Errors in measurementHard
Question
In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division $= 0.05\text{ }mm$, then the least count of the vernier callipers is $\_\_\_\_$ mm .
Options
A.0.002
B.0.05
C.0.02
D.0.005
Solution
$LC = 1MSD - 1MSD = 1MSD - \frac{48}{50}MSD$
$$= \frac{2}{50}MSD = \frac{2}{50} \times .05\text{ }mm = 0.002\text{ }mm$$
Create a free account to view solution
View Solution FreeMore Errors in measurement Questions
In an experiment, a set of reading are obtained $- 1.24\text{ }mm,1.25\text{ }mm,1.23\text{ }mm,1.21\text{ }mm$. The exp...If the percentage error in the measurement of radius of a sphere is 2%, then the maximum percentage errors in the measur...If error in measuring diamete r of a circle is 4%, the error in the radius of the circle would be...Accuracy and precision are __(i)___and these are respectively linked with ___(ii)___ & ___(iii)___ .Fill the blanks abov...Four persons measure the length of a rod as $20.00\text{ }cm,19.75\text{ }cm,17.01\text{ }cm$ and 18.25 cm . The relativ...