To solve this problem, we need to understand the relationship between the least count, the number of divisions on the circular scale, and the pitch of the screw gauge.$\\ \\$The least count of a screw gauge is given by the formula:$\\$Least count$= \frac{\mathrm{Pitch}}{\mathrm{Number\ of\ divisions\ on\ the\ circular\ scale}} \\ \\$Given:$\\$• Least count$= 0.01$ mm$\\$• Number of divisions on the circular scale$= 50 \\ \\$Let the pitch of the screw gauge be $P. \\$Then, according to the formula:$\\ 0.01 = \frac{P}{50} \\ \\$To find $P,$ we multiply both sides by 50:$\\ P = 0.01 \times 50 \\ P = 0.5$ mm$\\ \\$Therefore, the pitch of the screw gauge is $0.5$ mm, which corresponds to Option 2.