Vernier Caliper: GRADE A+ COMPLETE WITH ALL THE
ANSWER 2022 LATEST
1. The main scale of a Vernier caliper has 1 mm markings. The Vernier scale includes 20
equal divisions that correspond to the 18 divisions of the main scale. The least count
possible with these Vernier calipers is
(A). 0.02 mm
(B). 0.05 mm
(C). 0.1 mm
(D). 0.2 mm
Solution. For the given Vernier calipers, one main scale division (MSD) is 1 MSD = 1
mm.
Given 20 Vernier scale divisions (VSD) are equal to 18 MSD, we get 1 VSD = 18/20 = 0.9
mm.
The least count (LC) is given by LC = 1 MSD − 1 VSD = 1.0 − 0.9 = 0.1 mm
2. Vernier calipers are used to measure a cylinder's diameter with no zero error. It is
determined that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the
main scale. The 50 divisions on the Vernier scale are comparable to 2.45 cm. One of the
main scale divisions and the 24th division of the Vernier scale line up perfectly. The
cylinder's diameter is
(A). 5.112 cm
(B). 5.124 cm
(C). 5.136 cm
(D). 5.148 cm
Solution. Given data, one main scale division (MSD) and one Vernier scale division
(VSD) are 1 MSD = 5.15 cm − 5.10 cm = 0.05 cm
, 1 VSD = 2.45/50 cm = 0.049 cm.
The least count of the given Vernier calipers is LC = 1 MSD − 1 VSD = 0.001 cm.
For the given measurement, main scale reading (MSR) is 5.10 cm and the Vernier scale
reading (VSR) is 24.
Hence, the diameter D of the cylinder is D = MSR + VSR × LC = 5.10 + 24 × 0.001 = 5.124
cm.
3. M divisions on the main scale of a Vernier calipers coincide with (M + 1) divisions on its
Vernier scale. If each division on the main scale is of b units, determine the least count of
instrument
Solution. Given, a main scale division of the Vernier calipers is 1 MSD = b
Since (M + 1) Vernier scale divisions are equal to M main scale divisions,
We get 1 VSD = M/M + 1 MSD = Mb/M + 1
The least count is given by LC = 1 MSD − 1 VSD = b/M + 1
4. The Vernier scale, which has an instrument's 25 divisions, corresponds to the main scale's
24th division. 20 identical segments make up 1 cm on the main scale. The instrument's
least count is:
(A) 0.002 cm
(B) 0.05 cm
(C) 0.001 cm
(D) 0.02 cm
Solution. 1 MSD = 1/20 = 0.05 cm
Given as 25 VSD = 24 MSD
So, VSD = 24/25 MSD
ANSWER 2022 LATEST
1. The main scale of a Vernier caliper has 1 mm markings. The Vernier scale includes 20
equal divisions that correspond to the 18 divisions of the main scale. The least count
possible with these Vernier calipers is
(A). 0.02 mm
(B). 0.05 mm
(C). 0.1 mm
(D). 0.2 mm
Solution. For the given Vernier calipers, one main scale division (MSD) is 1 MSD = 1
mm.
Given 20 Vernier scale divisions (VSD) are equal to 18 MSD, we get 1 VSD = 18/20 = 0.9
mm.
The least count (LC) is given by LC = 1 MSD − 1 VSD = 1.0 − 0.9 = 0.1 mm
2. Vernier calipers are used to measure a cylinder's diameter with no zero error. It is
determined that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the
main scale. The 50 divisions on the Vernier scale are comparable to 2.45 cm. One of the
main scale divisions and the 24th division of the Vernier scale line up perfectly. The
cylinder's diameter is
(A). 5.112 cm
(B). 5.124 cm
(C). 5.136 cm
(D). 5.148 cm
Solution. Given data, one main scale division (MSD) and one Vernier scale division
(VSD) are 1 MSD = 5.15 cm − 5.10 cm = 0.05 cm
, 1 VSD = 2.45/50 cm = 0.049 cm.
The least count of the given Vernier calipers is LC = 1 MSD − 1 VSD = 0.001 cm.
For the given measurement, main scale reading (MSR) is 5.10 cm and the Vernier scale
reading (VSR) is 24.
Hence, the diameter D of the cylinder is D = MSR + VSR × LC = 5.10 + 24 × 0.001 = 5.124
cm.
3. M divisions on the main scale of a Vernier calipers coincide with (M + 1) divisions on its
Vernier scale. If each division on the main scale is of b units, determine the least count of
instrument
Solution. Given, a main scale division of the Vernier calipers is 1 MSD = b
Since (M + 1) Vernier scale divisions are equal to M main scale divisions,
We get 1 VSD = M/M + 1 MSD = Mb/M + 1
The least count is given by LC = 1 MSD − 1 VSD = b/M + 1
4. The Vernier scale, which has an instrument's 25 divisions, corresponds to the main scale's
24th division. 20 identical segments make up 1 cm on the main scale. The instrument's
least count is:
(A) 0.002 cm
(B) 0.05 cm
(C) 0.001 cm
(D) 0.02 cm
Solution. 1 MSD = 1/20 = 0.05 cm
Given as 25 VSD = 24 MSD
So, VSD = 24/25 MSD
