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(A) The work done $W$ in blowing a soap bubble of radius $r$ is given by $W = T 	imes \Delta A$, where $T$ is the surface tension and $\Delta A$ is t

Vedclass · Accepted Answer

slightly less than $4$ times

Vedclass · Answer

(A) The work done $W$ in blowing a soap bubble of radius $r$ is given by $W = T 	imes \Delta A$, where $T$ is the surface tension and $\Delta A$ is the change in surface area.Since a soap bubble has two surfaces (inner and outer), the total surface area is $A = 2 	imes (4\pi r^2) = 8\pi r^2$.Thus, $W = T 	imes 8\pi r^2$.For the first bubble of radius $R$, $W_1 = 8\pi R^2 T_1$.For the second bubble of radius $2R$, $W_2 = 8\pi (2R)^2 T_2 = 32\pi R^2 T_2$.When the solution is heated, the surface tension $T$ decreases, so $T_2 The ratio is $\frac{W_2}{W_1} = \frac{32\pi R^2 T_2}{8\pi R^2 T_1} = 4 	imes \frac{T_2}{T_1}$.Since $T_2 Therefore, the work required is slightly less than $4$ times the work required for the first bubble.

$A$ soap bubble of radius $R$ is blown. After heating the solution, a second bubble of radius $2R$ is blown. The work required to blow the second bubble in comparison to that required for the first bubble is

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