The work done in blowing a soap bubble of vol | vedclass

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

Question

(A) The volume of a soap bubble is given by $V = \frac{4}{3} \pi r^3$,which implies $r = (\frac{3V}{4\pi})^{1/3}$.The work done $W$ in blowing a soap

Vedclass · Accepted Answer

$2^{2/3} W$

Vedclass · Answer

(A) The volume of a soap bubble is given by $V = \frac{4}{3} \pi r^3$,which implies $r = (\frac{3V}{4\pi})^{1/3}$.The work done $W$ in blowing a soap bubble of radius $r$ is equal to the surface tension $T$ multiplied by the change in surface area. Since a soap bubble has two surfaces (inner and outer),the total surface area is $2 	imes (4 \pi r^2) = 8 \pi r^2$.Thus,$W = 8 \pi r^2 T$.Substituting the expression for $r$,we get $W = 8 \pi (\frac{3V}{4\pi})^{2/3} T$.This shows that $W \propto V^{2/3}$.For a bubble of volume $2V$,the new work $W'$ is given by $\frac{W'}{W} = (\frac{2V}{V})^{2/3} = 2^{2/3}$.Therefore,$W' = 2^{2/3} W$.

The work done in blowing a soap bubble of volume $V$ is $W$. The work required to blow a soap bubble of volume $2V$ is [where $T$ is the surface tension of the soap solution].

Similar Questions

Energy needed in breaking a liquid drop of radius $R$ into $n$ smaller drops,each of radius $r$,is [where $T$ is the surface tension of the liquid].

$A$ mercury drop of radius $1\, cm$ is sprayed into $10^{6}$ drops of equal size. The energy expressed in joule is (surface tension of mercury is $460 \times 10^{-3}\, N/m$)

How much work is required to form a bubble of radius $10 \ cm$ from a liquid with surface tension $\frac{3}{100} \ N/m$?

Energy required to form a soap bubble of diameter $20 \, cm$ will be (Surface tension for soap solution is $30 \, dynes/cm$)

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module