The Vernier constant of Vernier callipers is 0 . 1 mm and it has zero error of - 0 . 05 cm . While measuring diameter of a sphere, the main scale reading is 1 . 7 cm and coinciding vernier division is 5 . The corrected diameter will be _____ × 10 - 2 cm .

Step-by-step solution

Vernier constant is the least count of the vernier V . C = 0 . 1   mm = 0 . 01   cm zero error = - 0 . 05   cm ∴ correction = 0 . 05   cm M . S . R . = 1 . 7   cm V . S . R . = 5   V . C . = 5 × 0 . 01 = 0 . 05   cm The diameter of the sphere will be, diameter = MSR + VSR + correction 1 . 7 + 0 . 05 + 0 . 05 = 1 . 8   cm = 180 × 10 - 2   cm