A student measure length of an object is 2.70 mm. Find percentage error in this measurement
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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.
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