JEE PYQ: Units & Measurements - Question ID a3ab01814c4a (JEE Main 2024)

Core Formula & Concept:

In metrology, the least count (LC) of a vernier calliper is the smallest length that can be measured with the instrument. It is defined as the difference between the length of one main-scale division ($MSD$) and one vernier-scale division ($VSD$).

Mathematically, $LC = MSD - VSD$

When $N$ divisions on the vernier scale coincide with $(N-1)$ divisions on the main scale, the length of one vernier division is: $VSD = \frac{(N-1) \cdot MSD}{N}$

Substituting this into the least count formula gives: $LC = MSD - \frac{(N-1) \cdot MSD}{N} = \frac{MSD}{N}$

Thus, the least count can also be expressed as: $LC = \frac{MSD}{N}$

Step-by-Step Derivation:

Given:

• Number of vernier divisions, $N = 20$
• These 20 vernier divisions coincide with the $19^{\text{th}}$ main-scale division, meaning $20 \ VSD = 19 \ MSD$.
• Least count, $LC = 0.1 \ \text{mm}$

Step 1: Express the relationship between $VSD$ and $MSD$.
Since 20 vernier divisions coincide with 19 main-scale divisions, $20 \ VSD = 19 \ MSD$ Thus, $VSD = \frac{19}{20} MSD$

Step 2: Use the least count formula.
The least count is: $LC = MSD - VSD$ Substitute $VSD$ from Step 1: $LC = MSD - \frac{19}{20} MSD = \frac{1}{20} MSD$

Step 3: Solve for $MSD$.
Given $LC = 0.1 \ \text{mm}$, we have: $0.1 = \frac{1}{20} MSD$ Multiply both sides by 20: $MSD = 0.1 \times 20 = 2 \ \text{mm}$

Conclusion:
One main-scale division is equal to 2 mm, which corresponds to option B.

Common Traps & Exam Tip:

Trap 1: Misinterpreting the coincidence condition.
Students often confuse whether $N$ vernier divisions coincide with $N$ or $(N-1)$ main-scale divisions. In this question, 20 vernier divisions coincide with 19 main-scale divisions, so the correct relationship is $20 \ VSD = 19 \ MSD$, not $20 \ VSD = 20 \ MSD$.

Trap 2: Incorrectly applying the least count formula.
Some students mistakenly use $LC = \frac{MSD}{N-1}$ instead of $LC = \frac{MSD}{N}$. The correct formula is derived from $LC = MSD - VSD$, leading to $LC = \frac{MSD}{N}$ when $N \ VSD = (N-1) \ MSD$.

Exam Tip:
Always verify the coincidence condition first. If $N$ vernier divisions coincide with $(N-1)$ main-scale divisions, the least count is $LC = \frac{MSD}{N}$. This is a standard result and can save time in exams.