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JEE Main 2022 · 26 Jun · Shift 1, Physics question 57

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 contant will be ______ ×10−2\times10^{-2} mm.

Answer and worked solution

Official answer: 5

Main scale division 120\frac1{20} cm =0.5=0.5 mm. Vernier division: 10 VSD = 9 MSD, so 1 VSD =0.45=0.45 mm.

Vernier constant 0.050.05 mm =5×10−2=5\times10^{-2} mm.

Answer: 5

Topics:Vernier callipers and screw gaugeLeast count
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