Answer:
\[Y=\,\frac{FL}{\pi
{{r}^{2}}\Delta L}\] \[\therefore \] \[\Delta L=\,\frac{FL}{\pi
{{r}^{2}}Y}\]
or \[\Delta
L=\,\frac{800\,\times \,\,9.1}{3.14\,\times \,25\,\,\times \,{{10}^{-6}}\times
\,2\times \,{{10}^{11}}}\]
\[=4.64\times
{{10}^{4}}m\]
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