JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 3)
The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 $$\mu $$m diameter of a wire is :
500
100
200
50
توضیح
Least count = $${{Pitch} \over {Number\,\,of\,\,division\,\,on\,\,circular\,scale}}$$
5 $$ \times $$ 10$$-$$6 = $${{{{10}^{ - 3}}} \over N}$$
N = 200
5 $$ \times $$ 10$$-$$6 = $${{{{10}^{ - 3}}} \over N}$$
N = 200