In a vernier calliper,when both jaws touch ea | vedclass

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(C) The zero error is given as $0.04 	ext{ mm} = 0.004 	ext{ cm}$. Since the zero of the vernier scale shifts to the left,the zero error is negative

Vedclass · Accepted Answer

$20$

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(C) The zero error is given as $0.04 	ext{ mm} = 0.004 	ext{ cm}$. Since the zero of the vernier scale shifts to the left,the zero error is negative.The formula for zero error is: $	ext{Zero Error} = -(	ext{n} 	imes 	ext{Least Count})$,where $n$ is the coinciding division.Given $50 	ext{ VSD} = 49 	ext{ MSD}$,the Least Count $(LC)$ is $1 	ext{ MSD} - 1 	ext{ VSD} = 1 	ext{ MSD} - \frac{49}{50} 	ext{ MSD} = \frac{1}{50} 	ext{ MSD}$.Let $1 	ext{ MSD} = x 	ext{ cm}$. Then $LC = \frac{x}{50} 	ext{ cm}$.The $4^{	ext{th}}$ division coincides,so the zero error is $4 	imes LC = 4 	imes \frac{x}{50} = 0.004 	ext{ cm}$.Solving for $x$: $\frac{4x}{50} = 0.004 \implies 4x = 0.2 \implies x = 0.05 	ext{ cm}$.The number of main scale divisions in $1 	ext{ cm}$ is $N = \frac{1 	ext{ cm}}{x} = \frac{1}{0.05} = 20$.

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