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JEE Main 20146 April
Exam
JEE Main
Year
2014
Date
6 April
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2014 · 6 April · Physics · Mechanical Properties of Fluids

On heating water, bubbles being formed at the bottom of the vessel...

On heating water, bubbles being formed at the bottom of the vessel detatch and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r << R, and the surface tension of water is T, value of r just before bubbles detatch is :
(density of water is ρw{ρ}_{\text{w}})

On heating water, bubbles being formed at the bottom of the vessel detatch and rise. Take the bubbles to be spheres of

Options

  1. AR22ρwg3T{\text{R}}^{2} \sqrt{\dfrac{{2ρ}_{\text{w}} \text{g}}{3 T}}
  2. B{\text{R}}^{2} \sqrt{\dfrac{{&rho;}_{\text{w}} \text{g}}{6 T}}
  3. C{\text{R}}^{2} \sqrt{\dfrac{{&rho;}_{\text{w}} \text{g}}{T}}
  4. D{\text{R}}^{2} \sqrt{\dfrac{{3 &rho;}_{\text{w}} \text{g}}{T}}

Correct answer

A. R22ρwg3T{\text{R}}^{2} \sqrt{\dfrac{{2ρ}_{\text{w}} \text{g}}{3 T}}

Step-by-step solution

When the bubble gets detached,

Buoyant force = force due to surface tension
On heating water, bubbles being formed at the bottom of the vessel detatch and rise. Take the bubbles to be spheres of

∫ \text{T} \times dl \text\sin \theta = \dfrac{4}{3} \pi {\text{R}}^{3} {ρ}_{\text{w}} \text{g}

⇒T×2πr×rR=43πR3ρwg⇒ \text{T} \times 2 \pi \text{r} \times \dfrac{\text{r}}{\text{R}} = \dfrac{4}{3} \pi {\text{R}}^{3} {ρ}_{\text{w}} \text{g}
⇒r2=2R4ρwg3T⇒ {\text{r}}^{2} = \dfrac{2 {\text{R}}^{4} {ρ}_{\text{w}} \text{g}}{3T}
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