The excess pressure inside the first soap bub | vedclass

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Question

(A) The excess pressure inside a soap bubble of radius $r$ is given by $P = \frac{4T}{r}$, where $T$ is the surface tension.Given that the excess pres

Vedclass · Accepted Answer

$1$:$27$

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(A) The excess pressure inside a soap bubble of radius $r$ is given by $P = \frac{4T}{r}$, where $T$ is the surface tension.Given that the excess pressure in the first bubble $(P_1)$ is three times that in the second bubble $(P_2)$:$P_1 = 3P_2$$\frac{4T}{r_1} = 3 	imes \frac{4T}{r_2}$$\frac{1}{r_1} = \frac{3}{r_2} \implies r_2 = 3r_1$The volume of a spherical soap bubble is $V = \frac{4}{3}\pi r^3$.The ratio of the volumes is:$\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \left(\frac{r_1}{r_2}ight)^3$Substituting $r_2 = 3r_1$:$\frac{V_1}{V_2} = \left(\frac{r_1}{3r_1}ight)^3 = \left(\frac{1}{3}ight)^3 = \frac{1}{27}$Thus, the ratio of the volume of the first soap bubble to the volume of the second soap bubble is $1:27$.

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

The excess pressure inside the first soap bubble is three times that inside the second soap bubble. The ratio of the volume of the first soap bubble to the volume of the second soap bubble is

Similar Questions

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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