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JEE Advanced 2015 Paper 1
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© examsnet.com
Question : 10
Total: 60
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 division 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 pinch 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 o f the screw gauge is 0.1 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 o f the screw gauge is 0.005 mm
Validate
Solution:
For vernier callipers,
1 main scale division=
1
8
cm
1 vernier scale division =
1
10
cm
So least count=
1
40
c
m
For screw gauge.
pitch(p)=2 main scale division
So least count=
p
100
So option (1) and (2) are correct.
© examsnet.com
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