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1 cm of 1 cm of main scale of a vernier calipers is divided into 10 divisions the least count of the calipersis 0.05 CM then the vernier scale must have
Asked by amarateja3 | 09 Jan, 2024, 09:26: AM

Given least count 0.05 cm = 0.5 mm is unrealistic value.

( approximately we can measure 0.5 mm without vernier )

Normally main scale smallest division is 1 mm .

If vernier has 10 divisions , then least count is 0.1 mm = 0.01 cm

If vernier has 20 divisions , then least count is 0.05 mm = 0.005 cm .

Hence , it is assumed that least count of vernier is 0.005 cm

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If there are n divisions in vernier scale and n divisions of vernier scal coinicide with (n-1) divisions of main scale ,

then leasct count LC is

where MSD is 1 main scale division.

In given vernier Caliper , 1 cm is divided by 10 divisions.

MSD = 1 cm /10 = 0.1 cm

If n divisions of vernier coincide with (n-1) divisions of main scale divisions and leat count is 0.05 cm , then

from above expression, we get ,

n = 0.1/0.005 = 20

Answered by Thiyagarajan K | 09 Jan, 2024, 04:05: PM
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