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The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is $0.5 \mathrm{~mm}$, calculate the minimum inaccuracy in the measurement of distance.
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Verified Answer
As we know that inaccuracy will be measured by difference of 1 MSD and 1 VSD
where MSD $=$ main scale division and
VSD $=$ vernier scale division.
As given that $50 \mathrm{VSD}=49 \mathrm{MSD}$
$$
\begin{aligned}
&1 \mathrm{MSD}=\frac{50}{49} \mathrm{VSD} \\
&\begin{aligned}
1 \mathrm{VSD} &=\frac{49}{50} \mathrm{MSD} \\
\text { Inaccuracy } &=1 \mathrm{MSD}-1 \mathrm{VSD} \\
&=1 \mathrm{MSD}-\frac{49}{50} \mathrm{MSD}=\frac{1}{50} \mathrm{MSD}
\end{aligned}
\end{aligned}
$$
Given, Main scales division (1 MSD) $=0.5 \mathrm{~mm}$ Hence, minimum inaccuracy in the measurement of distance
$$
=\frac{1}{50} \times 0.5 \mathrm{~mm}=\frac{1}{100}=0.01 \mathrm{~mm}
$$
where MSD $=$ main scale division and
VSD $=$ vernier scale division.
As given that $50 \mathrm{VSD}=49 \mathrm{MSD}$
$$
\begin{aligned}
&1 \mathrm{MSD}=\frac{50}{49} \mathrm{VSD} \\
&\begin{aligned}
1 \mathrm{VSD} &=\frac{49}{50} \mathrm{MSD} \\
\text { Inaccuracy } &=1 \mathrm{MSD}-1 \mathrm{VSD} \\
&=1 \mathrm{MSD}-\frac{49}{50} \mathrm{MSD}=\frac{1}{50} \mathrm{MSD}
\end{aligned}
\end{aligned}
$$
Given, Main scales division (1 MSD) $=0.5 \mathrm{~mm}$ Hence, minimum inaccuracy in the measurement of distance
$$
=\frac{1}{50} \times 0.5 \mathrm{~mm}=\frac{1}{100}=0.01 \mathrm{~mm}
$$
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