Air (density rho) is being blown on a soap fi | vedclass

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Question

(D) The bubble detaches from the film when the force exerted by the dynamic pressure exceeds the force due to surface tension along the circumference

Vedclass · Accepted Answer

$\sqrt{\frac{8 T}{\rho R}}$

Vedclass · Answer

(D) The bubble detaches from the film when the force exerted by the dynamic pressure exceeds the force due to surface tension along the circumference of the pipe.The force due to dynamic pressure is given by $F_{	ext{dynamic}} = P_{	ext{dynamic}} 	imes A = (\frac{1}{2} ho v^2) 	imes (\pi R^2)$.The force due to surface tension acts along the circumference of the pipe. Since a soap film has two surfaces,the total surface tension force is $F_{	ext{surface tension}} = 2 	imes (T 	imes 2 \pi R) = 4 \pi R T$.For the bubble to detach,the dynamic force must be at least equal to the surface tension force:$\frac{1}{2} ho v^2 	imes \pi R^2 = 4 \pi R T$Simplifying the equation:$\frac{1}{2} ho v^2 R = 4 T$$v^2 = \frac{8 T}{ho R}$$v = \sqrt{\frac{8 T}{ho R}}$Thus,the minimum speed at which the bubble is formed is $v = \sqrt{\frac{8 T}{ho R}}$.

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