JEE Challenger
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PhyMedMain 2026 • 2nd Apr - Shift 1

Calculate Error in Youngs Modulus Measurement Using Screw Gauge and Scale

The diameter of a wire measured by a screw gauge of least count 0.001 cm0.001\text{ cm} is 0.08 cm0.08\text{ cm}. The length measured by a scale of least count 0.1 cm0.1\text{ cm} is 150 cm150\text{ cm}. When a weight of 100 N100\text{ N} is applied to the wire, the extension in length is 0.5 cm0.5\text{ cm}, measured by a micrometer of least count 0.001 cm0.001\text{ cm}. The error in the measured Young's modulus is α×109 N/m2\alpha \times 10^9\text{ N/m}^2. The value of α\alpha is _______. (Ignore the contribution of the load to Young's modulus error calculation)

Options

A

1.31.3

B

1.651.65

Correct
C

0.130.13

D

0.250.25

Topics & Concepts

Units and MeasurementsMeasuring InstrumentsElasticity

Step-by-Step Solution

To find the error in the measured Young's modulus, we first express Young's modulus YY in terms of the given parameters:

Y=StressStrain=F/AΔL/L=F⋅L(πD24)⋅ΔL=4FLπD2ΔLY = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L} = \frac{F \cdot L}{\left(\frac{\pi D^2}{4}\right) \cdot \Delta L} = \frac{4FL}{\pi D^2 \Delta L}

Given values:

  • Load applied, F=100 NF = 100\text{ N} (error in load is neglected)
  • Original length, L=150 cm=1.5 mL = 150\text{ cm} = 1.5\text{ m}
  • Least count of scale for LL, ΔLmeas=0.1 cm\Delta L_{\text{meas}} = 0.1\text{ cm}
  • Diameter, D=0.08 cm=8×10−4 mD = 0.08\text{ cm} = 8 \times 10^{-4}\text{ m}
  • Least count of screw gauge for DD, ΔD=0.001 cm\Delta D = 0.001\text{ cm}
  • Extension, ΔL=0.5 cm=5×10−3 m\Delta L = 0.5\text{ cm} = 5 \times 10^{-3}\text{ m}
  • Least count of micrometer for ΔL\Delta L, δ(ΔL)=0.001 cm\delta(\Delta L) = 0.001\text{ cm}

Step 1: Calculate the value of Young's Modulus (YY)

Y=4×100×1.5π×(8×10−4)2×(5×10−3)Y = \frac{4 \times 100 \times 1.5}{\pi \times (8 \times 10^{-4})^2 \times (5 \times 10^{-3})}

Y=600π×(64×10−8)×(5×10−3)Y = \frac{600}{\pi \times (64 \times 10^{-8}) \times (5 \times 10^{-3})}

Y=6003.2π×10−9=187.5π×109≈59.683×109 N/m2Y = \frac{600}{3.2 \pi \times 10^{-9}} = \frac{187.5}{\pi} \times 10^9 \approx 59.683 \times 10^9\text{ N/m}^2

Step 2: Calculate the relative error in Young's Modulus (ΔYY\frac{\Delta Y}{Y})

Taking the natural logarithm and differentiating Y=4FLπD2ΔLY = \frac{4FL}{\pi D^2 \Delta L}, the fractional error is given by:

ΔYY=ΔLmeasL+2ΔDD+δ(ΔL)ΔL\frac{\Delta Y}{Y} = \frac{\Delta L_{\text{meas}}}{L} + 2\frac{\Delta D}{D} + \frac{\delta(\Delta L)}{\Delta L}

Calculating each fractional error term:

  1. ΔLmeasL=0.1 cm150 cm=11500\frac{\Delta L_{\text{meas}}}{L} = \frac{0.1\text{ cm}}{150\text{ cm}} = \frac{1}{1500}
  2. 2ΔDD=2×0.001 cm0.08 cm=280=1402\frac{\Delta D}{D} = 2 \times \frac{0.001\text{ cm}}{0.08\text{ cm}} = \frac{2}{80} = \frac{1}{40}
  3. δ(ΔL)ΔL=0.001 cm0.5 cm=1500\frac{\delta(\Delta L)}{\Delta L} = \frac{0.001\text{ cm}}{0.5\text{ cm}} = \frac{1}{500}

Summing these terms:

ΔYY=11500+140+1500=2+75+63000=833000≈0.027667\frac{\Delta Y}{Y} = \frac{1}{1500} + \frac{1}{40} + \frac{1}{500} = \frac{2 + 75 + 6}{3000} = \frac{83}{3000} \approx 0.027667

Step 3: Calculate the absolute error in Young's Modulus (ΔY\Delta Y)

ΔY=Y×(ΔYY)\Delta Y = Y \times \left(\frac{\Delta Y}{Y}\right)

ΔY=(59.683×109 N/m2)×833000\Delta Y = \left(59.683 \times 10^9\text{ N/m}^2\right) \times \frac{83}{3000}

ΔY≈1.6513×109 N/m2\Delta Y \approx 1.6513 \times 10^9\text{ N/m}^2

Thus, comparing with ΔY=α×109 N/m2\Delta Y = \alpha \times 10^9\text{ N/m}^2, we get:

α=1.65\alpha = 1.65

Calculate Error in Youngs Modulus Measurement Using Screw Gauge and Scale | Physics PYQ Solution - JEE Challenger