Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$: In a Vernier calliper, if a positive zero error exists, then while taking measurements, the reading taken will be more than the actual reading.
Reason $(R)$: The zero error in a Vernier calliper might have happened due to a manufacturing defect or due to rough handling.
In the light of the above statements, choose the correct answer from the options given below:

  • A
    Both $(A)$ and $(R)$ are correct and $(R)$ is the correct explanation of $(A)$.
  • B
    Both $(A)$ and $(R)$ are correct but $(R)$ is not the correct explanation of $(A)$.
  • C
    $(A)$ is true but $(R)$ is false.
  • D
    $(A)$ is false but $(R)$ is true.

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Similar Questions

$N$ divisions on the main scale of a vernier calliper coincide with $(N + 1)$ divisions of the vernier scale. If each division of the main scale is $a$ units,then the least count of the instrument is:

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:

In a vernier calliper,when both jaws touch each other,the zero of the vernier scale shifts towards the left and its $4^{\text{th}}$ division coincides exactly with a certain division on the main scale. If $50$ vernier scale divisions $(VSD)$ are equal to $49$ main scale divisions $(MSD)$ and the zero error in the instrument is $0.04 \text{ mm}$,then how many main scale divisions are there in $1 \text{ cm}$?

In a screw gauge,$5$ complete rotations of the screw cause it to move a linear distance of $0.25\, cm$. There are $100$ circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of $4$ main scale divisions and $30$ circular scale divisions. Assuming negligible zero error,the thickness of the wire is (in $, cm$)

The main scale of a vernier calliper has $n$ divisions per $cm$. $n$ divisions of the vernier scale coincide with $(n-1)$ divisions of the main scale. The least count of the vernier calliper is,

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