JEE Main2012PhysicsMechanical Properties of FluidsActual
A large number of droplets, each of radius, r coalesce to form a bigger drop of radius, R . An engineer designs a machine so that the energy released in this process is converted into the kinetic energy of the drop. Velocity of the drop is ( T= surface tension, = density)
Options
- A[ T ( 1 r - 1 R ) ]^ 1 / 2
- B[ 6 T ( 1 r - 1 R ) ]^ 1 / 2
- C[ 3 T ( 1 r - 1 R ) ]^ 1 / 2
- D[ 2 T ( 1 r - 1 R ) ]^ 1 / 2
Correct answer
B. [ 6 T ( 1 r - 1 R ) ]^ 1 / 2
Step-by-step solution
When small droplets coelesce to form a bigger drop, energy released in this process is given by, 4 R^3 T [ 1 r - 1 R ] Where, R= radius of big drop r= radius of small drop T= surface tension According to question aligned & 1 2 m v^2=4 R^3 T [ 1 r - 1 R ] & 1 2 [ 4 3 R^3 ] v^2=4 R^3 T [ 1 r - 1 R ] V^2 & = 6 T [ 1 r - 1 R ] V= [ 6 T ( 1 r - 1 R ) ]^ 1 2 aligned