Question

In an instrument each centimetre of the main scale is divided in 20 equal parts .The instrument carries a vernier scale having 50 smallest and the length of 50 division of vernier scale is equal to the length of 49 division of the msin scale.Calculate the least count of the instrument。
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Answer 1

Answer:

0.001 mm

Explanation:

Least count is defined as the smallest value which can be measured by the instrument

Length of one main scale division 1 M.S.D.

= 1/ 20 cm

=0.05 cm=0.5 mm

Length of 49 divisions on main scale = Length of 50 divisions on Vernier scale

Value of smallest division on Vernier scale = 49/50 = 0.98 mm

Least count of vernier callipers

L.C.=1 M.S.D. – 0.98 × M.S.D

Least count = 0.05 – (0.98 x 0.05) = 0.001 mm

Answer 2

Explanation:

Given In an instrument each centimetre of the main scale is divided in 20 equal parts .The instrument carries a vernier scale having 50 smallest and the length of 50 division of vernier scale is equal to the length of 49 division of the msin scale.Calculate the least count of the instrument。

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• Now 1 main scale division = 1 / 20 cm = 0.05 cm = 0.5 mm
• Now 50 Vernier scale division will be 49 main scale division
• So 1 VSD will be 49 / 50 MSD
• This will be 49 / 50 x 0.5 mm
• Least count = 1 Main scale division – 1 Vernier scale division
•                    = 0.5 mm – 49 / 50 x 0.5  
•                   = 0.5 mm – 0.49 mm
•                    = 0.01 mm

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https://brainly.in/question/4919375