The work done in doubling the radius of a soa | vedclass

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Question

(A) soap bubble has two surfaces (inner and outer),so the work done $W$ is given by $W = 2 	imes T 	imes \Delta A$,where $T$ is the surface tension

Vedclass · Accepted Answer

$9.043	imes 10^3\,erg$

Vedclass · Answer

(A) soap bubble has two surfaces (inner and outer),so the work done $W$ is given by $W = 2 	imes T 	imes \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area.Initial radius $r_1 = 2\,cm$.Final radius $r_2 = 2 	imes r_1 = 4\,cm$.Change in surface area $\Delta A = 4\pi r_2^2 - 4\pi r_1^2 = 4\pi(r_2^2 - r_1^2)$.$\Delta A = 4\pi(4^2 - 2^2) = 4\pi(16 - 4) = 4\pi(12) = 48\pi\,cm^2$.Work done $W = 2 	imes 30\,dyne/cm 	imes 48\pi\,cm^2$.$W = 60 	imes 48\pi = 2880\pi\,erg$.Using $\pi \approx 3.14159$,$W \approx 2880 	imes 3.14159 \approx 9047.7\,erg$.Rounding to the given options,$W \approx 9.043 	imes 10^3\,erg$.

The work done in doubling the radius of a soap bubble of radius $2\,cm$ will be ( Surface Tension of soap solution $= 30\,dyne/cm$ ).

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