KCSE 2025 Physics P1 Q1 — Micrometer Zero Error (Actual Diameter)

KCSE 2025 Form 4 Measurement

Published

The Question

“A micrometer screw gauge has a zero error of +0.02 mm. When used to measure the diameter of a steel ball it reads 0.31 mm. Determine the actual diameter of the steel ball.”

1

Know how a positive zero error behaves

A zero error is what the instrument shows when the jaws are fully closed and it should read zero. A positive zero error of +0.02 mm means the gauge always reads 0.02 mm too high, so the true value is found by subtracting the error from the reading.

actual=reading−zero error\text{actual} = \text{reading} - \text{zero error}
2

Substitute the values

Put the recorded reading of 0.31 mm and the zero error of +0.02 mm into the correction. Both are already in millimetres, so no conversion is needed.

actual=0.31−0.02 mm\text{actual} = 0.31 - 0.02 \ \text{mm}
3

Do the subtraction

Take 0.02 mm away from 0.31 mm to obtain the corrected diameter of the steel ball.

0.29 mm0.29\ \mathrm{mm}

Final Result

The actual diameter of the steel ball is 0.29 mm.

Why this method works

Every measuring instrument can carry a systematic offset: the micrometer here does not truly start at zero, it starts at +0.02 mm. Because that same 0.02 mm is added to every single reading, the fault is systematic rather than random — it never averages out, so it must be corrected by calculation. The rule follows directly from the sign: a positive error means the scale is shifted up, so the reading overstates the true length and you subtract the error. Had the error been negative the scale would sit low, the reading would understate the length, and you would instead add the error back on.

✓

Reverse the logic: if the true diameter is 0.29 mm and the gauge always adds 0.02 mm, it would display 0.29 + 0.02 = 0.31 mm, exactly the reading given.