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 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 1

Correct Answer - (d)
Since, the measured value is 3.50 cm, the least count of meassuring instrument must be 0.01 cm. it is so for the given vernier calliper. Least count for vernier calliper = Vernier constant `= 1 MSD -1 VSD = 1 MSD - (9//10) MSD`
` = (1)/(10) (MSD) = (1)/(10)xx0.1 cm = 0.01 cm`