In a screw gauge when the circular scale is given five complete rotations it moves linearly by 2.5 mm

Question

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. [2026]

Smart Achievers · Accepted Answer

(4)

$MSD= \frac{2.5}{5} mm = 0.5 mm = pitch$

$L.C. = \frac{MSD}{pitch} = \frac{0.5}{100} mm = 5 \times 10^{- 3} mm$

Solution

Q.
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. [2026]

1 $1 \times 10^{- 2}$

2 $1 \times 10^{- 3}$

3 $5 \times 10^{- 2}$

4 $5 \times 10^{- 3}$

Ans.
(4)

$MSD= \frac{2.5}{5} mm = 0.5 mm = pitch$

$L.C. = \frac{MSD}{pitch} = \frac{0.5}{100} mm = 5 \times 10^{- 3} mm$

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Solution

Q. 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. [2026]

1 1×10-2

2 1×10-3

3 5×10-2

4 5×10-3

Ans. (4) MSD=2.55 mm=0.5 mm=pitch L.C.=MSDpitch=0.5100 mm=5×10-3 mm