Question

if in a vernier callipers 10 VSD coincides with 8 MSD, then the least count of Vernier calliper is [ given 1MSD=1mm]

Answer 1

Given,

1 MSD = 1 mm

10th VSD coincides with the 8th MSD.

Therefore, 1 VSD = 8/10 MSD = 8/10 × 1 mm = 0.8 mm

Thus, Least count = 1 MSD – 1 VSD = 1 mm – 0.8 mm = 0.2 mm

Answer 2

0.2 mm or 2 * $10^{-4}$m

Explanation:

Given that,

1 VSD = 8/10 MSD = 0.8 MSD

and

1MSD = 1mm

To find,

Least Count = ?

Procedure:

Least Count = 1 MSD - 1 VSD

= (1 - 0.8)

= 0.2 * 1 mm

= 0.2 mm

Therefore, the least count of Vernier calliper is 0.2mm 0r 0.0002 m or

2 *  $10^{-4}$m.

Learn more: Vernier Calliper

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