JEE PYQ: Units & Measurements - Question ID 5f6fccef78b7 (JEE Main 2026)

ID: 5f6fccef78b7JEE Main 2026Single Correct MCQ

In a screw gauge when the circular scale is given five complete rotations it moves linearly by 2.5 mm . If the circular scale has 100 divisions, the least count of screw gauge is ____\_\_\_\_ mm.

Select Option

Step-by-step Explanation

Core Formula & Concept:

In a screw gauge (also called a micrometer screw), two scales work together:

  • The main scale (linear scale) is marked along the sleeve.
  • The circular scale (thimble scale) is marked on the rotating thimble.

The pitch of the screw is the linear distance moved by the screw per complete rotation of the thimble. The least count (LC) is the smallest distance that can be measured with the instrument. It is given by:

Least Count=PitchNumber of divisions on circular scale\text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} Step-by-Step Derivation:

Step 1: Determine the pitch of the screw.

The question states that when the circular scale is given five complete rotations, it moves linearly by 2.5 mm. Therefore, the linear distance moved per complete rotation (pitch) is:

Pitch=Total linear movementNumber of rotations=2.5 mm5=0.5 mm\text{Pitch} = \frac{\text{Total linear movement}}{\text{Number of rotations}} = \frac{2.5\ \text{mm}}{5} = 0.5\ \text{mm}

Step 2: Use the number of divisions on the circular scale.

The circular scale has 100 divisions. The least count is the pitch divided by the number of divisions on the circular scale:

Least Count=PitchNumber of divisions=0.5 mm100=0.005 mm\text{Least Count} = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.5\ \text{mm}}{100} = 0.005\ \text{mm}

Step 3: Express the least count in scientific notation.

0.005 mm=5×103 mm0.005\ \text{mm} = 5 \times 10^{-3}\ \text{mm}.

This matches option D.

Common Traps & Exam Tip:

Trap 1: Students often confuse the total linear movement with the pitch. They might directly divide 2.5 mm by 100 divisions, skipping the pitch calculation. This leads to an incorrect least count of 2.5×1022.5 \times 10^{-2} mm.

Trap 2: Misinterpreting the number of rotations. If a student mistakenly takes 1 rotation instead of 5, the pitch becomes 2.5 mm, leading to a least count of 2.5×1022.5 \times 10^{-2} mm.

Exam Tip: Always calculate the pitch first by dividing the total linear movement by the number of complete rotations. Then, divide the pitch by the number of circular scale divisions to find the least count.

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