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

Options

A. A screw gauge having 100 division in the circular scale and pitch as 1mm
B. A screw gauge having 50 division in the circular scale and pitch as 1mm.
C. A meter scale.
D. A vernier calliper where 10 division in vernier scale match with 9 division in main scale and main scale has 10 division in 1 cm.

Answer

(d)

Explanation

As the measured value is 3.50 cm, the least count must be = 0.01 cm = 0.1 mm For given screw gauge in option (a), least count `= (1 mm)/(100) = 0.01mm`
Least count in option (b) `= (1mm)/(50) = 0.02 mm`
Least count in option (c ) =1 mm
Least count in option (d) = 1 MSD - 1 VSD
`=1MSD - (9)/(10)MSD`
`=(1)/(10) MSD = (1)/(10)xx 1mm = 0.1 mm`
Thus vernier calliper is usedc for the measurement. Hence option (d) is true.