A $5\, m$ long aluminium wire ($Y = 7 \times {10^{10}}N/{m^2})$ of diameter $3\, mm$ supports a $40\, kg$ mass. In order to have the same elongation in a copper wire $(Y = 12 \times {10^{10}}N/{m^2})$ of the same length under the same weight, the diameter should now be, in $mm.$
A$1.75$
B$1.5$
C$2.5$
D$5$
Diffcult
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C$2.5$
c (c) $l = \frac{{FL}}{{\pi {r^2}Y}} \Rightarrow {r^2} \propto \frac{1}{Y}$ $(F,L$ and $l$ are constant$)$
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