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

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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 $0.03 \,N/m$).

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