A vernier calipers has $1 \mathrm{~mm}$ marks on the main scale. It has 20 equal divisions on the vernier…
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
$0.02 \mathrm{~mm}$
$0.05 \mathrm{~mm}$
$0.1 \mathrm{~mm}$
$0.2 \mathrm{~mm}$
Solution
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)