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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 :A meter scaleA vernier calliper where the 10 divisions in vernier scale matches with 9 divisions in main scale and main scale has 10 divisions in 1 cmA screw gauge having 100 divisions in the circular scale and pitch is 1 mmA screw gauge having 50 divisions in the circular scale and pitch is 1 mm

Question

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

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Solution

The student used a Vernier Calliper where the 10 divisions in the Vernier scale matches with 9 divisions in the main scale and the main scale has 10 divisions in 1 cm.

Here's why:

  1. A meter scale typically has a least count of 1 mm or 0.1 cm, so it cannot measure up to two decimal places (i.e., 3.50 cm).

  2. A Vernier Calliper where the 10 divisions in the Vernier scale matches with 9 divisions in the main scale and the main scale has 10 divisions in 1 cm has a least count of 0.1 mm or 0.01 cm. This can measure up to two decimal places.

  3. A screw gauge having 100 divisions in the circular scale and pitch is 1 mm has a least count of 0.01 mm or 0.001 cm. This can measure up to three decimal places.

  4. A screw gauge having 50 divisions in the circular scale and pitch is 1 mm also has a least count of 0.02 mm or 0.002 cm. This can measure up to three decimal places.

Therefore, the student must have used the Vernier Calliper to measure the length of the rod as 3.50 cm.

This problem has been solved

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