If 450 text{ erg} of work is done in blowing | vedclass

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(C) The work done in blowing a soap bubble of radius $r$ is given by $W_1 = T 	imes \Delta A 	imes 2$, where $T$ is the surface tension and $\Delta

Vedclass · Accepted Answer

$3600$

Vedclass · Answer

(C) The work done in blowing a soap bubble of radius $r$ is given by $W_1 = T 	imes \Delta A 	imes 2$, where $T$ is the surface tension and $\Delta A$ is the change in surface area. Since a soap bubble has two surfaces, the area is $2 	imes (4\pi r^2)$.$W_1 = T 	imes 8\pi r^2 = 450 	ext{ erg}$.When the radius is increased to $3r$, the new work done $W_2$ is $T 	imes 8\pi (3r)^2 = T 	imes 8\pi (9r^2) = 9 	imes (T 	imes 8\pi r^2)$.Substituting the value of $W_1$, we get $W_2 = 9 	imes 450 = 4050 	ext{ erg}$.The additional work required is $W_{	ext{add}} = W_2 - W_1 = 4050 - 450 = 3600 	ext{ erg}$.

If $450 \text{ erg}$ of work is done in blowing a soap bubble of radius $r$, the additional work required to be done to blow it to a radius equal to $3r$ is (in $\text{ erg}$)

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