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JEE Advanced 2013 · Paper 1, Physics question 1

The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th^\text{th} division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is

Choose one option
Answer and worked solution

Official answer: B

Least count. 50 Vernier divisions = 2.45 cm, so one VSD = 0.049 cm; one MSD = 0.05 cm. LC = 0.05 − 0.049 = 0.001 cm.

Reading. 5.10+24×0.001=5.1245.10+24\times0.001=5.124 cm.

(B)

Topics:Vernier callipers and screw gauge
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