The work done in blowing a soap bubble of 10 | vedclass

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

Question

(B) soap bubble has two free surfaces (inner and outer). Work done $(W)$ is given by the formula: $W = S 	imes \Delta A 	imes 2$,where $S$ is the su

Vedclass · Accepted Answer

$75.36 	imes 10^{-4}$

Vedclass · Answer

(B) soap bubble has two free surfaces (inner and outer). Work done $(W)$ is given by the formula: $W = S 	imes \Delta A 	imes 2$,where $S$ is the surface tension and $\Delta A$ is the change in surface area. Given: $S = 0.03 \,N/m$,radius $r = 10 \,cm = 0.1 \,m$. The change in area $\Delta A = 4 \pi r^2$. Substituting the values: $W = 0.03 	imes (4 	imes 3.14 	imes (0.1)^2) 	imes 2$. $W = 0.03 	imes 4 	imes 3.14 	imes 0.01 	imes 2$. $W = 0.03 	imes 0.2512 	imes 2 = 0.015072 \,J$. $W = 75.36 	imes 10^{-4} \,J$.

The work done in blowing a soap bubble of $10 \,cm$ radius is ...... $J$ (surface tension of soap solution is $0.03 \,N/m$).

Similar Questions

$A$ soap bubble of radius $R$ is blown. After heating a solution,a second bubble of radius $2R$ is blown. The work required to blow the $2^{nd}$ bubble in comparison to that required for the $1^{st}$ bubble is

Two mercury drops (each of radius $r$) merge to form a bigger drop. The surface energy of the bigger drop, if $T$ is the surface tension, is

Two mercury drops of radii $r$ and $2r$ merge to form a bigger drop. The surface energy released in the process is nearly (Surface tension of mercury is $S$ and take $9^{2/3} = 4.326$). (in $\pi r^2 S$)

Explain surface energy and surface tension.

The work done in blowing a soap bubble of radius $R$ is $W_1$ and that of radius $2R$ is $W_2$. The ratio of $W_1$ to $W_2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module