13. A vernier calliper has a least count of 0.01 cm and
has a zero error of + 0.04 cm. Represent it with
the help of a diagram!
Answers
Answered by
0
Explanation:
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
Similar questions