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(B) Given that $50$ divisions of the vernier scale $(VSD)$ are equal to $49$ divisions of the main scale $(MSD)$.First,we find the value of one main s

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$7.023$

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(B) Given that $50$ divisions of the vernier scale $(VSD)$ are equal to $49$ divisions of the main scale $(MSD)$.First,we find the value of one main scale division $(MSD)$: $MSD = 7.05 \; cm - 7.00 \; cm = 0.05 \; cm$.The least count $(LC)$ of the vernier calliper is defined as $LC = 1 \; MSD - 1 \; VSD$.Since $50 \; VSD = 49 \; MSD$,we have $1 \; VSD = \frac{49}{50} \; MSD = 0.98 \; MSD$.Therefore,$LC = 1 \; MSD - 0.98 \; MSD = 0.02 \; MSD$.Substituting the value of $MSD$: $LC = 0.02 	imes 0.05 \; cm = 0.001 \; cm$.The main scale reading $(MSR)$ is the mark just before the vernier zero,which is $7.00 \; cm$.The vernier scale reading $(VSR)$ is the coinciding division multiplied by the least count: $VSR = 23 	imes 0.001 \; cm = 0.023 \; cm$.The total measured thickness is $MSR + VSR = 7.00 \; cm + 0.023 \; cm = 7.023 \; cm$.

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The circular scale of a micrometer has $200$ divisions and a pitch of $2 \, mm$. Find the measured value of the thickness of a thin sheet in $mm$.

Figure $1$ shows the configuration of the main scale and Vernier scale before measurement. Figure $2$ shows the configuration corresponding to the measurement of the diameter $D$ of a tube. The measured value of $D$ is (in $cm$)

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