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A student measured the diameter of a small steel ball using a screw gauge of least count 0.01 cm . The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauges has zero error of _ 0.04 cm , the correct diameter of the ball is

Answer»

Explanation:Given -MSR = 0.5 cmVSR = 25 LC = 0.001 cmZero ERROR = -0.0004 cm# Solution -Diameter of steel BALL measured by STUDENT -Measured diameter = MSR + VSR × LCMeasured diameter = 0.5 + 25 × 0.001Measured diameter = 0.525 cmTrue diameter of steel ball must be -TRUE diameter = Measured diamter - Zero errorTrue diameter = 0.525 - (-0.004)True diameter = 0.529 cmHope this HELPS you...



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