Question 40: Units , Dimensions and Error Analysis

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.

Accepted Answer

Given:
In $5$ complete rotations, the screw moves linearly by $2.5 \, 	ext{mm}$
Circular scale has $100$ divisions
First, find the pitch of the screw gauge.
Pitch means the linear distance moved in one complete rotation.
$ 	ext{Pitch} = \frac{2.5}{5} = 0.5 \, 	ext{mm} $
Now, least count is given by:
$ 	ext{Least Count} = \frac{	ext{Pitch}}{	ext{Number of circular scale divisions}} $
So,
$ 	ext{Least Count} = \frac{0.5}{100} = 0.005 \, 	ext{mm} $
$ 0.005 \, 	ext{mm} = 5 	imes 10^{-3} \, 	ext{mm} $
So, the correct answer is:
$ \boxed{5 	imes 10^{-3} \, 	ext{mm}} $
Option D

Answer

$1 	imes 10^{-2}$

Answer

$ 1 	imes 10^{-3}$

Answer

$5 	imes 10^{-2}$

Answer

$ 5 	imes 10^{-3}$

Units , Dimensions and Error Analysis

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.

Choose the correct answer: