Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

When a spherical drop carries a charge Q and has radius R, its electric potential (relative to infinity) is

$V\; =\; \backslash frac\{Q\}\{4\backslash pi\backslash varepsilon\_0\; R\}$

All 27 drops are identical, each having the same charge and radius. Let the charge of each small drop be q and its radius be r. Then the potential of a single drop is

$V\_0\; =\; \backslash frac\{q\}\{4\backslash pi\backslash varepsilon\_0\; r\}\; =\; 220\backslash \; \backslash text\{V\}$

When the 27 drops coalesce, the total charge adds linearly:

$Q\; =\; 27\backslash ,q$

The volume of a sphere is proportional to the cube of its radius. Because the volume of the big drop equals the sum of the volumes of the 27 small drops, we have

$\backslash frac\{4\}\{3\}\backslash pi\; R^\{3\}\; =\; 27\backslash left(\backslash frac\{4\}\{3\}\backslash pi\; r^\{3\}\backslash right)$

which simplifies to

$R^\{3\}\; =\; 27\backslash ,r^\{3\}$

Taking the cube root gives the radius of the big drop

$R\; =\; 27^\{1/3\}\backslash ,r\; =\; 3\backslash ,r$

Now substitute Q and R into the potential formula for the big drop:

$V\_\{\backslash text\{big\}\}\; =\; \backslash frac\{27\backslash ,q\}\{4\backslash pi\backslash varepsilon\_0\; (3\backslash ,r)\}\; =\; \backslash frac\{27\}\{3\}\backslash ,\backslash frac\{q\}\{4\backslash pi\backslash varepsilon\_0\; r\}$

Since the fraction \(\frac{q}{4\pi\varepsilon_0 r}\) is just the potential of one small drop (\(V_0 = 220\ \text{V}\)), we obtain

$V\_\{\backslash text\{big\}\}\; =\; 9\backslash ,V\_0\; =\; 9\; \backslash times\; 220\backslash \; \backslash text\{V\}\; =\; 1980\backslash \; \backslash text\{V\}$

Thus, the potential of the larger combined drop is 1980 V, which matches the given correct option.