In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in circular scale and pitch of screw gauge is 0.1 mm. When diameter of a sphere is measured, the reading of main scale is 5 mm and 50^ th division of circular scale coincides with the reference line of main scale. The diameter of sphere is _______ mm.

Step-by-step solution

Least count of the screw gauge is given by: LC = Pitch Number of divisions on circular scale LC = 0.1 100 = 0.001 mm When the two studs are in contact, the 5^ th division of the circular scale coincides with the reference line. This indicates a positive zero error. Zero error = +5 LC = +5 0.001 = +0.005 mm For the measurement of the sphere's diameter: Main scale reading (MSR) = 5 mm Circular scale reading (CSR) = 50 Measured diameter = MSR + CSR LC Measured diameter = 5 + 50 0.001 = 5.050 mm The true diameter is ob