Question

A vernier callipers has its main scale of $10cm$ divided into $200$ equal parts. Its vernier scale of $25$ divisions coincides with $12mm$ on the main scale. The least count of the instrument is:
A.$0.020cm$
B.$0.002cm$
C.$0.010cm$
D.$0.001cm$

Answer 1

The main two points which we have to keep while taking measurement using Vernier Caliper instrument is as following:
->The main scale contributes the main number(s) and one decimal place in the measurement reading.
For example, in reading 4.2 cm, 4 is the main number and 0.2 is the one decimal place number)
->The Vernier scale contributes the second decimal place to the reading.

Complete step-by-step solution:
To obtain the main scale reading the number which is immediate left of the zero on the Vernier scale is taken.
To obtain the Vernier scale reading we have to consider the alignment of the scale lines of the main scale and Vernier scale. The number on the main scale which coincides with the Vernier scale is taken as the main scale.
In order to obtain the final measurement reading, we will add the main scale reading and Vernier scale reading together.
10 cm of Main Scale $=$200 division Main scale
$\Rightarrow 1\text{MSD}=\dfrac{10cm}{200}=0.05cm$
It is given that 25 Vernier scale divisions coincide with 12 mm on the main scale.
Hence,
25 VSD = 12 mm on the main scale

\Rightarrow 25VSD=24MSD \\\ \Rightarrow 1VSD=\dfrac{25}{24}MSD=0.048~cm \\\ \end{array}$$ Least Count$=\left( 1MSD-1VSD \right)$ $\begin{aligned} & =\left( 0.050-0.048 \right)cm \\\ & =0.002cm \\\ \end{aligned}$ Therefore, the least count of the instrument is 0.002cm. **Option B is the right answer** **Note:** Main sources of error in the Vernier scale reading are following, ->Scale misreading (parallax effect), ->Excessive measuring force causing jaw tilt, ->Thermal expansion caused by a temperature difference between the caliper and workpiece.