JEE PYQ: Units & Measurements - Question ID d466415a421c (JEE Main 2022)

Core Formula & Concept:

In a screw gauge measurement, the diameter of a wire is determined by combining the main scale reading (MSR) and the circular scale reading (CSR). The key formulas are:

• Least Count (LC): $\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$ where Pitch is the distance moved by the screw on the main scale in one complete rotation.
• Circular Scale Reading (CSR): $\text{CSR} = \text{Number of divisions coinciding with pitch line} \times \text{LC}$
• Total Reading: $\text{Diameter} = \text{MSR} + \text{CSR} \pm \text{Error}$ The error is added if it is positive and subtracted if it is negative.

Step-by-Step Derivation:

Given data:

• Pitch = $0.5 \text{ mm}$ (distance moved in one rotation)
• Number of divisions on circular scale = $50$
• Main scale reading (MSR) = $2.5 \text{ mm}$
• Circular scale division coinciding with pitch line = $45$
• Negative error = $0.03 \text{ mm}$

Step 1: Calculate the Least Count (LC)

The least count is the smallest measurement that can be made with the screw gauge. $\text{LC} = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.5 \text{ mm}}{50} = 0.01 \text{ mm}$

Step 2: Calculate the Circular Scale Reading (CSR)

The circular scale reading is given by the number of divisions coinciding with the pitch line multiplied by the least count. $\text{CSR} = 45 \times 0.01 \text{ mm} = 0.45 \text{ mm}$

Step 3: Calculate the Observed Diameter

The observed diameter is the sum of the main scale reading and the circular scale reading. $\text{Observed Diameter} = \text{MSR} + \text{CSR} = 2.5 \text{ mm} + 0.45 \text{ mm} = 2.95 \text{ mm}$

Step 4: Apply the Error Correction

The instrument has a negative error of $0.03 \text{ mm}$. This means the measured value is $0.03 \text{ mm}$ larger than the actual value. To get the true diameter, we subtract the error from the observed diameter. $\text{True Diameter} = \text{Observed Diameter} - \text{Error} = 2.95 \text{ mm} - 0.03 \text{ mm} = 2.92 \text{ mm}$ Wait! This result ($2.92 \text{ mm}$) does not match any of the given options. Let's re-examine the error correction step carefully.

Re-evaluating Error Correction:

A negative error means the instrument reads higher than the actual value. Thus, the actual diameter is smaller than the observed diameter. However, the question states that the instrument has a negative error of $0.03 \text{ mm}$, which implies: $\text{Actual Diameter} = \text{Observed Diameter} - \text{Error}$ But the observed diameter is $2.95 \text{ mm}$, and subtracting $0.03 \text{ mm}$ gives $2.92 \text{ mm}$, which is not among the options. This suggests a possible misinterpretation of the error.

Alternatively, if the error is negative, it may mean the instrument reads lower than the actual value. In such cases, the actual diameter is larger than the observed diameter. Thus: $\text{Actual Diameter} = \text{Observed Diameter} + \text{Error} = 2.95 \text{ mm} + 0.03 \text{ mm} = 2.98 \text{ mm}$ This matches option C, which is the correct answer.

Conclusion: The correct interpretation is that a negative error means the instrument under-reads, so the actual diameter is larger than the observed diameter by the error magnitude.

Common Traps & Exam Tip:

Students often make the following mistakes in screw gauge questions:

1. Misinterpreting the Error Sign: A negative error does not always mean subtraction from the observed value. It depends on whether the instrument over-reads or under-reads. In this case, a negative error implies the instrument reads lower, so the actual value is higher.
2. Incorrect Least Count Calculation: Some students confuse pitch with least count or miscount the number of divisions on the circular scale.
3. Ignoring the Error: Forgetting to apply the error correction is a common oversight, leading to incorrect results.
4. Circular Scale Reading: Students may misread the circular scale division coinciding with the pitch line, especially if it is not clearly marked.

Exam Tip: Always double-check the error sign and its interpretation. If the error is negative, ask whether the instrument over-reads or under-reads before applying the correction.