A tube $A B$ of fine bore is connected to a manometer $M$ as shown. The stop cork $S$ controls the flow of air. $A B$ is dipped into and taken out of a liquid whose surface tension is $σ$ . A film of liquid is formed, which closes the end $B$ . On opening the stop cork for a while, air is forced into the tube and a bubble is formed at $B$ . 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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$4 r h ρ g$

$\frac{r h ρ g}{4}$

$r h ρ g$

$2 r h g ρ$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Pressure of air in the tube is higher than atmospheric pressure by $Δ P = ρ g h$
Excess pressure inside bubble is $Δ P = \frac{4 σ}{r}$
The pressure of air inside the bubble and the entire tube is same.
$∴ \frac{4 σ}{r} = ρ g h \Rightarrow σ = \frac{ρ g h r}{4}$