JEE Physics 2023 Question with Solution

Given: For the first wire, length is $L$, radius is $r$, applied force is $f$, and extension is $l$. For the second wire, length is $2L$, radius is $2r$, and applied force is $2f$.

Find: The increase in length of the second wire.

Use the extension formula:

$$\Delta L = \frac{FL}{YA}$$

where $Y$ is Young's modulus and $A = \pi r^2$.

For the first wire:

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

For the second wire:

$$\Delta L_2 = \frac{(2f)(2L)}{Y\pi (2r)^2}$$

Simplifying,

$$\Delta L_2 = \frac{4fL}{Y\pi \cdot 4r^2} = \frac{fL}{Y\pi r^2}$$

Comparing with the first wire,

$$\Delta L_2 = l$$

Therefore, the increase in length of the second wire is $l$. The correct option is A.

Given: Extension of a wire varies as force and length, and inversely as cross-sectional area.

Find: The extension of the second wire relative to the first.

Since

$$\Delta L \propto \frac{FL}{r^2}$$

for the same material, compare the two wires directly:

• force becomes $2f$
• length becomes $2L$
• radius becomes $2r$, so area factor becomes $4r^2$

Thus the net factor is

$$\frac{2 \times 2}{4} = 1$$

So the extension remains unchanged.

Therefore, the increase in length is $l$. The correct option is A.