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Units and Measurements question

2007 · Shift 2 · Q9
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Units and Measurements question

2007 · Shift 2 · Q9

JEE AdvancedPhysicsUnits and MeasurementsMCQ+3 / −1
A student performs an experiment to determine the Young's modulus of a wire, exactly 2 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 mm with an uncertainty of ±0.05  mm\pm0.05\;\mathrm{mm}±0.05mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ±0.01  mm\pm0.01\;\mathrm{mm}±0.01mm. Take g = 9.8 m/s2 (exact). The Young's modulus obtained from the reading is
  1. A
    (2.0  ±  0.3)×1011  N/m2\left(2.0\;\pm\;0.3\right)\times10^{11}\;\mathrm N/\mathrm m^2(2.0±0.3)×1011N/m2
  2. B
    (2.0  ±  0.2)×1011  N/m2\left(2.0\;\pm\;0.2\right)\times10^{11}\;\mathrm N/\mathrm m^2(2.0±0.2)×1011N/m2
  3. C
    (2.0  ±  0.1)×1011  N/m2\left(2.0\;\pm\;0.1\right)\times10^{11}\;\mathrm N/\mathrm m^2(2.0±0.1)×1011N/m2
  4. D
    (2.0  ±  0.05)×1011  N/m2\left(2.0\;\pm\;0.05\right)\times10^{11}\;\mathrm N/\mathrm m^2(2.0±0.05)×1011N/m2
View written solutionFree

Correct answer: B

  1. Formula for Young's modulus

For a wire,

Y=stressstrain=F/AΔL/L=FLA ΔLY = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\,\Delta L}Y=strainstress​=ΔL/LF/A​=AΔLFL​

For a circular wire, cross-sectional area

A=πd24A = \frac{\pi d^2}{4}A=4πd2​

So,

Y=4FLπd2ΔLY = \frac{4FL}{\pi d^2\Delta L}Y=πd2ΔL4FL​
  1. Substitute the measured values

Given:

  • L=2 mL = 2\,\text{m}L=2m (exact)
  • m=1.0 kgm = 1.0\,\text{kg}m=1.0kg, so F=mg=9.8 NF = mg = 9.8\,\text{N}F=mg=9.8N (exact)
  • d=0.4 mm=0.4×10−3 md = 0.4\,\text{mm} = 0.4\times10^{-3}\,\text{m}d=0.4mm=0.4×10−3m
  • ΔL=0.8 mm=0.8×10−3 m\Delta L = 0.8\,\text{mm} = 0.8\times10^{-3}\,\text{m}ΔL=0.8mm=0.8×10−3m

Thus,

Y=4(9.8)(2)π(0.4×10−3)2(0.8×10−3)Y = \frac{4(9.8)(2)}{\pi (0.4\times10^{-3})^2(0.8\times10^{-3})}Y=π(0.4×10−3)2(0.8×10−3)4(9.8)(2)​

Now,

(0.4×10−3)2=0.16×10−6=1.6×10−7(0.4\times10^{-3})^2 = 0.16\times10^{-6} = 1.6\times10^{-7}(0.4×10−3)2=0.16×10−6=1.6×10−7

Hence,

πd2ΔL=π(1.6×10−7)(0.8×10−3)=π(1.28×10−10)\pi d^2\Delta L = \pi (1.6\times10^{-7})(0.8\times10^{-3}) = \pi (1.28\times10^{-10})πd2ΔL=π(1.6×10−7)(0.8×10−3)=π(1.28×10−10)

And numerator,

4×9.8×2=78.44\times 9.8 \times 2 = 78.44×9.8×2=78.4

Therefore,

Y=78.4π×1.28×10−10approx1.95×1011 N/m2Y = \frac{78.4}{\pi \times 1.28\times10^{-10}} approx 1.95\times10^{11}\,\text{N/m}^2Y=π×1.28×10−1078.4​approx1.95×1011N/m2

which rounds to

Y≈2.0×1011 N/m2Y \approx 2.0\times10^{11}\,\text{N/m}^2Y≈2.0×1011N/m2
  1. Error calculation

Since

Y∝1d2ΔLY \propto \frac{1}{d^2\Delta L}Y∝d2ΔL1​

maximum fractional error is

ΔYY=2Δdd+Δ(ΔL)ΔL\frac{\Delta Y}{Y} = 2\frac{\Delta d}{d} + \frac{\Delta(\Delta L)}{\Delta L}YΔY​=2dΔd​+ΔLΔ(ΔL)​

Given:

Δdd=0.010.4=0.025=2.5%\frac{\Delta d}{d} = \frac{0.01}{0.4} = 0.025 = 2.5\%dΔd​=0.40.01​=0.025=2.5%

so

2Δdd=0.05=5%2\frac{\Delta d}{d} = 0.05 = 5\%2dΔd​=0.05=5%

Also,

Δ(ΔL)ΔL=0.050.8=0.0625=6.25%\frac{\Delta(\Delta L)}{\Delta L} = \frac{0.05}{0.8} = 0.0625 = 6.25\%ΔLΔ(ΔL)​=0.80.05​=0.0625=6.25%

Thus,

ΔYY=0.05+0.0625=0.1125=11.25%\frac{\Delta Y}{Y} = 0.05 + 0.0625 = 0.1125 = 11.25\%YΔY​=0.05+0.0625=0.1125=11.25%

Absolute error:

ΔY=0.1125×2.0×1011=0.225×1011 N/m2\Delta Y = 0.1125 \times 2.0\times10^{11} = 0.225\times10^{11}\,\text{N/m}^2ΔY=0.1125×2.0×1011=0.225×1011N/m2

Rounding suitably,

ΔY≈0.2×1011 N/m2\Delta Y \approx 0.2\times10^{11}\,\text{N/m}^2ΔY≈0.2×1011N/m2

So the result is

Y=(2.0±0.2)×1011 N/m2Y = (2.0 \pm 0.2)\times10^{11}\,\text{N/m}^2Y=(2.0±0.2)×1011N/m2
  1. Option check
  • A: ±0.3×1011\pm 0.3\times10^{11}±0.3×1011 ❌
  • B: ±0.2×1011\pm 0.2\times10^{11}±0.2×1011 ✅
  • C: ±0.1×1011\pm 0.1\times10^{11}±0.1×1011 ❌
  • D: ±0.05×1011\pm 0.05\times10^{11}±0.05×1011 ❌

Therefore, the correct option is B.

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