Question

Consider a Vernier callipers in which each 1cm on
main scale is divided into 8 equal divisions. If 5
divisions of the Vernier scale coincide with 4 divisions
on the main scale, then least count is​

Answer 1

Answer:

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.

Hope it helps you