A Vernier callipers has $1$ 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 callipers, the least count is?

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Solution

The correct option is D $0.2$ mm
$\begin{matrix} 1 MSD & = lmm 20 V S D & = 16 M S D 1 VSD & = \frac{16}{20} M S D = 0.8 M S D \end{matrix}$
$\begin{matrix} L \cdot C & = 1 M S D - 1 V S D = 1 - 0.8 = 0.2 m m \end{matrix}$

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A Vernier callipers has 1 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 callipers, the least count is?

The correct option is D 0.2 mm1 MSD=lmm20VSD=16MSD 1 VSD =1620MSD=0.8MSDL⋅C=1MSD−1VSD=1−0.8=0.2 mm

A Vernier callipers has $1$ 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 callipers, the least count is?

The correct option is D $0.2$ mm
$\begin{matrix} 1 MSD & = lmm 20 V S D & = 16 M S D 1 VSD & = \frac{16}{20} M S D = 0.8 M S D \end{matrix}$
$\begin{matrix} L \cdot C & = 1 M S D - 1 V S D = 1 - 0.8 = 0.2 m m \end{matrix}$