A $500 \mathrm{~W}$ heating unit is designed to operate on a $400 \mathrm{~V}$ line. If line voltage drops to $160 \mathrm{~V}$, the percentage drop in heat output will be:

The power $P$ consumed by a heating unit is given by the formula $P = \dfrac{V^2}{R}$, where $V$ is the voltage and $R$ is the resistance of the heating element.

Since the resistance $R$ is constant, the power is proportional to the square of the voltage, $P \propto V^2$.

Let $P_1 = 500 \text{ W}$ at $V_1 = 400 \text{ V}$. When the voltage drops to $V_2 = 160 \text{ V}$, the new power $P_2$ is given by:

$\dfrac{P_2}{P_1} = \left( \dfrac{V_2}{V_1} \right)^2$

Substituting the values:

$\dfrac{P_2}{500} = \left( \dfrac{160}{400} \right)^2 = \left( \dfrac{2}{5} \right)^2 = \dfrac{4}{25}$

$P_2 = 500 \times \dfrac{4}{25} = 20 \times 4 = 80 \text{ W}$

The drop in heat output is $P_1 - P_2 = 500 - 80 = 420 \text{ W}$.

The percentage drop in heat output is $\dfrac{P_1 - P_2}{P_1} \times 100 = \dfrac{420}{500} \times 100 = \dfrac{420}{5} = 84\%$.

Answer: 84%