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(D) The work done $W$ in blowing a soap bubble of radius $R$ is given by $W = T 	imes \Delta A$,where $\Delta A$ is the change in surface area. For a

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$\sqrt[3]{4} W$

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(D) The work done $W$ in blowing a soap bubble of radius $R$ is given by $W = T 	imes \Delta A$,where $\Delta A$ is the change in surface area. For a soap bubble,there are two surfaces,so $\Delta A = 2 	imes (4 \pi R^2) = 8 \pi R^2$.Thus,$W = 8 \pi R^2 T$.Since the volume of the bubble is $V = \frac{4}{3} \pi R^3$,we have $R^3 \propto V$,which implies $R \propto V^{1/3}$.Substituting this into the work equation: $W \propto R^2 \propto (V^{1/3})^2 = V^{2/3}$.Therefore,$\frac{W_2}{W_1} = \left( \frac{V_2}{V_1} ight)^{2/3}$.Given $V_1 = V$ and $V_2 = 2V$,we get $\frac{W_2}{W} = \left( \frac{2V}{V} ight)^{2/3} = 2^{2/3} = (2^2)^{1/3} = 4^{1/3} = \sqrt[3]{4}$.Hence,$W_2 = \sqrt[3]{4} W$.

Column-$I$	Column-$II$
$(a)$ Liquid drop in air	$(i)$ $\frac{4T}{R}$
$(b)$ Bubble of liquid in air	$(ii)$ $\frac{2T}{R}$
	$(iii)$ $\frac{2R}{T}$

If the work done in blowing a bubble of volume $V$ is $W$,then the work done in blowing the bubble of volume $2V$ from the same soap solution will be

Similar Questions

If the internal pressures of two soap bubbles are $1.01 \text{ atm}$ and $1.02 \text{ atm}$ respectively, find the ratio of their volumes.

$A$ soap bubble (surface tension $= T$) is charged to a maximum surface density of charge $= \sigma$. When it is just about to burst,its radius $R$ is given by:

Pressure inside two soap bubbles are $1.01$ and $1.02$ atmosphere, respectively. The ratio of their volumes is (in $ : 1$)

Formation of bubbles is in Column-$I$ and the pressure difference between them is given in Column-$II$. Match them appropriately.

Column-$I$ Column-$II$

$(a)$ Liquid drop in air $(i)$ $\frac{4T}{R}$

$(b)$ Bubble of liquid in air $(ii)$ $\frac{2T}{R}$

$(iii)$ $\frac{2R}{T}$

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