Explanation

(c) Let capacitance of each capacitor is $C.$
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Then equivalent capacitance in series is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{C''} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C^{''} = \frac{C}{2}$
Charge $Q = C^{''}V = \frac{C}{2}.15$ …(i)
When filled with dielectric
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C_{1} = 4C,\ \ C_{2} = C$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{C''} = \frac{1}{4C} + \frac{1}{C} = \frac{5}{4C}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C^{''} = \frac{4C}{5}$
Since, charge is conserved
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Q = C^{''}V^{''} = \frac{4C}{5}V''$ …(ii)
From Eqs. (i) and (ii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{C}{2} \times 15 = \frac{4C}{5} \times V''$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V^{''} = \frac{15 \times 5}{4 \times 2} = 9.4V$
$\approx 10\ V$