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A student measured the diameter of a small steel ball using a screw gauge of least count 0.0010.001 cm. The main scale reading is 55 mm and zero of circular scale division coincides with 2525 divisions above the reference level. If screw gauge has a zero error of −0.004-0.004 cm, the correct diameter of the ball is

1

0.0530.053 cm

2

0.5250.525 cm

3

0.5210.521 cm

4

0.5290.529 cm

Step-by-Step Solution

Diameter of the ball = MSR + CSR ×\times (Least count) −- Zero error = 0.50.5 cm + 25×0.001−(−0.004)25 \times 0.001 - (-0.004) = 0.5+0.025+0.0040.5 + 0.025 + 0.004 = 0.5290.529 cm

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