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 ______ $$\times 10^{-2}$$ cm.

We need to find the corrected diameter measured using Vernier callipers. Vernier constant (least count) = 0.1 mm = 0.01 cm, coinciding Vernier division = 5, Vernier scale reading = 5 × 0.01 = 0.05 cm.

Observed reading = Main scale reading + Vernier scale reading $$= 1.7 + 0.05 = 1.75 \text{ cm}$$.

Zero error = −0.05 cm (negative zero error). Corrected reading = Observed reading − Zero error $$= 1.75 - (-0.05) = 1.75 + 0.05 = 1.80 \text{ cm}$$.

$$1.80 \text{ cm} = 180 \times 10^{-2} \text{ cm}$$. The answer is 180 $$\times 10^{-2}$$ cm.