A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is ................ cm.

Properties of Matter PYQ — JEE Main 2021 · 26 Aug · Shift 1 · Q66 with Solution

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Correct answer: 2

For a soap bubble of radius $R$ , the excess pressure is

\Delta P = \frac{4T}{R}

where $T$ is the surface tension.

The smaller bubble is formed inside the bigger bubble.

So,

P_2 - P_{\text{atm}} = \frac{4T}{6}

P_1 - P_2 = \frac{4T}{3}

Therefore, pressure inside the smaller bubble relative to atmosphere is

P_1 - P_{\text{atm}} = (P_1 - P_2) + (P_2 - P_{\text{atm}})

P_1 - P_{\text{atm}} = \frac{4T}{3} + \frac{4T}{6}

P_1 - P_{\text{atm}} = 4T\left(\frac{1}{3}+\frac{1}{6}\right)=4T\cdot \frac{1}{2}=2T

Let the radius of the equivalent soap bubble be $R$ . Its excess pressure relative to atmosphere should be the same:

\frac{4T}{R} = 2T

Cancelling $T$ ,

\frac{4}{R} = 2

R = 2\,\text{cm}

The required radius is

\boxed{2}

cm.