While measuring diameter of a lead shot using screw gauge, the pitch scale reading and the head scale reading were found to be 4.0 mm and 50 respectively. If the least count of the instrument is 0.01 mm and zero error is -0.45 mm, then what is the diameter of the lead shot?

3.12 mm

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4.65 mm

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4.95 mm

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49.8 mm

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Solution

The correct option is C 4.95 mm
Main scale reading $M . S . R . = 4.0 m m$
Coinciding division $n = 50$
Least count of instrument $L . C . = 0.01 m m$
Reading of the lead shot $d_{1} = M . S . R . + n \times L . C . = 4.0 + 50 \times 0.01 = 4.50 m m$
Negative zero error $e r r o r = - 0.45 m m$
Diameter of lead shot $D = d_{1} - (e r r o r) = 4.50 - (- 0.45) = 4.95 m m$

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While measuring diameter of a lead shot using screw gauge, the pitch scale reading and the head scale reading were found to be 4.0 mm and 50 respectively. If the least count of the instrument is 0.01 mm and zero error is -0.45 mm, then what is the diameter of the lead shot?

The correct option is C 4.95 mmMain scale reading M.S.R.=4.0 mmCoinciding division n=50Least count of instrument L.C.=0.01 mmReading of the lead shot d1=M.S.R.+n×L.C.=4.0+50×0.01=4.50 mmNegative zero error error=−0.45 mmDiameter of lead shot D=d1−(error)=4.50−(−0.45)=4.95 mm