N Identical drops each of radius r combine into a big drop of radius R.The energy loss during this formation is 3E where E is the energy of the bigger drop.The value of 'N' is

A large number of liquid drops each of radius 'a' coalesce to form a single spherical drop of radius b. The energy released in the process is converted into kinetic energy of the big drops formed. The speed of big drop will be

A large number of droplets, each of radius a, coalesce to form a bigger drop of radius b . Assume that the energy released in the process is converted into the kinetic energy of the drop. The velocity of the drop is sigma = surface tension, rho = density)

A number of droplets, each of radius r , combine to form a drop of radius R . If T is the surface tension, the rise in temperature will be

n number of water droplets, each of radius r , coalese, to form a single drop of radius R . The rise in temperature d theta is

Two mercury drops each of radius r merge to form a bigger drop. Calculate the surface energy released.

Eight small drops, each of radius r and having same charge q are combined to form a big drop. The ratio between the potentials of the bigger drop and the smaller drop is

Two mercury drops (each of radius 'r' ) merge to from bigger drop. The surface energy of the bigger drop, if 'T' is the surface tension, is :

8000 identical water drops are combined to form a big drop then the ratio of the final surface energy to the initial surface energy of all the drops together is

The energy needed for breaking a drop of radius R into n drops each of radius r is

n charged drops, each of radius r and charge q , coalesce to from a big drop of radius R and charge Q . If V is the electric potential and E is the electric field at the surface of a drop, then.