Question

# A wire of length $\text{L}$and radius r is fixed at one end. When a stretching force $\text{F}$ is applied at free end, the elongation in the wire is $\text{l}$. When another wire of same material but of length $\text{2L}$ and radius $\text{2r}$ is also fixed at one end is stretched by a force $\text{2F}$ applied at free end, the elongation in the second wire will be?

A

$\frac{\text{l}}{4}$

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B

$\text{2l}$

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C

$\text{l}$

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D

$\frac{\text{l}}{2}$

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Solution

## The correct option is C $\text{l}$The explanation for the correct answer:Step 1: Given dataStretching force $\text{F}$Length $\text{2L}$Radius $\text{2r}$Step 2: Concept used$\begin{array}{r}\mathrm{Y}=\frac{\mathrm{FL}}{\mathrm{A\Delta L}}\\ \end{array}$Where,$\text{F}$ is the applied force($\mathrm{N}$)$\text{L}$ is the initial length($\mathrm{m}$)$\mathrm{A}$ is the area(${\mathrm{m}}^{2}$)$\text{Y}$ is young's modulus($\mathrm{Pa}$) Step 3: CalculationThe Young's modulus of a material is calculated as follows:Young's modulus is unchanged in the second scenario. Hence,$\mathrm{Y}=\frac{{\mathrm{F}}^{\mathrm{\prime }}{\mathrm{L}}^{\mathrm{\prime }}}{{\mathrm{A}}^{\mathrm{\prime }}{\mathrm{\Delta L}}^{\mathrm{\prime }}}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{F}×2\mathrm{L}}{\pi \left(2\mathrm{r}{\right)}^{2}×{\mathrm{\Delta L}}^{\mathrm{\prime }}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{FL}}{\pi {\mathrm{r}}^{2}{\mathrm{\Delta L}}^{\mathrm{\prime }}}$From the above formula $\mathrm{\Delta L}$$=\text{l}$Therefore, a correct answer is an option($\mathrm{C}$).

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