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. $\frac{\text{l}}{4}$
• B. $\text{2l}$
• C. $\text{l}$
• D. $\frac{\text{l}}{2}$

Answer 1

The correct option is C

$\text{l}$

The explanation for the correct answer:

Step 1: Given data

Stretching 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: Calculation

The Young's modulus of a material is calculated as follows:

Young's modulus is unchanged in the second scenario.

Hence,

$$\begin{gathered} \mathrm{Y}=\frac{{\mathrm{F}}^{\mathrm{\prime }}{\mathrm{L}}^{\mathrm{\prime }}}{{\mathrm{A}}^{\mathrm{\prime }}{\mathrm{\Delta L}}^{\mathrm{\prime }}} \\ =\frac{2\mathrm{F}\times 2\mathrm{L}}{\pi (2\mathrm{r}{)}^2\times {\mathrm{\Delta L}}^{\mathrm{\prime }}} \\ =\frac{\mathrm{FL}}{\pi {\mathrm{r}}^2{\mathrm{\Delta L}}^{\mathrm{\prime }}} \end{gathered}$$

From the above formula $\mathrm{\Delta L}$$=\text{l}$

Therefore, a correct answer is an option($\mathrm{C}$).