(C) The excess pressure inside a soap bubble of radius $R$ is given by $\Delta P = P_{in} - P_{out} = \frac{4T}{R}$,where $T$ is the surface tension.F

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air flows from the smaller bubble to the bigger

(C) The excess pressure inside a soap bubble of radius $R$ is given by $\Delta P = P_{in} - P_{out} = \frac{4T}{R}$,where $T$ is the surface tension.For two bubbles with radii $R_1$ and $R_2$ where $R_1 Since $R_1 \frac{4T}{R_2}$,which implies $P_1 > P_2$.Air always flows from a region of higher pressure to a region of lower pressure.Therefore,air flows from the smaller bubble to the bigger bubble.

Class 11 Physics Mechanical Properties of Fluids Excess Pressure and coalesce of Bubble and drop

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If two soap bubbles of different radii are connected by a tube,

A
air flows from the bigger bubble to the smaller bubble till the sizes become equal
B
air flows from bigger bubble to the smaller bubble till the sizes are interchanged
C
air flows from the smaller bubble to the bigger
D
there is no flow of air.

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Excess Pressure and coalesce of Bubble and drop

Two soap bubbles combine to form a single bubble. In this process,the change in volume and surface area are respectively $V$ and $A$. If $P$ is the atmospheric pressure,and $T$ is the surface tension of the soap solution,the following relation is true :

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The pressure inside two soap bubbles, $A$ and $B$, is $1.01 \,atm$ and $1.02 \,atm$ respectively. The ratio of their respective radii $(r_A : r_B)$ is (outside pressure $= 1 \,atm$).

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In a cylinder provided with a piston, air is under pressure $P_1$ at a constant temperature $t$. $A$ soap bubble with radius $r$ and surface tension $T$ is lying inside the cylinder. To reduce the radius of the soap bubble to half, the required air pressure inside the cylinder is

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$A$ large number of water drops,each of radius $r$,combine to form a single drop of radius $R$. If the surface tension is $T$ and the mechanical equivalent of heat is $J$,the rise in heat energy per unit volume will be:

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If two soap bubbles of different radii are connected by a tube,

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Two soap bubbles combine to form a single bubble. In this process,the change in volume and surface area are respectively $V$ and $A$. If $P$ is the atmospheric pressure,and $T$ is the surface tension of the soap solution,the following relation is true :

The pressure inside two soap bubbles, $A$ and $B$, is $1.01 \,atm$ and $1.02 \,atm$ respectively. The ratio of their respective radii $(r_A : r_B)$ is (outside pressure $= 1 \,atm$).

In a cylinder provided with a piston, air is under pressure $P_1$ at a constant temperature $t$. $A$ soap bubble with radius $r$ and surface tension $T$ is lying inside the cylinder. To reduce the radius of the soap bubble to half, the required air pressure inside the cylinder is

$A$ large number of water drops,each of radius $r$,combine to form a single drop of radius $R$. If the surface tension is $T$ and the mechanical equivalent of heat is $J$,the rise in heat energy per unit volume will be:

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