JEE PYQ: Units & Measurements - Question ID 95476e660a49 (JEE Main 2018)

In a screw gauge, the linear distance moved by the screw per complete rotation is called the pitch of the screw. The circular scale divisions allow us to measure fractions of this pitch. The key formulas are:

• Pitch ($p$): $p = \frac{\text{Linear distance moved}}{\text{Number of complete rotations}}$
• Least Count ($LC$): $LC = \frac{p}{\text{Number of circular scale divisions}}$
• Total Reading: $\text{Thickness} = (\text{Main scale reading} \times p) + (\text{Circular scale reading} \times LC)$

The main scale reading corresponds to full rotations (or pitch multiples), while the circular scale reading refines the measurement to fractions of the pitch.

Step 1: Calculate the pitch of the screw gauge.

Given that $5$ complete rotations move the screw by $0.25 \text{ cm}$: $p = \frac{0.25 \text{ cm}}{5} = 0.05 \text{ cm}.$

Step 2: Determine the least count (LC) of the screw gauge.

The circular scale has $100$ divisions. The least count is: $LC = \frac{p}{100} = \frac{0.05 \text{ cm}}{100} = 0.0005 \text{ cm}.$

Step 3: Compute the main scale contribution.

The main scale reading is $4$ divisions. Since each main scale division corresponds to one pitch ($0.05 \text{ cm}$): $\text{Main scale contribution} = 4 \times 0.05 \text{ cm} = 0.20 \text{ cm}.$

Step 4: Compute the circular scale contribution.

The circular scale reading is $30$ divisions. Each division corresponds to the least count ($0.0005 \text{ cm}$): $\text{Circular scale contribution} = 30 \times 0.0005 \text{ cm} = 0.015 \text{ cm}.$

Step 5: Calculate the total thickness of the wire.

Add the main scale and circular scale contributions: $\text{Thickness} = 0.20 \text{ cm} + 0.015 \text{ cm} = 0.2150 \text{ cm}.$

Students often make the following mistakes:

1. Misinterpreting the main scale reading: Some assume the main scale divisions correspond to the least count instead of the pitch. For example, they might multiply $4$ by $0.0005 \text{ cm}$ instead of $0.05 \text{ cm}$.
2. Incorrect pitch calculation: Dividing the linear distance by the number of circular scale divisions (e.g., $0.25 \text{ cm} / 100$) instead of the number of rotations.
3. Zero error confusion: The question specifies negligible zero error, but students sometimes overcomplicate the problem by assuming a zero error exists.
4. Unit mismatches: Forgetting to ensure all measurements are in the same unit (e.g., mixing cm and mm). Here, everything is in cm, so no conversion is needed.

Exam Tip: Always double-check the definitions of pitch and least count. Write down the formulas explicitly to avoid confusion between main scale and circular scale contributions.