From the figure given below find the least count of the screw guage and the diameter of the wire...pls fast I have my exam tomorrow ...
Dear student
(a) pitch of the screw gauge is 1 mm =0.1 cm
(b) Least count = smallest division on main scale / total number of division on circular scale
= 0.1 cm/50 =0.002 cm
(c) main scale reading a=4 mm= 0.4 cm
circular scale reading b = 47
Diameter of wire = a + (b x L.C) = 0.4+(47x 0.002) =0.494 cm
Regards
(a) pitch of the screw gauge is 1 mm =0.1 cm
(b) Least count = smallest division on main scale / total number of division on circular scale
= 0.1 cm/50 =0.002 cm
(c) main scale reading a=4 mm= 0.4 cm
circular scale reading b = 47
Diameter of wire = a + (b x L.C) = 0.4+(47x 0.002) =0.494 cm
Regards