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(A) The power $P$ dissipated by a heater coil connected to a constant voltage $V$ is given by $P = \frac{V^2}{R}$,where $R$ is the resistance of the c

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$2:1$

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(A) The power $P$ dissipated by a heater coil connected to a constant voltage $V$ is given by $P = \frac{V^2}{R}$,where $R$ is the resistance of the coil.Since $P \propto \frac{1}{R}$ and the resistance $R$ is directly proportional to the length $l$ of the wire $(R = \rho \frac{l}{A})$,we have $R \propto l$.Therefore,$P \propto \frac{1}{l}$.Let the original length be $l$ and the original resistance be $R$. The new length is $l' = \frac{l}{2}$,so the new resistance is $R' = \frac{R}{2}$.The power produced by the original coil is $P = \frac{V^2}{R}$.The power produced by the half coil is $P' = \frac{V^2}{R'} = \frac{V^2}{R/2} = 2 \frac{V^2}{R} = 2P$.Thus,the ratio of the heat produced by the half coil to that by the original coil is $P' : P = 2:1$.

$A$ heater coil is cut into two parts of equal length and one of them is used in the heater. The ratio of the heat produced by this half coil to that by the original coil is

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