Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5 mm$. The circular scale has 100 divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.
Measurement condition
Main scale reading
Circular scale reading
Two arms of gauge touching each other without wire
$0$ division
$4$ division
Attempt-$1$: With wire
$4$ division
$20$ division
Attempt-$2$: With wire
$4$ division
$16$ division
What are the diameter and cross-sectional area of the wire measured using the screw gauge?
Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5 mm$. The circular scale has 100 divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.
| Measurement condition | Main scale reading | Circular scale reading |
| Two arms of gauge touching each other without wire | $0$ division | $4$ division |
| Attempt-$1$: With wire | $4$ division | $20$ division |
| Attempt-$2$: With wire | $4$ division | $16$ division |
What are the diameter and cross-sectional area of the wire measured using the screw gauge?
- [IIT 2022]
- A
$2.22 \pm 0.02 mm , \pi(1.23 \pm 0.02) mm ^2$
- B
$2.22 \pm 0.01 mm , \pi(1.23 \pm 0.01) mm ^2$
- C
$2.14 \pm 0.02 mm , \pi(1.14 \pm 0.02) mm ^2$
- D
$2.14 \pm 0.01 mm , \pi(1.14 \pm 0.01) mm ^2$
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Asseretion $A:$ If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5\, {mm}$ and there are $50$ total divisions on circular scale, then least count is $0.001\, {cm}$.
Reason $R:$ Least Count $=\frac{\text { Pitch }}{\text { Total divisions on circular scale }}$
In the light of the above statements, choose the most appropriate answer from the options given below:
Asseretion $A:$ If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is $5\, {mm}$ and there are $50$ total divisions on circular scale, then least count is $0.001\, {cm}$.
Reason $R:$ Least Count $=\frac{\text { Pitch }}{\text { Total divisions on circular scale }}$
In the light of the above statements, choose the most appropriate answer from the options given below:
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The least count of the main scale of a screw gauge is $1\, mm$. The minimum number of divisions on its circular scale required to measure $5\,\mu m$ diameter of a wire is
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