- #1
harsh22902
- 14
- 2
- Homework Statement
- The main scale of vernier calipers reads in mm and its vernier scale is divided into 8 divisions which coincide with 5 divisions of the main scale. When the two jaws touch each other, the zero of the vernier coincides with the zero of main scale. When a rod is tightly placed along its length between the jaws, it is observed that the zero of vernier scale lies just left to 36th division of main scale and fourth division of vernier scale coincides with one main scale division. The measured value is (in cm) .
- Relevant Equations
- Reading = Main Scale Reading + (Vernier Scale Reading)(Least count)
Least Count = 1 MSR - 1VSR
My approach to this problem was pretty simple and straight.
There is no zero error here.
8 divisions of VS(Vernier scale)= 5 divisions on MS (Main Scale)
hence 1 VS division = 5/8 mm .
Now , MSR = 35 mm ( as zero of VS lies to the left of 36)
Least count = 1mm - 5/8 mm = 3/8 mm
Given in the question , VS division coinciding with 1 MS division = 4
Reading = 35 + 4*(3/8) mm = 36.5 mm
BUT the answer provided is 35.5 mm.
There is no zero error here.
8 divisions of VS(Vernier scale)= 5 divisions on MS (Main Scale)
hence 1 VS division = 5/8 mm .
Now , MSR = 35 mm ( as zero of VS lies to the left of 36)
Least count = 1mm - 5/8 mm = 3/8 mm
Given in the question , VS division coinciding with 1 MS division = 4
Reading = 35 + 4*(3/8) mm = 36.5 mm
BUT the answer provided is 35.5 mm.