Excess pressure of one soap bubble is four ti | vedclass

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Question

(a) $\Delta P = \frac{{4T}}{r}\; \Rightarrow \frac{{\Delta {P_1}}}{{\Delta {P_2}}} = 4$

$	herefore \;\frac{{{r_2}}}{{{r_1}}} = 4$ and $\frac{{{V_1

Vedclass · Accepted Answer

$1:64$

Vedclass · Answer

(a) $\Delta P = \frac{{4T}}{r}\; \Rightarrow \frac{{\Delta {P_1}}}{{\Delta {P_2}}} = 4$

$	herefore \;\frac{{{r_2}}}{{{r_1}}} = 4$ and $\frac{{{V_1}}}{{{V_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} ight)^3} = \frac{1}{{64}}$

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

Excess pressure of one soap bubble is four times more than the other. Then the ratio of volume of first bubble to another one is

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