Energy needed in breaking a drop of radius $R$ into $n$ drops of radii $r$ is given by

Question

When two soap bubbles of radii $ a $ and $ b $ ($ b>a $) coalesce, the radius of curvature of the common surface is:

Answer 1

To solve the problem, we need to calculate the energy required to break a large drop of radius $ R $ into $ n $ smaller drops, each of radius $ r $.

The energy change is related to the change in the surface area when the drop is broken down.

The surface area $ A $ of a single spherical drop is given by the formula:

$A = 4\pi R^2$

When the large drop is split into $ n $ smaller drops, each with radius $ r $, the total surface area becomes:

$A_{\text{new}} = n \times 4\pi r^2 = 4\pi nr^2$

The change in surface area, $\Delta A$, is:

$\Delta A = A_{\text{new}} - A = 4\pi nr^2 - 4\pi R^2 = 4\pi (nr^2 - R^2)$

The surface energy, or tension, associated with the surface area change is given by:

$\Delta U = T \times \Delta A$

where $ T $ is the surface tension of the liquid.

Substituting for $\Delta A$, the energy needed to break the drop is:

$\Delta U = 4\pi T (nr^2 - R^2)$

Therefore, the correct option is:

$4\pi T(nr^2 - R^2)$, which matches the provided correct answer.

Let's briefly rule out the other options:

$\frac{4}{3}\pi (r^3 n - R^2)$ does not involve surface areas correctly.
$4\pi T(R^2 - nr^2)$ has reversed terms and would imply negative energy, which isn't correct for breaking drops.
$4\pi T(nr^2 + R^2)$ suggests incorrect surface area calculations.

This confirms that the energy required is indeed given by $4\pi T(nr^2 - R^2)$.

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