AQA A-Level Physics Paper 1, November 2020: Question 23

1 mark · Medium difficulty · Multiple Choice

Determine the efficiency of an electric motor that lifts a load of weight W through a vertical height h in time t, given a potential difference V and current I.

Practise this question

Question

Multiple choice question 23 asking for the efficiency of an electric motor that lifts a load of weight W through a height h in time t, with potential difference V and current I. Four options are given: A (Wh / VIt), B (VI / Wht), C (Wht / VI), and D (VIt / Wh).
Question text

23 An electric motor lifts a load of weight W through a vertical height h in time t.

The potential difference across the motor is V and the current in it is I.

What is the efficiency of the motor?

[1 mark]

Wh

A

VIt

VI

B

Wht

Wht

C

VI

VIt

D

Wh

Mark scheme

Show the mark scheme Mark scheme indicating that the correct answer for question 23 is A.

23 A

How to answer it

Calculating Electric Motor Efficiency

What this question tests

This question assesses your ability to combine definitions of electrical power, mechanical work done (gravitational potential energy), and the general principle of efficiency. You must correctly identify input and output energies/powers within a dynamic system.

Question 2.3 [1 Mark]

Multiple Choice Solution & Breakdown

✅ Correct Answer: A

The correct option is A ( Wh / VIt ).

Mark Awarded: 1 / 1 for selecting option A.

💡 Key Knowledge

  • Efficiency definition: Efficiency = (Useful Output Energy / Total Input Energy) × 100% (or expressed as a ratio).
  • Useful Output Energy: Raising a load of weight W through a vertical height h gives gravitational potential energy, E_p = Wh (since weight W = mg ).
  • Total Input Energy: Electrical energy supplied to the motor is Power × time = VIt .

🧠 Exam Technique

For ratio-based multiple-choice questions involving efficiency, always write down the general formula first using words or standard symbols before looking at the options:

Efficiency = Useful Output / Total Input

Substitute the specific variables given in the stem ( Wh for output, VIt for input) to immediately match with the correct distractor.

❌ Common Errors

  • Inverting the ratio: Choosing option D ( VIt / Wh ) by accidentally putting input over output—a common slip under time pressure.
  • Misidentifying time dependence: Confusing energy with power and incorrectly including or omitting time t from either the numerator or denominator. Remember that electrical power is VI , so energy requires multiplying by time t .

Topics

Physics · 3.4 Mechanics and materials · 3.5 Electricity

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.