Consider a soap film on a rectangular frame of wire of area $3 \times 3 \,cm^2$. If the area of the soap film is increased to $5 \times 5 \,cm^2$, the work done in the process will be (surface tension of soap solution is $2.5 \times 10^{-2} \,N/m$).

  • A
    $9 \times 10^{-6} \,J$
  • B
    $16 \times 10^{-6} \,J$
  • C
    $40 \times 10^{-6} \,J$
  • D
    $80 \times 10^{-6} \,J$

Explore More

Similar Questions

$A$ liquid drop having surface energy $E$ is spread into $216$ droplets of the same size. The final surface energy of the droplets is (in $E$)

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 $W_1: W_2=$

$A$ soap bubble of radius $\frac{1}{\sqrt{\pi}} \text{ cm}$ is expanded to radius $\frac{3}{\sqrt{\pi}} \text{ cm}$. The surface tension of the soap solution is $25 \text{ dyne/cm}$. The work done during expansion in $\text{erg}$ is:

Twenty-seven droplets of water, each of radius $0.1 \,mm$, merge to form a single drop. Calculate the energy released during this process. (Take surface tension of water $T = 0.072 \,N/m$)

Find out the work done (in $\times 10^{-3} \; J$) to expand the soap bubble to radius $R = 5 \; cm$ (Surface tension of water $= 0.1 \; N/m$).

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