Errors in measurementMedium
Question
In a screw gauge when the circular scale is given five complete rotations it moves linearly by 2.5 mm . If the circular scale has 100 divisions, the least count of screw gauge is $\_\_\_\_$ mm.
Options
A.$1 \times 10^{- 2}$
B.$1 \times 10^{- 3}$
C.$5 \times 10^{- 2}$
D.$5 \times 10^{- 3}$
Solution
Sol. [Key Concept: Pitch $= \frac{linear\ travel}{rotations}$; least count $= \frac{pitch}{number\ of\ circular\ divisions}$.]
Pitch ⇒ $\frac{2.5\text{ }mm}{5} = 0.5\text{ }mm = 5 \times 10^{- 4}\text{ }m$
L.C. ⇒ $\frac{\text{~}\text{Pitch}\text{~}}{\text{~}\text{No. of divisions}\text{~}} = \frac{5 \times 10^{- 4}}{100}\text{ }m$
$${= 5 \times 10^{- 6}\text{ }m }{= 5 \times 10^{- 3}\text{ }mm }$$
Create a free account to view solution
View Solution FreeMore Errors in measurement Questions
A student performs an experiment to determine the Young′s modulus of a wire, exactly 2 m long, by Searle′s m...In a vernier, N divisions of vernier scale coincide with (N - 1) divisions of main scale (in which 1 division) represent...The length of a cylinder is measured with a meter rod having least count 0.1 cm. Its diameter is measured with vernier c...The period of oscillation of a simple pendulum in the experiment is recorded as 2.63s, 2.56s, 2.42s, 2.71s and 2.80s res...The length of the string of a simple pendulum is measured with a meter scale to be 63.5 cm, the radius of the bob plus t...