Question:
When a wire is stretched to double its length:
When a wire is stretched to double its length:
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For stretching problems:
- Volume remains constant.
- Use the relation \( r' = \frac{r}{\sqrt{\text{stretch ratio}}} \).
Updated On: Mar 26, 2025
- Its radius is halved
- Longitudinal strain is unity
- Stress is equal to two times Young’s modulus
- Young’s modulus is equal to thrice the energy per unit volume
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The Correct Option is A
Solution and Explanation
When a wire is stretched, its volume remains constant. Let the initial length be \( L \) and radius \( r \), and after stretching, the new length is \( 2L \).
Since volume is conserved:
\[ \pi r^2 L = \pi r'^2 (2L) \] \[ r'^2 = \frac{r^2}{2} \] \[ r' = \frac{r}{\sqrt{2}} \] Since \( \sqrt{2} \approx 1.414 \), the radius is slightly more than half but approximately close.
Since volume is conserved:
\[ \pi r^2 L = \pi r'^2 (2L) \] \[ r'^2 = \frac{r^2}{2} \] \[ r' = \frac{r}{\sqrt{2}} \] Since \( \sqrt{2} \approx 1.414 \), the radius is slightly more than half but approximately close.
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