JEE PYQ: Units & Measurements - Question ID 832ea2c4ca48 (JEE Main 2024)

Core Formula & Concept:

In a vernier calliper, the least count (LC) is the smallest length that can be measured accurately. It is defined as: $\text{Least Count} = \frac{\text{Value of 1 main scale division (MSD)}}{\text{Number of vernier scale divisions (VSD)}}$ However, when the vernier scale has a different number of divisions compared to the main scale, the least count is calculated as: $\text{Least Count} = 1 \text{ MSD} - 1 \text{ VSD}$ where $1 \text{ VSD} = \frac{\text{Total length of vernier scale}}{\text{Number of vernier divisions}}$.

A zero error occurs when the zero of the vernier scale does not coincide with the zero of the main scale when the jaws are closed. If the zero of the vernier scale shifts to the left, the zero error is positive (since the instrument reads a value larger than the actual measurement). The zero error is given by: $\text{Zero Error} = n \times \text{Least Count}$ where $n$ is the number of vernier divisions that coincide with a main scale division when the jaws are closed.

The question provides: - 50 vernier scale divisions (VSD) = 49 main scale divisions (MSD). - Zero error = $0.04 \text{ mm}$. - The $4^{\text{th}}$ vernier division coincides with a main scale division when the jaws are closed.

Step-by-Step Derivation:

Step 1: Calculate the Least Count (LC)

Given that 50 VSD = 49 MSD, let the length of 1 MSD be $x \text{ mm}$. Then: $$50 \text{ VSD} = 49 \text{ MSD} \\ \Rightarrow 1 \text{ VSD} = \frac{49}{50} \text{ MSD} = \frac{49}{50} x \text{ mm}$$ The least count is: $\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = x - \frac{49}{50} x = \frac{1}{50} x \text{ mm}$

Step 2: Relate Zero Error to Least Count

When the jaws are closed, the $4^{\text{th}}$ vernier division coincides with a main scale division. This means the zero of the vernier scale has shifted by 4 divisions to the left. The zero error is: $\text{Zero Error} = n \times \text{LC} = 4 \times \frac{1}{50} x \text{ mm}$ Given that the zero error is $0.04 \text{ mm}$: $$4 \times \frac{1}{50} x = 0.04 \\ \Rightarrow \frac{4x}{50} = 0.04 \\ \Rightarrow x = \frac{0.04 \times 50}{4} = 0.5 \text{ mm}$$ Thus, $1 \text{ MSD} = 0.5 \text{ mm}$.

Step 3: Calculate Number of Main Scale Divisions in 1 cm

We know that $1 \text{ cm} = 10 \text{ mm}$. If $1 \text{ MSD} = 0.5 \text{ mm}$, then the number of main scale divisions in $1 \text{ cm}$ is: $\text{Number of MSD} = \frac{10 \text{ mm}}{0.5 \text{ mm}} = 20$

Common Traps & Exam Tip:

1. Misinterpreting the Zero Error Direction: Students often confuse whether the zero error is positive or negative. Here, the zero shifts to the left, so the error is positive. If the zero shifts to the right, the error is negative.

2. Incorrect Least Count Calculation: Some students mistakenly take the least count as $\frac{1 \text{ MSD}}{50}$ instead of $1 \text{ MSD} - 1 \text{ VSD}$. Always remember that the least count is the difference between one main scale division and one vernier scale division.

3. Unit Confusion: Ensure all units are consistent (e.g., converting cm to mm or vice versa). Here, the zero error is given in mm, so the final answer must be consistent with mm.

4. Coinciding Division Misinterpretation: The question states that the $4^{\text{th}}$ vernier division coincides with a main scale division. This does not mean the zero error is $4 \text{ LC}$; it means the zero has shifted by 4 divisions, so the error is $4 \times \text{LC}$.