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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 zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of -0.004 cm the correct diameter of the ball is


Diameter of the ball

=MSR+CSR×( Least count )- Zero error 

=5 mm+25×0.001 cm-(-0.004)cm=0.5 cm+25×0.001 cm-(-0.004)cm=0.529 cm.

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