In a vernier callipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be _________ × 10⁻² mm.

Solution

Physics sir · 3 steps

Find one main scale division first, then subtract one vernier division from it.

1. One main division. 1 cm is split into 20 parts, so 1 MSD = (10 mm)/20 = 0.5 mm.
2. One vernier division. 10 vernier divisions span the same length as 9 main divisions. So 1 VSD = 9/10 MSD = 0.9 × 0.5 = 0.45 mm.

   VC = 1 MSD - 1 VSD = 0.5 mm - 0.45 mm = 0.05 mm

3. So. VC = 0.05 mm = 5 × 10⁻² mm, so the answer is 5.

The common slip: 10. Taking one main scale division as 1 mm instead of 0.5 mm gives a vernier constant of 0.1 mm, that is 10, not 5.

The idea to keep. Vernier constant is 1 MSD minus 1 VSD. If n VSD match (n − 1) MSD, it is just 1/n of one MSD. Always work out the real size of one MSD first.

Answer from the NTA final answer key. Solution by PrepWiser.