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Properties of Matter question

2024 · 8 Apr · Shift 1 · Q85
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Properties of Matter question

2024 · 8 Apr · Shift 1 · Q85

JEE MainPhysicsProperties of MatterNumerical+4 / −1
A liquid column of height 0.04 cm0.04 \mathrm{~cm}0.04 cm balances excess pressure of a soap bubble of certain radius. If density of liquid is 8×103 kg m−38 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}8×103 kg m−3 and surface tension of soap solution is 0.28 Nm−10.28 \mathrm{~Nm}^{-1}0.28 Nm−1, then diameter of the soap bubble is ‾cm\underline{\hspace{2cm}}\mathrm{cm}​cm. (if g=10 m s−2\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}g=10 m s−2 )
Numerical answer
View written solutionFree

Correct answer: 7

  1. Excess pressure of a soap bubble

For a soap bubble, the excess pressure is

ΔP=4Tr\Delta P = \frac{4T}{r}ΔP=r4T​

where TTT is surface tension and rrr is radius of the bubble.

  1. Pressure balanced by liquid column

A liquid column of height hhh balances this excess pressure, so

ρgh=4Tr\rho g h = \frac{4T}{r}ρgh=r4T​

Given:

  • h=0.04 cm=4×10−4 mh = 0.04\,\text{cm} = 4 \times 10^{-4}\,\text{m}h=0.04cm=4×10−4m
  • ρ=8×103 kg m−3\rho = 8 \times 10^3\,\text{kg m}^{-3}ρ=8×103kg m−3
  • T=0.28 N m−1T = 0.28\,\text{N m}^{-1}T=0.28N m−1
  • g=10 m s−2g = 10\,\text{m s}^{-2}g=10m s−2
  1. Substitute values
(8×103)(10)(4×10−4)=4(0.28)r(8 \times 10^3)(10)(4 \times 10^{-4}) = \frac{4(0.28)}{r}(8×103)(10)(4×10−4)=r4(0.28)​

First calculate the left side:

8×103×10×4×10−4=328 \times 10^3 \times 10 \times 4 \times 10^{-4} = 328×103×10×4×10−4=32

So,

32=1.12r32 = \frac{1.12}{r}32=r1.12​
  1. Solve for radius
r=1.1232=0.035 mr = \frac{1.12}{32} = 0.035\,\text{m}r=321.12​=0.035m
  1. Find diameter
d=2r=0.07 m=7 cmd = 2r = 0.07\,\text{m} = 7\,\text{cm}d=2r=0.07m=7cm

Final Answer

7\boxed{7}7​

The diameter of the soap bubble is 7 cm7\,\text{cm}7cm.

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