There are two Vernier calipers,both of which have $1 \ cm$ divided into $10$ equal divisions on the main scale. The Vernier scale of one of the calipers $(C_1)$ has $10$ equal divisions that correspond to $9$ main scale divisions. The Vernier scale of the other caliper $(C_2)$ has $10$ equal divisions that correspond to $11$ main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in $cm$) by calipers $C_1$ and $C_2$,respectively,are
- A$2.85$ and $2.82$
- B$2.87$ and $2.83$
- C$2.87$ and $2.86$
- D$2.87$ and $2.87$
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Diameter 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 ........... $cm$.
Assertion $A$: If in five complete rotations of the circular scale,the distance travelled on the main scale of the screw gauge is $5 \, mm$ and there are $50$ total divisions on the circular scale,then the least count is $0.001 \, cm$.
Reason $R$: $\text{Least Count} = \frac{\text{Pitch}}{\text{Total divisions on circular scale}}$
In the light of the above statements,choose the most appropriate answer from the options given below:
Reason $R$: $\text{Least Count} = \frac{\text{Pitch}}{\text{Total divisions on circular scale}}$
In the light of the above statements,choose the most appropriate answer from the options given below:
One main scale division of a vernier callipers is $a \ cm$ and $n^{\text{th}}$ division of the vernier scale coincides with $(n-1)^{\text{th}}$ division of the main scale. The least count of the callipers in $mm$ is
If $50$ Vernier divisions are equal to $49$ main scale divisions of a travelling microscope and one smallest reading of main scale is $0.5 \,mm$, the Vernier constant of travelling microscope is:
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
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