AQA AS Level Physics Paper 2, June 2022: Question 28

1 mark · Medium difficulty · Multiple Choice

Calculate the mass of water pumped to a height of 0.60 m each second given a solar energy incident rate of 1.5 W and an overall efficiency of 20%.

Practise this question

Question

A diagram shows a solar panel powering a water pump that lifts water through a vertical height of 0.60 m from a bottom container to a top container. Below the diagram, text states that solar energy is incident on the solar panel at a rate of 1.5 W, the overall efficiency is 20%, and asks what mass of water can be pumped into the top container each second, with four multiple-choice options A (5 g), B (50 g), C (100 g), and D (250 g).
Question text

28 A solar panel powers a pump for a water feature.

Solar energy is incident on the solar panel at a rate of 1.5 W.

Water from the bottom container is continually pumped through a vertical height

of 0.60 m to the top container.

The overall efficiency of the solar panel and the pump is 20%.

What mass of water can be pumped into the top container each second?

[1 mark]

A 5 g

B 50 g

C 100 g

D 250 g

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 28 is B (AO2), corresponding to 50 g.

28 B (AO2) 50 g

How to answer it

Solar-Powered Water Pump Efficiency

What this question tests

This question assesses your ability to combine the concepts of power, efficiency, and gravitational potential energy (GPE) in a practical, real-world context. You must relate input energy/power rates to useful output work per second, handle unit conversions for mass (grams to kilograms), and correctly apply standard physics equations under timed multiple-choice conditions.

Question 28 [1 mark]

Exam Breakdown & Step-by-Step Solution

✅ Correct Answer

B (50 g)

Marks awarded: 1 mark for selecting B.

💡 Key Knowledge

  • Power: Defined as energy transferred per second ( P = E / t ). Therefore, energy per second is numerically equal to power in Watts ( J s⁻¹ = W ).
  • Efficiency: Efficiency = (Useful output power / Total input power) × 100%
  • GPE: ΔEp = m g h

🧠 Exam Technique

In multiple-choice calculations, do not rush the arithmetic. Jot down each formula explicitly before substituting values to prevent substitution errors, and keep track of units (especially grams versus kilograms).

❌ Common Errors

  • Forgetting efficiency: Multiplying or dividing by the 20% factor in the wrong direction, or omitting it entirely.
  • Unit blindness: Finding the mass in kg ( 0.05 kg ) but failing to convert it into grams ( 50 g ), leading to option A ( 5 g ) or incorrect dimensional matching.

📐 Step-by-Step Calculation

  1. Find the useful output power:
    Input power = 1.5 W
    Efficiency = 20% = 0.20
    Useful Power = 1.5 W × 0.20 = 0.30 W (meaning 0.30 J of useful energy transferred per second).
  2. Relate power to gravitational potential energy per second:
    Useful power P = ΔEp / t = (m g h) / t
    Since we want the mass pumped each second, m / t is our unknown.
  3. Rearrange for mass per second ( m ):
    0.30 = (m × 9.81 × 0.60) / 1 (using g = 9.81 m s⁻² or 9.8 m s⁻² )
    m = 0.30 / (9.81 × 0.60) = 0.30 / 5.886 ≈ 0.0509 kg s⁻¹
    *(Note: If using g = 9.8 m s⁻² , 0.30 / (9.8 × 0.6) = 0.30 / 5.88 ≈ 0.051 kg s⁻¹ )*
  4. Convert kilograms to grams:
    0.050 kg × 1000 = 50 g .
    This matches option B.

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.