Creating small bubbles of soap!

A soap bubble of initial radius r 1 r_1 is blown at the end of a capillary tube of length \ell and cross sectional radius a a . It is then left so that the size of the air bubble gradually reduces and the new radius is r 2 r_2 . If the surface tension of the soap bubble is T T and the coefficient of viscosity of air is η \eta , then the time taken by the bubble to reduce to radius r 2 r_2 can be represented as

t = 6 η T a 4 x ( r 1 4 r 2 4 ) , t = \dfrac{6 \eta \ell}{T a^4 x} \left( {r_1}^4 - {r_2}^4 \right),

where all quantities are in SI units and x x is a positive integer. Viscosity of air and surface tension of soap solution is independent of temperature.

Evaluate the value of x x .


The answer is 3.

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1 solution

R G Staff
Dec 28, 2016

The pressure inside the bubble is at time t when the radius of the bubble is reduced to r will be P b u b b l e = P 0 + 4 T r {P_{bubble}} = {P_0} + \frac{{4T}}{r} . Now, through the capillary of length l, the air is flowing and we can use Poiseuillie's equation for it. That is, Q = d V d t = π Δ p a 4 8 η l Q = - \frac{{dV}}{{dt}} = \frac{{\pi \Delta p{a^4}}}{{8\eta l}} V = 4 3 π r 3 V = \frac{4}{3}\pi {r^3} Δ p = 4 T r \Delta p = \frac{{4T}}{r} Putting these values in the equation, 4 π r 2 d r d t = 4 π T a 4 8 η r l r 1 r 2 r 3 d r = T a 4 8 η l 0 t d t t = 2 η l T a 4 ( r 1 4 r 2 4 ) = 6 η l T a 4 x ( r 1 4 r 2 4 ) x = 3 \begin{gathered} - 4\pi {r^2}\frac{{dr}}{{dt}} = \frac{{4\pi T{a^4}}}{{8\eta rl}} \\ - \int_{{r_1}}^{{r_2}} {{r^3}dr} = \frac{{T{a^4}}}{{8\eta l}}\int_0^t {dt} \\ t = \frac{{2\eta l}}{{T{a^4}}}(r_1^4 - r_2^4) = \frac{{6\eta l}}{{T{a^4}x}}(r_1^4 - r_2^4) \\ x = 3 \\ \end{gathered}

Same way!!!

A Former Brilliant Member - 4 years, 5 months ago

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