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If r=5 10^{-4}m =10^3Kgm^{-3},g=10m/s^{2},T=0.11Nm^{-1} the ra

Question

If $r = 5 \times 1 0^{- 4} m ρ = 1 0^{3} K g m^{- 3}, g = 10 m / s^{2}, T = 0.11 N m^{- 1}$ the radius of the drop when it detached from the dropper is approximately:

1.4 \times 1 0^{- 3} m

3.3 \times 1 0^{- 3} m

2.0 \times 1 0^{- 3} m

4.1 \times 1 0^{- 3} m

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William Discussed

r = 5 \times 1 0^{- 4} m ρ = 1 0^{3} K g m^{- 3}, g = 10 m / s^{2}, T = 0.11 N m^{- 1}

the radius of the drop when it detached from the dropper is approximately:

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Text solutionVerified

m g = \int_{o}^{2 π r} T sin θ d s

m =

Mass of bubble that is getting detached.

T =

Surface tension.

∴ ρ \times \frac{4}{3} π R^{2} \times g = T \times 2 π r \times \frac{r}{R}

\frac{2}{3} \times ρ R^{3} g = T \frac{r ^{2}}{R}

\Rightarrow R^{4} = \frac{3 T . r ^{2}}{2 ρ g}

= \frac{3 \times 0.11 \times ( 5 \times 1 0 ^{- 4} ) ^{2}}{2 \times 1 0 ^{3} \times 10}

= \frac{3 \times 0.11 \times 25 \times 1 0 ^{- 8}}{2 \times 1 0 ^{4}}

R = 1.4 \times 1 0^{- 3} m

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Question Text	If $r = 5 \times 1 0^{- 4} m ρ = 1 0^{3} K g m^{- 3}, g = 10 m / s^{2}, T = 0.11 N m^{- 1}$ the radius of the drop when it detached from the dropper is approximately:
Answer Type	Text solution:1
Upvotes	150