When the tip of the screw of a screw gauge is in contact with the stud, 96th head, 96th head scale division coincides with the baseline. If the least count of this screw gauge is 0.01 mm then find out the correction to be made on the observation. (Take the number of divisions of the head scale as 100)

-0.04 mm

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0.4 mm

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0.04 mm

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0.09 mm

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Solution

The correct option is D 0.04 mm
$\begin{matrix} 96^{th} head scale division coinsidre with index Least count = 0.01 m m \begin{matrix} error = (100 - 96) \times L C = 4 \times 0.01 = 0.04 m m \end{matrix} correction, it is negative error so answer is = 0.04 m m \end{matrix}$

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The correct option is D 0.04 mm 96th head scale division coinsidre with index Least count =0.01 mm error =(100−96)×LC=4×0.01=0.04 mm correction, it is negative error so answer is =0.04 mm

The correct option is D 0.04 mm
$\begin{matrix} 96^{th} head scale division coinsidre with index Least count = 0.01 m m \begin{matrix} error = (100 - 96) \times L C = 4 \times 0.01 = 0.04 m m \end{matrix} correction, it is negative error so answer is = 0.04 m m \end{matrix}$