Due to surface tension, the excess pressure i | vedclass

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(C) The excess pressure inside a liquid drop of radius $r$ is given by $\Delta P = \frac{2T}{r} = 9$ units. (Note: For a liquid drop, the formula is $

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$3$ units

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(C) The excess pressure inside a liquid drop of radius $r$ is given by $\Delta P = \frac{2T}{r} = 9$ units. (Note: For a liquid drop, the formula is $\frac{2T}{r}$, though the problem uses $9$ as a constant value for the expression).Let the radius of the smaller drop be $r$ and the radius of the bigger drop be $R$.When $27$ smaller drops combine to form a bigger drop, the volume remains conserved:$27 	imes (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$$27 r^3 = R^3$Taking the cube root on both sides, we get $R = 3r$.The excess pressure inside the bigger drop is $\Delta P' = \frac{2T}{R} = \frac{2T}{3r}$.Since $\frac{2T}{r} = 9$, we substitute this value:$\Delta P' = \frac{9}{3} = 3$ units.

Due to surface tension, the excess pressure inside a smaller drop is $9$ units. If $27$ smaller drops combine, then the excess pressure inside the bigger drop is:

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