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Question

(D) soap bubble has two surfaces (inner and outer),so the change in surface area is $\Delta A = 2 	imes (4 \pi R^2) = 8 \pi R^2$.The work done $W$ is

Vedclass · Accepted Answer

$192 \pi 	imes 10^{-4} \, J$

Vedclass · Answer

(D) soap bubble has two surfaces (inner and outer),so the change in surface area is $\Delta A = 2 	imes (4 \pi R^2) = 8 \pi R^2$.The work done $W$ is given by $W = T 	imes \Delta A$,where $T$ is the surface tension.Given: $T = 60 	imes 10^{-3} \, N/m$ and $R = 0.2 \, m$.$W = 60 	imes 10^{-3} 	imes 8 \pi 	imes (0.2)^2$$W = 60 	imes 10^{-3} 	imes 8 \pi 	imes 0.04$$W = 480 \pi 	imes 10^{-3} 	imes 0.04$$W = 19.2 \pi 	imes 10^{-3} \, J = 192 \pi 	imes 10^{-4} \, J$.

The work done in blowing a soap bubble of radius $0.2 \, m$,given that the surface tension of the soap solution is $60 \times 10^{-3} \, N/m$,is:

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