The smallest division on the main scale of a | vedclass

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(C) Given: $10 	ext{ VSD} = 9 	ext{ MSD}$.Here, $	ext{MSD}$ is the Main Scale Division and $	ext{VSD}$ is the Vernier Scale Division.$1 	ext{ VSD

Vedclass · Accepted Answer

$3.15$

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(C) Given: $10 	ext{ VSD} = 9 	ext{ MSD}$.Here, $	ext{MSD}$ is the Main Scale Division and $	ext{VSD}$ is the Vernier Scale Division.$1 	ext{ VSD} = \frac{9}{10} 	ext{ MSD} = 0.9 	ext{ MSD}$.Least count $(LC)$ = $1 	ext{ MSD} - 1 	ext{ VSD} = (1 - 0.9) 	ext{ MSD} = 0.1 	ext{ MSD}$.Since $1 	ext{ MSD} = 0.1 	ext{ cm}$, $	ext{LC} = 0.1 	imes 0.1 	ext{ cm} = 0.01 	ext{ cm}$.From the left figure (zero error): The $6^{	ext{th}}$ division of the Vernier scale coincides with a main scale division. Since the zero of the Vernier scale is to the left of the zero of the main scale, the zero error is negative.Zero error = $-(10 - 6) 	imes 	ext{LC} = -4 	imes 0.01 	ext{ cm} = -0.04 	ext{ cm}$.From the right figure (observed reading): The main scale reading is $3.1 	ext{ cm}$. The $1^{	ext{st}}$ division of the Vernier scale coincides with a main scale division.Observed reading = $	ext{Main scale reading} + (	ext{Vernier coincidence} 	imes 	ext{LC}) = 3.1 	ext{ cm} + (1 	imes 0.01 	ext{ cm}) = 3.11 	ext{ cm}$.True diameter = $	ext{Observed reading} - 	ext{Zero error} = 3.11 	ext{ cm} - (-0.04 	ext{ cm}) = 3.15 	ext{ cm}$.

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