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Byju's Answer

Standard XII

Physics

Atmospheric Pressure

A tube of fin...

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 $σ$ . 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 $ρ$ be the density of manometer liquid and $r$ the radius of curvature of the bubble, then the surface tension $σ$ of the liquid is given by

ρ h r g

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2 ρ h r g

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4 ρ h r g

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\frac{r h ρ g}{4}

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Solution

The correct option is B $\frac{r h ρ g}{4}$
$p_{x} = p_{0} + ρ g h$ ......(i)
Now $p_{C} - p_{0} = \frac{4 σ}{r}$
This is the pressure drop in the soap bubble
$∵$ same air is filled we have $p_{x} = p_{y} = p_{C}$
Equating $p_{0}$ from both the above equations we get
$\Rightarrow p_{x} - ρ g h = p_{C} - \frac{4 σ}{r}$ .....(ii)
we get $σ = \frac{ρ g h r}{4}$

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The correct option is B rhρg4px=p0+ρgh ......(i)Now pC−p0=4σrThis is the pressure drop in the soap bubble∵ same air is filled we have px=py=pCEquating p0 from both the above equations we get⇒px−ρgh=pC−4σr.....(ii)we get σ=ρghr4