To determine the diameter of the wire using a screw gauge, we employ the formula:

$ \text{Total reading} = \text{MSR} + (\text{CSR} \times \text{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 $\frac{\text{Pitch}}{\text{Number of divisions on the circular scale}}$.

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{\text{Pitch}}{\text{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:

$ \text{Total reading} = \text{MSR} + (\text{CSR} \times \text{LC}) = 1 \mathrm{~mm} + (42 \times 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 \times 50 $

$ x = 71 $

Therefore, the value of $$x$$ is 71, which corresponds to Option B: 71.