Question 78: Units , Dimensions and Error Analysis

Question

While measuring diameter of wire using screw gauge the following readings were noted. Main scale reading is $$1 \mathrm{~mm}$$ and circular scale reading is equal to 42 divisions. Pitch of screw gauge is $$1 \mathrm{~mm}$$ and it has 100 divisions on circular scale. The diameter of the wire is $$\frac{x}{50} \mathrm{~mm}$$. The value of $$x$$ is :

Accepted Answer

To determine the diameter of the wire using a screw gauge, we employ the formula:
$ 	ext{Total reading} = 	ext{MSR} + (	ext{CSR} 	imes 	ext{LC}) $
where:
MSR (Main Scale Reading) is the value read directly from the main scale in mm.
CSR (Circular Scale Reading) is the number of divisions observed on the circular scale.
LC (Least Count) is the value of one division on the circular scale, calculated as $\frac{	ext{Pitch}}{	ext{Number of divisions on the circular scale}}$.
Given:
Main Scale Reading (MSR) = $1 \mathrm{~mm}$
Circular Scale Reading (CSR) = 42 divisions
Pitch of screw gauge = $1 \mathrm{~mm}$
Number of divisions on circular scale = 100
First, we find the Least Count (LC):
$ LC = \frac{	ext{Pitch}}{	ext{Number of divisions on the circular scale}} = \frac{1 \mathrm{~mm}}{100} = 0.01 \mathrm{~mm} $
Then we calculate the total measurement of the diameter of the wire:
$ 	ext{Total reading} = 	ext{MSR} + (	ext{CSR} 	imes 	ext{LC}) = 1 \mathrm{~mm} + (42 	imes 0.01 \mathrm{~mm}) = 1 \mathrm{~mm} + 0.42 \mathrm{~mm} = 1.42 \mathrm{~mm} $
Given that the diameter of the wire is also represented as $$\frac{x}{50} \mathrm{~mm}$$, we can equate this to our found total reading:
$ 1.42 \mathrm{~mm} = \frac{x}{50} \mathrm{~mm} $
Solving for $$x$$:
$ 1.42 = \frac{x}{50} $
$ x = 1.42 	imes 50 $
$ x = 71 $
Therefore, the value of $$x$$ is 71, which corresponds to Option B: 71.

Answer

42

Answer

71

Answer

21

Answer

142

Units , Dimensions and Error Analysis

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