JEE PYQ: Units & Measurements - Question ID 12c4b52c9c90 (JEE Main 2020)

A screw-gauge is a precision instrument used to measure small lengths. Its working principle relies on the pitch of the screw and the number of divisions on the circular scale. The key concepts and formulas are:

• Pitch ($p$): The distance moved by the screw along the main scale when it completes one full rotation. Mathematically, $p = \frac{\text{Distance moved on main scale}}{\text{Number of rotations}}$
• Least Count ($L$): The smallest length that can be measured with the screw-gauge. It is given by: $L = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$

The least count determines the precision of the instrument. Smaller the least count, higher the precision.

Step 1: Calculate the Pitch

Given: - Number of rotations = 6 - Distance moved on main scale = 3 mm

Step 2: Determine the Least Count

Given: - Number of divisions on circular scale = 50

Step 3: Convert Units to Match Options

The question provides options in both mm and cm. Convert $0.01 \text{ mm}$ to cm: $0.01 \text{ mm} = 0.01 \times 10^{-1} \text{ cm} = 0.001 \text{ cm}$

Step 4: Match with Given Options

The calculated least count is $0.001 \text{ cm}$, which corresponds to option D.

Students often make the following mistakes:

• Unit Confusion: Forgetting to convert mm to cm or vice versa, leading to incorrect option selection. Always double-check units.
• Incorrect Pitch Calculation: Dividing the total distance by the number of rotations incorrectly (e.g., dividing 6 by 3 instead of 3 by 6). Remember: pitch = distance / rotations.
• Misinterpreting Least Count Formula: Some students mistakenly multiply pitch by the number of divisions instead of dividing. The least count is always pitch divided by the number of circular scale divisions.

Exam Tip: Always write down the formulas clearly and substitute values with units. Cross-verify your final answer with the given options to avoid unit-related errors.