Answer

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**Hint**The change in potential energy is simply equal to the product of the surface tension and change in the total surface area of the mercury drops. The volume in the two states (large one drop and small big drop) are equal.

Formula used: In this solution we will be using the following formulae;

\[\Delta U = T4\pi {R^2}\left( {{N^{\dfrac{1}{3}}} - 1} \right)\], where \[\Delta U\] is the change in potential energy, \[R\] is the radius of the big drop of the liquid, \[N\] is the number of smaller drops, and \[T\] is the surface tension of the liquid.

\[V = \dfrac{4}{3}\pi {r^3}\], \[V\] is the volume of a sphere and \[r\] is the radius of the sphere. \[A = 4\pi {r^2}\] where \[A\] is the surface area of a sphere.

\[\Delta U = T\Delta A\] where \[\Delta A\] signifies the change in surface area of one big drop and that of the sum of many small drops.

**Complete Step-by-Step solution:**

Generally, for such a process, we the change in potential energy (which will be the energy expended) of the drops would be given by

\[\Delta U = T4\pi {R^2}\left( {{N^{\dfrac{1}{3}}} - 1} \right)\], where \[\Delta U\] is the change in potential energy, \[R\] is the radius of the big drop of the liquid, \[N\] is the number of smaller drops, and \[T\] is the surface tension of the liquid.

Hence, by inserting known values, we have

\[\Delta U = \left( {460 \times {{10}^{ - 3}}} \right)4\pi {\left( {0.01} \right)^2}\left( {{{\left( {{{10}^6}} \right)}^{\dfrac{1}{3}}} - 1} \right)\]

\[ \Rightarrow \Delta U = \left( {0.046} \right)4\pi {\left( {0.01} \right)^2}\left( {{{10}^2} - 1} \right)\]

Hence, by computation, we have

\[\Delta U = 0.057J\]

**Thus the correct option is A.**

**Note:**Generally, for such a process, the change potential energy (which will be the energy expended) of the drops would be given by

\[\Delta U = T\Delta A\]

Since, \[A = 4\pi {r^2}\] where \[A\] is the surface area of a sphere and \[r\] is the radius of the sphere.

Hence, we have

\[\Delta A = N4\pi {r^2} - 4\pi {R^2}\] where \[N\] is the number of drops and \[R\] is the radius of the big drop.

However, the volume is the same in both cases hence,

\[V = N\dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {R^3}\]

So, by cancellation, we have

\[N{r^3} = R\]

\[ \Rightarrow r = {N^{ - \dfrac{1}{3}}}R\]

So, we have that

\[\Delta A = N4\pi {\left( {{N^{ - \dfrac{1}{3}}}R} \right)^2} - 4\pi {R^2} = N4\pi {N^{ - \dfrac{2}{3}}}{R^2} - 4\pi {R^2}\]

\[ \Rightarrow \Delta A = 4\pi {R^2}\left( {{N^{1 - \dfrac{1}{3}}} - 1} \right) = 4\pi {R^2}\left( {{N^{\dfrac{1}{3}}} - 1} \right)\]

Then the change in potential is

\[\Delta U = T\left[ {4\pi {R^2}\left( {{N^{\dfrac{1}{3}}} - 1} \right)} \right]\]

\[ \Rightarrow \Delta U = T4\pi {R^2}\left( {{N^{\dfrac{1}{3}}} - 1} \right)\] which is the formula.

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