JEE Main 2024 · 30 January · evening · Single correct · +4 / −1
If 50 Vernier divisions are equal to 49 main scale divisions of a traveling microscope and one smallest reading of main scale is 0.5 mm, the Vernier constant of traveling microscope is
- (a) 0.01 mm
- (b) 0.01 cm
- (c) 0.1 mm
- (d) 0.1 cm
Answer and solution
Answer (a) 0.01 mm
Solution
Physics sir · 3 steps
Find how much shorter one Vernier division is than one main scale division. That gap is the Vernier constant.
- Size of one VSD. 50 Vernier divisions cover 49 main scale divisions, so 1 VSD = 49/50 MSD.
VC = 1 MSD - 1 VSD = (1 - 49/50)MSD = 1/50 MSD
- Put in numbers. One main scale division is 0.5 mm, so VC = 0.5/50 mm = 0.01 mm.
- So. The Vernier constant of the travelling microscope is 0.01 mm, which is option (a).
The idea to keep. Vernier constant is 1 MSD - 1 VSD. When n Vernier divisions match (n-1) main scale divisions, it is just MSD/n.
Answer from the NTA final answer key. Solution by PrepWiser.