A) 16
B) 2
C) 1/4
D) 4
Correct Answer: A
Solution :
\[\Delta l=\frac{FL}{\pi {{r}^{2}}Y},\] But \[V=\pi {{r}^{2}}L\] so \[L=\frac{V}{\pi {{r}^{2}}}\] \[=\frac{FV}{{{(\pi {{r}^{2}})}^{2}}Y}\propto \frac{1}{{{r}^{4}}}\] \[\frac{\Delta {{l}_{A}}}{\Delta {{l}_{B}}}={{\left( \frac{{{r}_{B}}}{{{r}_{A}}} \right)}^{4}}={{\left( \frac{{{r}_{B}}}{\frac{{{r}_{B}}}{2}} \right)}^{4}}=16\]You need to login to perform this action.
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