Work done in increasing the size of a soap bu | vedclass

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Question

(B) soap bubble has two surfaces (inner and outer). The work done $W$ in increasing the radius from $r_1$ to $r_2$ is given by $W = 2 	imes T 	imes

Vedclass · Accepted Answer

$0.4\,\pi \,mJ$

Vedclass · Answer

(B) soap bubble has two surfaces (inner and outer). The work done $W$ in increasing the radius from $r_1$ to $r_2$ is given by $W = 2 	imes T 	imes \Delta A$,where $\Delta A = 4\pi(r_2^2 - r_1^2)$.Thus,$W = 8\pi T (r_2^2 - r_1^2)$.Given: $T = 0.03\, N/m$,$r_1 = 3\, cm = 0.03\, m$,$r_2 = 5\, cm = 0.05\, m$.$W = 8 	imes \pi 	imes 0.03 	imes ((0.05)^2 - (0.03)^2)$.$W = 8 	imes \pi 	imes 0.03 	imes (0.0025 - 0.0009)$.$W = 8 	imes \pi 	imes 0.03 	imes 0.0016$.$W = 0.24 	imes \pi 	imes 0.0016 = 0.000384\pi\, J$.$W = 0.384\pi\, mJ \approx 0.4\pi\, mJ$.

Work done in increasing the size of a soap bubble from a radius of $3\, cm$ to $5\, cm$ is nearly (Surface tension of soap solution $= 0.03\, N/m$)

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