MHT CET202120 Sep 2021Evening ShiftPhysicsMechanical Properties of FluidsActual
Let ' R ₁ ' and ' R ₂ ' are radii of two mercury drops. A big mercury drop is formed from then under isothermal conditions. The radius of the resultant drop is
Options
- AR ₁^2+ R ₂^2
- B( R ₁^3+ R ₂^3 )^ 1 3
- CR ₁^2- R ₂^2
- DR₁+R₂ 2
Correct answer
B. ( R ₁^3+ R ₂^3 )^ 1 3
Step-by-step solution
The volume of the bigger drop will be equal to the sum of the volumes of the smaller drops aligned & 4 3 R ^3= 4 3 R ₁^3+ 4 3 R ₂^3 & R = ( R ₁^3+ R ₂^3 )^ 1 / 3 aligned