JEE Challenger
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PhyMedAdv 2022 • Paper 2

Calculation of Wire Diameter and Area Using Screw Gauge Measurements

The cross-sectional area of a given wire is evaluated with a screw gauge. The pitch associated with the main scale is 0.5 mm0.5\text{ mm}. The circular scale consists of 100100 divisions, and completing one full rotation of this circular scale advances the main scale by two divisions. The observations recorded are given in the table below.

Measurement conditionMain scale readingCircular scale readingTwo arms of gauge touching0 division4 divisionseach other without wireAttempt-1: With wire4 divisions20 divisionsAttempt-2: With wire4 divisions16 divisions\begin{array}{|l|l|l|} \hline \textbf{Measurement condition} & \textbf{Main scale reading} & \textbf{Circular scale reading} \\ \hline \text{Two arms of gauge touching} & 0\text{ division} & 4\text{ divisions} \\ \text{each other without wire} \\ \hline \text{Attempt-1: With wire} & 4\text{ divisions} & 20\text{ divisions} \\ \hline \text{Attempt-2: With wire} & 4\text{ divisions} & 16\text{ divisions} \\ \hline \end{array}

Determine the diameter as well as the cross-sectional area of the wire obtained from these screw gauge measurements.

Options

A

2.22±0.02 mm,π(1.23±0.02) mm22.22 \pm 0.02\text{ mm}, \quad \pi(1.23 \pm 0.02)\text{ mm}^2

B

2.22±0.01 mm,π(1.23±0.01) mm22.22 \pm 0.01\text{ mm}, \quad \pi(1.23 \pm 0.01)\text{ mm}^2

C

2.14±0.02 mm,π(1.14±0.02) mm22.14 \pm 0.02\text{ mm}, \quad \pi(1.14 \pm 0.02)\text{ mm}^2

Correct
D

2.14±0.01 mm,π(1.14±0.01) mm22.14 \pm 0.01\text{ mm}, \quad \pi(1.14 \pm 0.01)\text{ mm}^2

Topics & Concepts

Units and MeasurementsMeasuring Instruments

Step-by-Step Solution

To determine the diameter and the cross-sectional area of the wire from the given screw gauge observations, we follow these step-by-step calculations:

1. Calculation of Least Count (LC)

Given:

  • Pitch of the main scale (1 main scale division, MSD1\text{ main scale division, MSD}) = 0.5 mm0.5\text{ mm}
  • Main scale shift in one full rotation of circular scale = 2 divisions=2×0.5 mm=1.0 mm2\text{ divisions} = 2 \times 0.5\text{ mm} = 1.0\text{ mm}
  • Total circular scale divisions (NN) = 100100

The Least Count (LC\text{LC}) of the screw gauge is: LC=Distance advanced in one full rotationTotal circular scale divisions=1.0 mm100=0.01 mm\text{LC} = \frac{\text{Distance advanced in one full rotation}}{\text{Total circular scale divisions}} = \frac{1.0\text{ mm}}{100} = 0.01\text{ mm}

2. Determination of Zero Error

When the two arms touch without the wire:

  • Main Scale Reading (MSR)=0 divisions=0 mm\text{Main Scale Reading (MSR)} = 0\text{ divisions} = 0\text{ mm}
  • Circular Scale Reading (CSR)=4 divisions\text{Circular Scale Reading (CSR)} = 4\text{ divisions}

Zero Error=MSR+(CSR×LC)=0+(4×0.01 mm)=+0.04 mm\text{Zero Error} = \text{MSR} + (\text{CSR} \times \text{LC}) = 0 + (4 \times 0.01\text{ mm}) = +0.04\text{ mm}

3. Measurements of Wire Diameter

Attempt 1:

  • MSR=4 divisions=4×0.5 mm=2.0 mm\text{MSR} = 4\text{ divisions} = 4 \times 0.5\text{ mm} = 2.0\text{ mm}
  • CSR=20 divisions\text{CSR} = 20\text{ divisions}
  • Observed reading d1′=2.0 mm+(20×0.01 mm)=2.20 mmd_1' = 2.0\text{ mm} + (20 \times 0.01\text{ mm}) = 2.20\text{ mm}
  • Corrected diameter d1=d1′−Zero Error=2.20 mm−0.04 mm=2.16 mmd_1 = d_1' - \text{Zero Error} = 2.20\text{ mm} - 0.04\text{ mm} = 2.16\text{ mm}

Attempt 2:

  • MSR=4 divisions=4×0.5 mm=2.0 mm\text{MSR} = 4\text{ divisions} = 4 \times 0.5\text{ mm} = 2.0\text{ mm}
  • CSR=16 divisions\text{CSR} = 16\text{ divisions}
  • Observed reading d2′=2.0 mm+(16×0.01 mm)=2.16 mmd_2' = 2.0\text{ mm} + (16 \times 0.01\text{ mm}) = 2.16\text{ mm}
  • Corrected diameter d2=d2′−Zero Error=2.16 mm−0.04 mm=2.12 mmd_2 = d_2' - \text{Zero Error} = 2.16\text{ mm} - 0.04\text{ mm} = 2.12\text{ mm}

4. Mean Diameter and Error Analysis

  • Mean Diameter (dd): d=d1+d22=2.16+2.122=2.14 mmd = \frac{d_1 + d_2}{2} = \frac{2.16 + 2.12}{2} = 2.14\text{ mm}

  • Mean Absolute Error (Δd\Delta d): Δd=∣d1−d∣+∣d2−d∣2=∣2.16−2.14∣+∣2.12−2.14∣2=0.02+0.022=0.02 mm\Delta d = \frac{|d_1 - d| + |d_2 - d|}{2} = \frac{|2.16 - 2.14| + |2.12 - 2.14|}{2} = \frac{0.02 + 0.02}{2} = 0.02\text{ mm}

Thus, the measured diameter is: d=2.14±0.02 mmd = 2.14 \pm 0.02\text{ mm}

5. Calculation of Cross-Sectional Area and Error

  • Cross-Sectional Area (AA): A=πd24=π(2.14)24=4.5796π4=1.1449π≈π(1.14) mm2A = \frac{\pi d^2}{4} = \frac{\pi (2.14)^2}{4} = \frac{4.5796 \pi}{4} = 1.1449 \pi \approx \pi (1.14)\text{ mm}^2

  • Absolute Error in Area (ΔA\Delta A): Using standard fractional error propagation for A=πd24A = \frac{\pi d^2}{4}: ΔAA=2Δdd\frac{\Delta A}{A} = 2 \frac{\Delta d}{d} ΔA=2A(Δdd)=2(1.1449π)(0.022.14)≈0.0214π≈π(0.02) mm2\Delta A = 2 A \left(\frac{\Delta d}{d}\right) = 2 \left(1.1449 \pi\right) \left(\frac{0.02}{2.14}\right) \approx 0.0214 \pi \approx \pi (0.02)\text{ mm}^2

Thus, the cross-sectional area is: A=π(1.14±0.02) mm2A = \pi (1.14 \pm 0.02)\text{ mm}^2

Conclusion

  • Diameter: 2.14±0.02 mm2.14 \pm 0.02\text{ mm}
  • Area: π(1.14±0.02) mm2\pi(1.14 \pm 0.02)\text{ mm}^2

This corresponds to Option C.

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Calculation of Wire Diameter and Area Using Screw Gauge Measurements | Physics PYQ Solution - JEE Challenger