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

Answer 1

The correct option is D 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.

The measured distance by the student is 3.50 cm which means the least count of the measurement done is 0.01 cm.

A screw gauge having 100 divisions on a circular scale with a pitch of 1 mm will have a least count of (1 mm)/(100) i.e. 0.01 mm

similarly screw gauge having 50 divisions on a circular scale with a pitch of 1 mm will have a least count of (1 mm)/(50) i.e. 0.5 mm

A meter scale has a least count of 1 mm.

And the least count of a vernier caliper which has 10 divisions in 1 cm on main scale and 10 divisions in vernier scale match with 9 division in main scale can be given as

1 cm/(10*10) = 0.01 cm

Thus the student must have used the vernier caliper for the measurement.