Consider a Vernier callipers 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:
Text Solution
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(b,c)For vernier callipers,
1 main scale division =
cm
1 vernier scale division =
cm
So least count =
cm
For screw gauge,
pitch (p) = 2 main scale division
So least count 
So option & are correct.
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