
In a slide caliper, $\left ({m + 1} \right)$ number of vernier divisions is equal to m number of smallest main scale divisions. If $d$units are the magnitude of the smallest main scale division, then the magnitude of the vernier constant is.
A. ${\text { }}\dfrac{d}{{\left ({m + 1} \right)}}{\text {units}} $
B. ${\text { }}\dfrac{d}{m}{\text {units}}$
C. ${\text { }}\dfrac{{md}}{{\left ({m + 1} \right)}}{\text {units}} $
D. $\dfrac{{\left( {m + 1} \right)}}{m}{\text{ units}}$
Answer
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Hint: Here, we have to explain vernier scale division and main scale division and show that in a slide caliper, $\left( {m + 1} \right)$ number of vernier divisions is equal to m number of smallest main scale divisions. If $d$ units is the magnitude of the smallest main scale division.
Complete step by step solution:
As we know that, the Vernier caliper is an extremely precise measuring instrument; the reading error is $\dfrac{1}{{20}}$ mm. That is equal to $0.05{\text {mm}} $. Close the jaws lightly on the object to be measured and the object is kept perpendicular to the vernier caliper. A vernier scale is a visual aid to take an accurate measurement reading between two graduation markings on a linear scale by using mechanical interpolation, thereby increasing resolution and reducing measurement uncertainty by using Vernier acuity to reduce human estimation error.
On the main scale division, the distance between two successive marks on the main scale is equal to $1{\text {mm}}$. And the vernier scale division is, the distance between two successive marks on the vernier scale.
Thus, the $\left ({m + 1} \right)$ vernier division is equal to the $m$ main scale division.
One division on vernier scale $\dfrac{m}{{\left( {m + 1} \right)}}$ division on main scale
And the Vernier constant is given by
$ = \left ( {1 - \dfrac{m}{{m + 1}}} \right)d$
We take the lowest common multiple of the denominator that is $\left ({m + 1} \right)$
So, we can write $ = \dfrac{{m + 1 - m}} {{m + 1}} \times d$
$ = \dfrac{d}{{\left ({m + 1} \right)}}$
Hence, the correct option is (A).
Note: Be careful while calculating the error on the vernier scale and the main scale division during the experiment. Write the proper units and be careful on writing the decimal value.
Complete step by step solution:
As we know that, the Vernier caliper is an extremely precise measuring instrument; the reading error is $\dfrac{1}{{20}}$ mm. That is equal to $0.05{\text {mm}} $. Close the jaws lightly on the object to be measured and the object is kept perpendicular to the vernier caliper. A vernier scale is a visual aid to take an accurate measurement reading between two graduation markings on a linear scale by using mechanical interpolation, thereby increasing resolution and reducing measurement uncertainty by using Vernier acuity to reduce human estimation error.
On the main scale division, the distance between two successive marks on the main scale is equal to $1{\text {mm}}$. And the vernier scale division is, the distance between two successive marks on the vernier scale.
Thus, the $\left ({m + 1} \right)$ vernier division is equal to the $m$ main scale division.
One division on vernier scale $\dfrac{m}{{\left( {m + 1} \right)}}$ division on main scale
And the Vernier constant is given by
$ = \left ( {1 - \dfrac{m}{{m + 1}}} \right)d$
We take the lowest common multiple of the denominator that is $\left ({m + 1} \right)$
So, we can write $ = \dfrac{{m + 1 - m}} {{m + 1}} \times d$
$ = \dfrac{d}{{\left ({m + 1} \right)}}$
Hence, the correct option is (A).
Note: Be careful while calculating the error on the vernier scale and the main scale division during the experiment. Write the proper units and be careful on writing the decimal value.
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