A) \[2L\]
B) \[L\]
C) \[L/2\]
D) \[L/4\]
Correct Answer: B
Solution :
As, the Young's modulus, \[Y=\frac{FL}{\pi {{r}^{2}}l}\] \[\therefore \] \[l\propto \frac{FL}{{{r}^{2}}}\] \[\Rightarrow \] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\left( \frac{{{F}_{1}}}{{{F}_{2}}} \right)\left( \frac{{{L}_{1}}}{{{L}_{2}}} \right)\left( \frac{r_{2}^{2}}{r_{1}^{2}} \right)\] \[=\left( \frac{F}{2F} \right)\times \left( \frac{L}{2L} \right)\times \left( \frac{4{{r}^{2}}}{{{r}^{2}}} \right)=1:1\] or \[{{l}_{1}}={{l}_{2}}=L\]You need to login to perform this action.
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