The amount of work done in blowing a soap bub | vedclass

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

Question

(B) soap bubble has two surfaces (inner and outer). Therefore,the total surface area of a soap bubble of radius $r$ is $A = 2 	imes (4\pi r^2) = 8\pi

Vedclass · Accepted Answer

$2\pi(D^2 - d^2)T$

Vedclass · Answer

(B) soap bubble has two surfaces (inner and outer). Therefore,the total surface area of a soap bubble of radius $r$ is $A = 2 	imes (4\pi r^2) = 8\pi r^2$.Initial radius $r_1 = d/2$,so initial area $A_1 = 8\pi(d/2)^2 = 2\pi d^2$.Final radius $r_2 = D/2$,so final area $A_2 = 8\pi(D/2)^2 = 2\pi D^2$.The change in surface area is $\Delta A = A_2 - A_1 = 2\pi(D^2 - d^2)$.The work done $W$ is given by $W = T 	imes \Delta A$.Substituting the values,$W = T 	imes 2\pi(D^2 - d^2) = 2\pi(D^2 - d^2)T$.

The amount of work done in blowing a soap bubble such that its diameter increases from $d$ to $D$ is ($T=$ surface tension of solution).

Similar Questions

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

The work done in increasing the diameter of a soap bubble from $2 \ cm$ to $4 \ cm$ is (Surface tension of soap solution $= 3.5 \times 10^{-2} \ N/m$)

One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to the total initial surface energy is

The surface tension of a soap solution is $T$. Work done in blowing a soap bubble of diameter $2d$ is (in $\pi d^2 T$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module