The radii of two soap bubbles are $r_1$ and $r_2$. In isothermal condition,they combine with each other to form a single bubble. The radius of the resultant bubble is

  • A
    $R=\frac{r_1+r_2}{2}$
  • B
    $R=r_1(r_1+r_2)$
  • C
    $R=\sqrt{r_1^2+r_2^2}$
  • D
    $R=r_1+r_2$

Explore More

Similar Questions

$A$ vessel having a small hole in the bottom must hold water without leakage when water is poured to a height of $7 \text{ cm}$. What is the radius of the hole (in $\text{ mm}$)? [Surface tension of water is $0.07 \text{ N/m}$, angle of contact is $0^{\circ}$, and $g = 10 \text{ m/s}^2$]

There are two liquid drops of different radii. The excess pressure inside over the outside is

When a large bubble rises from the bottom of a water lake to its surface,its radius doubles. If the atmospheric pressure is equal to the pressure of a water column of height $H$,then the depth of the lake will be

Pressure inside two soap bubbles are $1.02 \ atm$ and $1.05 \ atm$ respectively. The ratio of their surface area is

The excess pressure inside the first soap bubble is three times that of a second soap bubble. What is the ratio of the volumes of the first bubble to the second bubble?

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