Question

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

Answer 1

Let the radius of small drop is r and radius of big drop is R
A/C to question,
two smaller drops are collapsed and formed a bigger drop.
∴ volume of two smaller drops = volume of bigger drop
⇒ 4/3 πr³ + 4/3 πr³ = 4/3 πR³
⇒ 2r³ = R³
⇒R = $\sqrt[3]{2}$r

Now, surface energy of each smaller drop = surface tension × surface area of smaller drop
= T × 4πr² [ assumed surface tension = T ]
surface energy of bigger drop = surface tension × surface area of bigger drop
= T × 4πR²
= T × 4π($\sqrt[3]{2}$r)²

Now, ratio of energy of bigger to smaller drop = T × 4π($\sqrt[3]{2}$r)²/T × 4πr²
= 2⅔ : 1

Hence, answer is 2⅔ : 1

Answer 2

Answer:

Hear you go

Explanation:

hope it help you