Question

The elastic potential energy stored in a steel wire of length 20 m stretched through 2 cm is 80 J. The cross sectional area of the wire is______ mm² . (Given, y = 2.0 x 10¹¹ Nm^-2)

Answer 1

Correct Answer: 40

The elastic potential energy stored in a stretched wire is given by: \[ U = \frac{1}{2} \times \frac{F \Delta L}{A Y}, \] where:

• \( U = 80 \, \text{J} \) (elastic potential energy)
• \( L = 20 \, \text{m} \) (length of the wire)
• \( \Delta L = 2 \, \text{cm} = 0.02 \, \text{m} \) (extension in the wire)
• \( Y = 2.0 \times 10^{11} \, \text{Nm}^{-2} \) (Young's modulus of steel)
• \( A \) is the cross-sectional area of the wire (to be calculated)

Rearranging the formula for \( A \): \[ A = \frac{F \Delta L}{2 U Y}. \] The force \( F \) is given by: \[ F = Y \times \frac{\Delta L}{L}. \] Substitute \( F \) back into the equation for \( A \): \[ A = \frac{Y \times \frac{\Delta L}{L} \times \Delta L}{2 U Y} = \frac{\Delta L^2}{2 U \times L}. \] Substitute the given values: \[ A = \frac{(0.02)^2}{2 \times 80 \times 20} = \frac{0.0004}{3200}. \] Simplify: \[ A = 1.25 \times 10^{-7} \, \text{m}^2. \] Convert \( \text{m}^2 \) to \( \text{mm}^2 \): \[ A = 1.25 \times 10^{-7} \times 10^{6} = 0.125 \, \text{mm}^2. \]