Easy MCQ +4 / -1 PYQ · JEE Mains 2024

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:

  1. A 2 times Correct answer
  2. B 4 times
  3. C 3 times
  4. D $\frac{3}{2}$ times

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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