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Question

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

Options

A.A meter scale
B.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
C.A screw gauge having 100 divisions in the circular scale and pitch as 1 mm
D.A screw gauge having 50 divisions in the circular scale and pitch as 1 mm

Solution

As measured value is 3.50 cm, the least count must be 0.01 cm = 0.1 mm
For vernier scale with 1 MSD = 1 mm and 9 MSD = 10 VSD,
Least count = 1 MSD - 1 VSD
        = 0.1 mm

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