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A student recorded the length of a rod as 3.50 cm. Which measuring instrument was most likely used?
- A metre scale
- A vernier caliper in which 10 vernier divisions coincide with 9 main-scale divisions, and the main scale is marked in cm with 10 divisions per cm
- A screw gauge with a pitch of 1 mm and 100 divisions on the circular scale
- A screw gauge with a pitch of 1 mm and 50 divisions on the circular scale
Correct answer: A vernier caliper in which 10 vernier divisions coincide with 9 main-scale divisions, and the main scale is marked in cm with 10 divisions per cm
Solution
A reading of 3.50 cm is precise to 0.01 cm (0.1 mm), so the instrument's least count must be 0.1 mm. For the vernier where 10 VSD = 9 MSD with 1 MSD = 1 mm, LC = MSD/10 = 0.1 mm, which exactly matches. The screw gauges (LC = 0.01 mm or 0.02 mm) would give three decimal places in cm, more precision than recorded.
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