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Properties of Matter question

2020 · 3 Sep · Shift 1 · Q44
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Properties of Matter question

2020 · 3 Sep · Shift 1 · Q44

JEE MainPhysicsProperties of MatterNumerical+4 / −1
When a long glass capillary tube of radius 0.015 cm is dipped in a liquid, the liquid rises to a height of 15 cm within it. If the contact angle between the liquid and glass to close to 0o, the surface tension of the liquid, in milliNewton m–1, is [ρ\rhoρ(liquid) = 900 kgm–3, g = 10 ms–2] (Give answer in closest integer) ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 101

  1. Use the capillary rise formula

For a liquid rising in a capillary tube,

h=2Tcos⁡θρgrh = \frac{2T\cos\theta}{\rho g r}h=ρgr2Tcosθ​

where:

  • h=15 cm=0.15 mh = 15\,\text{cm} = 0.15\,\text{m}h=15cm=0.15m
  • r=0.015 cm=1.5×10−4 mr = 0.015\,\text{cm} = 1.5\times 10^{-4}\,\text{m}r=0.015cm=1.5×10−4m
  • ρ=900 kg m−3\rho = 900\,\text{kg m}^{-3}ρ=900kg m−3
  • g=10 m s−2g = 10\,\text{m s}^{-2}g=10m s−2
  • θ≈0∘⇒cos⁡θ≈1\theta \approx 0^\circ \Rightarrow \cos\theta \approx 1θ≈0∘⇒cosθ≈1

So,

T=hρgr2T = \frac{h\rho g r}{2}T=2hρgr​

  1. Substitute the values

T=(0.15)(900)(10)(1.5×10−4)2T = \frac{(0.15)(900)(10)(1.5\times10^{-4})}{2}T=2(0.15)(900)(10)(1.5×10−4)​

First compute stepwise:

900×10=9000900\times 10 = 9000900×10=9000

0.15×9000=13500.15\times 9000 = 13500.15×9000=1350

1350×1.5×10−4=2025×10−4=0.20251350\times 1.5\times10^{-4} = 2025\times10^{-4} = 0.20251350×1.5×10−4=2025×10−4=0.2025

Now divide by 222:

T=0.20252=0.10125 N m−1T = \frac{0.2025}{2} = 0.10125\,\text{N m}^{-1}T=20.2025​=0.10125N m−1

  1. Convert to milliNewton per metre

0.10125 N m−1=101.25 mN m−10.10125\,\text{N m}^{-1} = 101.25\,\text{mN m}^{-1}0.10125N m−1=101.25mN m−1

Closest integer:

101\boxed{101}101​

  1. Comparison with stored answer

Stored correct answer = 101101101

This matches the derived answer.

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