A wire of length $L$ and radius $r$ is clamped at one end. If its other end is pulled by a force $F$, its length increases by $l$. If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant, the increase in length will become:
Solution
<p>$$\begin{aligned}
& Y=\frac{\text { stress }}{\text { strain }} \\
& Y=\frac{\frac{\mathrm{F}}{\frac{\pi \mathrm{r}^2}{\ell}}}{\frac{\ell}{\mathrm{L}}}
\end{aligned}$$</p>
<p>$$\mathrm{F}=\mathrm{Y} \pi \mathrm{r}^2 \times \frac{\ell}{\mathrm{L}} \quad \text{.... (i)}$$</p>
<p>$$\begin{aligned}
& \mathrm{Y}=\frac{\frac{\mathrm{F} / 2}{\pi \mathrm{r}^2 / 4}}{\frac{\Delta \ell}{\mathrm{L}}} \\
& \mathrm{F}=\mathrm{Y} \frac{\Delta \ell}{\mathrm{L}} \times 2 \times \frac{\pi \mathrm{r}^2}{4}
\end{aligned}$$</p>
<p>From (i)</p>
<p>$$\begin{aligned}
& \mathrm{Y} \pi \mathrm{r}^2 \frac{\ell}{\mathrm{L}}=\mathrm{Y} \frac{\Delta \ell}{\mathrm{L}} \frac{\pi \mathrm{r}^2}{2} \\
& \Delta \ell=2 \ell
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity
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