The work done in blowing a soap bubble of radius $R$ is $W_1$ at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius $2R$ is blown and the work done is $W_2$. Then:

  • A
    $W_2 = 0$
  • B
    $W_2 = 4 W_1$
  • C
    $W_2 < 4 W_1$
  • D
    $W_2 = W_1$

Explore More

Similar Questions

Energy needed in breaking a liquid drop of radius $R$ into $n$ smaller drops,each of radius $r$,is [where $T$ is the surface tension of the liquid].

If the work done in blowing a soap bubble of volume $V$ is $W$,then the work done in blowing a soap bubble of volume $2V$ will be

The potential energy of a molecule on the surface of a liquid compared to one inside the liquid is

If $W_1$ is the work done in increasing the radius of a soap bubble from $r$ to $2r$ and $W_2$ is the work done in increasing the radius of the soap bubble from $2r$ to $3r$,then $W_1: W_2=$

$A$ liquid drop having surface energy $E$ is spread into $512$ droplets of the same size. The final surface energy of the droplets is (in $E$)

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