If n main scale divisions coincide with (n + 1) vernier scale divisions. The least count of vernier callipers, when each centimetre on the main scale is divided into five equal parts, will be :

Units and Measurements PYQ — JEE Main 2022 · 30 Jun · Shift 1 · Q43 with Solution

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Correct answer: A

Find the value of one main scale division (MSD)

Each centimetre on the main scale is divided into $5$ equal parts.

Therefore, $1\,\text{MSD} = \frac{1\,\text{cm}}{5} = 0.2\,\text{cm} = 2\,\text{mm}$
Use the given relation between main scale and vernier scale

Given: $n\,\text{MSD} = (n+1)\,\text{VSD}$

Hence, $1\,\text{VSD} = \frac{n}{n+1}\,\text{MSD}$
Formula for least count

For a vernier calipers, $\text{Least Count} = 1\,\text{MSD} - 1\,\text{VSD}$

Substituting $1\,\text{VSD}$ : $\text{LC} = \text{MSD} - \frac{n}{n+1}\,\text{MSD}$

$\text{LC} = \left(1 - \frac{n}{n+1}\right)\text{MSD}$

$\text{LC} = \frac{1}{n+1}\,\text{MSD}$
Substitute $1\,\text{MSD} = 2\,\text{mm}$

$\text{LC} = \frac{1}{n+1}\times 2\,\text{mm}$

$\boxed{\text{LC} = \frac{2}{n+1}\,\text{mm}}$
Check options
- A: $\dfrac{2}{n+1}$ mm ✅
- B: $\dfrac{5}{n+1}$ mm ❌
- C: $\dfrac{1}{2n}$ mm ❌
- D: $\dfrac{1}{5n}$ mm ❌

So, the correct option is A.