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)

$16J$

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 $F^{\prime }=8F...\left(i\right)$
Work done is stretching a wire,
$W=\frac{1}{2}\times F\times \Delta L$
For sam extension
$W\propto F$
$\therefore \frac{W^{\prime }}{W}=\frac{F^{\prime }}{F}=8$ [Using (i)]
$W^{\prime }=8W=8\times 2J=16J$