Home
Class 12
PHYSICS
Two drops of equal radius coalesce to fo...

Two drops of equal radius coalesce to form a bigger drop. What is ratio of surface energy of bigger drop to smaller one?

A

`2^(1//2):1`

B

`1:1`

C

`2^(2//3):1`

D

None of these

Text Solution

Verified by Experts

Promotional Banner

Similar Questions

Explore conceptually related problems

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

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 :

Two drops of equal radius r coalesce to form a single drop under isothermal conditions . The radius of such a drop would be

When water droplets merge to form a bigger drop

When water droplets merge to form a bigger drop

Two small drops of mercury each of radius r form a single large drop. The ratio of surface energy before and after this change is

If n drops, each of capacitance C and charged to a potential V, coalesce to form a big drop, the ratio of the energy stored in the big drop to that in each small drop will be

If 5 small spherical droplet coalesce to form a bigger drop then temperature of bigger drop in comparison to smaller drops will

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)

Six small raindrops each of radius 1.5 mm, come down with a terminal velocity of 6 cm s^(-1) . They coalesce to form a bigger drop. What is the terminal velocity of the bigger drop?