Home
Class 11
PHYSICS
The radii of two soap bubbles are r(1) a...

The radii of two soap bubbles are `r_(1) and r_(2) (r_(2) lt r_(1))`. They meet to produce a double bubble. The radius of their common interface is

Text Solution

Verified by Experts


`becauser_(1)ltr_(2)thereforeP_(1)gtP_(2)` small part of bubbles is in equilibrium
`impliesP_(1)(DeltaA)-P_(2)(DeltaA)=(4T)/(r)DeltaAimplies(4T)/(r_(1))-(4T)/(r_(2))=(4T)/(r)impliesr=(r_(1)r_(2))/(t_(2)-t_(1))``
Promotional Banner

Topper's Solved these Questions

  • ELASTICITY, SURFACE TENSION AND FLUID MECHANICS

    ALLEN|Exercise Exercise 1 (Elasticity)|16 Videos
  • ELASTICITY, SURFACE TENSION AND FLUID MECHANICS

    ALLEN|Exercise Exercise 1 (Surface Tension)|21 Videos
  • CENTRE OF MASS

    ALLEN|Exercise EXERCISE-V B|19 Videos
  • ERROR AND MEASUREMENT

    ALLEN|Exercise Part-2(Exercise-2)(B)|22 Videos

Similar Questions

Explore conceptually related problems

The radii of two soap bubbles are r_(i) and r_(2) . In isothermal conditions, two meet together in vacuum. Then the radius kof the resultant bubble is given by

Under isothermal condition two soap bubbles of radii r_(1) and r_(2) coalesce to form a single bubble of radius r. The external pressure is p_(0) . Find the surface tension of the soap in terms of the given parameters.

Two soap bubbles of radii 2 cm and 4 cm join to form a double bubble in air, then radius of. curvature of interface is

The excess pressure inside a soap bubble of radius R is (S is the surface tension)

Two soap bubbles A and B have radii r_(1) and r_(2) respectively. If r_(1) lt r_(2) than the excess pressure inside

The radii of two planets are respectively R_(1) and R_(2) and their densities are respectively rho_(1) and rho_(2) .The ratio of the accelerations due to gravity at their surface is

A soap bubble of radius r is placed on another bubble of radius 2r . The radius of the surface common to both the bubbles is

Two soap bubbles are stuck together with an intermediate film separating them. Compute the radius of curvature of this film given that the radii of the bubbles in this arrangement are r_(1) and r_(2) respectively. If r_(1) gt r_(2) state clearly which way the intermediate film will bulge. For the case when r_(1)=r_(2)=2cm calculate he radius of the bubble formed by bursting the intermediate film. The volume of a spherical dome of radius R and height h is pib^(2)(3R-b)//3 .

Consider two concentric spherical metal shells of radii r_(1) and r_(2) (r_(2) gt r_(1)) . If the outer shell has a charge q and the inner one is grounded, then the charge on the inner shell is

Two soap bubble of different radii R_(1) and R_(2) (ltR_(1)) coalesce to form an interface of radius R as shown in figure. The correct value of R is .