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Question:

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

if the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.

if the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.

if the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.

if the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.

Solution:

Vernier Calipers:
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 - 5 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 = 1/5 M.S.D = 1/5 * (1/8 cm) = 1/40 cm = 1/40 * 10 mm = 0.25 mm
Screw Gauge:
Number of divisions on circular scale = 100
One complete rotation moves it by 2 divisions on the linear scale.
Pitch = 2 divisions * 1/8 cm/division = 1/4 cm = 2.5 mm
Least count of screw gauge = Pitch / Number of divisions on circular scale
Least count = 2.5 mm / 100 = 0.025 mm
If the pitch of the screw gauge is twice the least count of the Vernier calipers, then:
Pitch = 2 * 0.25 mm = 0.5 mm
Least count of screw gauge = 0.5 mm / 100 = 0.005 mm
If the least count of the linear scale of the screw gauge is twice the least count of the Vernier calipers, then:
Least count of linear scale = 2 * 0.25 mm = 0.5 mm
Least count of screw gauge = 0.5 mm / 100 = 0.005 mm