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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?
$9F$
$6F$
$F$
$4F$
Solution
$Young's\,modulus,\,Y = \frac{{Fl}}{{A\Delta l}}$
Since initial volume of wires are same
$\therefore $ Their areas of cross sections are $A$ and $3A$ and lengths are $3l$ and $l$ respectively.
For Wire $1,$
$\Delta l = \left( {\frac{F}{{AY}}} \right)3l\,$ $…(i)$
For wire $2$, let $F'$ force is applied
$\frac{{F'}}{{3A}} = Y\frac{{\Delta l}}{l}$
$ \Rightarrow \Delta l = \left( {\frac{{F'}}{{3AY}}} \right)l$
From eqns $(i)$ and $(ii),$
$\left( {\frac{F}{{AY}}} \right)3l = \left( {\frac{{F'}}{{3AY}}} \right)l \Rightarrow F' = 9F$
