JEE Physics 2023 Question with Solution

Given: The initial resistance of the wire is $R_i = 5 \, \Omega$ and the wire is stretched to $5$ times its original length.

Find: The new resistance after stretching.

The solution states that the volume of wire remains constant during stretching.

So,

Since the new length is $l_f = 5l$, we get

Using resistance relation,

For the stretched wire,

Therefore, the new resistance is $125 \, \Omega$. The correct option is A.

Given: The wire length becomes $5l$ while material remains the same, so $\rho$ is unchanged.

Find: How the resistance changes.

Since volume is constant,

Resistance depends directly on length and inversely on area:

After stretching, length becomes $5$ times and area becomes $\frac{1}{5}$ times. Hence,

So the resistance becomes $25$ times the original value.

Therefore, the correct option is A. Note that the raw options list labels $125 \, \text{ohm}$ as option C, but the solution explicitly concludes option A and computes $125 \, \Omega$.