
The least count of a Vernier calliper is found to be as,
$\begin{align}
& A.\dfrac{1MSD}{\text{number of VSD }\!\!'\!\!\text{ s}} \\
& B.1MSD\times \text{number of VSD }\!\!'\!\!\text{ s} \\
& C.\text{both A and B} \\
& \text{D}\text{.1MSD-number of VSD }\!\!'\!\!\text{ s} \\
\end{align}$
Answer
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Hint: A Vernier scale can be considered as a visual aid to make an accurate and precise measurement reading between two markings which are graduated on a linear scale by the use of mechanical interpolation. By increasing the resolution and reducing the uncertainty in measurement will be obtained by the use of a Vernier scale in order to reduce human estimation error.
Complete step by step solution:
In the field of physics, the least count of a measuring device is defined as the smallest and most appropriate value in the measured quantity which can be resolved on the scale of an instrument. The least count error is present with both systematic and random errors. Devices having a higher precision will reduce the least count error. By the repetition of the observations and by considering the arithmetic mean of the result, where the mean value would be very close to the true value of the measured value. A Vernier scale or a Vernier caliper will be having a least count of $0.1mm$. The least count of the Vernier caliper can be found by taking the ratio of the smallest reading in the main scale to the total number of divisions in the Vernier scale. The least count of a Vernier caliper will be the difference between the smallest reading of the main scale and the smallest reading of the Vernier scale.
$\dfrac{1MSD}{\text{number of VSD }\!\!'\!\!\text{ s}}=LC$
Therefore the correct answer will be option A.
Note: The least count will be in relation to the precision of an instrument. The device which can measure the smaller variations in a value relative to the measurement of another device is having a smaller least count value and hence will be more precise. Any measurement produced by the device, which can be assumed repeatable and which is not less than the resolution of the least count. The least count of a device will be inversely proportional to the precision of the device.
Complete step by step solution:
In the field of physics, the least count of a measuring device is defined as the smallest and most appropriate value in the measured quantity which can be resolved on the scale of an instrument. The least count error is present with both systematic and random errors. Devices having a higher precision will reduce the least count error. By the repetition of the observations and by considering the arithmetic mean of the result, where the mean value would be very close to the true value of the measured value. A Vernier scale or a Vernier caliper will be having a least count of $0.1mm$. The least count of the Vernier caliper can be found by taking the ratio of the smallest reading in the main scale to the total number of divisions in the Vernier scale. The least count of a Vernier caliper will be the difference between the smallest reading of the main scale and the smallest reading of the Vernier scale.
$\dfrac{1MSD}{\text{number of VSD }\!\!'\!\!\text{ s}}=LC$
Therefore the correct answer will be option A.
Note: The least count will be in relation to the precision of an instrument. The device which can measure the smaller variations in a value relative to the measurement of another device is having a smaller least count value and hence will be more precise. Any measurement produced by the device, which can be assumed repeatable and which is not less than the resolution of the least count. The least count of a device will be inversely proportional to the precision of the device.
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