Question 14: Units , Dimensions and Error Analysis

Question

In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in circular scale and pitch of screw gauge is 0.1 mm. When diameter of a sphere is measured, the reading of main scale is 5 mm and 50
th
division of circular scale coincides with the reference line of main scale. The diameter of sphere is ______ mm.

Accepted Answer

For a screw gauge, first we find the least count.
$$ 	ext{Least count}=\frac{	ext{Pitch}}{	ext{Number of circular scale divisions}} =\frac{0.1\ 	ext{mm}}{100}=0.001\ 	ext{mm} $$
Now let us find the zero error.
When the two studs are in contact, the zero of main scale reference line coincides with the $5^	ext{th}$ division of circular scale.
This means the screw gauge shows a reading of
$$ 5 	imes 0.001 = 0.005\ 	ext{mm} $$
So the instrument has a positive zero error of
$$ +0.005\ 	ext{mm} $$
Hence, zero correction is
$$ -0.005\ 	ext{mm} $$
Now for the sphere:
Main scale reading $= 5\ 	ext{mm}$
Circular scale reading $= 50$
So observed reading is
$$ 5 + 50 	imes 0.001 = 5 + 0.050 = 5.050\ 	ext{mm} $$
Applying zero correction:
$$ 	ext{Correct reading} = 5.050 - 0.005 = 5.045\ 	ext{mm} $$
Therefore, the diameter of the sphere is
$$ \boxed{5.045\ 	ext{mm}} $$
So, the correct option is
A
.

Answer

5.045

Answer

5.055

Answer

5.450

Answer

5.550

Units , Dimensions and Error Analysis

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