A metal wire length L, cross-sectional area A and Young's modulus Y is stretched by a variable force F. F is varying in such a way that F is always slightly greater than the elastic forces of resistance in the wire. When the elongation in the wire is l, up to this instant
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a
the work done by F is YAl22L
b
the work done by F is YAl2L
c
the elastic potential energy stored in wire is YAl22L
d
no energy is lost during elongation
answer is A.
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Detailed Solution
In this question, the applied force is not constant. Let us consider at any instant the wire elongates by x. Then work done by F to elongate it more by dx isdW=Fdx=YAxLdx[x≪L]W=∫0l YAxdxL=YAl22LAs internal restoring force and applied force are almost same (equal and opposite), numerical value of work done by these two forces would be equal.