Question

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

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

Answer 1

The Correct Option is 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. $\therefore\quad$ $F$ $\propto$ $\frac{r^{2}}{L}$ $\therefore\quad$ $\frac{F'}{F}$ $=\frac{\left(2r\right)^{2}}{\left(L /2\right)}$ $\times\frac{L}{r^{2}}=8$ or $F'$ $=8 F$ $\quad\ldots\left(i\right)$ Work done in stretching a wire, $W=\frac{1}{2}\times$ $F\times\Delta L$ For same extension $W$ $\propto$ $F$ $\therefore\quad$ $\frac{W'}{W}$ $=\frac{F'}{F}=8$ $\quad\left[Using \left(i\right)\right]$ $W'$ $=8W=8\times2 \,J$ $=16 \,J$