Caliper Measurements

For the measurement consider the bottom scale of the caliper, which is in metric units, and ignore the top scale, which is calibrated in inches. In the bottom scale, there are the fixed scale (above) and the sliding scale (below). The boldface numbers on the fixed scale are in centimetres, and the tick marks between the boldface numbers are in millimetres, like a simple ruler; we also have 50 tick marks on the sliding scale. One can find the uncertainty (which is equal to the "least count") dividing the smallest division on main scale ( $1.0\;\textrm{mm}$ ) with the total number of divisions on vernier scale (50), so we have $U= \frac{1.0\;\textrm{mm}}{50}=0.02\;\textrm{mm}$ . On a caliper the leftmost tick mark on the sliding scale will let we read from the fixed scale the number of whole millimetres. In the figure, we have that the leftmost tick mark on the main scale is between $12\;\textrm{mm}$ and $13\;\textrm{mm}$ , so the number of whole millemetres is 12. After that, we must examine the vernier to determine which of its divisions coincide or are most coincident with a division on the main scale. Then we must add the number of these divisions to the main scale reading. We see that the 40th tick mark (the 8 one) on the sliding scale is in coincidence with the one above it. So $M=12.00+0.80=12.80\; \textrm{mm}$ , considering the uncertainty of this caliper. In conclusion, the final answer, putting $M$ and $U$ side by side, is $1280002$ , because the measurement can be written as $(12.80 \pm 0.02)\;\textrm{mm}$ .