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3 Conservation of energy

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Question 82
F

A student calculates the efficiency of a lamp using this working.

efficiency=0.072240=0.00030 \text{efficiency}=\frac{0.072}{240}=0.00030 efficiency=2400.072​=0.00030

The useful light output was measured as 0.072 kJ. The total energy input was measured as 240 J.

Which correction is needed?

[1]
A

Use 240/72240/72240/72; the efficiency is 3.33.33.3.

B

Use 72/24072/24072/240; the efficiency is 0.300.300.30.

C

Use 168/240168/240168/240; the efficiency is 0.700.700.70.

D

Use 0.072/2400.072/2400.072/240; the efficiency is 0.000300.000300.00030.

Markscheme

3 Conservation of energy Questions

  1. GCSE
  2. /Physics
  3. /3 Conservation of energy

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