UPSC Combined Geo-Scientist, Mains · 2024 · Mains Paper III
Q1(a). Percentage relative error of a balance reading
Question
A weighing balance showed the weight of a substance 9.5 g, which is 0.5 g lower than its actual weight. Calculate % relative error.
Answer in brief
True mass = 10.0 g; relative error = (9.5 − 10.0)/10.0 × 100 = −5.0 %, so the balance reads 5 % low.
Key points
- True (actual) mass: xt = 9.5 + 0.5 = 10.0 g.
- Absolute error = measured value − true value, with its sign: E = 9.5 − 10.0 = −0.5 g.
- Relative error = absolute error as a percentage of the true value: Er = −0.5/10.0 × 100 = −5.0 % (−50 parts per thousand).
- The negative sign is kept: the result is lower than the true value; relative accuracy = 95 %.
- A balance that reads low is a determinate (systematic) error; it is removed by calibration with standard masses.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.

Coversrelative errorabsolute errordeterminate erroraccuracycalibration
All parts of Question 1
- 1(a) Percentage relative error of a balance reading5 marks
- 1(b) Partition ratio k′ and relative retention α5 marks
- 1(c) Transmittance from Beer’s law5 marks
- 1(d) Identifying a magnesium compound from its thermogram5 marks
- 1(e) Properties needed in a detector for X-ray fluorescence5 marks
- 1(f) Generation of the plasma in the ICP torch5 marks
- 1(g) Calorific value of a fuel5 marks
- 1(h) AAS and ICP compared: accuracy, interferences, detection limit5 marks