There are two Vernier calipers,both of which | vedclass

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(B) $C_1$: $1$ Main scale division $(MSD)$ $= 0.1 \ cm$.$1$ Vernier scale division $(VSD)$ $= 0.9/10 = 0.09 \ cm$.Least count $(LC)$ $= 1 	ext{ MSD}

Vedclass · Accepted Answer

$2.87$ and $2.83$

Vedclass · Answer

(B) $C_1$: $1$ Main scale division $(MSD)$ $= 0.1 \ cm$.$1$ Vernier scale division $(VSD)$ $= 0.9/10 = 0.09 \ cm$.Least count $(LC)$ $= 1 	ext{ MSD} - 1 	ext{ VSD} = 0.1 - 0.09 = 0.01 \ cm$.Main scale reading $(MSR)$ $= 2.8 \ cm$.Number of Vernier division matching main scale division $n = 7$.Reading $R_1 = 	ext{MSR} + n 	imes 	ext{LC} = 2.8 + 7 	imes 0.01 = 2.87 \ cm$.$C_2$: $1$ Main scale division $(MSD)$ $= 0.1 \ cm$.$1$ Vernier scale division $(VSD)$ $= 1.1/10 = 0.11 \ cm$.Least count $(LC)$ $= 1 	ext{ VSD} - 1 	ext{ MSD} = 0.11 - 0.1 = 0.01 \ cm$.Main scale reading $(MSR)$ $= 2.8 \ cm$.Since the size of the Vernier scale division is larger than the main scale division,the matching division is counted from the back. In the figure,the $7$th Vernier division is matching,so we consider $n = 10 - 7 = 3$.Reading $R_2 = 	ext{MSR} + n 	imes 	ext{LC} = 2.8 + 3 	imes 0.01 = 2.83 \ cm$.

Similar Questions

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In an experiment,the angles are required to be measured using an instrument where $29$ divisions of the main scale exactly coincide with $30$ divisions of the vernier scale. If the smallest division of the main scale is half a degree $(= 0.5^\circ)$,then the least count of the instrument is:

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