Define least count of vernier callipers. How do you determine it ?
Answer
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Hint:The Vernier caliper least counts formula is calculated by dividing the smallest reading of the main scale with the total number of divisions of the vernier scale.LC of vernier caliper is the difference between one smallest reading of the main scale and one smallest reading of vernier scale which is 0.1 mm 0r 0.01 cm.
Complete step by step answer:
Let us suppose that there are n divisions on vernier callipers of equal length to that of $$(n - 1)$$ divisions on the main scale and the value of 1 main scale division is $x$.
Then, Value of $n$ divisions on vernier calliper $ = (n - 1)x$
Similarly, value of 1 division on vernier calliper $ = \dfrac{{(n - 1)x}}{n}$
Now, using the definition of least count, it can be written as
${\text{L}}{\text{.C}}{\text{. = }}\dfrac{{(Value{\text{ of one main scale division)}}}}{{{\text{(Total no}}{\text{. of divisions on vernier scale)}}}}$
Therefore, we get the following result
${\text{Least count = x - }}\dfrac{{(n - 1)x}}{n}$
Taking LCM and solving the equation we get
$L.C. = \dfrac{{xn - nx + x}}{n} = \dfrac{x}{n}$
We know that, Value of 1 main scale division = $1\,mm$
And, total number of divisions on vernier scale $ = 10$
Therefore,
$L.C. = \dfrac{x}{n} \\
\Rightarrow L.C.= \dfrac{1}{{10}} \\
\Rightarrow L.C.= 0.1\,mm \\
\therefore L.C.= 0.01\,cm$
Additional Information: On the similar line there is also the lowest count error which is the error related to the instrument’s resolution. It can have the graduations at 1mm division scale spacing. A vernier scale can have a minimum count of 0.01mm, while a micrometre can have a minimum count of 0.001 mm based on which scale you are using and how accurate the scale is.
Note: The vernier calliper has two scales, one of them is called the main scale and the other one is called the vernier scale. The least count is defined as $L.C = 1MSD - 1VSD$ where MSD is the main scale divisions and VSD are the vernier scale divisions. The least count in vernier calliper is also known as vernier constant.
Complete step by step answer:
Let us suppose that there are n divisions on vernier callipers of equal length to that of $$(n - 1)$$ divisions on the main scale and the value of 1 main scale division is $x$.
Then, Value of $n$ divisions on vernier calliper $ = (n - 1)x$
Similarly, value of 1 division on vernier calliper $ = \dfrac{{(n - 1)x}}{n}$
Now, using the definition of least count, it can be written as
${\text{L}}{\text{.C}}{\text{. = }}\dfrac{{(Value{\text{ of one main scale division)}}}}{{{\text{(Total no}}{\text{. of divisions on vernier scale)}}}}$
Therefore, we get the following result
${\text{Least count = x - }}\dfrac{{(n - 1)x}}{n}$
Taking LCM and solving the equation we get
$L.C. = \dfrac{{xn - nx + x}}{n} = \dfrac{x}{n}$
We know that, Value of 1 main scale division = $1\,mm$
And, total number of divisions on vernier scale $ = 10$
Therefore,
$L.C. = \dfrac{x}{n} \\
\Rightarrow L.C.= \dfrac{1}{{10}} \\
\Rightarrow L.C.= 0.1\,mm \\
\therefore L.C.= 0.01\,cm$
Additional Information: On the similar line there is also the lowest count error which is the error related to the instrument’s resolution. It can have the graduations at 1mm division scale spacing. A vernier scale can have a minimum count of 0.01mm, while a micrometre can have a minimum count of 0.001 mm based on which scale you are using and how accurate the scale is.
Note: The vernier calliper has two scales, one of them is called the main scale and the other one is called the vernier scale. The least count is defined as $L.C = 1MSD - 1VSD$ where MSD is the main scale divisions and VSD are the vernier scale divisions. The least count in vernier calliper is also known as vernier constant.
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