Two soap bubbles of radii r and 2r are arrang | vedclass

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(D) The excess pressure inside a soap bubble of radius $R$ is given by $\Delta P = \frac{4T}{R}$,where $T$ is the surface tension of the soap solution

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The radius of the smaller bubble will decrease and that of the bigger bubble will increase

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(D) The excess pressure inside a soap bubble of radius $R$ is given by $\Delta P = \frac{4T}{R}$,where $T$ is the surface tension of the soap solution.For the smaller bubble of radius $r$,the pressure inside is $P_1 = P_{atm} + \frac{4T}{r}$.For the bigger bubble of radius $2r$,the pressure inside is $P_2 = P_{atm} + \frac{4T}{2r} = P_{atm} + \frac{2T}{r}$.Since $P_1 > P_2$,when the valve is opened,air flows from the region of higher pressure (smaller bubble) to the region of lower pressure (bigger bubble).Consequently,the radius of the smaller bubble decreases and the radius of the bigger bubble increases.

Two soap bubbles of radii $r$ and $2r$ are arranged as shown in the diagram. The valve is now opened. Which one of the following will result?

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