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(D) The initial power dissipated is given by $W = \frac{V^2}{R} \quad ...(i)$When the wire is cut into two equal halves,the resistance of each half be

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$4 W$

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(D) The initial power dissipated is given by $W = \frac{V^2}{R} \quad ...(i)$When the wire is cut into two equal halves,the resistance of each half becomes $R' = \frac{R}{2}$.When these two halves are connected in parallel,the equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R} + \frac{2}{R} = \frac{4}{R}$So,$R_{eq} = \frac{R}{4}$.The new power dissipation $W'$ with the same potential difference $V$ is:$W' = \frac{V^2}{R_{eq}} = \frac{V^2}{R/4} = 4 \left( \frac{V^2}{R} \right)$Substituting equation $(i)$ into this expression,we get:$W' = 4W$.

When a potential difference $V$ is applied across a wire of resistance $R$,it dissipates energy at a rate $W$. If the wire is cut into two halves and these halves are connected in parallel across the same supply,the energy dissipation rate will become:

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