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JEE Advanced 2013 Paper 1
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© examsnet.com
Question : 3
Total: 60
The diameter of a cylinder is measured using a vernier calipers with no zero error. It is found that the zero of the vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The vernier scale has 50 division equivalent to 2.45 cm. The
24
t
h
division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is
5.112 cm
5.124 cm
5.136cm
5.148cm
Validate
Solution:
50
V
S
D
=
2.45
c
m
1
V
S
D
=
2.45
50
c
m
=
0.049
c
m
Least count of vernier =1MSD-1 VSD= 0.05 cm - 0.049 cm=0.001cm
Thickness of the object = Main scale reading + vernier scale reading × least count = 5.10+ (24) (0.001) = 5.124 cm.
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