A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it? (a) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm. (b) A screw gauge having 50 divisions in the circular scale and pitch as 1 mm. (c) A meter scale. (d) A vernier calliper where the 10 divisions in vernier scale matches with 9 divisions in main scale and main scale has 10 divisions in 1 cm.

Answer: (d). 3.50 cm is written to 0.01 cm, so the instrument's least count is 0.01 cm = 0.1 mm. Screw gauges in (a), (b) have LC 0.01 mm and 0.02 mm (would give 3 decimals in cm); a metre scale has LC 1 mm. Vernier in (d): MSD = 1 mm, 10 VSD = 9 MSD ⇒ LC = 0.1 mm. So the answer is (d).

A student measured the length of a rod and wrote it as 3.50 cm. Which… | JEE Main 2014

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Least Count, Vernier Callipers & Screw Gauge

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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Answer: (d)

3.50 cm is written to 0.01 cm, so the instrument's least count is $0.01$ cm $= 0.1$ mm.
Screw gauges in (a), (b) have LC $0.01$ mm and $0.02$ mm (would give 3 decimals in cm); a metre scale has LC 1 mm.
Vernier in (d): MSD $= 1$ mm, $10$ VSD $= 9$ MSD $\Rightarrow$ LC $= 0.1$ mm.
So the answer is (d).

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