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 vermier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm

B. A screw gausge having 100 divisions in the circular scale and pitch as 1 mm

C. A meter scale

D. A screw gauge have 50 divisions in the circular scale and pitch as 1 mm

Answer 1

The correct option is (A) A vermier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm

If student measure 3.50 cm it means that there in an uncertianly of order 0.01 cm.

L.C of V.C = 1 MSD−1 VSD

= \(\frac{1}{10} [1 - \frac{9}{10}] = \frac{1}{100} cm\)