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 25 t h division of circular scale coincides with reference line. If screw gauge has a zero error of - 0 . 004 cm , the correct diameter of the ball is

Question

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 25 t h division of circular scale coincides with reference line. If screw gauge has a zero error of - 0 . 004 cm , the correct diameter of the ball is, 0 . 053 cm, 0 . 525 cm, 0 . 521 cm, 0 . 529 cm

MARKS App · Accepted Answer

Diameter of the steel ball = main scale reading + (circular scale reading × Least count) - zero error = MSR + ( CSR × L . C ) − zero error = 0.5 cm + 25 × 0.001 − ( − 0.004 ) = 0.5 + 0.025 + 0.004 = 0.529 cm

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