99 Percentile Qs Bank for JEE MainPhysicsExperimental Physics
A student measured the length of a rod and wrote it as 3 . 50   cm . Which instrument did he use to measure it?
Options
- AA meter scale.
- BA vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scal
- CA screw gauge having 100 divisions in the circular scale and pitch as 1   mm .
- DA screw gauge having 50 divisions in the circular scale and pitch as 1   mm .
Correct answer
B. A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scal
Step-by-step solution
As per the given data the vernier has a least count of 0 . 01   cm , which is consistent with the information given in the question. The measured value of the length of the rod = 3 . 50   cm So, the least count of the measuring instrument must be 0 . 01   cm = 0 . 1   mm For the vernier scale 10   MSD = 1   cm = 10   mm → 1 MSD = 1   mm also 9   MSD = 10   VSD ∴ Least count (L.C.) = 1   MSD - 1   VSD = ( 1 - 0 . 9 )   mm = 0 . 1   mm