Question

2 Wires of the same material have their lengths in the ratio 1:3 and diameters in the ratio 2:1 . If they are stretched by the same force, What is their ratio of elongations?

Answer 1

Dear Student,

$$\begin{gathered} \frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}=\frac{1}{3} \\ \frac{{\mathrm{d}}_1}{{\mathrm{d}}_2}=\frac{2}{1} \\ \mathrm{Also},\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{2}{1}....\left[\mathrm{because}\mathrm{ratio}\mathrm{of}\mathrm{diameter}\mathrm{and}\mathrm{radius}\mathrm{are}\mathrm{same}\right] \\ \mathrm{We}\mathrm{know}\mathrm{that}, \\ \mathrm{Y}=\frac{\mathrm{Fl}}{{\mathrm{\pi r}}^2∆\mathrm{l}} \\ \mathrm{or},∆\mathrm{l}\mathrm{\alpha }\frac{\mathrm{l}}{{\mathrm{r}}^2} \\ \mathrm{for}\mathrm{same}\mathrm{force}\mathrm{and}\mathrm{young}\text{'}\mathrm{s}\mathrm{modulus} \\ ∆{\mathrm{l}}_1=\frac{{\mathrm{l}}_1}{{\left({\mathrm{r}}_1\right)}^2} \\ ∆{\mathrm{l}}_2=\frac{{\mathrm{l}}_2}{{\left({\mathrm{r}}_2\right)}^2} \\ \frac{∆{\mathrm{l}}_1}{∆{\mathrm{l}}_2}=\frac{\frac{{\mathrm{l}}_1}{{\left({\mathrm{r}}_1\right)}^2}}{\frac{{\mathrm{l}}_2}{{\left({\mathrm{r}}_2\right)}^2}} \\ \frac{∆{\mathrm{l}}_1}{∆{\mathrm{l}}_2}=\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}\times {\left(\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}\right)}^2 \\ \frac{∆{\mathrm{l}}_1}{∆{\mathrm{l}}_2}=\frac{1}{3}\times {\left(\frac{1}{2}\right)}^2 \\ \frac{∆{\mathrm{l}}_1}{∆{\mathrm{l}}_2}=\frac{1}{3}\times \frac{1}{4} \\ \frac{∆{\mathrm{l}}_1}{∆{\mathrm{l}}_2}=\frac{1}{12} \end{gathered}$$