NEETPhysicsUnits and MeasurementMCQ+4 / −1
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
- A0.521 cm
- B0.525 cm
- C0.053 cm
- D0.529 cm
View written solutionFree
Correct answer: D
Diameter of the ball
= MSR + CSR × (Least count) – Zero error
= 5 mm + 25 × 0.001 cm – (–0.004) cm
= 0.5 cm + 25 × 0.001 cm – (–0.004) cm = 0.529 cm.
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