Question

A wire is stretched by 0.01 m by a certain force F. Another wire of same material whose diameter and length are double to the original wire is stretched by the same force. Then its elongation will be

• A. 0.005 m
• B. 0.01 m
• C. 0.02 m
• D. 0.002 m

Answer 1

Answer: (A)

0.005 m

Answer 2

$l = \frac{FL}{\pi r^{2}Y}$ ∴ $l \propto \frac{L}{r^{2}}$ [As F and Y are constants]

$\frac{l_{2}}{l_{1}} = \left( \frac{L_{2}}{L_{1}} \right)\left( \frac{r_{1}}{r_{2}} \right)^{2} = (2) \times \left( \frac{1}{2} \right)^{2} = \frac{1}{2}$⇒ $l_{2} = \frac{l_{1}}{2} = \frac{0.01}{2} = 0.005m$.