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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

71721.jpg


A
ρhrg
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B
2ρhrg
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C
4ρhrg
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D
rhρg4
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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 $$\sigma  = \dfrac{\rho ghr}{4}$$

139939_71721_ans.png

Physics

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