Question

Which of the following is the most accurate instrument?
A) A scale whose least count is \[0.1cm\]
B) A Vernier caliper in which $10$ vernier divisions coincide with $9$ main scale divisions whose main scale is millimetre scale.
C) A microscope in which $50$ Vernier divisions coincide with $49$ main scale divisions whose main scale is millimetre scale.
D) A microscope in which $50$ Vernier divisions coincide with $49$ main scale divisions and in the main scale there are $20$ divisions in $1cm$.

Answer 1

Hint: Finding the least count (L.C.) of a Vernier caliper, tells us about the accuracy of the instrument. In the given question, find the least count of all 4 options in the same unit and compare them. The Vernier caliper with the lowest value of least count is the one with the highest accuracy.

Formula used:
The L.C. is given by:
${\text{L}}{\text{.C}}{\text{. = 1 m}}{\text{.s}}{\text{.d}}{\text{.}} - {\text{1 v}}{\text{.s}}{\text{.d}}{\text{.}}$
Where the value of v.s.d. must be written in terms of m.s.d

Complete step by step solution:
In order to find the accuracy of a Vernier caliper, we should know its least count. So we analyse all the four given options-
A) Here, the least count is already given to us, which is ${\text{0}}{\text{.1cm}}$.

B) Here, the value of m.s.d. is given as-
$1m.s.d. = 0.1cm$ , as it is a millimetre scale. $\{ 1cm = 10mm\} $
We also know that,
$10v.s.d. = 9m.s.d.$
So, $1v.s.d. = \dfrac{9}{{10}}m.s.d.$
$L.C. = 1m.s.d. - 1v.s.d.$
On substituting the value of $1v.s.d$we get-
$L.C. = 1m.s.d. - \dfrac{9}{{10}}m.s.d.$
Therefore,$L.C. = \dfrac{1}{{10}}m.s.d.$
Now, we put the value of $1m.s.d.$ here-
$L.C. = \dfrac{1}{{10}} \times 0.1{\text{ }}cm$
$L.C. = 0.01cm$

C) Here, the value of m.s.d. is given as-
$1m.s.d. = 0.1cm$ (Millimetre scale)
Also,
$50v.s.d. = 49m.s.d.$
$1v.s.d. = \dfrac{{49}}{{50}}m.s.d.$
Therefore L.C. is given by-
$L.C. = 1m.s.d. - 1v.s.d.$
\[L.C. = 1m.s.d. - \dfrac{{49}}{{50}}m.s.d.\]
$L.C. = \dfrac{1}{{50}}m.s.d.$
$L.C. = \dfrac{1}{{50}} \times 0.1{\text{ }}cm$
$L.C. = 0.002cm$

D) Here, it is said that there are 20 divisions in $1{\text{ cm}}$, which means,
$1m.s.d. = \dfrac{1}{{20}}cm = 0.05cm$
And $50v.s.d. = 49m.s.d.$
$1v.s.d. = \dfrac{{49}}{{50}}m.s.d.$
Therefore, L.C. is-
$L.C. = 1m.s.d. - \dfrac{{49}}{{50}}m.s.d.$
$L.C. = \dfrac{1}{{50}} \times 0.05 = 0.001{\text{ }}cm$

It is clear that $0.001cm$ is the lowest least count, thus (D) is the correct option.

Note: All of the values of the least count should be compared only when they are in the same unit, which is centimeters here. Other units like millimeters can also be used but are less common in the case of Vernier calipers.