99 Percentile Qs Bank for JEE MainPhysicsMechanical Properties of Fluids
Under isothermal conditions, two soap bubbles of radii ' r₁ ' and ' r₂ ' combine to forms a single soap bubble of radius ' R '. The surface tension of soap solution is ( P = outside pressure)
Options
- AP ( R ^3+ r ₁^3+ r ₂^3 ) 4 ( r ₁^2- r ₂^2+ R ^2 )
- BP ( R ^2+ r ₁^2+ r ₂^2 ) 4 ( r ₁^2- r ₂^2+ R ^2 )
- CP (R^3-r₁^3-r₂^3 ) 4 (r₁^2+r₂^2-R^2 )
- DP ( R ^2- r ₁^2- r ₂^2 ) 4 ( r ₁^3+ r ₂^3- R ^3 )
Correct answer
C. P (R^3-r₁^3-r₂^3 ) 4 (r₁^2+r₂^2-R^2 )
Step-by-step solution
Pressure inside the first bubble = P + 4 ~T r ₁ Pressure inside the second bubble = P + 4 ~T r ₂ Using the formula PV = nR ( = absolute temp ) ( P + 4 ~T r ₁ ) 4 3 r ₁^3= n ₁ R ( R is molar gas constant) ( P + 4 ~T r ₂ ) 4 3 r ₂^3= n ₂ R and ( P + 4 ~T R ) 4 3 R ^3= ( n ₁+ n ₂ ) R ( P + 4 ~T R ) 4 3 R ^3= ( P + 4 ~T r ₁ ) 4 3 r ₁^3+ ( P + 4 ~T r ₂ ) 4 r ₂^3 3 ( P + 4 ~T R ) R ^3= ( P + 4 ~T r ₁ ) r ₁^3+ ( P + 4 ~T r ₂ ) r ₂^3 on solving: T = P ( R ^3- r ₁^3- r ₂^3 ) 4 ( r ₁^2+ r ₂^2- R ^2 )