One end of a horizontal thick copper wire of length $2 L$ and radius $2 R$ is welded to an end of another horizontal thin copper wire of length $L$ and radius $R$. When the arrangement is stretched by a applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is :
A$0.25$
B$0.50$
C$2.00$
D$4.00$
IIT 2013, Medium
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C$2.00$
c $Y=\frac{\left(\frac{F}{A}\right)}{\frac{\Delta \ell_1}{L}} $ ...............$(i)$
$Y=\frac{\left(\frac{F}{4 A }\right)}{\frac{\Delta \ell_2}{2 L}} $ .................$(ii)$
$\frac{\Delta \ell_1}{\Delta \ell_2}=2$
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