Question

If the length of a vernier scale having 25 divisions correspond to 23 main scale divisions, then given that, \[M.S.D{\text{ }} = {\text{ }}1mm\] the least count of the vernier calipers is:
A.80mm
B.0.04mm
C.0.08cm
D.None of the above

Answer 1

Hint: We will first calculate the value of V.S.D from the equation given in the problem, i.e. \[25{\text{ }}V.S.D{\text{ }} = {\text{ }}23{\text{ }}M.S.D\] . Then calculate the value of least count by using the equation \[L.C = 1M.S.D - 1V.S.D\]

Complete answer:
The Vernier caliper gives a direct reading of the distance measured with high accuracy and precision. Vernier calipers are functionally identical, with different ways of reading the results. These calipers compromise a calibrated scale with a fixed jaw, and another jaw, with a pointer, that slides along the scale.
Vernier calipers can measure internal dimensions. This probe is slender and can get into deep grooves that may prove difficult for other measuring tools. The simplest method is to read the position of the pointer directly on the scale.

The Vernier scales may include metric measurements on the lower part of the scale and inch measurements on the upper or vice-versa. A Vernier scale allows more accurate interpolation and is the universal practice. Vernier calipers are commonly used in industry to provide a precision to 0.01mm.
M.S.D - The gap between two successive marks on the main scale is one main scale division (M.S.D).
V.S.D - The distance between two consecutive marks on the main scale is one main scale division (M.S.D).
Now we are given in the problem, \[M.S.D{\text{ }} = {\text{ }}1mm\]
\[25{\text{ }}V.S.D{\text{ }} = {\text{ }}23{\text{ }}M.S.D\]
\[ \Rightarrow V.S.D{\text{ }} = {\text{ }}\dfrac{{23}}{{25}}{\text{ M}}{\text{.S}}{\text{.D}}\]
Now we know that the least count of a verniers calipers is given by the equation
\[L.C = 1M.S.D - 1V.S.D\]
\[ \Rightarrow L.C = 1M.S.D - \dfrac{{23}}{{25}}{\text{ M}}{\text{.S}}{\text{.D}}\]
\[ \Rightarrow L.C = \dfrac{2}{{25}}{\text{ M}}{\text{.S}}{\text{.D}}\]
\[ \Rightarrow L.C = \dfrac{2}{{25}}{\text{ }} \times {\text{1mm}}\]
\[ \Rightarrow L.C = 0.8mm = 0.08cm\]

Hence, option C is the correct option.

Note:
In measurement science, the smallest and most precise value in the calculated quantity that can be resolved on the instrument's scale is the least count of a measuring instrument. The least count is related to the precision of an instrument; an instrument that can calculate smaller changes in a value compared to another instrument.