Question

If the work done in stretching a wire by $1mm$ is $2J$, the work necessary for stretching another wire of same material but with double radius of cross-section and half the length by $1mm$ is

• A. $16J$
• B. $8J$
• C. $4J$
• D. $\frac{1}{4}$ $J$

Answer 1

Answer: (A)

Stretching force, $F=$ $\frac{Y\pi r^2\Delta L}{L}$
where the symbols have their usual meanings.
Both the wires are of same material, so $Y$ will be equal,
extension in both the wires is same, so $\Delta L$ will be equal.
$\therefore$ $F$ $\propto$ $\frac{r^2}{L}$
$\therefore$ $\frac{F^{\prime }}{F}$ $=\frac{{\left(2r\right)}^2}{\left(L/2\right)}$ $\times \frac{L}{r^2}=8$
or $F^{\prime }$ $=8F$ $\dots \left(i\right)$
Work done in stretching a wire,
$W=\frac{1}{2}\times$ $F\times \Delta L$
For same extension
$W$ $\propto$ $F$
$\therefore$ $\frac{W^{\prime }}{W}$ $=\frac{F^{\prime }}{F}=8$ $\left[Using\left(i\right)\right]$
$W^{\prime }$ $=8W=8\times 2J$ $=16J$