in a horizontal circle. The wire is 4 m long with a diameter of 2 mm, and Young’s modulus is \( 2 \times 10^{11} \) Nm\(^2\). The strain in the wire is:

• A. \( 2.3 \times 10^{-4} \)
• B. \( 2.3 \times 10^{-5} \)
• C. \( 4.6 \times 10^{-4} \)
• D. \( 6.9 \times 10^{-4} \)

Hint:

For problems involving strain, use Young’s modulus and the stress formula.

Answer 1

The Correct Option is A

Step 1: Calculating the Tension in the Wire Centripetal force provides tension: \[ T = \frac{m v^2}{r} = \frac{4 \times 12^2}{4} = 144 N \] Step 2: Calculating Strain Strain is given by: \[ \text{Strain} = \frac{\text{Stress}}{\text{Young’s Modulus}} \] Stress: \[ \text{Stress} = \frac{T}{A} = \frac{144}{\pi (1 \times 10^{-3})^2} = \frac{144}{\pi \times 10^{-6}} \] \[ = 4.6 \times 10^7 \] \[ \text{Strain} = \frac{4.6 \times 10^7}{2 \times 10^{11}} \] \[ = 2.3 \times 10^{-4} \] Thus, the correct answer is option (1).

Answer 2

Given:
Mass of stone, \( m = 4 \, \text{kg} \)
Speed, \( v = 12 \, \text{m/s} \)
Length of wire, \( L = 4 \, \text{m} \)
Diameter of wire, \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
Young's modulus, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)

Step 1: Calculate the tension \( T \) in the wire due to circular motion:
\[ T = \frac{m v^2}{r} = \frac{4 \times 12^2}{4} = \frac{4 \times 144}{4} = 144 \, \text{N} \]

Step 2: Calculate the cross-sectional area \( A \) of the wire:
\[ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(1 \times 10^{-3}\right)^2 = \pi \times 10^{-6} \, \text{m}^2 \]

Step 3: Calculate the strain \( \varepsilon \) using Young's modulus formula:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{T/A}{\varepsilon} \implies \varepsilon = \frac{T}{Y A} \]

Step 4: Substitute values:
\[ \varepsilon = \frac{144}{2 \times 10^{11} \times \pi \times 10^{-6}} = \frac{144}{2 \times 10^{11} \times 3.1416 \times 10^{-6}} = \frac{144}{6.2832 \times 10^{5}} \approx 2.29 \times 10^{-4} \]

Therefore, the strain in the wire is:
\[ \boxed{2.3 \times 10^{-4}} \]