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(C) The power dissipated in a resistor is given by $P = \frac{V^2}{R}$.Initially,for a wire of resistance $R$ with potential $V_0$ applied,the power d

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$16$

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(C) The power dissipated in a resistor is given by $P = \frac{V^2}{R}$.Initially,for a wire of resistance $R$ with potential $V_0$ applied,the power dissipation is $P_1 = \frac{V_0^2}{R}$.When the wire is cut into two equal halves,the resistance of each half becomes $R' = \frac{R}{2}$.Now,a potential $V_0$ is applied across each half. The power dissipated in one half is $P' = \frac{V_0^2}{R'} = \frac{V_0^2}{R/2} = \frac{2V_0^2}{R}$.The total power dissipation across both wires is $P_2 = P' + P' = 2 	imes \frac{2V_0^2}{R} = \frac{4V_0^2}{R}$.Calculating the ratio $P_2 : P_1 = \frac{4V_0^2/R}{V_0^2/R} = 4$.Given the ratio is $\sqrt{x} : 1$,we have $\sqrt{x} = 4$.Squaring both sides,$x = 16$.

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