JEE Physics 2024 Question with Solution

Given: The original resistance of the wire is $R$. The wire is cut into two equal halves and the two halves are connected in parallel.

Find: The equivalent resistance of the parallel combination.

For a uniform wire, resistance is directly proportional to length. Therefore, when the wire is cut into two equal halves, the resistance of each half becomes

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

Now these two resistances, each equal to $R/2$, are connected in parallel. Hence,

$$\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{1}{R/2} + \frac{1}{R/2} = \frac{2}{R} + \frac{2}{R} = \frac{4}{R}$$

So,

$$R_{eq} = \frac{R}{4}$$

Therefore, the equivalent resistance is $R/4$. The correct option is B.

Given: A wire of resistance $R$ is cut into two equal parts.

Find: The equivalent resistance after connecting the two parts in parallel.

The resistance of a uniform wire is

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

If the wire is cut into two equal halves, the length of each piece becomes $L/2$ while the material and cross-sectional area remain unchanged. So the resistance of each half is

$$R_{half} = \rho \frac{L/2}{A} = \frac{1}{2}\rho \frac{L}{A} = \frac{R}{2}$$

Now two equal resistors of value $R/2$ are connected in parallel:

$$R_{eq} = \frac{\left(\frac{R}{2}\right)\left(\frac{R}{2}\right)}{\left(\frac{R}{2}\right)+\left(\frac{R}{2}\right)}$$$$R_{eq} = \frac{R^2/4}{R} = \frac{R}{4}$$

Therefore, the equivalent resistance of the combination is $R/4$. The correct option is B.

The solution contains an incorrect heading marking option C, but the working clearly gives $R/4$, so B is the correct answer.