Q136 / 160 · TS EAMCET 2016 · Physics
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
- Correct answerA$6.67 \times 10^3$
- B$3.34 \times 10^3$
- C$2.23 \times 10^3$
- D$4.5 \times 10^3$
Solution
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,
$R=6.67 \times 10^{-3} \mathrm{~m}$
Hence, the correct answer is option: 'A'.