A soap film is made by dropping a circular frame of radius b in soap solution. A bubble is formed by blowing air with speed v in the form of a cylinder. A ir stops after striking surface of soap bubble. Density of air is $ρ$ . The radius of bubble is R>b. The radius R of the bubble when the soap bubble separates from the ring is $\frac{βT}{{ρv}^{2}}$ . Find the value of β . (where, surface tension of liquid is T)

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answer is 4.

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Detailed Solution

Mass of air entering in 1s is $m = {πb}^{2} vρ$
Change in momentum in 1sec is $= mv - 0 = {πb}^{2} {ρv}^{2}$
Thus, $F = {πb}^{2} {ρv}^{2}$
Pressure due to this force is $Δp = \frac{F}{A} = \frac{{πb}^{2} {ρv}^{2}}{{πb}^{2}} = {ρv}^{2}$
When this excess pressure equals to $\frac{4 T}{R}$ , the bubble gets separated
$∴ {ρv}^{2} = \frac{4 T}{R} \Rightarrow= \frac{4 T}{{ρv}^{2}}; So, β = 4$