$A$ screw gauge with a pitch of $0.5 \ mm$ and a circular scale with $50$ divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement,it is found that when the two jaws of the screw gauge are brought in contact,the $45^{th}$ division coincides with the main scale line and the zero of the main scale is barely visible. What is the thickness of the sheet (in $mm$) if the main scale reading is $0.5 \ mm$ and the $25^{th}$ division coincides with the main scale line?
- A$0.70$
- B$0.50$
- C$0.75$
- D$0.80$
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Similar Questions
The smallest division on the main scale of a vernier callipers is $1 \ mm$,and $10$ vernier divisions coincide with $9$ main scale divisions. While measuring the diameter of a sphere,the zero mark of the vernier scale lies between $2.0 \ cm$ and $2.1 \ cm$ of the main scale,and the fifth division of the vernier scale coincides with a main scale division. The diameter of the sphere is:
Easy
View SolutionArea of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5 \text{ mm}$. 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?
| 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?
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 ....... $mm$
MediumAIEEE 2008
View SolutionThe pitch of the screw gauge is $1\, mm$ and there are $100$ divisions on the circular scale. When nothing is put in between the jaws,the zero of the circular scale lies $8$ divisions below the reference line. When a wire is placed between the jaws,the first linear scale division is clearly visible while $72^{nd}$ division on the circular scale coincides with the reference line. The radius of the wire is.........$mm$
DifficultJEE MAIN 2021
View SolutionThe vernier scale used for measurement has a positive zero error of $0.2\, mm$. If while taking a measurement it was noted that the '$0$' on the vernier scale lies between $8.5\, cm$ and $8.6\, cm$ and the vernier coincidence is $6$,then the correct value of measurement is ............. $cm$. (Least count $= 0.01\, cm$)
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