JEE Main 2022 · 29 Jun · Shift 2 · Physics Q51

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PhysicsInteger+4 / -1Question 51 of 90

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 ______ $\times10^{-2}$ cm.

Your answer

Answer and worked solution

Official answer: 180

Reading. $1.7+5\times0.01=1.75$ cm. Correction for zero error $-0.05$ cm: $1.75+0.05=1.80$ cm $=180\times10^{-2}$ cm.

Answer: 180

Topics:Vernier callipers and screw gauge

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JEE Main 2022 · 29 Jun · Shift 2, Physics question 51