The excess pressure inside an air bubble of radius $r$ just below the surface of the water is $P_{1}$ . The excess pressure inside a drop of the same radius just outside the surface is $P_{2}$ . If $T$ is surface tension then:

P_{1}

2 P_{2}

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P_{1}

P_{2}

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P_{2}

2 P_{1}

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P_{2}

= 0,

P_{1}

\neq

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Solution

The correct option is B $P_{1}$ = $2 P_{2}$
Since, an air bubble has 2 air-liquid interface, therefore the excess pressure inside a bubble is a result of 2 surfaces rather than just 1 in case of drop.

Excess pressure:

Bubble $= \frac{4 T}{R}$

Drop $= \frac{2 T}{R}$

Therefore, $P_{1} = 2 P_{2}$

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The excess pressure inside an air bubble of radius r just below the surface of the water is P1. The excess pressure inside a drop of the same radius just outside the surface is P2. If T is surface tension then:

The correct option is B P1 = 2P2Since, an air bubble has 2 air-liquid interface, therefore the excess pressure inside a bubble is a result of 2 surfaces rather than just 1 in case of drop.Excess pressure:Bubble =4TRDrop =2TRTherefore, P1=2P2

The excess pressure inside an air bubble of radius $r$ just below the surface of the water is $P_{1}$ . The excess pressure inside a drop of the same radius just outside the surface is $P_{2}$ . If $T$ is surface tension then:

The correct option is B $P_{1}$ = $2 P_{2}$
Since, an air bubble has 2 air-liquid interface, therefore the excess pressure inside a bubble is a result of 2 surfaces rather than just 1 in case of drop.

Excess pressure:

Bubble $= \frac{4 T}{R}$

Drop $= \frac{2 T}{R}$

Therefore, $P_{1} = 2 P_{2}$