A condenser having a capacity of $6\,\mu F$ is charged to $100\, V$ and is then joined to an uncharged condenser of $14\,\mu F$ and then removed. The ratio of the charges on $6\,\mu F$ and $14\,\mu F$ and the potential of $6\,\mu F$ will be
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(c) Let ${q_1},\,{q_2}$ be the charges on two condensers
 $V = \frac{{{q_1}}}{6} = \frac{{{q_2}}}{{14}}$

$==>$ $\frac{{{q_1}}}{{{q_2}}} = \frac{6}{{14}} = \frac{3}{7}$
Also ${q_1} + {q_2} = 600$

$==>$ ${q_1} + \frac{{14}}{6}{q_1} = 600$

$==>$ ${q_1} = \frac{{600}}{{20}} \times 6$
 $V = \frac{{{q_1}}}{6} = \frac{{600}}{{20}} = 30\,volt$

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