The capacitance of a parallel plate capacitor in which a metal plate of thickness $$t$$ is inserted is given by
$$C = \frac{{{\varepsilon _0}A}}{{d - t}}.\,{\text{Here}}\,\,t \to 0\,\therefore C = \frac{{{\varepsilon _0}A}}{d}$$

To get a capacitance of 2 $$\mu F$$ arrangement of capacitors of capacitance $$1\mu F$$ as shown in figure 8 capacitors of $$1\mu F$$  in parallel with four such branches in series ie., 32 such capacitors are required.

$$\frac{1}{{{C_{eq}}}} = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\,\,\,\,\,\,\,\,\,\,\,\therefore \frac{1}{{{C_{eq}}}} = 2\mu F$$

The two condensers in the circuit are in parallel order, hence $$C' = C + \frac{C}{2} = \frac{{3C}}{2}$$
The work done in charging the equivalent capacitor is stored in the form of potential energy.
Hence, $$W = U = \frac{1}{2}C{V^2}$$
So, for the equivalent capacitor $${C'}$$
$$\eqalign{ & = \frac{1}{2}\left( {\frac{{3C}}{2}} \right){V^2}\,\,\left( {C' = \frac{{3C}}{2}} \right) \cr & = \frac{3}{4}C{V^2} \cr} $$

Capacitance of spherical conductor $$ = 4\pi {\varepsilon _0}a$$
where $$a$$ is radius of conductor.
Therefore, $$C = \frac{1}{{9 \times {{10}^9}}} \times 1 = \frac{1}{9} \times {10^{ - 9}}$$
$$\eqalign{ & = 0.11 \times {10^{ - 9}}F \cr & = 1.1 \times {10^{ - 10}}F \cr} $$

When a parallel plate air capacitor connected to a cell of emf $$V,$$ then charge stored will be $$q = CV \Rightarrow V = \frac{q}{C}$$
Also energy stored is $$U = \frac{1}{2}C{V^2} = \frac{{{q^2}}}{{2C}}$$
As the battery is disconnected from the capacitor the charge will not be destroyed, i.e. $$q' = q$$  with the introduction of dielectric in the gap of capacitor the new capacitance will be $$C' = CK \Rightarrow V' = \frac{q}{{C'}} = \frac{q}{{CK}}$$
The new energy stored will be
$$\eqalign{ & U' = \frac{{{q^2}}}{{2CK}} \Rightarrow \Delta U = U' - U \cr & = \frac{{{q^2}}}{{2C}}\left( {\frac{1}{K} - 1} \right) \cr & = \frac{1}{2}C{V^2}\left( {\frac{1}{K} - 1} \right) \cr} $$
Hence, option (D) is incorrect

Before introducing a slab capacitance of plates
$${C_1} = \frac{{{\varepsilon _0}A}}{3}$$
If a slab of dielectric constant $$K$$ is introduced between plates then

$$C = \frac{{K{\varepsilon _0}A}}{d}\,{\text{then}}\,{{C'}_1} = \frac{{{\varepsilon _0}A}}{{2.4}}$$
$${C_1}$$ and $${{C'}_1}$$ are in series hence,
$$\eqalign{ & \frac{{{\varepsilon _0}A}}{3} = \frac{{k\frac{{{\varepsilon _0}A}}{3}.\frac{{{\varepsilon _0}A}}{{2.4}}}}{{k\frac{{{\varepsilon _0}A}}{3} + \frac{{{\varepsilon _0}A}}{{2.4}}}} \cr & 3\,k = 2.4\,k + 3 \cr & 0.6\,k = 3 \cr} $$
Hence, the dielectric constant of slap is given by, $$k = \frac{{30}}{6} = 5$$

I
When the capacitors are joined in series,
$${U_{{\text{series}}}} = \frac{1}{2}\frac{{{C_1}}}{{{n_1}}}{\left( {4V} \right)^2}$$
II
When the capacitors are joined in parallel,
$${U_{{\text{parallel}}}} = \frac{1}{2}\left( {{n_2}{C_2}} \right){V^2}$$
Given, $${U_{{\text{series}}}} = {U_{{\text{parallel}}}}$$
So, $$\frac{1}{2}\frac{{{C_1}}}{{{n_1}}}{\left( {4V} \right)^2} = \frac{1}{2}\left( {{n_2}{C_2}} \right){V^2}$$
$$ \Rightarrow {C_2} = \frac{{16{C_1}}}{{{n_2}{n_1}}}$$

Volume of 8 small drops = Volume of big drop
$$8 \times \frac{4}{3}\pi {R^3} = \frac{4}{3}\pi {R^3} \Rightarrow R = 2r$$
As capacity is proportional to $$r,$$ hence capacity becomes 2 times.

$$C = \frac{{2 \times 2}}{{2 + 2}} + 2 = 3\,\mu F$$

As we know that the energy stored in the capacitor is given by,
$$\eqalign{ & U = \frac{1}{2}C{V^2}\left[ {_{V = \,\,{\text{voltage across the plate}}}^{C = \,\,{\text{capacitance of capacitor}}}} \right] \cr & U = \frac{1}{2}\left( {\frac{{A{\varepsilon _0}}}{d}} \right){\left( {Ed} \right)^2}\,\,\,\,\left( {\because C = \frac{{{\varepsilon _0}A}}{d}\,\,{\text{and}}\,\,V = Ed} \right) \cr & U = \frac{1}{2}\frac{{{\varepsilon _0}A}}{d}{\left( {Ed} \right)^2},\,\,U = \frac{1}{2}{\varepsilon _0}{E^2}Ad \cr} $$