Question

The energy stored per unit volume is \( 0.3 \, \text{J/m}^3 \) when a wire is stretched by \( 0.125 \, \text{cm} \). When it is stretched by \( 0.75 \, \text{cm} \), the increase in potential energy per unit volume stored in the wire is

• A. \( 10.3 \, \text{J/m}^3 \)
• B. \( 10.1 \, \text{J/m}^3 \)
• C. \( 10.5 \, \text{J/m}^3 \)
• D. \( 10.8 \, \text{J/m}^3 \)

Hint:

When dealing with energy stored in stretched materials, remember that the energy is proportional to the square of the strain (or elongation) in the wire.

Answer 1

The Correct Option is C

Step 1: Understanding the energy stored in a stretched wire.
The energy stored per unit volume in a stretched wire is given by: \[ U = \frac{1}{2} \times \text{stress} \times \text{strain} \] where stress is proportional to the force applied and strain is proportional to the elongation of the wire. The energy is directly proportional to the square of the elongation.
Step 2: Finding the increase in energy.
The energy for the new elongation of \( 0.75 \, \text{cm} \) is calculated by comparing the energy with the previous elongation. The increase in potential energy is: \[ \Delta U = 10.5 \, \text{J/m}^3 \] Step 3: Conclusion.
Thus, the increase in energy stored per unit volume is \( 10.5 \, \text{J/m}^3 \), corresponding to option (C).