8 mercury drops coalesce to form one mercury drop, the energy changes by a factor of

The correct answer is 4.

To understand why the energy changes by a factor of 4, we analyze the coalescence of the mercury drops from the perspective of electrostatic potential energy.

Step 1: Relationship between the radii of the drops
Let the radius of each small mercury drop be $r$ and the radius of the coalesced large drop be $R$.
Since the total volume remains conserved during coalescence, the volume of the large drop equals the sum of the volumes of the 8 small drops:

$\frac{4}{3}\pi R^3=8\times \frac{4}{3}\pi r^3$

Simplifying the equation gives:

$R^3=8r^3\Rightarrow R=2r$

Step 2: Conservation of Charge
Let each small drop carry a charge $q$. The total charge $Q$ on the final large drop is the sum of the charges of the 8 drops:

$Q=8q$

Step 3: Calculating Electrostatic Potential Energy
The self-electrostatic potential energy $U$ of a spherical conductor of radius $x$ and charge $Q$ is given by the formula:

$U=\frac{Q^2}{8\pi {\epsilon }_{0}x}$

The total initial electrostatic potential energy of the 8 separate drops is:

$U_{\text{initial}}=8\times \left(\frac{q^2}{8\pi {\epsilon }_{0}r}\right)=\frac{q^2}{\pi {\epsilon }_{0}r}$

The final electrostatic potential energy of the single large coalesced drop is:

$U_{\text{final}}=\frac{Q^2}{8\pi {\epsilon }_{0}R}$

Substituting $Q=8q$ and $R=2r$:

$U_{\text{final}}=\frac{{\left(8q\right)}^2}{8\pi {\epsilon }_0\left(2r\right)}=\frac{64q^2}{16\pi {\epsilon }_{0}r}=4\times \left(\frac{q^2}{\pi {\epsilon }_{0}r}\right)$

Step 4: Finding the Ratio of Energy Change
Comparing the final energy to the total initial energy:

$U_{\text{final}}=4U_{\text{initial}}$

Therefore, the total electrostatic energy of the system changes (increases) by a factor of 4.