JEE Physics 2023 Question with Solution

Given: The length of the copper wire is increased by $20\%$, so $L_2 = 1.2L_1$.

Find: The percentage increase in resistance.

The resistance of a wire is given by

$$R = \rho \frac{L}{A}$$

where $L$ is the length and $A$ is the cross-sectional area.

When a wire is stretched, its volume remains constant:

$$L_1 A_1 = L_2 A_2$$

As $L_2 = 1.2L_1$, the area becomes

$$A_2 = \frac{A_1}{1.2}$$

The new resistance is

$$R_2 = \rho \frac{L_2}{A_2} = \rho \frac{1.2L_1}{A_1/1.2} = 1.44R_1$$

Therefore, the percentage increase in resistance is

$$\%\,\text{Increase} = \frac{R_2 - R_1}{R_1} \times 100 = \frac{1.44R_1 - R_1}{R_1} \times 100 = 44\%$$

Therefore, the percentage increase in resistance is $44\%$.

Given: The wire is stretched so that its length becomes $1.2$ times the original length.

Find: The percentage increase in resistance.

If volume remains constant, then resistance varies as

$$R \propto L^2$$

So,

$$\frac{R_2}{R_1} = \left(1.2\right)^2 = 1.44$$

Hence the increase is

$$(1.44 - 1) \times 100 = 44\%$$

Therefore, the correct numerical answer is $44$.