JEE Physics 2024 Question with Solution

Given: A wire of length $L$ and radius $r$ is stretched by force $F$ and its extension is $l$.

Find: The new extension when force becomes $\frac{F}{2}$ and radius becomes $\frac{r}{2}$, with length unchanged.

Using Young's modulus relation, the elongation of the wire is

$$l = \frac{FL}{AY}$$

where cross-sectional area of the wire is

$$A = \pi r^2$$

So initially,

$$l = \frac{FL}{\pi r^2 Y}$$

Now the new radius is

$$r' = \frac{r}{2}$$

and the new force is

$$F' = \frac{F}{2}$$

Hence the new area is

$$A' = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4}$$

Therefore, the new elongation is

$$l' = \frac{F'L}{A'Y} = \frac{\frac{F}{2} \cdot L}{\frac{\pi r^2}{4} \cdot Y}$$ $$l' = \frac{FL \cdot 4}{2\pi r^2 Y} = 2\frac{FL}{\pi r^2 Y} = 2l$$

Therefore, the increase in length becomes $2$ times the original increase in length. The correct option is D.

Given: Extension of a wire depends on applied force and cross-sectional area.

Find: How the extension changes when $F \to \frac{F}{2}$ and $r \to \frac{r}{2}$.

Since

$$l \propto \frac{F}{r^2}$$

for constant $L$ and $Y$, we compare directly:

$$\frac{l'}{l} = \frac{\frac{F}{2}}{\left(\frac{r}{2}\right)^2} \cdot \frac{r^2}{F}$$ $$\frac{l'}{l} = \frac{\frac{F}{2}}{\frac{r^2}{4}} \cdot \frac{r^2}{F} = 2$$

So,

$$l' = 2l$$

Therefore, the increase in length becomes $2$ times the original value. The correct option is D.