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To solve this problem, let's analyze the efficiency of the electric motor in terms of the input electrical energy and the output mechanical energy.

1. **Input Power (Electrical Power)**:  
   The electric motor has a potential difference \( V \) and a current \( I \). Therefore, the electrical power input to the motor is given by:
   \[
   \text{Electrical Power} = VI
   \]
   Over a time \( t \), the total electrical energy supplied to the motor is:
   \[
   \text{Electrical Energy} = VI \times t
   \]

2. **Output Power (Useful Mechanical Power)**:  
   The motor lifts a mass \( m \) to a height \( h \) at a steady speed. The useful work done against gravity to lift the mass is:
   \[
   \text{Mechanical Work} = mgh
   \]

3. **Efficiency Calculation**:  
   The efficiency \( \eta \) of the motor is the ratio of the useful output energy to the input energy:
   \[
   \eta = \frac{\text{Mechanical Work}}{\text{Electrical Energy}}
   \]
   Substituting the values, we get:
   \[
   \eta = \frac{mgh}{VI \times t}
   \]

Now we can check the answer choices to find the correct expression:

- **Option D**: \(\frac{mgh}{VIt}\)

This matches our derived formula for the efficiency, so the correct answer is:
\[
\boxed{\text{D}}
\]

Would you like further clarification on any part of this solution, or do you have other questions?

Here are five related questions you might find interesting:
1. How would the efficiency change if the motor had to lift the mass at a faster speed?
2. What happens to the efficiency if the current \( I \) increases but the voltage \( V \) stays the same?
3. How can we calculate the total energy lost due to inefficiency in this scenario?
4. What other factors could affect the efficiency of an electric motor beyond electrical and mechanical energy?
5. How does the efficiency formula apply if the motor lifts an object vertically and then holds it in place?

**Tip:** Always remember that efficiency is a ratio, often expressed as a percentage, and represents how well energy input is converted to useful work.

An electric motor with potential difference V and current I lifts a mass m through a height h in time t at a steady speed v. The efficiency of the motor is given by?

Find the efficiency of an electric motor lifting a mass m through a height h in time t. This problem covers electrical power, mechanical work, and efficiency calculations. Suitable for students in Grades 11-12.

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