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EEJ MAIN Physics QUESTION #7103
Question 1
A metallic wire elongates by $0.04\text{ m}$ under force $F$. If both length and diameter are doubled, find the elongation under the same force $F$ in meters [cite: 482, 484].
  • $0.01\text{ m}$
  • $0.02\text{ m}$✔️
  • $0.04\text{ m}$
  • $0.08\text{ m}$
Correct Answer Explanation

Extension $\Delta L = \frac{FL}{AY} = \frac{FL}{\pi (d/2)^2 Y} \propto \frac{L}{d^2}$ [cite: 482, 484].

  • New extension $\Delta L' \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2} \Delta L$
  • $\Delta L' = \frac{1}{2} \times 0.04 = 0.02\text{ m}$[cite: 484].