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Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by $\Delta l$ on applying a force F, how much force is needed to stretch the second wire by the same amount?
Explanation
$\Delta l \propto Fl/A$; with $l \propto 1/A$ at fixed volume, $F \propto A^2$.
Detailed Solution
Same volume, so wire 1 has area A and length 3l; wire 2 has area 3A and length l.
For wire 1: $\Delta l = \left(\frac{F}{AY}\right)3l$ ...(i)
For wire 2: $\frac{F'}{3A} = Y\frac{\Delta l}{l} \Rightarrow \Delta l = \left(\frac{F'}{3AY}\right)l$ ...(ii)
From equations (i) and (ii): $\left(\frac{F}{AY}\right)3l = \left(\frac{F'}{3AY}\right)l$
$\Rightarrow F' = 9F$