Question

A screw gauge has a positive error of 4 divisions. When this screw gauge holds a sphere, the main scale reading is 4 mm and the head scale coinciding division is 24. If it's least count is 0.01 mm, find out the volume of the sphere.​

Answer 1

Answer:

Positive error   =4×0.01=0.04 mm

Diameter of sphere   d=M.S.R+n×L.C. − zero error

Or   d=4+24×0.01−0.04=4.20 mm

Volume of sphere   V=34π(2d)3

⟹ V=34π×(24.20)3=38.808 mm3