A) 16 J
B) 8 J
C) 4 J
D) \[\frac{1}{4}J\]
Correct Answer: A
Solution :
: Stretching force,\[F=\frac{Y\pi {{r}^{2}}\Delta l}{l}\]where, Vis Young?s modules. \[\Delta l\]is change in length of the wire and / is original length of the wire. As both wires are of same material, Y will be equal, extension in both the wires is same, so \[\Delta l\] will be equal. \[\therefore \]\[F\propto \frac{{{r}^{2}}}{l}\] \[\therefore \]\[\frac{F'}{F}=\frac{{{(2r)}^{2}}}{(l/2)}\times \frac{l}{{{r}^{2}}}=8\]or\[F'=8F\] ...(i) \[\because \] Work done in stretching a wire,\[W=\frac{1}{2}\times F\times \Delta l\] For same extension \[W\propto F\] \[\therefore \]\[\frac{W'}{W}=\frac{F'}{F}=8\] (Using (i)) \[W'=8W=8\times 2J=16J\]
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