Question 10

Step 1: Calculate the Least Count (LC) of the screw gauge. $\text{LC} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}} = \frac{0.5 \text{ mm}}{100} = 0.005 \text{ mm}$

Step 2: Calculate the measured diameter of Wire-1 and determine the zero error. $\text{Measured Reading}_1 = \text{LS} + (\text{CS} \times \text{LC}) = 0.5 + (42 \times 0.005) = 0.5 + 0.21 = 0.71 \text{ mm}$ Given actual diameter of Wire-1 = 0.650 mm. $\text{Zero Error} (e) = \text{Measured Reading} - \text{Actual Reading} = 0.71 - 0.65 = 0.06 \text{ mm}$

Step 3: Calculate the measured diameter of Wire-2 and find the true diameter $d$. $\text{Measured Reading}_2 = 1.5 + (95 \times 0.005) = 1.5 + 0.475 = 1.975 \text{ mm}$ $\text{True diameter } d = \text{Measured Reading}_2 - e = 1.975 - 0.06 = 1.915 \text{ mm}$

Step 4: Convert the diameter to $\mu m$. $d = 1.915 \times 1000 = 1915 \mu m$

Final Answer: 1915