Question

(a) A vernier scale has 10 divisions. It slides over the main scale, whose pitch is 1.0 mm. If the number of divisions on the left hand of zero of the vernier scale on the main scale is 56 and the 8th vernier scale division coincides with the main scale, calculate the length in centimetres.
(b) If the above instrument has a negative error of 0.07 cm, calculate the corrected length.

Answer 1

No. of divisions on vernier scale = 10
Pitch = 1.0 mm

Pitch = `"Pitch"/"No. of divisions on the vernier scale"`

L.C. = `1.0/10` mm

L.C. = 0.1 mm

L.C. = 0.01 cm

There is 56 number of main scale division on the left hand of zero of the vernier scale.

⇒ Main scale reading = 56 mm = 5.6 cm
Vernier scale reading coinciding with the main scale = 8th

(a) Length recorded = Main scale reading + L.C. × V.S.D.
= 5.6 + 0.01 × 8
= 5.6 + 0.08
= 5.68 cm

(b) Negative error = −0.07 cm
⇒ Correction = −(−0.07) = +0.07 cm
∴ Corrected length = Observed reading + Correction
= 5.68 + (+0.07)
= 5.68 + 0.07
= 5.75 cm