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Capacitance

166033 Four capacitors with capacitance $C_{1}=1 \mu \mathrm{F}$, $\mathrm{C}_{2}=1.5 \mu \mathrm{F}, \quad \mathrm{C}_{3}=2.5 \mu \mathrm{F}$ and $\mathrm{C}_{4}=0.5 \mu \mathrm{F}$ are connected as shown in figure to a $30 \mathrm{~V}$ sources. The potential difference between points a and $b$ is

1 $5 \mathrm{~V}$

2 $-9 \mathrm{~V}$

3 $10 \mathrm{~V}$

4 $-13 \mathrm{~V}$

Explanation:

: According to question,

Given, $\mathrm{C}_{1}=1 \mu \mathrm{F}, \mathrm{C}_{2}=1.5 \mu \mathrm{F}, \mathrm{C}_{3}=2.5 \mu \mathrm{F}, \mathrm{C}_{4}=0.5 \mu \mathrm{F}$ Equivalent capacitance of $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$, are in series,
$\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$
$\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{1}+\frac{1}{1.5}=\frac{2.5}{1.5}=\frac{5}{3}$
$\mathrm{C}^{\prime} =\frac{3}{5} \mu \mathrm{F}$
Equivalent capacitance of $\mathrm{C}_{3}$ and $\mathrm{C}_{4}$, are in series,
$\frac{1}{C^{\prime \prime}} =\frac{1}{2.5}+\frac{1}{0.5}=\frac{6}{2.5}=\frac{12}{5}$
$\mathrm{C}^{\prime \prime} =\frac{5}{12} \mu \mathrm{F}$
Total capacitance $\left(\mathrm{C}_{\mathrm{eq}}\right)=\mathrm{C}^{\prime}+\mathrm{C}^{\prime \prime}$
$=\frac{3}{5}+\frac{5}{12}=\frac{61}{60} \mu \mathrm{F}$
Total charge $\left(\mathrm{Q}_{\text {total }}\right)=\mathrm{C}_{\mathrm{eq}} \times \mathrm{V}$
$=\frac{61}{60} \times 30=\frac{61}{2} \mu \mathrm{C}$
This charge is distributed in upper and lower of the circuit,
$\frac{\mathrm{Q}^{\prime}}{\mathrm{C}^{\prime}}=\frac{\mathrm{Q}^{\prime \prime}}{\mathrm{C}^{\prime \prime}}=\mathrm{V} \text { ( parallel) }$
$\frac{\mathrm{Q}^{\prime}}{\mathrm{Q}^{\prime \prime}}=\frac{\mathrm{C}^{\prime}}{\mathrm{C}^{\prime \prime}}=\frac{3 / 5}{5 / 12}=\frac{36}{25}$
$\mathrm{Q}^{\prime}=\frac{36 \mathrm{Q}^{\prime \prime}}{25}$
From equation (i), we get-
$\because \quad Q^{\prime}+Q^{\prime \prime}=\frac{61}{2}$
$\frac{36 Q^{\prime \prime}}{25}+Q^{\prime \prime}=\frac{61}{2}$
$\frac{61 Q^{\prime \prime}}{25}=\frac{61}{2} \Rightarrow Q^{\prime \prime}=\frac{25}{2} \mu F$
$\therefore \quad Q^{\prime}=\frac{61}{2}-\frac{25}{2}=\frac{36}{2}=18 \mu \mathrm{C}$
If potential difference $\left(V^{\prime}\right)$ across capacitor $C_{1}$
$\mathrm{V}^{\prime}=\frac{\mathrm{Q}^{\prime}}{\mathrm{C}_{1}}=\frac{18}{1}=18 \mathrm{~V}$
$\mathrm{~V}^{\prime \prime}=\frac{\mathrm{Q}^{\prime \prime}}{\mathrm{C}_{3}}=\frac{25 / 2}{5 / 2}=5 \mathrm{~V}$
So,
$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{a}}=18 \mathrm{~V}$
$\mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{b}}=5 \mathrm{~V}$
$\left(\mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{b}}\right)-\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{a}}\right)=5-18$
$\mathrm{~V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}=-13 \mathrm{~V}$

[CG PET -2016

Capacitance

166034 In given circuit when switch $S$ has been closed then charge on capacitor $A$ and $B$ respectively are

1 $3 \mathrm{q}, 6 \mathrm{q}$

2 $6 \mathrm{q}, 3 \mathrm{q}$

3 $4.5 \mathrm{q}, 4.5 \mathrm{q}$

4 $5 \mathrm{q}, 4 \mathrm{q}$

[CG PET- 2008]

Capacitance

166035 Two capacitors $C_{1}$ and $C_{2}=2 C_{1}$ are connected in a circuit with a switch between them as shown in the figure. Initially the switch is open and $C_{1}$ holds charge $Q$. The switch is closed. At steady state, the charge on each capacitor will be

1 Q, 2Q

2 $\frac{\mathrm{Q}}{3}, \frac{2 \mathrm{Q}}{3}$

3 $\frac{3 \mathrm{Q}}{2}, 3 \mathrm{Q}$

4 $\frac{2 \mathrm{Q}}{3}, \frac{4 \mathrm{Q}}{3}$

[BITSAT-2005]

Capacitance

166036 $n$ identical drops, each of capacitance $C$ and charged to a potential $\mathrm{V}$, coalesce to form a bigger drop. Then the ratio of the energy stored in the big drop to that in each small drop is

1 $\mathrm{n}^{5 / 3}: 1$

2 $\mathrm{n}^{4 / 3}: 1$

3 $\mathrm{n}: 1$

4 $n^{3}: 1$

5 $\mathrm{n}^{2 / 3}: 1$

[Kerala CEE - 2010]

Capacitance

166033 Four capacitors with capacitance $C_{1}=1 \mu \mathrm{F}$, $\mathrm{C}_{2}=1.5 \mu \mathrm{F}, \quad \mathrm{C}_{3}=2.5 \mu \mathrm{F}$ and $\mathrm{C}_{4}=0.5 \mu \mathrm{F}$ are connected as shown in figure to a $30 \mathrm{~V}$ sources. The potential difference between points a and $b$ is

1 $5 \mathrm{~V}$

2 $-9 \mathrm{~V}$

3 $10 \mathrm{~V}$

4 $-13 \mathrm{~V}$

Explanation:

: According to question,

Given, $\mathrm{C}_{1}=1 \mu \mathrm{F}, \mathrm{C}_{2}=1.5 \mu \mathrm{F}, \mathrm{C}_{3}=2.5 \mu \mathrm{F}, \mathrm{C}_{4}=0.5 \mu \mathrm{F}$ Equivalent capacitance of $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$, are in series,
$\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$
$\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{1}+\frac{1}{1.5}=\frac{2.5}{1.5}=\frac{5}{3}$
$\mathrm{C}^{\prime} =\frac{3}{5} \mu \mathrm{F}$
Equivalent capacitance of $\mathrm{C}_{3}$ and $\mathrm{C}_{4}$, are in series,
$\frac{1}{C^{\prime \prime}} =\frac{1}{2.5}+\frac{1}{0.5}=\frac{6}{2.5}=\frac{12}{5}$
$\mathrm{C}^{\prime \prime} =\frac{5}{12} \mu \mathrm{F}$
Total capacitance $\left(\mathrm{C}_{\mathrm{eq}}\right)=\mathrm{C}^{\prime}+\mathrm{C}^{\prime \prime}$
$=\frac{3}{5}+\frac{5}{12}=\frac{61}{60} \mu \mathrm{F}$
Total charge $\left(\mathrm{Q}_{\text {total }}\right)=\mathrm{C}_{\mathrm{eq}} \times \mathrm{V}$
$=\frac{61}{60} \times 30=\frac{61}{2} \mu \mathrm{C}$
This charge is distributed in upper and lower of the circuit,
$\frac{\mathrm{Q}^{\prime}}{\mathrm{C}^{\prime}}=\frac{\mathrm{Q}^{\prime \prime}}{\mathrm{C}^{\prime \prime}}=\mathrm{V} \text { ( parallel) }$
$\frac{\mathrm{Q}^{\prime}}{\mathrm{Q}^{\prime \prime}}=\frac{\mathrm{C}^{\prime}}{\mathrm{C}^{\prime \prime}}=\frac{3 / 5}{5 / 12}=\frac{36}{25}$
$\mathrm{Q}^{\prime}=\frac{36 \mathrm{Q}^{\prime \prime}}{25}$
From equation (i), we get-
$\because \quad Q^{\prime}+Q^{\prime \prime}=\frac{61}{2}$
$\frac{36 Q^{\prime \prime}}{25}+Q^{\prime \prime}=\frac{61}{2}$
$\frac{61 Q^{\prime \prime}}{25}=\frac{61}{2} \Rightarrow Q^{\prime \prime}=\frac{25}{2} \mu F$
$\therefore \quad Q^{\prime}=\frac{61}{2}-\frac{25}{2}=\frac{36}{2}=18 \mu \mathrm{C}$
If potential difference $\left(V^{\prime}\right)$ across capacitor $C_{1}$
$\mathrm{V}^{\prime}=\frac{\mathrm{Q}^{\prime}}{\mathrm{C}_{1}}=\frac{18}{1}=18 \mathrm{~V}$
$\mathrm{~V}^{\prime \prime}=\frac{\mathrm{Q}^{\prime \prime}}{\mathrm{C}_{3}}=\frac{25 / 2}{5 / 2}=5 \mathrm{~V}$
So,
$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{a}}=18 \mathrm{~V}$
$\mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{b}}=5 \mathrm{~V}$
$\left(\mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{b}}\right)-\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{a}}\right)=5-18$
$\mathrm{~V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}=-13 \mathrm{~V}$

[CG PET -2016

Capacitance

166034 In given circuit when switch $S$ has been closed then charge on capacitor $A$ and $B$ respectively are

1 $3 \mathrm{q}, 6 \mathrm{q}$

2 $6 \mathrm{q}, 3 \mathrm{q}$

3 $4.5 \mathrm{q}, 4.5 \mathrm{q}$

4 $5 \mathrm{q}, 4 \mathrm{q}$

[CG PET- 2008]

Capacitance

166035 Two capacitors $C_{1}$ and $C_{2}=2 C_{1}$ are connected in a circuit with a switch between them as shown in the figure. Initially the switch is open and $C_{1}$ holds charge $Q$. The switch is closed. At steady state, the charge on each capacitor will be

1 Q, 2Q

2 $\frac{\mathrm{Q}}{3}, \frac{2 \mathrm{Q}}{3}$

3 $\frac{3 \mathrm{Q}}{2}, 3 \mathrm{Q}$

4 $\frac{2 \mathrm{Q}}{3}, \frac{4 \mathrm{Q}}{3}$

[BITSAT-2005]

Capacitance

166036 $n$ identical drops, each of capacitance $C$ and charged to a potential $\mathrm{V}$, coalesce to form a bigger drop. Then the ratio of the energy stored in the big drop to that in each small drop is

1 $\mathrm{n}^{5 / 3}: 1$

2 $\mathrm{n}^{4 / 3}: 1$

3 $\mathrm{n}: 1$

4 $n^{3}: 1$

5 $\mathrm{n}^{2 / 3}: 1$

[Kerala CEE - 2010]

Capacitance

166033 Four capacitors with capacitance $C_{1}=1 \mu \mathrm{F}$, $\mathrm{C}_{2}=1.5 \mu \mathrm{F}, \quad \mathrm{C}_{3}=2.5 \mu \mathrm{F}$ and $\mathrm{C}_{4}=0.5 \mu \mathrm{F}$ are connected as shown in figure to a $30 \mathrm{~V}$ sources. The potential difference between points a and $b$ is

1 $5 \mathrm{~V}$

2 $-9 \mathrm{~V}$

3 $10 \mathrm{~V}$

4 $-13 \mathrm{~V}$

Explanation:

: According to question,

Given, $\mathrm{C}_{1}=1 \mu \mathrm{F}, \mathrm{C}_{2}=1.5 \mu \mathrm{F}, \mathrm{C}_{3}=2.5 \mu \mathrm{F}, \mathrm{C}_{4}=0.5 \mu \mathrm{F}$ Equivalent capacitance of $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$, are in series,
$\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$
$\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{1}+\frac{1}{1.5}=\frac{2.5}{1.5}=\frac{5}{3}$
$\mathrm{C}^{\prime} =\frac{3}{5} \mu \mathrm{F}$
Equivalent capacitance of $\mathrm{C}_{3}$ and $\mathrm{C}_{4}$, are in series,
$\frac{1}{C^{\prime \prime}} =\frac{1}{2.5}+\frac{1}{0.5}=\frac{6}{2.5}=\frac{12}{5}$
$\mathrm{C}^{\prime \prime} =\frac{5}{12} \mu \mathrm{F}$
Total capacitance $\left(\mathrm{C}_{\mathrm{eq}}\right)=\mathrm{C}^{\prime}+\mathrm{C}^{\prime \prime}$
$=\frac{3}{5}+\frac{5}{12}=\frac{61}{60} \mu \mathrm{F}$
Total charge $\left(\mathrm{Q}_{\text {total }}\right)=\mathrm{C}_{\mathrm{eq}} \times \mathrm{V}$
$=\frac{61}{60} \times 30=\frac{61}{2} \mu \mathrm{C}$
This charge is distributed in upper and lower of the circuit,
$\frac{\mathrm{Q}^{\prime}}{\mathrm{C}^{\prime}}=\frac{\mathrm{Q}^{\prime \prime}}{\mathrm{C}^{\prime \prime}}=\mathrm{V} \text { ( parallel) }$
$\frac{\mathrm{Q}^{\prime}}{\mathrm{Q}^{\prime \prime}}=\frac{\mathrm{C}^{\prime}}{\mathrm{C}^{\prime \prime}}=\frac{3 / 5}{5 / 12}=\frac{36}{25}$
$\mathrm{Q}^{\prime}=\frac{36 \mathrm{Q}^{\prime \prime}}{25}$
From equation (i), we get-
$\because \quad Q^{\prime}+Q^{\prime \prime}=\frac{61}{2}$
$\frac{36 Q^{\prime \prime}}{25}+Q^{\prime \prime}=\frac{61}{2}$
$\frac{61 Q^{\prime \prime}}{25}=\frac{61}{2} \Rightarrow Q^{\prime \prime}=\frac{25}{2} \mu F$
$\therefore \quad Q^{\prime}=\frac{61}{2}-\frac{25}{2}=\frac{36}{2}=18 \mu \mathrm{C}$
If potential difference $\left(V^{\prime}\right)$ across capacitor $C_{1}$
$\mathrm{V}^{\prime}=\frac{\mathrm{Q}^{\prime}}{\mathrm{C}_{1}}=\frac{18}{1}=18 \mathrm{~V}$
$\mathrm{~V}^{\prime \prime}=\frac{\mathrm{Q}^{\prime \prime}}{\mathrm{C}_{3}}=\frac{25 / 2}{5 / 2}=5 \mathrm{~V}$
So,
$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{a}}=18 \mathrm{~V}$
$\mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{b}}=5 \mathrm{~V}$
$\left(\mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{b}}\right)-\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{a}}\right)=5-18$
$\mathrm{~V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}=-13 \mathrm{~V}$

[CG PET -2016

Capacitance

166034 In given circuit when switch $S$ has been closed then charge on capacitor $A$ and $B$ respectively are

1 $3 \mathrm{q}, 6 \mathrm{q}$

2 $6 \mathrm{q}, 3 \mathrm{q}$

3 $4.5 \mathrm{q}, 4.5 \mathrm{q}$

4 $5 \mathrm{q}, 4 \mathrm{q}$

[CG PET- 2008]

Capacitance

166035 Two capacitors $C_{1}$ and $C_{2}=2 C_{1}$ are connected in a circuit with a switch between them as shown in the figure. Initially the switch is open and $C_{1}$ holds charge $Q$. The switch is closed. At steady state, the charge on each capacitor will be

1 Q, 2Q

2 $\frac{\mathrm{Q}}{3}, \frac{2 \mathrm{Q}}{3}$

3 $\frac{3 \mathrm{Q}}{2}, 3 \mathrm{Q}$

4 $\frac{2 \mathrm{Q}}{3}, \frac{4 \mathrm{Q}}{3}$

[BITSAT-2005]

Capacitance

166036 $n$ identical drops, each of capacitance $C$ and charged to a potential $\mathrm{V}$, coalesce to form a bigger drop. Then the ratio of the energy stored in the big drop to that in each small drop is

1 $\mathrm{n}^{5 / 3}: 1$

2 $\mathrm{n}^{4 / 3}: 1$

3 $\mathrm{n}: 1$

4 $n^{3}: 1$

5 $\mathrm{n}^{2 / 3}: 1$

[Kerala CEE - 2010]

Capacitance

166033 Four capacitors with capacitance $C_{1}=1 \mu \mathrm{F}$, $\mathrm{C}_{2}=1.5 \mu \mathrm{F}, \quad \mathrm{C}_{3}=2.5 \mu \mathrm{F}$ and $\mathrm{C}_{4}=0.5 \mu \mathrm{F}$ are connected as shown in figure to a $30 \mathrm{~V}$ sources. The potential difference between points a and $b$ is

1 $5 \mathrm{~V}$

2 $-9 \mathrm{~V}$

3 $10 \mathrm{~V}$

4 $-13 \mathrm{~V}$

Explanation:

: According to question,

Given, $\mathrm{C}_{1}=1 \mu \mathrm{F}, \mathrm{C}_{2}=1.5 \mu \mathrm{F}, \mathrm{C}_{3}=2.5 \mu \mathrm{F}, \mathrm{C}_{4}=0.5 \mu \mathrm{F}$ Equivalent capacitance of $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$, are in series,
$\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$
$\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{1}+\frac{1}{1.5}=\frac{2.5}{1.5}=\frac{5}{3}$
$\mathrm{C}^{\prime} =\frac{3}{5} \mu \mathrm{F}$
Equivalent capacitance of $\mathrm{C}_{3}$ and $\mathrm{C}_{4}$, are in series,
$\frac{1}{C^{\prime \prime}} =\frac{1}{2.5}+\frac{1}{0.5}=\frac{6}{2.5}=\frac{12}{5}$
$\mathrm{C}^{\prime \prime} =\frac{5}{12} \mu \mathrm{F}$
Total capacitance $\left(\mathrm{C}_{\mathrm{eq}}\right)=\mathrm{C}^{\prime}+\mathrm{C}^{\prime \prime}$
$=\frac{3}{5}+\frac{5}{12}=\frac{61}{60} \mu \mathrm{F}$
Total charge $\left(\mathrm{Q}_{\text {total }}\right)=\mathrm{C}_{\mathrm{eq}} \times \mathrm{V}$
$=\frac{61}{60} \times 30=\frac{61}{2} \mu \mathrm{C}$
This charge is distributed in upper and lower of the circuit,
$\frac{\mathrm{Q}^{\prime}}{\mathrm{C}^{\prime}}=\frac{\mathrm{Q}^{\prime \prime}}{\mathrm{C}^{\prime \prime}}=\mathrm{V} \text { ( parallel) }$
$\frac{\mathrm{Q}^{\prime}}{\mathrm{Q}^{\prime \prime}}=\frac{\mathrm{C}^{\prime}}{\mathrm{C}^{\prime \prime}}=\frac{3 / 5}{5 / 12}=\frac{36}{25}$
$\mathrm{Q}^{\prime}=\frac{36 \mathrm{Q}^{\prime \prime}}{25}$
From equation (i), we get-
$\because \quad Q^{\prime}+Q^{\prime \prime}=\frac{61}{2}$
$\frac{36 Q^{\prime \prime}}{25}+Q^{\prime \prime}=\frac{61}{2}$
$\frac{61 Q^{\prime \prime}}{25}=\frac{61}{2} \Rightarrow Q^{\prime \prime}=\frac{25}{2} \mu F$
$\therefore \quad Q^{\prime}=\frac{61}{2}-\frac{25}{2}=\frac{36}{2}=18 \mu \mathrm{C}$
If potential difference $\left(V^{\prime}\right)$ across capacitor $C_{1}$
$\mathrm{V}^{\prime}=\frac{\mathrm{Q}^{\prime}}{\mathrm{C}_{1}}=\frac{18}{1}=18 \mathrm{~V}$
$\mathrm{~V}^{\prime \prime}=\frac{\mathrm{Q}^{\prime \prime}}{\mathrm{C}_{3}}=\frac{25 / 2}{5 / 2}=5 \mathrm{~V}$
So,
$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{a}}=18 \mathrm{~V}$
$\mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{b}}=5 \mathrm{~V}$
$\left(\mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{b}}\right)-\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{a}}\right)=5-18$
$\mathrm{~V}_{\mathrm{a}}-\mathrm{V}_{\mathrm{b}}=-13 \mathrm{~V}$

[CG PET -2016

Capacitance

166034 In given circuit when switch $S$ has been closed then charge on capacitor $A$ and $B$ respectively are

1 $3 \mathrm{q}, 6 \mathrm{q}$

2 $6 \mathrm{q}, 3 \mathrm{q}$

3 $4.5 \mathrm{q}, 4.5 \mathrm{q}$

4 $5 \mathrm{q}, 4 \mathrm{q}$

[CG PET- 2008]

Capacitance

166035 Two capacitors $C_{1}$ and $C_{2}=2 C_{1}$ are connected in a circuit with a switch between them as shown in the figure. Initially the switch is open and $C_{1}$ holds charge $Q$. The switch is closed. At steady state, the charge on each capacitor will be

1 Q, 2Q

2 $\frac{\mathrm{Q}}{3}, \frac{2 \mathrm{Q}}{3}$

3 $\frac{3 \mathrm{Q}}{2}, 3 \mathrm{Q}$

4 $\frac{2 \mathrm{Q}}{3}, \frac{4 \mathrm{Q}}{3}$

[BITSAT-2005]

Capacitance

166036 $n$ identical drops, each of capacitance $C$ and charged to a potential $\mathrm{V}$, coalesce to form a bigger drop. Then the ratio of the energy stored in the big drop to that in each small drop is

1 $\mathrm{n}^{5 / 3}: 1$

2 $\mathrm{n}^{4 / 3}: 1$

3 $\mathrm{n}: 1$

4 $n^{3}: 1$

5 $\mathrm{n}^{2 / 3}: 1$

[Kerala CEE - 2010]

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