Pressure inside a soap bubble is greater than the pressure outside by an amount: (given: $R =$ Radius of bubble,$S =$ Surface tension of bubble)

  • A
    $\frac{4 S}{R}$
  • B
    $\frac{4 R}{S}$
  • C
    $\frac{S}{R}$
  • D
    $\frac{2 S}{R}$

Explore More

Similar Questions

The excess pressure inside a soap bubble of radius $2 \ cm$ is $50 \ dyne/cm^2$. The surface tension is

Two water drops each of radius $r$ coalesce to form a bigger drop. If $T$ is the surface tension,the surface energy released in this process is

Two soap bubbles of radii $r_1$ and $r_2$ in vacuum coalesce under isothermal conditions. The resulting bubble has a radius equal to

The excess pressure inside a spherical drop of water $A$ is four times that of another drop $B$. Then the ratio of mass of drop $A$ to that of drop $B$ is

Two soap bubbles of radii $r$ and $2r$ are arranged as shown in the diagram. The valve is now opened. Which one of the following will result?

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