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

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 isNEET 2018

TestHub · Accepted Answer

Diameter of the steel ball = main scale reading + (circular scale reading × Least count) - zero error

$= MSR + (CSR \times L . C) -$ zero error

$= 0.5 cm + 25 \times 0.001 - (- 0.004)$

$= 0.5 + 0.025 + 0.004$

$= 0.529 cm$

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