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JEE Main Physics Class 11 Physical world and Measurements Part 2 Questions
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© examsnet.com
Question : 73
Total: 100
In an experiment to find out the diameter of wire using screw gauge, the following observations were noted :
(A) Screw moves
0.5
mm
on main scale in one complete rotation
(B) Total divisions on circular scale
=
50
(C) Main scale reading is
2.5
mm
(D)
45
th
division of circular scale is in the pitch line
(E) Instrument has
0.03
mm
negative error
Then the diameter of wire is :
[29-Jul-2022-Shift-1]
2.92
mm
2.54
mm
2.98
mm
3.45
mm
Validate
Solution:
MSR
=
2.5
mm
CSR
=
45
×
0.5
50
mm
=
0.45
mm
Diameter reading
=
MSR
+
CSR
−
zero error
=
2.5
+
0.45
−
(
−
0.03
)
=
2.98
mm
© examsnet.com
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