JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 9)
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:
2 times
4 times
3 times
$$\frac{3}{2}$$ times
Explanation
$$\begin{aligned} & Y=\frac{\text { stress }}{\text { strain }} \\ & Y=\frac{\frac{\mathrm{F}}{\frac{\pi \mathrm{r}^2}{\ell}}}{\frac{\ell}{\mathrm{L}}} \end{aligned}$$
$$\mathrm{F}=\mathrm{Y} \pi \mathrm{r}^2 \times \frac{\ell}{\mathrm{L}} \quad \text{.... (i)}$$
$$\begin{aligned} & \mathrm{Y}=\frac{\frac{\mathrm{F} / 2}{\pi \mathrm{r}^2 / 4}}{\frac{\Delta \ell}{\mathrm{L}}} \\ & \mathrm{F}=\mathrm{Y} \frac{\Delta \ell}{\mathrm{L}} \times 2 \times \frac{\pi \mathrm{r}^2}{4} \end{aligned}$$
From (i)
$$\begin{aligned} & \mathrm{Y} \pi \mathrm{r}^2 \frac{\ell}{\mathrm{L}}=\mathrm{Y} \frac{\Delta \ell}{\mathrm{L}} \frac{\pi \mathrm{r}^2}{2} \\ & \Delta \ell=2 \ell \end{aligned}$$
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