Question

The work to be done to produce a strain of \(10^{-3}\) in a steel wire of mass 2.96 kg and density 7.4 gcm\(^{-3}\) is (Young's modulus of steel = \(2 \times 10^{11}\) Nm\(^{-2}\))

• A. 0.04 kJ
• B. 0.04 J
• C. 100 kJ
• D. 400 J

Hint:

Remember the formulas for elastic potential energy. The energy per unit volume is \( \frac{1}{2}(\text{Stress})(\text{Strain}) \), which can also be written as \( \frac{1}{2} Y (\text{Strain})^2 \) or \( \frac{(\text{Stress})^2}{2Y} \). To get the total energy (work done), multiply the energy density by the total volume of the material.

Answer 1

The Correct Option is A

The work done in stretching a wire is equal to the elastic potential energy stored in it.
The energy stored per unit volume (energy density) is given by \( u = \frac{1}{2} \times \text{Stress} \times \text{Strain} \).
Young's modulus is defined as \( Y = \frac{\text{Stress}}{\text{Strain}} \), so Stress = \( Y \times \text{Strain} \).
Substituting this into the energy density formula: \( u = \frac{1}{2} \times (Y \times \text{Strain}) \times \text{Strain} = \frac{1}{2} Y (\text{Strain})^2 \).
The total work done (total energy stored) is \( W = u \times \text{Volume} \).
We are given: Strain = \(10^{-3}\) and \( Y = 2 \times 10^{11} \) Nm\(^{-2}\).
First, let's find the volume of the wire.
Mass \( m = 2.96 \) kg.
Density \( \rho = 7.4 \text{ gcm}^{-3} \). We must convert this to SI units (kg/m\(^3\)).
\( \rho = 7.4 \frac{\text{g}}{\text{cm}^3} \times \frac{1 \text{ kg}}{1000 \text{ g}} \times \left(\frac{100 \text{ cm}}{1 \text{ m}}\right)^3 = 7.4 \times \frac{1}{1000} \times 10^6 \frac{\text{kg}}{\text{m}^3} = 7400 \) kg/m\(^3\).
Volume \( V = \frac{\text{Mass}}{\text{Density}} = \frac{2.96}{7400} = \frac{296}{740000} = \frac{4}{10000} = 4 \times 10^{-4} \) m\(^3\).
Now, calculate the total work done.
\( W = \left( \frac{1}{2} Y (\text{Strain})^2 \right) \times V \).
\( W = \frac{1}{2} (2 \times 10^{11}) (10^{-3})^2 \times (4 \times 10^{-4}) \).
\( W = (10^{11}) (10^{-6}) (4 \times 10^{-4}) = 4 \times 10^{11-6-4} = 4 \times 10^1 = 40 \) J.
The question asks for the answer in kJ.
\( 40 \text{ J} = 0.040 \text{ kJ} \).