⇒Electrostatics

Correct Option is (B)

Energy stored in a charged capacitor is

$\mathrm{U}=\frac{{\mathrm{Q}}^2}{2\mathrm{C}}\Rightarrow \mathrm{U}\propto {\mathrm{Q}}^2$

As per given condition, when, ${\mathrm{Q}}_2=({\mathrm{Q}}_1+3)$,

$\begin{array}{ll}\therefore & {\mathrm{U}}_2=44\mathrm{\%}\text{ of }{\mathrm{U}}_1=\frac{144}{100}{\mathrm{U}}_1\\ \therefore & \frac{{\mathrm{Q}}_2^2}{2\mathrm{C}}=\frac{144}{100}\times \frac{{\mathrm{Q}}_1^2}{2\mathrm{C}}\\ \therefore & {\mathrm{Q}}_1=\frac{10}{12}{\mathrm{Q}}_2\\ \therefore & {\mathrm{Q}}_1=\frac{5}{6}({\mathrm{Q}}_1+3)\\ \therefore & 6{\mathrm{Q}}_1=5{\mathrm{Q}}_1+15\\ \therefore & {\mathrm{Q}}_1=15\mathrm{C}\end{array}$

Correct Option is (B)

In the given circuit, $C_3$ and $C_4$ are in series and ${\mathrm{C}}_3={\mathrm{C}}_4=8\mu \mathrm{F}$

$\therefore \frac{1}{{\mathrm{C}}_{\mathrm{S}}}=\frac{1}{{\mathrm{C}}_3}+\frac{1}{{\mathrm{C}}_4}$

$\begin{array}{rl}\therefore {\mathrm{C}}_{\mathrm{S}}& =\frac{{\mathrm{C}}_3^2}{2{\mathrm{C}}_3}\\ {\mathrm{C}}_{\mathrm{S}}& =\frac{{\mathrm{C}}_3}{2}\\ \therefore {\mathrm{C}}_{\mathrm{S}}& =4\mu \mathrm{F}\text{.... (i)}\end{array}$

${\mathrm{C}}_5$ and ${\mathrm{C}}_6$ are in parallel and ${\mathrm{C}}_5={\mathrm{C}}_6=4\mu \mathrm{F}$

${\mathrm{C}}_{\mathrm{P}}={\mathrm{C}}_5+{\mathrm{C}}_6$

$\therefore {\mathrm{C}}_{\mathrm{P}}=8\mu \mathrm{F}$

$\therefore$ Equivalent circuit is as shown in figure.

Now, ${\mathrm{C}}_2$ and ${\mathrm{C}}_{\mathrm{P}}$ are in series and their combination in parallel with ${\mathrm{C}}_{\mathrm{S}}$

$\begin{array}{rl}\therefore {\mathrm{C}}_{\mathrm{E}}& =\frac{{\mathrm{C}}_2{\mathrm{C}}_{\mathrm{p}}}{{\mathrm{C}}_2{\mathrm{C}}_{\mathrm{p}}}+{\mathrm{C}}_{\mathrm{s}}\\ & {\mathrm{C}}_{\mathrm{E}}=\frac{(8)(8)}{16}+4\\ \therefore & {\mathrm{C}}_{\mathrm{E}}=8\mu \mathrm{F}\end{array}$

Now, ${\mathrm{C}}_1$ and ${\mathrm{C}}_{\mathrm{E}}$ are in series,

$\begin{array}{rl}& \therefore C=\frac{C_1{C}_E}{C_1+C_E}\\ & \therefore C=\frac{(8)(8)}{16}\\ & \therefore C=4\mu F\end{array}$

Correct Option is (D)

$\begin{array}{rl}& \frac{1}{C_s}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}\\ \\ & \therefore C_s=\frac{C}{3}\end{array}$

Now, ${\mathrm{C}}_{\mathrm{s}}$ and $\mathrm{C}$ are connected in parallel,

$\therefore {\mathrm{C}}_{\mathrm{net}}={\mathrm{C}}_{\mathrm{s}}+\mathrm{C}=\frac{\mathrm{C}}{3}+\mathrm{C}=\frac{4\mathrm{C}}{3}$

Correct Option is (B)

$\begin{array}{rl}V& =20-(-20)=40\text{ volt }\\ Q& =40C\\ C& =\frac{Q}{V}\\ \therefore C& =\frac{40}{40}\\ \therefore C& =1F\end{array}$

Correct Option is (C)

We know, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}$

As both the charged spheres are at the same potential, the charge on both spheres is

$\begin{array}{rl}& {\mathrm{Q}}_1={\mathrm{C}}_1\mathrm{V}\\ & {\mathrm{Q}}_2={\mathrm{C}}_2\mathrm{V}\end{array}$

The charge densities of both spheres are

${\sigma }_1=\frac{{\mathrm{Q}}_1}{{\mathrm{A}}_1}=\frac{{\mathrm{C}}_1\mathrm{V}}{4\pi {\mathrm{r}}_1^2}$

Similiarly,

${\sigma }_2=\frac{{\mathrm{Q}}_2}{{\mathrm{A}}_2}=\frac{{\mathrm{C}}_2\mathrm{V}}{4\pi {\mathrm{r}}_2^2}$

Taking the ratios,

$\begin{array}{rl}\frac{{\sigma }_2}{{\sigma }_1}& =\frac{C_2{r}_1^2}{C_1{r}_2^2}\\ \frac{{\sigma }_2}{{\sigma }_1}& =\frac{2\times {10}^{-6}\times (0.03{)}^2}{3\times {10}^{-6}\times (0.02{)}^2}=\frac{3}{2}\\ \therefore \frac{{\sigma }_1}{{\sigma }_2}& =\frac{2}{3}\end{array}$