Verified Solution • 2025–26 Session moderate

QJEEM22Ph5401: JEE Main 2022 Physics - Class 11 JEEM22Ph5401 Solution: In a vernier callipers, each cm on the main scale is divided…

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 _________ ×\times 10 −- 2 mm.

Step-by-Step Verified Solution

1 MSD = 120{1 \over {20}} cm

∵\because 10 VSD = 9 MSD

1 VSD = 910{9 \over {10}} ×\times 120{1 \over {20}} cm = 9200{9 \over {200}} ×\times 10 mm = 0.45 mm

Now, 1 MSD = 120{1 \over {20}} ×\times 10 mm = 0.50 mm

LC = (0.50 −- 0.45) mm = 0.05 mm

= 5 ×\times 10 −- 2 mm

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