Question 10

NUMERICALMEDIUM

Two thin wires, Wire-1 of diameter 0.650 mm and Wire-2 of unknown diameter dd are given. To obtain the value of dd, 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.

WireLS (mm)CS (Readings)Wire-10.542Wire-21.595\begin{array}{|l|c|c|} \hline \text{Wire} & \text{LS (mm)} & \text{CS (Readings)} \\ \hline \text{Wire-1} & 0.5 & 42 \\ \text{Wire-2} & 1.5 & 95 \\ \hline \end{array}

The value of dd (in μm\mu m) is:

Correct Answer: 1915

Detailed Solution

Step 1: Calculate the Least Count (LC) of the screw gauge. LC=PitchNumber of circular scale divisions=0.5 mm100=0.005 mm\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. Measured Reading1=LS+(CS×LC)=0.5+(42×0.005)=0.5+0.21=0.71 mm\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. Zero Error(e)=Measured Reading−Actual Reading=0.71−0.65=0.06 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 dd. Measured Reading2=1.5+(95×0.005)=1.5+0.475=1.975 mm\text{Measured Reading}_2 = 1.5 + (95 \times 0.005) = 1.5 + 0.475 = 1.975 \text{ mm} True diameter d=Measured Reading2−e=1.975−0.06=1.915 mm\text{True diameter } d = \text{Measured Reading}_2 - e = 1.975 - 0.06 = 1.915 \text{ mm}

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

Final Answer: 1915

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