Solution

Question\n\nThere are $50$ divisions on the circular scale of a screw gauge. If the head (thimble) of the screw is given $10$ revolutions, then the spindle advances by $5\,\mathrm{mm}$. There is also zero error as the $2$nd division of the circular scale coincides with the datum line and zero of circular scale is below the datum line. What is the thickness of a glass slab as measured by the described screw gauge shown in the figure?\n\nContent coming soon. Write the full solution here, or use the part files (p-i.mdx, p-ii.mdx, …) for each part.