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A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of –0.004 cm, the correct diameter of the ball is
Explanation
Subtract the (negative) zero error from the observed reading.
Detailed Solution
Diameter of the ball = MSR + CSR × (Least count) – Zero error
$= 0.5 + 25\times0.001 - (-0.004)$
$= 0.5 + 0.025 + 0.004$
$= 0.529$ cm
