JEE MAIN - Physics (2020 - 6th September Morning Slot - No. 18)
A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead of the
pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of 0.5mm is noticed on the pitch scale. The nature of zero error involved and the least
count of the screw gauge, are respectively :
Positive, 0.1 mm
Positive, 0.1 $$\mu $$m
Positive, 10 $$\mu $$m
Negative, 2 $$\mu $$m
توضیح
Least count of screw gauge
= $${{0.5} \over {50}}$$
= 1 $$ \times $$ 10-5 m
= 10 $$\mu $$m
Zero error in positive.
= $${{0.5} \over {50}}$$
= 1 $$ \times $$ 10-5 m
= 10 $$\mu $$m
Zero error in positive.