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Two identical springs of copper and stee...

Two identical springs of copper and steel are equally strectched. On which more work will have to be done?

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Work done in stretching a wire is given by `W = (1)/(2)FxxDeltal`
[Where F is applied force and `Delta l` is extension in the wire]
As springs of steel and copper are equally stretched. Therefore, for same force (F),
`"W"prop Deltal " …..(i)"`
Young's modulus (Y) `=(F)/(A) xx (l)/(Deltal)implies Deltal = (F)/(A) xx (l)/(Y)`
As both springs are identical, `" "Delta l prop (1)/(Y) " ......(ii)"`
From Eqs. (i) and (ii), we get `" "W prop (1)/(Y)`
`therefore" "(W_("steel"))/(W_("copper"))=(Y_("copper"))/(Y_("steel")) lt 1" "("As, " Y_("steel") gt Y_("copper"))`
`implies " "W_("steel")lt W_("copper")`
Therefore, more work will be done for stretching copper spring.
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