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.

Pitch of the screw gauge is the linear distance moved in one complete rotation.

\(\text{Pitch} = \dfrac{2.5}{5} = 0.5 \text{ mm}\)

Least count of the screw gauge is given by the ratio of the pitch to the total number of divisions on the circular scale.

\(\text{Least count} = \dfrac{\text{Pitch}}{\text{Number of divisions on circular scale}}\)

\(\text{Least count} = \dfrac{0.5}{100} = 0.005 \text{ mm}\)

\(\text{Least count} = 5 \times 10^{-3} \text{ mm}\)

Answer: \(5 \times 10^{-3}\)