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
A$0.521 \,cm$
B$\;$$0.525\, cm$
C$0.529\, cm$
D$\;$ $0.053\, cm$
NEET 2018, Medium
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C$0.529\, cm$
c Diameter of the ball
$= MSR \,+ \,CSR \times$ (Least count) $- Zero\, error$
$= 5\ mm + 25 \times 0.001\ cm - (-0.004)\ cm$
$= 0.5\ cm + 25 \times 0.001\ cm - (-0.004)\ cm = 0.529\ cm.$
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