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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Correct Answer

(A) A vernier calliper where the \(10\) divisions in vernier scale matches with \(9\) division in main scale and main scale has \(10\) divisions in \(1\; cm\).

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Detailed explanation

If student measures \(3.50 \mathrm{cm},\) it means that there is an uncertainly of order \(0.01 \mathrm{cm} .\)

For vernier scale with \(1 \mathrm{MSD}=\frac{1}{10} \mathrm{cm}\) and \(9 \mathrm{MSD}=10 \mathrm{VSD}\)

\(LC\) of vernier calliper \(=1 \mathrm{MSD}-1 \mathrm{VSD}\)

\(=\frac{1}{10}\left(1-\frac{9}{10}\right)=\frac{1}{100} \mathrm{cm}=0.01 \mathrm{cm}\)

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