Calculate the work done, if a wire is loaded by $'Mg'$ weight and the increase in length is $'l'$
A$Mgl$
B$Zero$
C$Mgl/2$
D$2Mgl$
Medium
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C$Mgl/2$
c (c) Work done $=$ $\frac{1}{2}Fl = \frac{{Mgl}}{2}$
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