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?

(a) 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,

L.C. = 1/10 mm

$\Rightarrow \text{L.C.}=0.1\text{mm}$

$\Rightarrow \text{L.C.}=0.01\text{cm}$

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

As we know,

Vernier scale reading = Vernier scale division × L.C.

Vernier scale reading = 4 × 0.01cm

⇒ Vernier scale reading = 0.04cm

$\Rightarrow \text{Vernier scale reading }=0.04\text{cm}$

Hence, the Vernier scale reading is 0.04 cm.

and

Main scale reading of the vernier callipers = 1.8 cm

$\text{∴ Total reading}=\text{main scale reading}+\text{vernier scale reading}$

$\text{Total reading }=1.8\text{cm}+0.04\text{cm}$

$\Rightarrow \text{Total reading }=1.84\text{cm}$

Hence, the length is 1.84 cm.

(b) zero error is -0.02cm,

Substituting the values in the formula above we get,

$\text{Correct reading }=1.84\; –(-0.02)$

$\text{Correct reading }=1.84+0.02$

$\text{Correct reading }=1.86\text{cm}$

Hence, the correct length is 1.86 cm.