JEE Main | 2014Errors in measurementMedium
Question
A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?
Options
A.A meter scale
B.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
C.A screw gauge having 100 divisions in the circular scale and pitch as 1 mm
D.A screw gauge having 50 divisions in the circular scale and pitch as 1 mm
Solution
As measured value is 3.50 cm, the least count must be 0.01 cm = 0.1 mm
For vernier scale with 1 MSD = 1 mm and 9 MSD = 10 VSD,
Least count = 1 MSD - 1 VSD
= 0.1 mm
For vernier scale with 1 MSD = 1 mm and 9 MSD = 10 VSD,
Least count = 1 MSD - 1 VSD
= 0.1 mm
Create a free account to view solution
View Solution FreeMore Errors in measurement Questions
In an experiment the values of two spring constants were measured as $k_{1} = (10 \pm 0.2)N/m$ and $k_{2} = (20 \pm 0.3)...Significant figure in 3400 are -...The external and internal radius of a hollow cylinder are measured to be (4.23 ± 0.01) cm and (3.89 ± 0.01)cm....When both jaws of vernier callipers touch each other, zero mark of the vernier scale is right to zero mark of main scale...An experiment measures quantities a, b and c, and X is calculated from X = ab2/c3. If the percentage error in a, b and c...