A large number of liquid drops each of radius | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) When $n$ drops of radius $r$ coalesce to form a single drop of radius $R$,the volume is conserved: $n(\frac{4}{3}\pi r^3) = \frac{4}{3}\pi R^3$,so

Vedclass · Accepted Answer

$\sqrt{\frac{6T}{\rho}\left(\frac{1}{r} - \frac{1}{R}\right)}$

Vedclass · Answer

(D) When $n$ drops of radius $r$ coalesce to form a single drop of radius $R$,the volume is conserved: $n(\frac{4}{3}\pi r^3) = \frac{4}{3}\pi R^3$,so $n = (R/r)^3$.The energy released during the process is equal to the decrease in surface area multiplied by the surface tension $T$:$E = T 	imes (A_{initial} - A_{final}) = T 	imes (n \cdot 4\pi r^2 - 4\pi R^2)$.Substituting $n = R^3/r^3$:$E = 4\pi T (\frac{R^3}{r^3} \cdot r^2 - R^2) = 4\pi T R^3 (\frac{1}{r} - \frac{1}{R})$.This energy is converted into the kinetic energy of the big drop:$K.E. = \frac{1}{2} M v^2 = E$,where $M$ is the mass of the big drop,$M = ho \cdot V = ho \cdot \frac{4}{3}\pi R^3$.Equating the two:$\frac{1}{2} (\frac{4}{3}\pi R^3 ho) v^2 = 4\pi T R^3 (\frac{1}{r} - \frac{1}{R})$.$\frac{2}{3} \pi R^3 ho v^2 = 4\pi T R^3 (\frac{1}{r} - \frac{1}{R})$.$v^2 = \frac{4 	imes 3}{2} \frac{T}{ho} (\frac{1}{r} - \frac{1}{R}) = \frac{6T}{ho} (\frac{1}{r} - \frac{1}{R})$.$v = \sqrt{\frac{6T}{ho} (\frac{1}{r} - \frac{1}{R})}$.

$A$ large number of liquid drops each of radius $r$ coalesce to form a single drop of radius $R$. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given,surface tension of liquid $T$,density $\rho$):

Similar Questions

The excess pressure inside one soap bubble is four times that of another. What is the ratio of the volume of the first bubble to the second bubble?

Pressure inside two soap bubbles is $1.01 \, atm$ and $1.03 \, atm$. The ratio between their volumes is (Pressure outside the soap bubble is $1 \, atm$). (in $ : 1$)

$A$ drop of liquid of diameter $2.8\, mm$ breaks up into $125$ identical drops. The change in energy is nearly ........ $erg$ (Surface Tension of liquid = $75\, dynes/cm$).

Two soap bubbles with radii $r_1$ and $r_2$ $(r_1 > r_2)$ come in contact. Their common surface has a radius of curvature $r$. Find $r$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module