n droplets of equal sixe of radus r calesce to from a bigger drop of radius r the energy liberated is equal to ------------
To determine the energy released when n droplets of equal size with radius r coalesce to form a bigger drop with radius R, we can use the principle of energy conservation.
First, let's calculate the initial total volume of the droplets. The volume of each droplet can be calculated using the formula for the volume of a sphere: V = (4/3)πr³. Since all the droplets have the same size, the total initial volume will be n times the volume of a single droplet:
Initial total volume (V_initial) = n * (4/3)πr³
When the droplets coalesce, they combine to form a larger drop. Hence, the final volume of the bigger drop can be calculated using the formula for the volume of a sphere: V = (4/3)πR³.
Now, according to the law of conservation of volume, the total volume of the droplets before and after coalescence should remain the same. Therefore, we can equate the initial and final volumes:
V_initial = V_final
n * (4/3)πr³ = (4/3)πR³
Next, let's solve this equation for the final radius, R. Divide both sides of the equation by (4/3)π to isolate R³:
n * r³ = R³
Taking the cube root of both sides, we find:
R = (n * r)^(1/3)
Now we have the final radius of the bigger drop, R, in terms of the initial radius, r, and the number of droplets, n.
To calculate the energy liberated during this coalescence process, we can use the formula:
Energy liberated = Surface area of the bigger drop * Surface tension
The surface area of a sphere can be calculated using the formula: A = 4πR²
Therefore,
Energy liberated = 4πR² * Surface tension
where Surface tension is the property of the liquid and is denoted by the symbol γ.
Hence, the energy liberated when n droplets of equal size with radius r coalesce to form a bigger drop with radius R is given by the equation:
Energy liberated = 4π [(n * r)^(1/3)]² * γ