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 ( $σ$ surface tension, $ρ$ density)

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

${[\frac{σ}{ρ} (\frac{1}{a} - \frac{1}{b})]}^{1 / 2}$

${[\frac{6 σ}{ρ} (\frac{1}{a} - \frac{1}{b})]}^{1 / 2}$

${[\frac{3 σ}{ρ} (\frac{1}{a} - \frac{1}{b})]}^{1 / 2}$

${[\frac{2 σ}{ρ} (\frac{1}{a} - \frac{1}{b})]}^{1 / 2}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let say number of droplets are n.

$Energy released = [n \times 4 π a^{2} - 4 π b^{2}] σ$

$Now, n \times \frac{4}{3} π a^{3} = \frac{4}{3} π b^{3} or n = \frac{b^{3}}{a^{3}}$

Therefore, energy released is

$\begin{array}{l} = [\frac{b^{3}}{a^{3}} \times 4 π a^{2} - 4 π b^{2}] σ = 4 π b^{2} [\frac{b}{a} - 1] σ \\ Now, K.E of big drop=energy released \\ \Rightarrow \frac{1}{2} (\frac{4}{3} π b^{3}) ρ v^{2} = 4 π b^{2} [\frac{b}{a} - 1] σ \\ ⇒ v = {[\frac{6 σ}{ρ} (\frac{1}{a} - \frac{1}{b})]}^{1 / 2} \end{array}$