Physics, asked by summerkhiangte8098, 9 months ago

Consider a Vernier callipers in which each 1cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions in its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then :

Answers

Answered by Fatimakincsem
1

Hence if the pitch of the screw gauge is twice the least count of vernier calipers the least count of screw gauge is 0.005 mm.

Explanation:

1 cm is divided into 8 divisions, 1 main scale division is given as   1 / 8   cm

L.C. of Vernier is given as 1 M.S.D − 1 V.S.D

Given 5 Vernier scale coincides with 4 main scale divisions:

5 V.S.D = 4 M.S.D

L.C = 1 M.S.D −  4 / 5  M.S.D

L.C = 1 M.S.D −   4  / 5   M.S.D =  1 / 5 M.S.D =  1 / 40   cm

Given pitch of Screw gauge is 2 times L.C. of V.S

p = 2 ×  1 / 40   ​cm = 1 /  20 cm

The least count of screw gauge = Pitch / No. of divisions on circular scale

No. of divisions on circular scale = 100 = 1 /20 / 100 = 0.0005 cm

No. of divisions on circular scale = 0.005 mm

Hence if the pitch of the screw gauge is twice the least count of vernier calipers the least count of screw gauge is 0.005 mm.

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