In a vernier calliper, when both jaws touch each other, zero of the vernier scale shifts towards left and its 4th division coincides exactly with a certain division on main scale. If 50 vernier scale divisions equal to 49 main scale divisions and zero error in the instrument is 0.04 mm then how many main scale divisions are there in 1 cm ?
- 5
- 40
- 10
- 20
Answer
BONUS
Solution
50 VSD = 49 MSD LC = 1 MSD - 1 VSD = 150 MSD
vernier zero lies to the left of main-scale zero, the zero error is negative
4th VSD coincides ⇒ zero error = -(50 - 4) LC = -46 LC
46 150 MSD = 0.04 mm
1 MSD = 0.04 5046 = 123 mm
MSD in 1 cm = 10 23 = 230
(4 LC = 0.04 mm, giving 20, would hold only for a positive zero error.)
230 is not among the options, no option is correct ⇒ question dropped
Answer: Bonus (full marks to all who attempted)
Note: The official NTA answer key gives option (4). Our faculty team finds that no option is fully correct, so this question should be dropped (bonus: full marks to everyone who attempted it).