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A vernier calipers has $1 \mathrm{~mm}$ marks on the main scale. It has 20 equal divisions on the vernier scale which match with 16 main scale divisions. For this vernier calipers, the least count is
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The correct answer is:
$0.2 \mathrm{~mm}$
$0.2 \mathrm{~mm}$
Least count of vernier calipers
$$
\begin{aligned}
& \text { LC }=1 \mathrm{MSD}-1 \text { VSD } \\
& =\frac{\text { Smallest division on main scale }}{\text { Number of divisions on vernier scale }} \\
& 20 \text { divisions of vernier scale } \\
& =16 \text { divisions of main scale } \\
& \therefore \quad 1 \mathrm{VSD}=\frac{16}{20} \mathrm{~mm}=0.8 \mathrm{~mm} \\
& \therefore \quad \text { LC }=1 \mathrm{MSD}-1 \mathrm{VSD} \\
& =1 \mathrm{~mm}-0.8 \mathrm{~mm} \\
& \therefore \quad=0.2 \mathrm{~mm} \\
&
\end{aligned}
$$
$\therefore$ The correct option is (d).
$$
\begin{aligned}
& \text { LC }=1 \mathrm{MSD}-1 \text { VSD } \\
& =\frac{\text { Smallest division on main scale }}{\text { Number of divisions on vernier scale }} \\
& 20 \text { divisions of vernier scale } \\
& =16 \text { divisions of main scale } \\
& \therefore \quad 1 \mathrm{VSD}=\frac{16}{20} \mathrm{~mm}=0.8 \mathrm{~mm} \\
& \therefore \quad \text { LC }=1 \mathrm{MSD}-1 \mathrm{VSD} \\
& =1 \mathrm{~mm}-0.8 \mathrm{~mm} \\
& \therefore \quad=0.2 \mathrm{~mm} \\
&
\end{aligned}
$$
$\therefore$ The correct option is (d).
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