The one division of main scale of vernier callipers reads $1\,mm$ and $10$ divisions of Vernier scale is equal to the $9$ divisions on main scale. When the two jaws of the instrument touch each other the $zero$ of the Vernier lies to the right of $zero$ of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws,the $zero$ of the Vernier scale lies in between $4.1\,cm$ and $4.2\,cm$ and $6^{\text{th}}$ Vernier division coincides with a main scale division. The diameter of the bob will be $.............10^{-2}\,cm$
- A$413$
- B$411$
- C$141$
- D$412$
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In a vernier callipers,$(N+1)$ divisions of vernier scale coincide with $N$ divisions of main scale. If $1 \text{ MSD}$ represents $0.1 \text{ mm}$,the vernier constant (in $\text{cm}$) is:
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View Solution$A$ vernier callipers has $20$ divisions on the vernier scale, which coincide with $19$ divisions on the main scale. The least count of the instrument is $0.1 \,mm$. One main scale division is equal to $...$ $mm$.
The vernier scale of a travelling microscope has $50$ divisions which coincide with $49$ main scale divisions. If each main scale division is $0.5 \ mm$,calculate the minimum inaccuracy in the measurement of distance. (in $mm$)
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View SolutionAssertion $A$: 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 $0.001 \, cm$.
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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:
$A$ student measured the diameter of a wire using a screw gauge with the least count $0.001\, cm$ and listed the measurements. The measured value should be recorded as (in $, cm$)
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