Biology, asked by shrushti2383, 1 year ago

Workdone in spraying a liquid droplet of radius r into n droplet

Answers

Answered by rahul286741
0
If the droplet is sprayed into multiple droplets, the total volume is unchanged. Thus, we can write

n(4/3)πr³ = (4/3)πR³

where R is the radius of the initial drop and r is the radius of the droplets.

We can divide the equation by (4/3)π and we get

nr³ = R³

or

r = R(1/n)¹/³

The total surface area of the droplets is n4πr².

We rewrite it by using the last equation

n4πr² = n4πR²(1/n)²/³ = 4πR²n¹/³

4πR² is the surface area of the original drop, so the total area has increased by a factor of n¹/³.

The surface energy is the work required to increase the surface area of a substance by unit area. Then, the surface energy is proportional to surface area.

Hence, the surface energy has increased by a factor n¹/³ and is n¹/³u

The original surface energy was u, so the increase in surface energy is
n¹/³u – u = [n¹/³ – 1]u

If n = 1000 then n¹/³ = 10

and [n¹/³ – 1]u = [10 – 1]u = 9u

Similar questions