The surface tension of a liquid is $5 \, N/m$. If a thin film of the area $0.02 \, m^2$ is formed on a loop,then its surface energy will be

  • A
    $5 \times 10^2 \, J$
  • B
    $2.5 \times 10^{-2} \, J$
  • C
    $2 \times 10^{-1} \, J$
  • D
    $5 \times 10^{-1} \, J$

Explore More

Similar Questions

$A$ drop of mercury of radius $2\, mm$ is split into $8$ identical droplets. Find the increase in surface energy in $\mu J$. (Surface tension of mercury is $0.465\;J/m^2$)

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:

The surface energy required to create a liquid surface of area $0.04 \text{ m}^2$ for a liquid of surface tension $75 \text{ N/m}$ will be ....... $\text{J}$

$A$ liquid drop having surface energy $E$ is spread into $729$ droplets of same size. The final surface energy of the droplets is

The surface tension of a soap solution is $3.5 \times 10^{-2} \, N m^{-1}$. The amount of work done required to increase the radius of a soap bubble from $10 \, cm$ to $20 \, cm$ is $..... \times 10^{-4} \, J$.

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