If r=5\times 10^{-4}m\quad \rho=10^3Kgm^{-3},g=10m/s^{2},T=0.11Nm^ | Filo

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Class 11 Physics Mechanics Mechanical Properties of Fluids

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:

(a)

1.4 \times 1 0^{- 3} m

(b)

3.3 \times 1 0^{- 3} m

(c)

2.0 \times 1 0^{- 3} m

(d)

4.1 \times 1 0^{- 3} m

Correct answer: (a)

Solution:

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 . 1 1 \times ( 5 \times 1 0 ^{- 4} ) ^{2}}{2 \times 1 0 ^{3} \times 1 0}

= \frac{3 \times 0 . 1 1 \times 2 5 \times 1 0 ^{- 8}}{2 \times 1 0 ^{4}}

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

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