A soap bubble of radius 3, {cm} is formed ins | vedclass

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Question

Excess pressure inside the smaller soap bubble

$\Delta {P}=\frac{4 {S}}{{r}_{1}}+\frac{4 {S}}{{r}_{2}}\ldots 	ext { (i) }$

The excess pressure

Vedclass · Accepted Answer

$2$

Vedclass · Answer

Excess pressure inside the smaller soap bubble

$\Delta {P}=\frac{4 {S}}{{r}_{1}}+\frac{4 {S}}{{r}_{2}}\ldots 	ext { (i) }$

The excess pressure inside the equivalent soap bubble

$\Delta {P}=\frac{4 {S}}{{R}_{{eq}}} \ldots 	ext { (ii) }$

From $ (i) \,\& \,(ii)$

$\frac{4 {S}}{{R}_{{eq}}}=\frac{4 {S}}{{T}_{1}}+\frac{4 {S}}{{T}_{2}}$

$\frac{1}{{R}_{{eq}}}=\frac{1}{{T}_{1}}+\frac{1}{{I}_{2}}$

$=\frac{1}{6}+\frac{1}{3}$

${R}_{{eq}}=2 \,{cm}$

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

A soap bubble of radius $3\, {cm}$ is formed inside the another soap bubble of radius $6\, {cm}$. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is .......... ${cm}$

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