Question

11.The elongation of a steel wire is stretched by a force e.If a wire of the same material of double the length and half the diameter is subjected to double the force, its elongation will be

Answer 1

Hi,

$$\begin{gathered} \mathrm{Young}\mathrm{modulus}\mathrm{of}\mathrm{steel}\mathrm{wire}\mathrm{is}\mathrm{given}\mathrm{by} \\ Y=\frac{F}{A}\times \frac{l}{∆l} \\ \Rightarrow Y=\frac{4\times F}{\mathrm{\pi }{d}^2}\times \frac{l}{∆l}.....\left(1\right) \\ \mathrm{If}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{wire}\mathrm{is}\mathrm{doubled}\mathrm{and}\mathrm{diameter}\mathrm{of}\mathrm{wire}\mathrm{is}\mathrm{halfed},\mathrm{Then} \\ \mathrm{Y}=\frac{16\times \mathit{2}F}{{\mathrm{\pi d}}^2}\times \frac{2l}{∆l\mathit{\text{'}}}.....\left(2\right) \\ \mathrm{From}\mathrm{equation}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{have} \\ \frac{4F}{{\mathrm{\pi d}}^2}\times \frac{l}{∆l}=\frac{32F}{\pi {\mathrm{d}}^2}\times \frac{2l}{∆l\mathit{\text{'}}} \\ \Rightarrow \frac{1}{∆\mathrm{l}}=\frac{16}{∆\mathrm{l}\text{'}} \\ \Rightarrow ∆l\mathit{\text{'}}=16∆l \\ \mathrm{Hence},\mathrm{elongation}\mathrm{will}\mathrm{increase}\mathrm{by}16\mathrm{times}. \end{gathered}$$