JEE MAIN - Physics (2022 - 29th June Evening Shift - No. 19)
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.
Answer
180
Explanation
Since zero error is negative, we will add 0.05 cm.
$$\Rightarrow$$ Corrected reading = 1.7 cm + 5 $$\times$$ 0.1 mm + 0.05 cm
= 180 $$\times$$ 10$$-$$2 cm
Comments (0)
