Identical springs of steel and copper are equally stretched. On which, more work will have to be done?
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Work done in stretching a wire, W = \(\frac{1}{2}\)Fx∆l.
[F = applied force, ∆l = extension in wire]
Spring are equally stretched, therefore, for same force (F).
W ∝ ∆l (i)
Y(Young’s modulus)
Y = \(\frac{F}{A}\) × \(\frac{l}{∆l}\) or ∆l = \(\frac{F}{A}\) × \(\frac{l}{Y}\)
Both springs are identical
∆l ∝ \(\frac{l}{Y}\) (ii)
From (i) & (ii)
W ∝ \(\frac{l}{W}\) or \(\frac{Y_{steel}}{W_{copper}}\) = \(\frac{Y_{copper}}{Y_{steel}}\) < 1 (∴ Ysteel > Ycopper)
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