Fundamentholfundamenthol
NEET2018Paper 2018Physics
Q.

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

  1. A

    0.521 cm

  2. B

    0.053 cm

  3. C

    0.525 cm 

  4. D

    0.529 cm 

Solution

Diameter of the ball

= MSR + CSR × (Least count) – Zero error

= 0.5 cm + 25 × 0.001 – (–0.004)

= 0.5 + 0.025 + 0.004

= 0.529 cm

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