Properties Of Matter Ques 86
- A soap bubble is being blown at the end of very narrow tube of radius $b$. Air (density $\rho$ ) moves with a velocity $v$ inside the tube and comes to rest inside the bubble. The surface tension of the soap solution is $T$. After sometime the bubble, having grown to radius $r$ separates from the tube. Find the value of $r$. Assume that $r»b$ so, that you can consider the air to be falling normally on the bubble’s surface.
(2003, 4M)
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Answer:
Correct Answer: 86.$\frac{4 T}{\rho v^{2}}$
Solution:
Formula:
- Surface Tension force $=2 \pi b \times 2 T \sin \theta$
Mass of the air per second entering the bubble $=\rho A v$
Momentum of air per second
$=$ Force due to air $=\rho A v^{2}$
The bubble will separate from the tube when force due to moving air becomes equal to the surface tension force inside the bubble.
$2 \pi b \times 2 T \sin \theta=\rho A v^{2}$
putting $\sin \theta=\frac{b}{r}, A=\pi b^{2}$ and solving, we get
$$ r=\frac{4 T}{\rho v^{2}} $$
BETA
