If W_1 is the work done in increasing the rad | vedclass

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(A) The work done in increasing the radius of a soap bubble from $r_1$ to $r_2$ is given by the formula $W = T 	imes \Delta A$, where $T$ is the surf

Vedclass · Accepted Answer

$3: 5$

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(A) The work done in increasing the radius of a soap bubble from $r_1$ to $r_2$ is given by the formula $W = T 	imes \Delta A$, where $T$ is the surface tension and $\Delta A$ is the change in surface area. Since a soap bubble has two surfaces, the change in area is $\Delta A = 2 	imes (4\pi r_2^2 - 4\pi r_1^2) = 8\pi(r_2^2 - r_1^2)$.For $W_1$ (from $r$ to $2r$): $W_1 = 8\pi T ((2r)^2 - r^2) = 8\pi T (4r^2 - r^2) = 8\pi T (3r^2) = 24\pi T r^2$.For $W_2$ (from $2r$ to $3r$): $W_2 = 8\pi T ((3r)^2 - (2r)^2) = 8\pi T (9r^2 - 4r^2) = 8\pi T (5r^2) = 40\pi T r^2$.Therefore, the ratio $W_1: W_2 = (24\pi T r^2) : (40\pi T r^2) = 24:40 = 3:5$.

If $W_1$ is the work done in increasing the radius of a soap bubble from $r$ to $2r$ and $W_2$ is the work done in increasing the radius of the soap bubble from $2r$ to $3r$, then $W_1: W_2=$

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