Question

What change in surface energy will be noticed when a drop of radius r 1000 r surface tension s

Answer 1

The change in the surface energy, when a drop of radius R is broken into 1000 drops each of radius r is $36\pi R^2s$

• Surface area of radius R = 1000 * surface area of radius r
• ∴
• $\dfrac{4}{3}\pi R^3 = 1000 * \dfrac{4}{3}\pi r^3\\\\R^3 = 1000r^3\\\\r = \dfrac{R}{10}$
• Now, from given, we have,
• surface energy = surface tension * increase in surface area
• $= s [(1000)4\pi r^2 - (4\pi R^2)]\\\\=4\pi s[1000r^2-R^2]\\\\=4\pi s[\frac{R^2}{100} * 1000 - R^2]\\\\=4\pi s[10R^2-R^2]\\\\=4\pi s(9R^2)\\\\=36\pi s R^2$

Answer 2

The change in surface energy will be S.E = 36 πR^2 . T

Explanation:

Surface Energy:

It is defined as the amount of work done in increasing the area of the liquid against surface tension.

4/3 πR^3 = 1000  4/3 π r^3

r^3 = R^3 / 1000

As we know that

S.E = Surface tension x surface area

S.E = 1000 4πr^2 . T - (4πR^2) T

S.E = 1000 4π R^2 / 100 x T - 4πR^2 . T

S.E = 4πR^2 . T (10 - 1)

S.E = 4πR^2 . T (9)

S.E = 36 πR^2 . T

Thus the change in surface energy will be S.E = 36 πR^2 . T

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