The energy needed for breaking a liquid drop | vedclass

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

Question

(A) The volume of the large drop is equal to the sum of the volumes of the '$n$' small droplets.$V_{large} = n 	imes V_{small}$$\frac{4}{3} \pi R^3 =

Vedclass · Accepted Answer

$4\pi T R^2 [\frac{R}{r} - 1]$

Vedclass · Answer

(A) The volume of the large drop is equal to the sum of the volumes of the '$n$' small droplets.$V_{large} = n 	imes V_{small}$$\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3$$R^3 = n r^3 \implies n = \frac{R^3}{r^3}$The energy required is equal to the increase in surface area multiplied by the surface tension '$T$'.$E = T 	imes (	ext{Final Surface Area} - 	ext{Initial Surface Area})$$E = T 	imes (n 	imes 4\pi r^2 - 4\pi R^2)$Substitute $n = \frac{R^3}{r^3}$:$E = 4\pi T (\frac{R^3}{r^3} 	imes r^2 - R^2)$$E = 4\pi T (\frac{R^3}{r} - R^2)$$E = 4\pi T R^2 (\frac{R}{r} - 1)$

The energy needed for breaking a liquid drop of radius '$R$' into '$n$' droplets each of radius '$r$' is [$T$ = surface tension of the liquid]

Similar Questions

The potential energy of a molecule on the surface of a liquid compared to one inside the liquid is

If the surface tension of a liquid is $T$,the gain in surface energy for an increase in liquid surface by $A$ is

$A$ big water drop is formed by the combination of $n$ small water droplets of equal radii. The ratio of the surface energy of $n$ droplets to the surface energy of the big drop is

Surface tension of a soap bubble is $2.0 \times 10^{-2} \; N m^{-1}$. Work done to increase the radius of the soap bubble from $3.5 \; cm$ to $7 \; cm$ will be $......... \times 10^{-4} \; J$ [Take $\pi = \frac{22}{7}$]

If $1000$ droplets of water of surface tension $0.07\,N/m$,each having the same radius $1\,mm$,combine to form a single drop,the released surface energy in the process is: (Take $\pi = \frac{22}{7}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module