An electric motor lifts a load of weight $W$ through a vertical height $h$ in time $t$ - AQA - A-Level Physics - Question 23 - 2020 - Paper 1

The efficiency ( \eta) of the motor can be defined as the ratio of useful output power to the total input power.

1. Calculate the useful output power: The useful output power ( \text{Power}_{\text{out}}) is the gravitational potential energy gained by the load per unit time:

   $\text{Power}_{\text{out}} = \frac{W \cdot h}{t}$

2. Calculate the total input power: The electrical power supplied to the motor can be calculated as:

   $\text{Power}_{\text{in}} = V \cdot I$

3. Calculate the efficiency:

   $\eta = \frac{\text{Power}_{\text{out}}}{\text{Power}_{\text{in}}} = \frac{\frac{W \cdot h}{t}}{V \cdot I} = \frac{W \cdot h}{V \cdot I \cdot t}$

   From this relationship, we can see that the efficiency of the motor is given by:

   $\eta = \frac{Wh}{VI}$

   Thus, the answer is Option C: $\frac{Wh}{VI}$.