$A$ student performs an experiment of measuring the thickness of a slab with a vernier calliper whose $50$ divisions of the vernier scale are equal to $49$ divisions of the main scale. He noted that the zero of the vernier scale is between $7.00 \; cm$ and $7.05 \; cm$ mark of the main scale and the $23^{rd}$ division of the vernier scale exactly coincides with the main scale. The measured value of the thickness of the given slab using the calliper will be (in $; cm$)
- A$7.23$
- B$7.023$
- C$7.073$
- D$7.73$
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The diameter of a wire is measured with a screw gauge having a least count of $0.01\;mm$. Which of the following correctly expresses the diameter?
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View SolutionIdentify the physical quantity that cannot be measured using a spherometer:
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View SolutionWhich devices are used for the measurement of lengths of different orders?
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View SolutionIn an experiment to find out the diameter of a wire using a screw gauge,the following observations were noted:
$(A)$ Screw moves $0.5 \ mm$ on the main scale in one complete rotation.
$(B)$ Total divisions on the circular scale $= 50$.
$(C)$ Main scale reading is $2.5 \ mm$.
$(D)$ $45^{\text{th}}$ division of the circular scale is in the reference line.
$(E)$ Instrument has $0.03 \ mm$ negative zero error.
Then the diameter of the wire is: (in $mm$)
$(A)$ Screw moves $0.5 \ mm$ on the main scale in one complete rotation.
$(B)$ Total divisions on the circular scale $= 50$.
$(C)$ Main scale reading is $2.5 \ mm$.
$(D)$ $45^{\text{th}}$ division of the circular scale is in the reference line.
$(E)$ Instrument has $0.03 \ mm$ negative zero error.
Then the diameter of the wire is: (in $mm$)
Difficult
View SolutionDiameter of a steel ball is measured using a Vernier callipers which has divisions of $0.1\,cm$ on its main scale $(MS)$ and $10$ divisions of its vernier scale $(VS)$ match $9$ divisions on the main scale. Three such measurements for a ball are given as:
$S$.No. $MS\;(cm)$ $VS$ divisions $(1)$ $0.5$ $8$ $(2)$ $0.5$ $4$ $(3)$ $0.5$ $6$
If the zero error is $-0.03\,cm,$ then the mean corrected diameter is ........... $cm$.
| $S$.No. | $MS\;(cm)$ | $VS$ divisions |
| $(1)$ | $0.5$ | $8$ |
| $(2)$ | $0.5$ | $4$ |
| $(3)$ | $0.5$ | $6$ |
If the zero error is $-0.03\,cm,$ then the mean corrected diameter is ........... $cm$.
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