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AIEEE 2008 Solved Paper
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© examsnet.com
Question : 39
Total: 104
Two full turns of the circular scale of a screw gauge cover a distance of
1
mm
on its main scale. The total number of divisions on the circular scale is 50 . Further, it is found that the screw gauge has a zero error of
−
0.03
mm
while measuring the diameter of a thin wire, a student notes the main scale reading of
3
mm
and the number of circular scale divisions in line with the main scale as 35 . The diameter of the wire is
3.32
mm
3.73
mm
3.67
mm
3.38
mm
Validate
Solution:
Least count of screw gauge
=
0.5
50
mm
=
0.01
mm
Main scale reading
=
3
mm
Vernier scale reading
=
35
∴
Reading
=
[Main scale reading
+
circular scale reading
×
L.C]
−
(zero error)
=
[
3
+
35
×
0.01
]
−
(
−
0.03
)
=
3.38
mm
© examsnet.com
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