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A soap bubble of radius $1.0 \mathrm{~cm}$ is formed inside another soap bubble of radius $2.0 \mathrm{~cm}$. The radius of an another soap bubble which has the same pressure difference as that between the inside of the smaller and outside of large soap bubble, in metres is
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Verified Answer
The correct answer is:
$6.67 \times 10^3$
Given, radius of first soap bubble $\left(R_1\right)=2.0 \mathrm{~cm}$
Radius of second soap bubble $\left(R_2\right)=1.0 \mathrm{~cm}$
We know that,
Radius of another bubble having same pressure difference,
or,
$$
R=\frac{R_1 R_2}{R_1+R_2}
$$
$R=\frac{2.0 \times 1.0 \times 10^{-4}}{(2.0+1.0) \times 10^{-2}}$
or,
$$
R=\frac{2 \times 10^{-4}}{3 \times 10^{-2}}
$$
or,
$R=6.67 \times 10^{-3} \mathrm{~m}$
Radius of second soap bubble $\left(R_2\right)=1.0 \mathrm{~cm}$
We know that,
Radius of another bubble having same pressure difference,
or,
$$
R=\frac{R_1 R_2}{R_1+R_2}
$$
$R=\frac{2.0 \times 1.0 \times 10^{-4}}{(2.0+1.0) \times 10^{-2}}$
or,
$$
R=\frac{2 \times 10^{-4}}{3 \times 10^{-2}}
$$
or,
$R=6.67 \times 10^{-3} \mathrm{~m}$
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