The head scale of a screw gauge contains 100 divisions and it's pitch is 1mm. what is the least count of the screw gauge​

0.01 mm

Explanation:

The least count of the screw gauge = pitch/number of divisions on circular scale

Pitch = 1 mm and number of divisions = 100

Substituting the values of the variables into the equation, we have

least count of the screw gauge = pitch/number of divisions on circular scale =  1 mm/100 = 0.01 mm

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge