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

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answer is 180.

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Detailed Solution

Measured diameter = MSR + VSR × VC
$= 1 .7 + 0 .01 \times 5$
= 1.75
Corrected = Measured – Error
= 1.75 – (–0.05)
= 1.80 cm
$= 180 \times 10^{- 2} cm$
hence answer is 180.

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