Answer:
Here,\[\therefore \]
\[\frac{{{D}_{cu}}}{{{D}_{iron}}}=\]
As, \[\sqrt{\frac{{{Y}_{iron}}}{{{Y}_{CU}}}}\]
\[=\sqrt{\frac{190\times
{{10}^{9}}}{110\times {{10}^{9}}}}=\sqrt{\frac{19}{11}}=1\cdot 31\]\[0\cdot
065c{{m}^{2}}.\]
\[{{Y}_{steel}}=2\times
{{10}^{11}}N{{m}^{-2}}.\]
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