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Properties of Matter question

2016 · 10 Apr · Shift 1 · Q47
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Properties of Matter question

2016 · 10 Apr · Shift 1 · Q47

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A thin 1 m long rod has a radius of 5 mm. A force of 50 π\piπ kN is applied at one end to determine its Young’s modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is 0.01 mm, which of the following statements is false ?
  1. A
    Δγγ{{\Delta \gamma } \over \gamma }γΔγ​ gets minimum contribution from the uncertainty in the length.
  2. B
    The figure of merit is the largest for the length of the rod.
  3. C
    The maximum value of γ\gammaγ that can be determined is 2 ×\times× 1014 N/m2
  4. D
    Δγγ{{\Delta \gamma } \over \gamma }γΔγ​ gets its maximum contribution from the uncertainty in strain
View written solutionFree

Correct answer: NO OPTION IS FALSE; A, B, C AND D ARE ALL TRUE.

  1. Young’s modulus formula

For a rod under longitudinal stress,

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

For a circular cross-section,

A=πr2A = \pi r^2A=πr2

So,

γ=FLπr2ΔL\gamma = \frac{F L}{\pi r^2 \Delta L}γ=πr2ΔLFL​

Given:

  • L=1 mL = 1\,\text{m}L=1m
  • r=5 mm=5×10−3 mr = 5\,\text{mm} = 5 \times 10^{-3}\,\text{m}r=5mm=5×10−3m
  • F=50π kN=50π×103 NF = 50\pi\,\text{kN} = 50\pi \times 10^3\,\text{N}F=50πkN=50π×103N
  • least count in all length measurements =0.01 mm=10−5 m= 0.01\,\text{mm} = 10^{-5}\,\text{m}=0.01mm=10−5m
  • force is exact
  1. Relative error in Young’s modulus

Since

γ=FLπr2ΔL\gamma = \frac{F L}{\pi r^2 \Delta L}γ=πr2ΔLFL​

and FFF is exact, the maximum fractional error is

Δγγ=ΔLrodL+2Δrr+Δ(ΔL)ΔL\frac{\Delta \gamma}{\gamma} = \frac{\Delta L_{\text{rod}}}{L} + 2\frac{\Delta r}{r} + \frac{\Delta (\Delta L)}{\Delta L}γΔγ​=LΔLrod​​+2rΔr​+ΔLΔ(ΔL)​

where:

  • ΔLrod\Delta L_{\text{rod}}ΔLrod​ is uncertainty in rod length measurement,
  • Δr\Delta rΔr is uncertainty in radius,
  • Δ(ΔL)\Delta(\Delta L)Δ(ΔL) is uncertainty in extension.

Each direct length measurement has uncertainty 10−5 m10^{-5}\,\text{m}10−5m.

  1. Contribution from uncertainty in rod length
ΔLrodL=10−51=10−5\frac{\Delta L_{\text{rod}}}{L} = \frac{10^{-5}}{1} = 10^{-5}LΔLrod​​=110−5​=10−5

This is very small.

So statement A:

Δγγ\dfrac{\Delta \gamma}{\gamma}γΔγ​ gets minimum contribution from the uncertainty in the length.

is true.

  1. Contribution from uncertainty in radius
2Δrr=2×10−55×10−3=2×2×10−3=4×10−32\frac{\Delta r}{r} = 2\times \frac{10^{-5}}{5\times 10^{-3}} = 2\times 2\times 10^{-3} = 4\times 10^{-3}2rΔr​=2×5×10−310−5​=2×2×10−3=4×10−3
  1. Contribution from uncertainty in strain

Since strain involves extension ΔL\Delta LΔL, its contribution is

Δ(ΔL)ΔL\frac{\Delta(\Delta L)}{\Delta L}ΔLΔ(ΔL)​

This will be largest when ΔL\Delta LΔL is smallest measurable, i.e. of order least count.

To determine the maximum Young’s modulus measurable, the extension should be minimum detectable:

ΔLmin⁡=0.01 mm=10−5 m\Delta L_{\min} = 0.01\,\text{mm} = 10^{-5}\,\text{m}ΔLmin​=0.01mm=10−5m

Then,

Δ(ΔL)ΔL=10−510−5=1\frac{\Delta(\Delta L)}{\Delta L} = \frac{10^{-5}}{10^{-5}} = 1ΔLΔ(ΔL)​=10−510−5​=1

This is much larger than the previous contributions:

  • from length: 10−510^{-5}10−5
  • from radius: 4×10−34\times 10^{-3}4×10−3
  • from strain: 111

Hence statement D is true.

  1. Figure of merit

Figure of merit of an instrument is inverse of least count effect, i.e. larger measured quantity compared to least count means better precision.

Compare:

  • rod length =1 m=1000 mm= 1\,\text{m} = 1000\,\text{mm}=1m=1000mm
  • radius =5 mm= 5\,\text{mm}=5mm
  • extension is very small

Since the rod length is the largest quantity measured with the same least count, its relative error is minimum, so figure of merit is largest for rod length.

Thus statement B is true.

  1. Maximum value of Young’s modulus measurable

Use minimum detectable extension:

ΔLmin⁡=10−5 m\Delta L_{\min} = 10^{-5}\,\text{m}ΔLmin​=10−5m

Now,

γmax⁡=FLπr2ΔLmin⁡\gamma_{\max} = \frac{F L}{\pi r^2 \Delta L_{\min}}γmax​=πr2ΔLmin​FL​

Substitute values:

F=50π×103,L=1,r=5×10−3,ΔLmin⁡=10−5F = 50\pi\times 10^3, \quad L=1, \quad r=5\times 10^{-3}, \quad \Delta L_{\min}=10^{-5}F=50π×103,L=1,r=5×10−3,ΔLmin​=10−5 γmax⁡=50π×103×1π(5×10−3)2(10−5)\gamma_{\max} = \frac{50\pi\times 10^3 \times 1}{\pi (5\times 10^{-3})^2 (10^{-5})}γmax​=π(5×10−3)2(10−5)50π×103×1​

Cancel π\piπ:

γmax⁡=50×10325×10−6×10−5\gamma_{\max} = \frac{50\times 10^3}{25\times 10^{-6}\times 10^{-5}}γmax​=25×10−6×10−550×103​ (5×10−3)2=25×10−6(5\times 10^{-3})^2 = 25\times 10^{-6}(5×10−3)2=25×10−6

So,

γmax⁡=50×10325×10−11=2×103×1011=2×1014 N/m2\gamma_{\max} = \frac{50\times 10^3}{25\times 10^{-11}} = 2\times 10^3 \times 10^{11} = 2\times 10^{14}\,\text{N/m}^2γmax​=25×10−1150×103​=2×103×1011=2×1014N/m2

Thus statement C is also true.

  1. Conclusion
  • A: true
  • B: true
  • C: true
  • D: true

Therefore, none of the given statements is false.

So the stored answer C is incorrect.

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