Practical Physics
Two thin wires, Wire-1 of diameter 0.650 mm and Wire-2 of unknown diameter d are given. To obtain the value of d, the diameters of the two wires are measured with a screw gauge. The screw gauge has a pitch of 0.5 mm and there are 100 divisions on the circular scale (CS). The smallest division on the linear scale (LS) is 0.5 mm. The table shows the readings of LS and CS for the measurements. The value of d (in µm) is :
| LS (mm) | CS | |
|---|---|---|
| Wire-1 | 0.5 | 42 |
| Wire-2 | 1.5 | 95 |
Explanation:
Least count of the screw gauge is
$ \text{Least count} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{0.5\ \text{mm}}{100} = 0.005\ \text{mm} $
Now, observed diameter of Wire-1:
$ \text{Diameter of Wire-1} = \text{LS} + \text{CS} \times \text{LC} $
$ = 0.5 + 42 \times 0.005 $
$ = 0.5 + 0.21 = 0.71\ \text{mm} $
But the actual diameter of Wire-1 is given as
$ 0.650\ \text{mm} $
So, zero error of the screw gauge is
$ \text{Zero error} = 0.71 - 0.650 = 0.060\ \text{mm} $
This is a positive zero error, so zero correction will be
$ -0.060\ \text{mm} $
Now, observed diameter of Wire-2:
$ = 1.5 + 95 \times 0.005 $
$ = 1.5 + 0.475 = 1.975\ \text{mm} $
Applying zero correction:
$ d = 1.975 - 0.060 = 1.915\ \text{mm} $
Convert into $\mu$m:
$ 1\ \text{mm} = 1000\ \mu\text{m} $
So,
$ d = 1.915 \times 1000 = 1915\ \mu\text{m} $
Hence, the value of $d$ is
$ \boxed{1915\ \mu\text{m}} $