In the given circuit, charge $Q_2$ on the $2\ μF$ capacitor changes as $C$ is varied from $1\ μF$ to $3\ μF$. $Q_2$ as a function of '$C$' is given properly by: (figures are drawn schematically and are not to scale)
JEE MAIN 2015, Diffcult
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From figure, $Q_{2}=\frac{2}{2+1} Q=\frac{2}{3} Q$
$Q = E\left( {\frac{{C \times 3}}{{C + 3}}} \right)$
$ {\therefore Q_{2}=\frac{2}{3}\left(\frac{3 C E}{C+3}\right)=\frac{2 C E}{C+3}}$
Therefore graph $d$ correctly dipicts.
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