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

Engineering

Physics

Vernier Calipers and Travelling Microscope

Question

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.

A screw gauge having 100 divisions in the circular scale and pitch as 1mm.

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

A meter scale.

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The student measured 3.50 cm, which has three significant figures with precision to 0.01 cm (0.1 mm). This precision indicates a vernier caliper or screw gauge was used, not a meter scale (precision ~0.1 cm).

For vernier caliper: Least Count (LC) = $\frac{1 MSD}{10}$ = $\frac{0.1 cm}{10}$ = 0.01 cm. This matches the measurement's precision.

For screw gauges: LC = $\frac{Pitch}{No. of divisions}$ . Options give LC=0.01mm (0.001cm) or 0.02mm (0.002cm), which are more precise than 0.01 cm.

Final Answer: 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.