Assertion A : If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is 5 mm and there are 50 total divisions on circular scale, then least count is 0.001 cm.
Reason R :
Least Count = ${{Pitch} \over {Total\,divisions\,on\,circular\,scale}}$
In the light of the above statements, choose the most appropriate answer from the options given below :
