A soap bubble assumes a spherical surface. Wh | vedclass

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

Vedclass · Accepted Answer

The pressure of air inside the bubble is less than the atmospheric pressure; that is why the atmospheric pressure has compressed it equally from all sides to give it a spherical shape.

Vedclass · Answer

(C) For a soap bubble, the excess pressure inside is given by $\Delta P = P_{in} - P_{out} = \frac{4T}{R}$, where $T$ is the surface tension and $R$ is the radius of the bubble. This implies that $P_{in} = P_{out} + \frac{4T}{R}$. Since $\frac{4T}{R} > 0$, the pressure inside the bubble is always greater than the atmospheric pressure. Therefore, statement $(C)$ is wrong because it incorrectly claims that the internal pressure is less than the atmospheric pressure.

Column-$I$	Column-$II$
$(a)$ Liquid drop in air	$(i)$ $\frac{4T}{R}$
$(b)$ Bubble of liquid in air	$(ii)$ $\frac{2T}{R}$
	$(iii)$ $\frac{2R}{T}$

$A$ soap bubble assumes a spherical surface. Which of the following statements is wrong?

Similar Questions

The radius of one soap bubble is four times that of another. What is the ratio of the excess pressure inside the two bubbles?

The excess pressure inside a spherical water drop $A$ is four times that of another water drop $B$. Then, the ratio of the mass of water drop $A$ to that of drop $B$ is

Formation of bubbles is in Column-$I$ and the pressure difference between them is given in Column-$II$. Match them appropriately.

Column-$I$ Column-$II$

$(a)$ Liquid drop in air $(i)$ $\frac{4T}{R}$

$(b)$ Bubble of liquid in air $(ii)$ $\frac{2T}{R}$

$(iii)$ $\frac{2R}{T}$

Two small drops of mercury,each of radius $R$,coalesce to form a single large drop. The ratio of the total surface energies before and after the change is:

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