JEE PYQ: Units & Measurements - Question ID 2551024b4ccc (JEE Main 2022)

ID: 2551024b4cccJEE Main 2022Single Correct MCQ

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 :

Select Option

Step-by-step Explanation

Core Formula & Concept:

In vernier callipers, the least count (LC) is the smallest measurement that can be accurately read. It is determined by the difference between one main scale division (MSD) and one vernier scale division (VSD). The key formula is:

Least Count (LC)=1 MSD1 VSD\text{Least Count (LC)} = 1 \text{ MSD} - 1 \text{ VSD}

Here, the main scale is divided such that each centimetre is split into 5 equal parts. This means:

1 MSD=1 cm5=2 mm1 \text{ MSD} = \frac{1 \text{ cm}}{5} = 2 \text{ mm}

The problem states that n main scale divisions coincide with (n + 1) vernier scale divisions. This implies:

n×MSD=(n+1)×VSDn \times \text{MSD} = (n + 1) \times \text{VSD}

From this, we can express the length of one vernier scale division (VSD) in terms of the main scale division (MSD).

--- Step-by-Step Derivation:

Step 1: Express VSD in terms of MSD

Given that n MSDs coincide with (n + 1) VSDs:

n×MSD=(n+1)×VSDn \times \text{MSD} = (n + 1) \times \text{VSD} VSD=nn+1×MSD\Rightarrow \text{VSD} = \frac{n}{n + 1} \times \text{MSD}

Step 2: Compute the Least Count (LC)

Using the formula for least count:

LC=1 MSD1 VSD\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} =MSDnn+1×MSD= \text{MSD} - \frac{n}{n + 1} \times \text{MSD} =MSD(1nn+1)= \text{MSD} \left(1 - \frac{n}{n + 1}\right) =MSD((n+1)nn+1)= \text{MSD} \left(\frac{(n + 1) - n}{n + 1}\right) =MSD(1n+1)= \text{MSD} \left(\frac{1}{n + 1}\right)

Step 3: Substitute the value of MSD

Since each centimetre on the main scale is divided into 5 equal parts:

1 MSD=1 cm5=2 mm1 \text{ MSD} = \frac{1 \text{ cm}}{5} = 2 \text{ mm}

Substituting this into the least count expression:

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

Step 4: Match with the given options

The derived least count is:

2n+1 mm\frac{2}{n + 1} \text{ mm}

This matches Option A.

--- Common Traps & Exam Tip:

1. Misinterpreting the Main Scale Division: Many students mistakenly assume that each main scale division is 1 mm instead of 2 mm (since 1 cm is divided into 5 parts). This leads to incorrect calculations. Always verify the value of 1 MSD from the given scale division.

2. Incorrect Least Count Formula: Some students confuse the least count formula with: LC=1 MSDn\text{LC} = \frac{1 \text{ MSD}}{n} This is incorrect. The correct formula is the difference between 1 MSD and 1 VSD.

3. Unit Confusion: The question asks for the least count in mm, but students sometimes compute it in cm and forget to convert. Always double-check the units in the final answer.

Exam Tip: When solving vernier callipers problems, always:

  • First, determine the value of 1 main scale division (MSD).
  • Use the coincidence condition to find the value of 1 vernier scale division (VSD).
  • Compute the least count as the difference between 1 MSD and 1 VSD.

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