JEE Challenger
More from Mechanical Properties of Solids
PhyMedMarks to AllAdv 2007 • Paper 2

Determining Youngs Modulus of a Wire with Uncertainty Using Searles Method

A student performs an experiment to determine the Young's modulus of a wire, exactly 2 m2\text{ m} long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm0.8\text{ mm} with an uncertainty of ±0.05 mm\pm 0.05\text{ mm} at a load of exactly 1.0 kg1.0\text{ kg}. The student also measures the diameter of the wire to be 0.4 mm0.4\text{ mm} with an uncertainty of ±0.01 mm\pm 0.01\text{ mm}. Take g=9.8 m/s2g = 9.8\text{ m/s}^2 (exact). The Young's modulus obtained from the reading is

Official Notice: Marks Awarded to All

This question was dropped / full marks were awarded to all candidates in the official answer key by the exam conducting body due to an ambiguity or error in the question or options.

Options

A

(2.0±0.3)×1011 N/m2(2.0 \pm 0.3) \times 10^{11}\text{ N/m}^2

B

(2.0±0.2)×1011 N/m2(2.0 \pm 0.2) \times 10^{11}\text{ N/m}^2

C

(2.0±0.1)×1011 N/m2(2.0 \pm 0.1) \times 10^{11}\text{ N/m}^2

D

(2.0±0.05)×1011 N/m2(2.0 \pm 0.05) \times 10^{11}\text{ N/m}^2

Topics & Concepts

Mechanical Properties of SolidsElasticityMeasuring Instruments

Step-by-Step Solution

To determine the Young's modulus YY of the wire and its uncertainty, we use the formula from Searle's method:

Y=StressStrain=F/Al/L=MgL(πd24)l=4MgLπd2lY = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{l/L} = \frac{M g L}{\left(\frac{\pi d^2}{4}\right) l} = \frac{4MgL}{\pi d^2 l}

where:

  • Length of the wire, L=2 mL = 2\text{ m} (exact)
  • Load, M=1.0 kgM = 1.0\text{ kg} (exact)
  • Acceleration due to gravity, g=9.8 m/s2g = 9.8\text{ m/s}^2 (exact)
  • Extension, l=0.8 mm=0.8×10−3 ml = 0.8\text{ mm} = 0.8 \times 10^{-3}\text{ m}, with Δl=0.05 mm\Delta l = 0.05\text{ mm}
  • Diameter, d=0.4 mm=0.4×10−3 md = 0.4\text{ mm} = 0.4 \times 10^{-3}\text{ m}, with Δd=0.01 mm\Delta d = 0.01\text{ mm}

1. Calculation of the Value of YY:

Substituting the given values:
Y=4×1.0×9.8×2π×(0.4×10−3)2×(0.8×10−3)Y = \frac{4 \times 1.0 \times 9.8 \times 2}{\pi \times (0.4 \times 10^{-3})^2 \times (0.8 \times 10^{-3})}

Y=78.4π×0.16×10−6×0.8×10−3=78.40.40212×10−9≈1.95×1011 N/m2≈2.0×1011 N/m2Y = \frac{78.4}{\pi \times 0.16 \times 10^{-6} \times 0.8 \times 10^{-3}} = \frac{78.4}{0.40212 \times 10^{-9}} \approx 1.95 \times 10^{11}\text{ N/m}^2 \approx 2.0 \times 10^{11}\text{ N/m}^2

2. Calculation of Uncertainty ΔY\Delta Y:

Since MM, gg, and LL are exact quantities, the relative error in YY depends only on the measurements of dd and ll:

ΔYY=2(Δdd)+Δll\frac{\Delta Y}{Y} = 2\left(\frac{\Delta d}{d}\right) + \frac{\Delta l}{l}

Substituting the fractional uncertainties:
Δdd=0.01 mm0.4 mm=0.025\frac{\Delta d}{d} = \frac{0.01\text{ mm}}{0.4\text{ mm}} = 0.025

Δll=0.05 mm0.8 mm=0.0625\frac{\Delta l}{l} = \frac{0.05\text{ mm}}{0.8\text{ mm}} = 0.0625

Thus, the relative uncertainty in YY is:
ΔYY=2(0.025)+0.0625=0.05+0.0625=0.1125\frac{\Delta Y}{Y} = 2(0.025) + 0.0625 = 0.05 + 0.0625 = 0.1125

Now, calculating the absolute uncertainty ΔY\Delta Y:
ΔY=0.1125×Y≈0.1125×(2.0×1011 N/m2)=0.225×1011 N/m2\Delta Y = 0.1125 \times Y \approx 0.1125 \times (2.0 \times 10^{11}\text{ N/m}^2) = 0.225 \times 10^{11}\text{ N/m}^2

(Using the unrounded value Y≈1.95×1011 N/m2Y \approx 1.95 \times 10^{11}\text{ N/m}^2, ΔY≈0.219×1011 N/m2\Delta Y \approx 0.219 \times 10^{11}\text{ N/m}^2.)

3. Reporting with Significant Figures:

  • If standard rounding off to one significant figure is applied:
    ΔY≈0.2×1011 N/m2  ⟹  Y=(2.0±0.2)×1011 N/m2\Delta Y \approx 0.2 \times 10^{11}\text{ N/m}^2 \implies Y = (2.0 \pm 0.2) \times 10^{11}\text{ N/m}^2
    which corresponds to Option (B).

  • If the conservative convention of experimental error analysis is followed (where uncertainties are rounded up to the nearest upper digit so as not to underestimate the experimental error):
    ΔY≈0.3×1011 N/m2  ⟹  Y=(2.0±0.3)×1011 N/m2\Delta Y \approx 0.3 \times 10^{11}\text{ N/m}^2 \implies Y = (2.0 \pm 0.3) \times 10^{11}\text{ N/m}^2
    which corresponds to Option (A).

Therefore, both (A) and (B) are correct answers.

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