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Step-by-Step Solution
Step 1: Understand the Problem
We have 27 identical small spherical drops of mercury, each charged such that they have the same potential V (10 V here, though the numerical voltage value does not affect the ratio we seek). These drops combine to form a single larger spherical drop. We want to find how many times bigger the electrostatic potential energy of the combined large drop is, compared to the electrostatic potential energy of one of the smaller drops.
Step 2: Relate Volumes to Find the Radius of the Big Drop
When 27 smaller drops of radius r merge, their total volume becomes the volume of the bigger drop. Hence,
27 \left(\frac{4}{3}\pi\,r^3 \right) = \frac{4}{3}\pi\,R^3 \\
\implies 27\,r^3 = R^3 \\
\implies R = 3r.
Step 3: Express the Potential Energy of One Small Drop
The self-energy (electrostatic potential energy) of a single charged spherical drop of radius r and charge q is given by:
U_1 = \frac{3}{5} \,\frac{k\,q^2}{r},
where k is the Coulomb constant.
Step 4: Total Charge on the Big Drop
Since there are 27 small drops, each carrying charge q, the total charge on the final big drop is:
Q = 27\,q.
Step 5: Express the Potential Energy of the Big Drop
The self-energy of the bigger drop of radius R and charge Q is:
U = \frac{3}{5}\,\frac{k\,Q^2}{R}.
Substitute Q = 27\,q and R = 3r :
U = \frac{3}{5}\,\frac{k\,(27\,q)^2}{3\,r}
= \frac{3}{5}\,\frac{k\,(27^2)\,q^2}{3\,r}
= \frac{27^2}{3} \left( \frac{3}{5}\,\frac{k\,q^2}{r} \right).
Step 6: Simplify the Expression
U = \frac{27^2}{3} \,U_1
= \frac{729}{3} \,U_1
= 243 \,U_1.
Step 7: Conclude the Ratio
The potential energy of the bigger drop is therefore 243 times that of a single smaller drop. Hence, the required ratio is 243.
