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n identical drops of a liquid each having charge q and radius r coalesce. Match the unfilled places in COLUMN-I with their respective fills given in COLUMN-II (E=electric field, V=electric potential, U=electric potential energy, $σ$ =surface charge density)
COLUMN-I COLUMN-II
A) $E_{b i g} = .... E_{s m a l l}$ p) $n^{1 / 3}$
B) $V_{b i g} = .... V_{s m a l l}$ q)    $n^{\frac{5}{3}}$
C) $U_{b i g} = .... U_{s m a l l}$ r)    $n^{2 / 3}$
D)    $σ_{b i g} = ..... σ_{s m a l l}$ s) $n^{- 1 / 3}$

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a→p, b→r, c→q, d→p

a→p, b→r, c→q, d→s

a→s, b→p, c→r, d→q

a→s, b→s, c→r, d→q

answer is A.

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Detailed Solution

n drops coalesce, then radius ( R ) of bigger drop having charge nq is

$n (\frac{4}{3} π r^{3}) = \frac{4}{3} π R^{3}$

$\Rightarrow R^{3} = n r^{3} \Rightarrow R = n^{\frac{1}{3}} r$

Now, $E_{b i g} = \frac{n q}{4 π ε_{0} R^{2}} \Rightarrow E_{b i g} = \frac{n q}{4 π ε_{0} n^{\frac{2}{3}} r^{2}} = n^{\frac{1}{3}} E_{s m a l l}$

SO, $A \to (p)$

$S i m i l a r l y, V_{b i g} = \frac{n q}{4 π ε_{0} R} = n^{\frac{2}{3}} V_{s m a l l}$

$S o, B \to (r) S i n c e, C_{b i g} = 4 π ε_{0} R = n^{\frac{1}{3}} C_{s m a l l} S o, C \to (p) S i n c e, σ_{b i g} = \frac{n q}{4 π R^{2}} = n^{\frac{1}{3}} σ_{s m a l l} S o, D \to (p)$

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