  Question

A tube of fine bore AB is connected to a manometer M as shown. The stop cock S controls the flow of air. AB is dipped into a liquid whose surface tension is $$\sigma$$. On opening the stop cock for a while, a bubble is formed at B and the manometer level is recorded, showing a difference $$h$$ in the levels in the two arms. If $$\rho$$ be the density of manometer liquid and $$r$$ the radius of curvature of the bubble, then the surface tension $$\sigma$$ of the liquid is given by A
ρhrg  B
2ρhrg  C
4ρhrg  D
rhρg4  Solution

The correct option is B $$\dfrac{rh\rho g}{4}$$$$p_{x}=p_{0}+\rho gh$$ ......(i)Now $$p_{C}-p_{0}=\dfrac{4\sigma }{r}$$This is the pressure drop in the soap bubble$$\because$$ same air is filled we have $$p_x=p_y=p_C$$Equating $$p_0$$ from both the above equations we get$$\Rightarrow p_{x}-\rho gh=p_C-\dfrac{4\sigma}{r}$$.....(ii)we get $$\sigma = \dfrac{\rho ghr}{4}$$ Physics

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