Correct option is (C)

1. Pitch of the screw gauge: 1 mm


2. Number of divisions on the circular scale: 100 divisions


3. Zero error: The zero of the circular scale lies 5 divisions below the reference line, indicating a positive zero error.


4. Measurement data:

• 4 linear scale divisions are visible.


• 60 divisions on the circular scale coincide with the reference line.

Step-by-Step Calculation

1. Least Count of the screw gauge:

$\text{Least Count}=\frac{\text{Pitch}}{\text{Number of Divisions on Circular Scale}}=\frac{1\text{ mm}}{100}=0.01\text{ mm}$

1. Main Scale Reading (MSR):

The linear scale shows 4 divisions, so the main scale reading is:

$\text{MSR}=4\text{ mm}$

1. Circular Scale Reading (CSR):

60 divisions coincide with the reference line, so the circular scale reading is:

$\text{CSR}=60\times \text{Least Count}=60\times 0.01\text{ mm}=0.60\text{ mm}$

1. Zero Error:

The zero error is 5 divisions below the reference line, indicating a positive zero error:

$\text{Zero Error}=+5\times \text{Least Count}=+5\times 0.01\text{ mm}=+0.05\text{ mm}$

1. Total Reading without considering zero error:

$\text{Total Reading (without zero error)}=\text{MSR}+\text{CSR}=4\text{ mm}+0.60\text{ mm}=4.60\text{ mm}$

1. Corrected Reading considering zero error:

Since the zero error is positive, we subtract it from the total reading:

$\text{Corrected Reading}=\text{Total Reading (without zero error)}-\text{Zero Error}=4.60\text{ mm}-0.05\text{ mm}=4.55\text{ mm}$

Conclusion

The diameter of the wire is:

Option C: 4.55 mm

Correct option is (B)

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.