The main scale of a vernier caliper reads in | vedclass

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(C) The main scale division $(MSD)$ is $1 	ext{ mm}$.Given that $8$ vernier scale divisions $(VSD)$ coincide with $5$ main scale divisions $(MSD)$.Th

Vedclass · Accepted Answer

$3.65$

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(C) The main scale division $(MSD)$ is $1 	ext{ mm}$.Given that $8$ vernier scale divisions $(VSD)$ coincide with $5$ main scale divisions $(MSD)$.Therefore,$1 	ext{ VSD} = \frac{5}{8} 	ext{ MSD} = \frac{5}{8} 	ext{ mm} = 0.625 	ext{ mm}$.The least count $(LC)$ of the vernier caliper is $1 	ext{ MSD} - 1 	ext{ VSD} = 1 	ext{ mm} - 0.625 	ext{ mm} = 0.375 	ext{ mm}$.The zero of the vernier scale is to the left of the $36^{th}$ division of the main scale,so the main scale reading $(MSR)$ is $35 	ext{ mm}$.The fourth division of the vernier scale coincides with the main scale,so the vernier scale reading $(VSR)$ is $4 	imes 	ext{LC} = 4 	imes 0.375 	ext{ mm} = 1.5 	ext{ mm}$.The total reading is $	ext{MSR} + 	ext{VSR} = 35 	ext{ mm} + 1.5 	ext{ mm} = 36.5 	ext{ mm}$.Converting to centimeters,$36.5 	ext{ mm} = 3.65 	ext{ cm}$.

Similar Questions

$A$ specially designed Vernier calliper has the main scale least count of $1 \,mm$. On the Vernier scale,there are $10$ equal divisions and they match with $11$ main scale divisions. Then,the least count of the Vernier calliper is ........... $mm$.

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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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