A  500 W heating unit is designed to operate from a 115 volt line. If the line voltage drops to 110  volt. The percentage drop in heat output will be

$$\begin{gathered} we know P=\frac{V^2}{R} \\ \begin{array}{l}\therefore \frac{P^1}{P}=\frac{V^{1^2}}{V^2}\\ \Rightarrow \frac{P^1}{P}={\left(\frac{110}{115}\right)}^2={\left(1-\frac{5}{115}\right)}^2\\ u\sin g binomial approx\\ \Rightarrow \frac{P^1}{P} = 1-\frac{10}{115}=1-\frac{2}{23}\\ \Rightarrow \frac{P^1}{P} -1= -\frac{2}{23}\\ multiplying bothsides with 100 we get\\ \Rightarrow \left(\frac{P^1}{P}-1\right)\times 100=-\frac{200}{23}=-8.6\%\\ \end{array} \end{gathered}$$