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(A) The resistance of the original wire is $R = \rho \frac{L}{A} = \rho \frac{L}{\pi (d/2)^2} = \frac{4 \rho L}{\pi d^2}$.The power consumed is $P = \

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$(B, D)$

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(A) The resistance of the original wire is $R = ho \frac{L}{A} = ho \frac{L}{\pi (d/2)^2} = \frac{4 ho L}{\pi d^2}$.The power consumed is $P = \frac{V^2}{R}$. The time taken to heat the water is $t = 4 	ext{ minutes}$.For the new wires,each has length $L$ and diameter $2d$. The resistance of each new wire is $R' = ho \frac{L}{\pi (d)^2} = \frac{ho L}{\pi d^2} = \frac{R}{4}$.If the two new wires are connected in series,the equivalent resistance is $R_s = R' + R' = 2R' = \frac{R}{2}$.The power in series is $P_s = \frac{V^2}{R_s} = \frac{V^2}{R/2} = 2P$. Since $P \propto 1/t$,the time taken is $t_s = \frac{t}{2} = \frac{4}{2} = 2 	ext{ minutes}$.If the two new wires are connected in parallel,the equivalent resistance is $R_p = \frac{R' 	imes R'}{R' + R'} = \frac{R'}{2} = \frac{R/4}{2} = \frac{R}{8}$.The power in parallel is $P_p = \frac{V^2}{R_p} = \frac{V^2}{R/8} = 8P$. The time taken is $t_p = \frac{t}{8} = \frac{4}{8} = 0.5 	ext{ minutes}$.Thus,$(B)$ and $(D)$ are correct.

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