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JEE Main Physics Class 11 Physical world and Measurements Part 1 Questions
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Question : 55
Total: 100
The pitch and the number of divisions, on the circular scale for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 division below the mean line.
The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of the sheet is:
[9 Jan. 2019 II]
5.755 mm
5.950 mm
5.725 mm
5.740 mm
Validate
Solution:
Least count of screw gauge,
LC
=
Pitch
No. of division
=
0.5
×
10
–
3
=
0.5
×
10
–
2
mm
+
ve error
=
3
×
0.5
×
10
–
2
mm
=
1.5
×
10
–
2
mm
=
0.015
mm
Reading = MSR + CSR – (+ve error)
=
5.5
mm
+
(
48
×
0.5
×
10
–
2
)
−
0.015
= 5.5 + 0.24 – 0.015 = 5.725 mm
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