A student measured the diameter of a small steel ball using a screw gauge of least count $0.001 c m$ . The main scale reading is $5 m m$ and zero of circular scale division coincides with $25$ divisions above the reference level. If screw gauge has a zero error of $- 0.004 c m$ , the correct diameter of the ball is

0.521 c m

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0.525 c m

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0.053 c m

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0.529 c m

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Solution

The correct option is D $0.529 c m$
Reading of screw gauge
$=$ MSR + VSR × LC + zero error
$= 0.5 c m + 25 \times 0.001 c m + 0.004 c m$
$= 0.529 c m$

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