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(B) The pressure inside a soap bubble is given by $P = P_0 + \frac{4T}{r}$,where $P_0 = 8 \ N/m^2$ is the external pressure,$T = 0.04 \ N/m$ is the su

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$6$

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(B) The pressure inside a soap bubble is given by $P = P_0 + \frac{4T}{r}$,where $P_0 = 8 \ N/m^2$ is the external pressure,$T = 0.04 \ N/m$ is the surface tension,and $r$ is the radius.For bubble $A$ $(r_A = 0.02 \ m)$: $P_A = 8 + \frac{4 	imes 0.04}{0.02} = 8 + 8 = 16 \ N/m^2$.For bubble $B$ $(r_B = 0.04 \ m)$: $P_B = 8 + \frac{4 	imes 0.04}{0.04} = 8 + 4 = 12 \ N/m^2$.Using the ideal gas law $PV = nRT$,and assuming temperature $T$ is constant,$n = \frac{PV}{RT}$.For bubble $A$: $n_A = \frac{P_A V_A}{RT} = \frac{16 	imes \frac{4}{3} \pi (0.02)^3}{RT}$.For bubble $B$: $n_B = \frac{P_B V_B}{RT} = \frac{12 	imes \frac{4}{3} \pi (0.04)^3}{RT}$.Taking the ratio $\frac{n_B}{n_A} = \frac{12 	imes (0.04)^3}{16 	imes (0.02)^3} = \frac{12}{16} 	imes (2)^3 = \frac{3}{4} 	imes 8 = 6$.

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

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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