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
- 0.521 cm
- 0.525 cm
- 0.053 cm
- 0.529 cm
Answer
(4) 0.529 cm
Solution
Observed reading = MSR + (CSR × LC)
= 0.5 cm + 25 0.001 cm
= 0.525 cm
Correct reading = observed reading − zero error
= 0.525 - (-0.004)
= 0.529 cm
Answer: (4)