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A student measured the length of a rod and wrote it as 3.50 cm which instrument did he use to measure it?

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A vernier calipers where the 10 divisions in vernier scale matches with 9 divisions in main scale which has 10 divisions in 1 cm

A meter scale.

A screw gauge having 100 divisions in the circulas scale and pitch is 1 mm.

A screw gauge having 50 division in the circular scale and pitch as 1 mm

answer is C.

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Detailed Solution

The measurement is 3.50cm, so the least count of the instrument used must be 0.01cm.

$For vernier scale with 1 MSD = 1 cm$

For Vernier scale with 1MSD=1cm and 9MSD=10VSD.

LC of Vernier calliper =1MSD-1VSD.

$= \frac{1}{10} (1 - \frac{9}{10}) = \frac{1}{100} cm$

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