In an experiment to find out the diameter of wire using screw gauge, the following observation were noted: (a) Screw moves 0 . 5 mm on main scale in one complete rotation (b) Total divisions on circular scale = 50 (c) Main scale reading is 2 . 5 mm (d) 45 th division of circular scale is in the pitch line (e) Instrument has 0 . 03 mm negative error Then the diameter of wire is :

Step-by-step solution

From the data given in the question, Pitch of the screw gauge = 0 . 5   mm . Total divisions on the circular scale n = 50 . Therefore, the least count of the screw gauge is LC = pitch n = 0 . 5 50 = 0 . 01   mm Now, MSR = 2 . 5   mm , CSR = 45 Diameter reading = MSR + LC × CSR - zero error ⇒ d = 2 . 5 + 0 . 45 - - 0 . 03 ⇒ d = 2 . 98   mm