MHT CET202526 Apr 2025Morning ShiftPhysicsMechanical Properties of FluidsActual
Let R₁, R₂ and R₃ be the radii of three mercury drops. A big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is
Options
- A(R₁^3+R₂^3+R₃^3 )^ 1 3
- B(R₁^2+R₂^3-R₃^3 )^ 1 3
- C(R₁^3+R₂^3+R₃^3 )
- D(R₁+R₂+R₃ )^3
Correct answer
A. (R₁^3+R₂^3+R₃^3 )^ 1 3
Step-by-step solution
Volume Conservation Principle Under isothermal conditions with constant density, the total volume remains conserved during the coalescence of three mercury drops. The volume of each spherical drop is V = 4 3 R^3 . The combined volume of the small drops with radii R₁ , R₂ , and R₃ equals the volume of the resultant drop with radius R : 4 3 R^3 = 4 3 R₁^3 + 4 3 R₂^3 + 4 3 R₃^3 Simplifying, we factor out the common term: R^3 = R₁^3 + R₂^3 + R₃^3 Taking the cube root of both sides, the radius of the big drop is: R = (R