An air bubble of radius $r$ in water is at depth $h$ below the water surface. If $P$ is the atmospheric pressure, and $d$ and $T$ are the density and surface tension of water respectively, then the pressure inside the bubble will be:

  • A
    $P + hdg - (4T/r)$
  • B
    $P + hdg + (2T/r)$
  • C
    $P + hdg - (2T/r)$
  • D
    $P + hdg + (4T/r)$

Explore More

Similar Questions

When a big drop of water is formed from $n$ small drops of water,the energy loss is $3E$,where $E$ is the energy of the bigger drop. The radius of the bigger drop is $R$ and that of the smaller drop is $r$. Then the value of $n$ is:

Formation of bubbles is in Column-$I$ and the pressure difference between them is given in Column-$II$. Match them appropriately.
Column-$I$ Column-$II$
$(a)$ Liquid drop in air $(i)$ $\frac{4T}{R}$
$(b)$ Bubble of liquid in air $(ii)$ $\frac{2T}{R}$
$(iii)$ $\frac{2R}{T}$

$Assertion :$ Smaller drops of liquid resist deforming forces better than the larger drops.
$Reason :$ Excess pressure inside a drop is directly proportional to its surface area.

If two soap bubbles $A$ and $B$ of radii $r_1$ and $r_2$ respectively are kept in vacuum at constant temperature,then the ratio of masses of air inside the bubbles $A$ and $B$ is

$A$ spherical liquid drop of radius $R$ is divided into eight equal droplets. If surface tension is $T$,then the work done in this process will be

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo