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

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(B) soap bubble has two surfaces (inner and outer). The work done $W$ in changing the radius of a soap bubble from $r_1$ to $r_2$ is given by:$W = 2 \

Vedclass · Accepted Answer

$0.4$

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(B) soap bubble has two surfaces (inner and outer). The work done $W$ in changing the radius of a soap bubble from $r_1$ to $r_2$ is given by:$W = 2 	imes T 	imes \Delta A = 2 	imes T 	imes 4\pi(r_2^2 - r_1^2) = 8\pi T(r_2^2 - r_1^2)$Given: $T = 0.03\, Nm^{-1}$,$r_1 = 3\, cm = 0.03\, m$,$r_2 = 5\, cm = 0.05\, m$.Substituting the values:$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 $= 0.03\, Nm^{-1}$) (in $pi \,mJ$)

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