Question

A metal wire of natural length 50cm and cross-sectional area 4.0 mm² is fixed at end. A mass of 2.4kg is hung from the other end of the wire. If the elastic pot energy of the wire is 1.8x10^-4J,then its Young's modulus is:(Take g=10 ms^-2)

• A. 1.6x10¹¹ Nm²
• B. 2.4x10¹¹ Nm²
• C. 3.2x10¹¹ Nm²
• D. 1.8x10¹¹ Nm²
• E. 2.0x10¹¹ Nm²

Answer 1

Answer: (E)

Given:

• Natural length of wire, \( L_0 = 50 \, \text{cm} = 0.5 \, \text{m} \)
• Cross-sectional area, \( A = 4.0 \, \text{mm}^2 = 4.0 \times 10^{-6} \, \text{m}^2 \)
• Mass hung, \( m = 2.4 \, \text{kg} \) (force \( F = mg = 24 \, \text{N} \))
• Elastic potential energy, \( U = 1.8 \times 10^{-4} \, \text{J} \)
• Acceleration due to gravity, \( g = 10 \, \text{ms}^{-2} \)

Step 1: Relate Energy to Extension

Elastic potential energy in a stretched wire is given by:

\[ U = \frac{1}{2} F \Delta L \]

Solve for extension \( \Delta L \):

\[ \Delta L = \frac{2U}{F} = \frac{2 \times 1.8 \times 10^{-4}}{24} = 1.5 \times 10^{-5} \, \text{m} \]

Step 2: Calculate Young's Modulus (Y)

Young's modulus is defined as:

\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L_0} \]

Substitute values:

\[ Y = \frac{24/(4.0 \times 10^{-6})}{1.5 \times 10^{-5}/0.5} = \frac{6 \times 10^6}{3 \times 10^{-5}} = 2.0 \times 10^{11} \, \text{Nm}^{-2} \]

Conclusion:

The Young's modulus of the wire is \( 2.0 \times 10^{11} \, \text{Nm}^{-2} \).

Answer: \(\boxed{E}\)

Answer 2

1. Define variables and given information:

• L = 50 cm = 0.5 m (natural length of the wire)
• A = 4.0 mm² = 4.0 × 10⁻⁶ m² (cross-sectional area)
• m = 2.4 kg (mass hung from the wire)
• g = 10 m/s² (acceleration due to gravity)
• EPE = 1.8 × 10⁻⁴ J (elastic potential energy)
• Y = ? (Young's modulus)

2. Calculate the force applied by the mass:

The force (F) applied by the hanging mass is equal to its weight:

\[F = mg = (2.4 \, kg)(10 \, m/s^2) = 24 \, N\]

3. Relate elastic potential energy to Young's modulus:

The elastic potential energy (EPE) stored in a stretched wire is given by:

\[EPE = \frac{1}{2} \frac{F^2 L}{AY}\]

where:

• F is the force applied
• L is the original length
• A is the cross-sectional area
• Y is Young's modulus

4. Rearrange the EPE equation and solve for Y:

We can rearrange the EPE equation to solve for Young's modulus (Y):

\[Y = \frac{1}{2} \frac{F^2 L}{A(EPE)}\]

5. Substitute the values and calculate Y:

\[Y = \frac{1}{2} \frac{(24 \, N)^2 (0.5 \, m)}{(4.0 \times 10^{-6} \, m^2)(1.8 \times 10^{-4} \, J)}\]

\[Y = \frac{1}{2} \frac{576 \times 0.5}{7.2 \times 10^{-10}} = \frac{288}{7.2 \times 10^{-10}} = 40 \times 10^{10} = 2.0 \times 10^{11} \, N/m^2\]