Question

Two wires A and B are made of same material. The wire A has a length l and diameter r while the wire B has a length 2l and diameter r/2. If the two wires are stretched by the same force, the elongation in A divided by the elongation in B is
(a) 1/8
(b) 1/4
(c) 4
(d) 8

Answer 1

Correct option: (a) 1/8

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{Young}\text{'}\mathrm{s}\mathrm{modulus}\mathrm{of}\mathrm{the}\mathrm{wire}\text{'}\mathrm{s}\mathrm{material}\mathrm{be}\mathrm{Y}. \\ \mathrm{Here}: \\ \mathrm{Force}=\mathrm{F} \\ {\mathrm{A}}_1={\mathrm{\pi r}}^2 \\ {\mathrm{L}}_1=\mathrm{l} \\ {\mathrm{A}}_2=\mathrm{\pi }{\left(\frac{\mathrm{r}}{2}\right)}^2=\frac{{\mathrm{\pi r}}^2}{4} \\ {\mathrm{L}}_2=2\mathrm{l} \\ \mathrm{Let}\mathrm{the}\mathrm{elongation}\mathrm{in}\mathrm{A}\mathrm{be}\mathrm{x}\mathrm{and}\mathrm{that}\mathrm{in}\mathrm{B}\mathrm{be}\mathrm{y}. \\ \mathrm{Since}\mathrm{the}\mathrm{Young}\text{'}\mathrm{s}\mathrm{modulus}\mathrm{for}\mathrm{both}\mathrm{the}\mathrm{wires}\mathrm{is}\mathrm{the}\mathrm{same}: \\ \mathrm{Y}=\frac{{\displaystyle \raisebox{1ex}{$\mathrm{F}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{A}}_1$}\right.}}{{\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$l$}\right.}}=\frac{{\displaystyle \raisebox{1ex}{$\mathrm{F}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{A}}_2$}\right.}}{{\displaystyle \raisebox{1ex}{$\mathrm{y}$}\!\left/ \!\raisebox{-1ex}{$2\mathrm{l}$}\right.}} \\ \Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=\frac{{\mathrm{A}}_2}{2{\mathrm{A}}_1} \\ \Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=\frac{1}{8} \end{gathered}$$