Question

The number of circular divisions on the shown screw gauge is $50$. It moves $0.5mm$ on main scale for one complete rotation. Main scale reading is $2$. The diameter (in $mm$) of the ball is

Answer 1

First figure the zero error of the screw gauge is:
$ZE=\frac{5\times 0.5}{50}=0.05mm$

least count of circular scale is,
$=\frac{0.5}{50}=0.01mm$

Circular scale reading
$=25\times 0.01=0.25mm$

Main scale reading$=2\times 0.5=1mm$
Actual measurement
$=MSR+CSR-ZE$
$=1+0.25-0.05=1.20mm$

Final answer:$1.20mm$