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 :
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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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