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In a Vernier calipers, when both jaws touch each other, zero of the Vernier scale is shifted to the right of zero of the main scale and $$7^{th}$$ Vernier division coincides with a main scale reading. If the value of 1 main scale division is 1 mm and there are 10 Vernier scale divisions, then the Vernier caliper has
When the Vernier zero is to the right of the main scale zero → positive zero error.
Given:
1 MSD = 1 mm
10 VSD
Least count:
LC = 1 MSD − 1 VSD
Since 10 VSD = 9 MSD ⇒ 1 VSD = 0.9 mm
So:
LC = 1 − 0.9 = 0.1 mm
Now, 7th Vernier division coincides:
Zero error = 7 × LC = 7 × 0.1 = 0.7 mm
Since shift is to the right → positive error
final answer:
+0.7 mm
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