in a verniercallipers with no zero error, 7" division of main scale coincides with 10
- division of vernier scale when nothing is kept between the calipers (Take I MSD =
Imm) When a rod of length, is kept between the calipers, it is found that zero of the
vernier lies between the 9 and 10" divisions of main scale and 4 division of the
vernier coincides with 12 division of main scale.
A) 92mm
B) / -10.2mm
Least count 0.3mm
D) vernier constant = 0.3mm
Answers
Numerical problems based on Vernier callipers
Question 1.
Diameter of a steel ball is measured using a vernier callipers which has divisions of 0.1cm on its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main scale. Three such measurements for a ball are given as :
Q11 JEE 2015
If the zero error is – 0.03 cm, then mean corrected diameter is:
(1) 0.53 cm
(2) 0.56 cm
(3) 0.59 cm
(4) 0.52 cm
Solution —
MSD = 0.1 cm
VSD = 9/10 MSD = 0.9 × 0.1= 0.09 cm
Least count = MSD – VSD = 0.1 – 0.09 = 0.01 cm
Total reading = Main Scale Reading + Vernier coincidence × Least Count
1st reading = 0.5 + 8 × 0.01 = 0.58 cm
2nd reading = 0.5 + 4 × 0.01 = 0.54 cm
3rd reading = 0.5 + 6 × 0.01 = 0.56 cm
Average reading = (0.58 + 0.54 + 0.56)/3 = 0.56 cm
Corrected reading = reading – zero error = 0.56 – (-0.03) = 0.56 + 0.03 = 0.59 cm