Question

Problem from IIT JEE 2015
....
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 calipers, 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,

A. If the pitch of the screw gauge is twice the least count of the Vernier calipers,

the least count of the screw gauge is 0.01 mm.

B. If the pitch of the screw gauge is twice the least count of the Vernier calipers,

the least count of the screw gauge is 0.005 mm.

C. If the least count of the linear scale of the screw gauge is twice the least count

of the Vernier calipers, the least count of the screw gauge is 0.01 mm.

D. If the least count of the linear scale of the screw gauge is twice the least count

of the Vernier calipers, the least count of the screw gauge is 0.005 mm.​

Answer 1

Answer:

means

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Answer 2

Answer:

Hope it helps

Explanation:

Vernier Calipers:

1 cm is divided into 8 divisions, 1 main scale division is given as 81cm

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

Given 5 Vernier scale coincides with 4 main scale divisions:

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

L.C=1M.S.D−54M.S.D

L.C=1M.S.D−54M.S.D=51M.S.D=401cm 

Checking option A and B: 

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

p=2×401cm=201cm

The least count of screw gauge is no.ofdivisionsoncircularscalepitch

No. of divisions on circular scale =100=100201=0.0005 cm=0.005 mm

Hence, option B is correct.

Checking Option C and D:

Given least count of the linear scale of the screw gauge is twice the least count of the Vernier calipers.

One complete rotation of the circular scale moves it by two divisions on the linear scale, so pitch is 2 times linear scale division of screw gauge,

Pitch =2 × linear scale division of screw gauge =2×2 × L.C of V.S=4× L.C of V.S =4×401=10