(D) The energy of a drop is given by $U = T \times A$, where $T$ is surface tension and $A$ is surface area.Energy of $n$ small drops: $U_i = n \times

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(D) The energy of a drop is given by $U = T \times A$, where $T$ is surface tension and $A$ is surface area.Energy of $n$ small drops: $U_i = n \times (4\pi r^2 T)$.Energy of the big drop: $E = 4\pi R^2 T$.Energy loss is given as $\Delta U = U_i - E = 3E$.Substituting the values: $n(4\pi r^2 T) - 4\pi R^2 T = 3(4\pi R^2 T)$.$n(4\pi r^2 T) = 4\pi R^2 T + 12\pi R^2 T$.$n(4\pi r^2 T) = 16\pi R^2 T$.$n = \frac{16\pi R^2 T}{4\pi r^2 T} = 4\frac{R^2}{r^2}$.

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

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When a big drop of water is formed from $n$ small drops of water, the energy loss is $3E$, where $E$ is the energy of the bigger drop. If $R$ is the radius of the bigger drop and $r$ is the radius of the smaller drop, then the number of smaller drops $(n)$ is:

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A
$\frac{4R}{r^2}$
B
$\frac{4R}{r}$
C
$\frac{2R^2}{r}$
D
$\frac{4R^2}{r^2}$

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Excess Pressure and coalesce of Bubble and drop

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When a big drop of water is formed from $n$ small drops of water, the energy loss is $3E$, where $E$ is the energy of the bigger drop. If $R$ is the radius of the bigger drop and $r$ is the radius of the smaller drop, then the number of smaller drops $(n)$ is:

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

$64$ small drops of mercury,each of radius $r$ and charge $q$,coalesce to form a single large drop. What is the ratio of the surface charge density of a small drop to that of the large drop?

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