In a vernier callipers, there are 10 divisions on the vernier scale and 1 cm on the main scale is divided in 10 parts. While measuring a length, the zero of the vernier lies just ahead of 1.8 cm mark and 4th division of vernier coincides with a main scale division.

1. Find the length

2. If zero of vernier callipers is -0.02cm, what is the correct length?

(i) As we know,

$\text{Least count (L.C.)} = \dfrac{\text{ Value of 1 main scale div (x)}}{\text{total no. of div on vernier (n)}}$

Given,

Total number of divisions on the vernier scale (n) = 10

Value of one main scale division (x) = 1 mm

Coinciding division = 4

Substituting the values in the formula above we get,

$\text{L.C.} = \dfrac{1}{\text{10}} \\[0.5em] \Rightarrow \text{L.C.} = 0.1\text{mm} \\[0.5em] \Rightarrow \text{L.C.} = 0.01\text{cm} \\[0.5em]$

Hence, least count of the vernier callipers is 0.01 cm.

As we know,

$\text{Vernier scale reading} = \text{Vernier scale division} \times \text{L.C.} \\[0.5em] \text{Vernier scale reading} = 4 \times 0.01 \text{cm} \\[0.5em] \Rightarrow \text{Vernier scale reading} = 0.04 \text{cm} \\[0.5em]$

Hence, Vernier scale reading is 0.04 cm.

and

Main scale reading of the vernier callipers = 1.8 cm

Therefore,

$\text{Total reading} = \text{main scale reading} + \text{vernier scale reading} \\[0.5em] \text{Total reading} = 1.8 \text{cm} + 0.04 \text{cm} \\[0.5em] \Rightarrow \text{Total reading}= 1.84\text{cm} \\[0.5em]$

Hence, the length is 1.84 cm.

(b) As we know,

$\text{Correct reading} = \text {observed reading} – \text{zero error}$

Given,

zero error is -0.02cm,

Substituting the values in the formula above we get,

$\text{Correct reading} = 1.84 – (-0.02) \\[0.5em] \text{Correct reading} = 1.84 + 0.02 \\[0.5em] \text{Correct reading} = 1.86 \text{cm} \\[0.5em]$

Hence, the correct length is 1.86 cm.