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Q.

A student measured the diameter of a small steel ball using a screw guage of least count 0.001cm. The main scale reading is 5mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw guage has a zero error of –0.004 cm, the correct diameter of the ball is

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a

0.525 cm

b

0.053 cm

c

0.529 cm

d

0.521 cm

answer is C.

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Detailed Solution

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MSR+(LC×VSR )+Zero correction
=0.5+25 × 0.001+0.004=0.529

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