Physics Q26 – JEE-MAIN 2026

Home Physics Units & MeasurementsQuestion #551

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.

Clue 1 of 3: Understanding Screw Gauge Mechanics

The least count of a screw gauge depends on its pitch and the number of divisions on the circular scale.

🧠 Full Solution Path

Step 1: Calculate the Pitch of the Screw Gauge✦ Active

The pitch is the linear distance moved by the screw for one complete rotation. Given that the screw moves $2.5$ mm for $5$ rotations.

Pitch (P) = \frac{Distance moved}{Number of rotations} = \frac{2.5 mm}{5} = 0.5 mm

Step 2: Calculate the Least Count of the Screw Gauge○ Expand

The least count is the smallest measurement that can be made with the instrument. It is calculated by dividing the pitch by the total number of divisions on the circular scale. Given Pitch ( $P$ ) = $0.5$ mm and Number of divisions ( $N$ ) = $100$ .

Least Count (L C) = \frac{Pitch}{Number of divisions} = \frac{0.5 mm}{100} = 0.005 mm

Step 3: Express the Least Count in Scientific Notation○ Expand

The calculated least count is $0.005$ mm. Converting this to scientific notation gives:

L C = 0.005 mm = 5 \times 10^{- 3} mm

This value matches option 4.

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Physics Question 26 – JEE-MAIN 2026

Clue 1 of 3: Understanding Screw Gauge Mechanics

🧠 Full Solution Path