Question

(d) A vernier used for measuring length has least count of 0.01 cm and positive error of seve
vernier divisions. If the final length, as measured from vernier, is 2.58 cm, find the correct
length​

Answer 1

Answer:

L.C. = 0.01 cm

Error = + 0.02 cm ⇒ Positive zero

error = + 0.02 cm

main scale reading (MSR) = 3.60 cm 8

th

vernier scale coincides with main scale

⇒ vernier coincidence (VC) = 8

We know that

Correct diameter = [main scale reading+ (L.C × V.C)] - (error)

= [3.60 + (0.01 × 8)] - [+ 0.002]

= (3.60 + 0.08)- (+0.02)

= 3.68 - 0.02

= 3.66 cm

∴correctradius=

2

diameter

=

2

3.66

=1.83cm