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(B) The power $P$ of a heater is given by $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.Since $V$ is constant,$P \propto \fra

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Increase by about $11\%$

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(B) The power $P$ of a heater is given by $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.Since $V$ is constant,$P \propto \frac{1}{R}$.The resistance $R$ of a wire is given by $R = ho \frac{l}{A}$,so $R \propto l$.Therefore,$P \propto \frac{1}{l}$.Let the initial length be $l_1 = 100$ and the final length be $l_2 = 100 - 10 = 90$.Using the proportionality $P_1 l_1 = P_2 l_2$,we get:$P_2 = P_1 	imes \frac{l_1}{l_2} = P_1 	imes \frac{100}{90} = P_1 	imes 1.111$.The percentage change in power is $\frac{P_2 - P_1}{P_1} 	imes 100 = (1.111 - 1) 	imes 100 = 11.1\%$.Thus,the power increases by about $11\%$.

If the length of the filament of a heater is reduced by $10\%$,the power of the heater will

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