JEE Challenger
More from Mechanical Properties of Solids
PhyMedAdv 2014 • Paper 1

Maximum Percentage Error in Youngs Modulus from Searles Experiment

During a Searle's experiment, the zero mark of the Vernier scale lies between 3.20×10−2 m3.20 \times 10^{-2}\text{ m} and 3.25×10−2 m3.25 \times 10^{-2}\text{ m} on the main scale. The 20th20^{\text{th}} division of the Vernier scale aligns perfectly with one of the main scale marks. When an extra load of 2 kg2\text{ kg} is attached to the wire, the zero of the Vernier scale remains between 3.20×10−2 m3.20 \times 10^{-2}\text{ m} and 3.25×10−2 m3.25 \times 10^{-2}\text{ m} of the main scale, but now the 45th45^{\text{th}} Vernier division coincides with a main scale division. The length of the thin wire is 2 m2\text{ m} and its cross-sectional area is 8×10−7 m28 \times 10^{-7}\text{ m}^2. The least count of the Vernier scale is 1.0×10−5 m1.0 \times 10^{-5}\text{ m}. The maximum percentage error in calculating the Young's modulus of the wire is

Official Numerical Answer4

Topics & Concepts

Mechanical Properties of SolidsElasticityMeasuring Instruments

Step-by-Step Solution

To find the maximum percentage error in calculating the Young's modulus of the wire, we use the formula for Young's modulus (YY):

Y=F⋅LA⋅ΔLY = \frac{F \cdot L}{A \cdot \Delta L}

where:

  • FF is the force applied due to the load,
  • LL is the original length of the wire,
  • AA is the cross-sectional area of the wire,
  • ΔL\Delta L is the elongation of the wire.

Step 1: Calculate the elongation (ΔL\Delta L)

The initial reading R1R_1 of the Vernier scale is given by: R1=MSR+20×LCR_1 = \text{MSR} + 20 \times \text{LC}

The final reading R2R_2 after applying the additional load of 2 kg2\text{ kg} is given by: R2=MSR+45×LCR_2 = \text{MSR} + 45 \times \text{LC}

Since the Main Scale Reading (MSR=3.20×10−2 m\text{MSR} = 3.20 \times 10^{-2}\text{ m}) remains the same in both cases, the elongation ΔL\Delta L is:

ΔL=R2−R1=(45−20)×LC=25×LC\Delta L = R_2 - R_1 = (45 - 20) \times \text{LC} = 25 \times \text{LC}

Given the least count LC=1.0×10−5 m\text{LC} = 1.0 \times 10^{-5}\text{ m}:

ΔL=25×1.0×10−5 m=2.5×10−4 m\Delta L = 25 \times 1.0 \times 10^{-5}\text{ m} = 2.5 \times 10^{-4}\text{ m}

Step 2: Determine the maximum percentage error in YY

Since the parameters FF, LL, and AA are given as exact values without specified measurement uncertainties, the error in Young's modulus YY arises purely from the error in the measurement of elongation ΔL\Delta L.

The maximum error in measuring elongation using this scale is equal to the least count of the instrument: δ(ΔL)=LC=1.0×10−5 m\delta(\Delta L) = \text{LC} = 1.0 \times 10^{-5}\text{ m}

The relative error in Young's modulus YY is: ΔYY=δ(ΔL)ΔL\frac{\Delta Y}{Y} = \frac{\delta(\Delta L)}{\Delta L}

Therefore, the maximum percentage error is: Percentage Error=(δ(ΔL)ΔL)×100%=(1.0×10−5 m25×10−5 m)×100%=125×100%=4%\text{Percentage Error} = \left( \frac{\delta(\Delta L)}{\Delta L} \right) \times 100\% = \left( \frac{1.0 \times 10^{-5}\text{ m}}{25 \times 10^{-5}\text{ m}} \right) \times 100\% = \frac{1}{25} \times 100\% = 4\%

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