In a vernier callipers, 50 vernier scale divisions are equal to 4… — Solution | JEE Physics

MCQEasyJEE 2026Vernier Calipers & Screw Gauge

In a vernier callipers, $50$ vernier scale divisions are equal to $48$ main scale divisions. If one main scale division = $0.05 \, \text{mm}$ , then the least count of the vernier callipers is _____ $\text{mm}$ .

A
$0.02$
B
$0.005$
C
$0.002$
D
$0.05$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: $50$ vernier scale divisions are equal to $48$ main scale divisions, and $1 \, \text{MSD} = 0.05 \, \text{mm}$ .

Find: The least count of the vernier callipers.

Step 1: Relation between vernier and main scale.

50 \, \text{VSD} = 48 \, \text{MSD}

So,

1 \, \text{VSD} = \frac{48}{50} \, \text{MSD}

Step 2: Define least count.

\text{Least Count} = 1 \, \text{MSD} - 1 \, \text{VSD}

Step 3: Substitute values.

\text{LC} = 1 - \frac{48}{50} = \frac{2}{50} \, \text{MSD}

Step 4: Convert into mm.

\text{LC} = \frac{2}{50} \times 0.05 = 0.002 \, \text{mm}

Therefore, the least count is $0.002 \, \text{mm}$ . The correct option is C.

Common mistakes

Using $\frac{48}{50} \, \text{MSD}$ directly as the least count is incorrect because that quantity is $1 \, \text{VSD}$ , not the difference between one main scale division and one vernier scale division. Always use $\text{LC} = 1 \, \text{MSD} - 1 \, \text{VSD}$ .
Forgetting to multiply by $0.05 \, \text{mm}$ at the end gives the least count only in terms of MSD, not in millimetres. After finding $\frac{2}{50} \, \text{MSD}$ , convert it into mm using the value of $1 \, \text{MSD}$ .

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Answer

Step-by-step solution

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