JEE Physics 2024 Question with Solution

Given: Potential difference $V$ is applied across a wire of resistance $R$, and the initial energy dissipation rate is $W$.

Find: The new energy dissipation rate when the wire is cut into two equal halves and the halves are connected in parallel across the same supply.

For the original wire,

$$W = \frac{V^2}{R}$$

Since resistance of a uniform wire is directly proportional to its length, cutting the wire into two equal halves makes the resistance of each half

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

When these two equal resistances are connected in parallel,

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

Hence,

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

Now the new power dissipation is

$$P' = \frac{V^2}{R_{eq}} = \frac{V^2}{R/4} = \frac{4V^2}{R}$$

Using $W = \frac{V^2}{R}$,

$$P' = 4W$$

Therefore, the energy dissipation rate becomes $4W$. Hence, the correct option is D.

Given: Original resistance is $R$ and original power is $W = \frac{V^2}{R}$.

Find: New power after cutting the wire into two equal halves and connecting them in parallel.

Each half has resistance $R/2$. Two equal resistors of resistance $R/2$ in parallel have equivalent resistance equal to half of $R/2$, that is,

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

At constant supply voltage, power is inversely proportional to resistance:

$$P \propto \frac{1}{R}$$

So reducing resistance from $R$ to $R/4$ increases power by a factor of $4$:

$$P' = 4W$$

Therefore, the energy dissipation rate becomes $4W$. Hence, the correct option is D.