JEE Main 2026 — Apr 4, Shift 1 — Physics solution

JEE Main2026Apr 4, Shift 1Physics

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.

A
$1 \times 1 0^{- 2}$
B
$1 \times 1 0^{- 3}$
C
$5 \times 1 0^{- 2}$
D
$5 \times 1 0^{- 3}$

Solution

Pitch is the linear distance travelled per full rotation:

$Pitch = \frac{2.5 mm}{5} = 0.5 mm .$

Least count $= \frac{Pitch}{Number of circular divisions} = \frac{0.5}{100} = 5 \times 1 0^{- 3}$ mm.

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