Mechanical Properties Of Solids Question 5
Question 5 - 2024 (29 Jan Shift 2)
A wire of length $L$ and radius $r$ is clamped at one end. If its other end is pulled by a force $\mathrm{F}$, 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) 3 times
(2) $3 / 2$ times
(3) 4 times
(4) 2 times
Show Answer
Answer: (4)
Solution:
$\mathrm{Y}=\frac{\text { stress }}{\text { strain }}$
$\mathrm{Y}=\frac{\frac{\mathrm{F}}{\frac{\ell}{2}}}{\frac{\ell}{\mathrm{L}}}$
$\mathrm{F}=\mathrm{Y} \pi \mathrm{r}^{2} \times \frac{\ell}{\mathrm{L}}$.
$\mathrm{Y}=\frac{\frac{\mathrm{Fr} / 2}{\frac{\Delta \ell}{2}}}{\mathrm{~L}}$
$\mathrm{F}=\mathrm{Y} \frac{\Delta \ell}{\mathrm{L}} \times 2 \times \frac{\pi \mathrm{r}^{2}}{4}$
From (i)
$\mathrm{Y} \pi \mathrm{r}^{2} \frac{\ell}{\mathrm{L}}=\mathrm{Y} \frac{\Delta \ell}{\mathrm{L}} \frac{\pi \mathrm{r}^{2}}{2}$
$\Delta \ell=2 \ell$
