Question

There are two Vernier calipers both of which have $1$ cm divided into $10$ equal divisions on the main scale. The Vernier scale of one of the calipers ($C_1$) has $10$ equal divisions that correspond to $9$ main scale divisions. The Vernier scale of the other caliper ($C_2$) has 10 equal divisions that corresponds to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers $C_1$ and $C_2$, respectively, are:

• A. $2.87$ and $2.87$
• B. $2.87$ and $2.83$
• C. $2.85$ and $2.82$
• D. $2.87$ and $2.86$

Answer 1

The correct option is B $2.87$ and $2.83$

MSD = $\frac{1}{10}$ cm = 0.1 cm

For first vernier caliper, 10 VSD = 9 MSD

⇒ VSD = $\frac{9}{10}$ MSD = $0.9\times 0.1$ = 0.09 cm

Reading

= main scale reading upto coinciding main scale division − n × VSD

= $3.5-7\times 0.09$ = 3.5 – 0.63 = 2.87 cm

For second vernier caliper, 10 VSD = 11 MSD

⇒ VSD = $\frac{11}{10}$ MSD = $1.1\times 0.1$ = 0.11 cm

Reading =$3.6-7\times 0.11$ = 3.6 – 0.77 = 2.83 cm