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

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Question

(A) soap bubble has two surfaces (inner and outer), so the work done $W$ in changing its radius from $r_1$ to $r_2$ is given by $W = T 	imes \Delta A

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$0.4$

Vedclass · Answer

(A) soap bubble has two surfaces (inner and outer), so the work done $W$ in changing its radius from $r_1$ to $r_2$ is given by $W = T 	imes \Delta A 	imes 2$, where $\Delta A = 4\pi(r_2^2 - r_1^2)$.Thus, $W = 8\pi T(r_2^2 - r_1^2)$.Given: $T = 0.03 \ Nm^{-1}$, $r_1 = 3 \ cm = 0.03 \ m$, $r_2 = 5 \ cm = 0.05 \ m$.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$$W = 0.24 \pi 	imes 0.0016 = 0.000384 \pi \ J$$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 $5 \ cm$ in millijoules is nearly (surface tension of soap solution $= 0.03 \ Nm^{-1}$) (in $\pi$)

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