If the excess pressure inside the soap bubble of radius $r_{1}$ in air is equal to the excess pressure inside air bubble of radius $r_{2}$ inside the soap solution, then $r_{1} : r_{2}$ is:

2 : 1

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Solution

The correct option is A $2 : 1$
Excess pressure inside the soap bubble is given by
$P_{1} = \frac{4 S}{r_{1}}$
Excess pressure inside the air bubble is given by
$P_{2} = \frac{2 S}{r_{2}}$
Now $P_{1} = P_{2}$
$∴ \frac{4 S}{r_{1}} = \frac{2 S}{r_{2}}$
$∴ \frac{r_{1}}{r_{2}} = \frac{2}{1} = 2 : 1$

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If the excess pressure inside the soap bubble of radius r1 in air is equal to the excess pressure inside air bubble of radius r2 inside the soap solution, then r1:r2 is:

The correct option is A 2:1Excess pressure inside the soap bubble is given by P1=4Sr1Excess pressure inside the air bubble is given by P2=2Sr2Now P1=P2∴4Sr1=2Sr2∴r1r2=21=2:1

If the excess pressure inside the soap bubble of radius $r_{1}$ in air is equal to the excess pressure inside air bubble of radius $r_{2}$ inside the soap solution, then $r_{1} : r_{2}$ is: