JEE PYQ: Units & Measurements - Question ID 427dd72478ff (JEE Main 2024)

In metrology, the least count of a Vernier calliper is the smallest measurement that can be accurately read with the instrument. It is determined by the difference between the value of one main-scale division and one Vernier-scale division.

Key formula: $\text{Least Count (LC)} = \text{Value of 1 main-scale division} - \text{Value of 1 Vernier-scale division}$

When $n$ divisions on the main scale coincide with $(n+1)$ divisions on the Vernier scale, the least count simplifies to: $\text{LC} = \frac{\text{Value of 1 main-scale division}}{n+1}$

1. Identify given data: - 10 divisions on the main scale coincide with 11 divisions on the Vernier scale. - Each main-scale division = 5 units. 2. Compute the value of one Vernier-scale division: Since 10 main-scale divisions = 11 Vernier-scale divisions, $10 \times 5 = 11 \times \text{(Vernier division value)}$ $\Rightarrow \text{Vernier division value} = \frac{10 \times 5}{11} = \frac{50}{11} \text{ units}$ 3. Determine the least count: $\text{LC} = \text{Main-scale division} - \text{Vernier-scale division}$ $\text{LC} = 5 - \frac{50}{11} = \frac{55 - 50}{11} = \frac{5}{11} \text{ units}$ 4. Match with the options: The calculated least count $\frac{5}{11}$ matches option A.

- Students often confuse the number of coinciding divisions. They mistakenly use 10 Vernier divisions matching 11 main-scale divisions, leading to an incorrect least count. - Another frequent error is forgetting to subtract the Vernier division value from the main-scale division value, instead directly dividing the main-scale value by the number of Vernier divisions. - Always verify the coincidence statement carefully: “10 main = 11 Vernier” implies the Vernier scale is slightly finer.