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Multiple Choice Questions

One and More than One Correct

1. Given, [JEE 2006]

$R_1 = 1\,\Omega$, $C_1 = 2\,\mu\text{F}$

$R_2 = 2\,\Omega$, $C_2 = 4\,\mu\text{F}$

The time constants (in $\mu\text{s}$) for the circuits I, II, III are respectively

a. $18, 8/9, 4$

b. $18, 4, 8/9$

c. $4, 8/9, 18$

d. $8/9, 18, 4$

Answer: (d)

Solution:

$$\tau_1 = R_{\text{eq}} C_{\text{eq}} = \left(\frac{R_1 R_2}{R_1 + R_2}\right) \left(\frac{C_1 C_2}{C_1 + C_2}\right) = \left(\frac{1 \times 2}{1 + 2}\right) \left(\frac{2 \times 4}{2 + 4}\right) = \frac{2}{3} \times \frac{4}{3} = \frac{8}{9}\,\mu\text{s}$$ $$\tau_2 = (R_1 + R_2)(C_1 + C_2) = (1 + 2)(2 + 4) = 3 \times 6 = 18\,\mu\text{s}$$ $$\tau_3 = (R_1 + R_2) \left(\frac{C_1 C_2}{C_1 + C_2}\right) = (1 + 2) \left(\frac{2 \times 4}{2 + 4}\right) = 3 \times \frac{4}{3} = 4\,\mu\text{s}$$

2. A circuit is connected as shown in the figure with the switch S open. When the switch is closed the total amount of charge that flows from Y to X is; [JEE 2007 P-I]

a. $0$

b. $54\,\mu\text{C}$

c. $27\,\mu\text{C}$

d. $81\,\mu\text{C}$

Answer: (c)

Solution:

Initial charge distribution (when switch S is open):

Charges on capacitors are $+9\,\mu\text{C}$ and $+18\,\mu\text{C}$.

Final charge distribution (when switch S is closed):

Charges on capacitors are $+18\,\mu\text{C}$ and $+18\,\mu\text{C}$.

Total amount of charge that flows from Y to X is $27\,\mu\text{C}$.

3. A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric constant $K = 2$. The level of liquid is $\frac{d}{3}$ initially. Suppose the liquid level decreases at a constant speed V, the time constant as a function of time t is: [JEE 2008 P-II]

a. $\frac{6 \varepsilon_0 R}{5d + 3Vt}$

b. $\frac{(15d + 9Vt) \varepsilon_0 R}{22d^2 - 2dVt - 9V^2t^2}$

c. $\frac{6 \varepsilon_0 R}{5d - 3Vt}$

d. $\frac{(15d - 9Vt) \varepsilon_0 R}{22d^2 + 2dVt - 9V^2t^2}$

Answer: (a)

Solution:

$$C_{\text{equivalent}} = \frac{\varepsilon_0 \cdot 2}{\frac{d}{3} - Vt} + \frac{\varepsilon_0}{d - \left(\frac{d}{3} - Vt\right)} = \frac{2\varepsilon_0}{\frac{d}{3} - Vt} + \frac{\varepsilon_0}{\frac{2d}{3} + Vt}$$ $$\tau = C_{\text{equivalent}} R = \frac{6 \varepsilon_0 R}{5d + 3Vt}$$

4. A $2\,\mu\text{F}$ capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is: [JEE 2011 P-I]

a. $0\%$

b. $20\%$

c. $75\%$

d. $80\%$

Answer: (d)

Solution:

$$U_i = \frac{1}{2} (2) V^2$$ $$\Delta U_{\text{loss}} = \frac{1}{2} \frac{(2)(8)}{2 + 8} V^2$$ $$\eta_{\text{lost}} = \frac{\Delta U_{\text{loss}}}{U_i} = \frac{4}{5} = 80\%$$

5. In the given circuit, a charge of $+80\,\mu\text{C}$ is given to the upper plate of the $4\,\mu\text{F}$ capacitor. Then in the steady state, the charge on the upper plate of the $3\,\mu\text{F}$ capacitor is: [JEE 2012 P-II]

a. $+32\,\mu\text{C}$

b. $+40\,\mu\text{C}$

c. $+48\,\mu\text{C}$

d. $+80\,\mu\text{C}$

Answer: (c)

Solution:

Charge on capacitor of $3\,\mu\text{F}$:

$$Q = \frac{3}{2 + 3} \times 80 = 48\,\mu\text{C}$$

6. Two capacitors $C_1$ and $C_2$ are charged to $120\,\text{V}$ and $200\,\text{V}$ respectively. It is found that by connecting them together the potential on each one can be made zero. Then [JEE Main 2013]

a. $5C_1 = 3C_2$

b. $3C_1 = 5C_2$

c. $C_1 + C_2 = 0$

d. $9C_1 = 4C_2$

Answer: (b, d)

Solution:

Potential $= 0$ on connecting them together i.e., $Q = 0$ i.e., $V_1 C_1 = V_2 C_2$ (capacitance is positive but they are connected with opposite polarity)

$$120 C_1 = 200 C_2 \implies 3C_1 = 5C_2$$

7. In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch $S_1$ is pressed first to fully charge the capacitor $C_1$ and then released. The switch $S_2$ is then pressed to charge the capacitor $C_2$. After some time, $S_2$ is released and then $S_3$ is pressed. After some time, [JEE Adv. 2013 P-I]

a. the charge on the upper plate of $C_1$ is $2CV_0$

b. the charge on the upper plate of $C_1$ is $CV_0$

c. the charge on the upper plate of $C_1$ is $0$

d. the charge on the upper plate of $C_2$ is $-CV_0$

Answer: (a, d)

Solution:

After switch $S_1$ is closed, $C_1$ is charged by $2CV_0$, when switch $S_2$ is closed, $C_1$ and $C_2$ both have plate charge $CV_0$. When $S_3$ is closed, then upper plate of $C_2$ becomes charged by $-CV_0$ and lower plate by $+CV_0$.

8. A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is $C_1$. When the capacitor is charged, the plate area covered by the dielectric gets charge $Q_1$ and the rest of the area gets charge $Q_2$. The electric field in the dielectric is $E_1$ and that in the other portion is $E_2$. Choose the correct option/options, ignoring edge effects. [JEE Adv. 2014 P-I]

a. $\frac{E_1}{E_2} = 1$

b. $\frac{E_1}{E_2} = \frac{1}{K}$

c. $\frac{Q_1}{Q_2} = \frac{3}{K}$

d. $\frac{C}{C_1} = \frac{2 + K}{K}$

Answer: (b)

Solution:

As $E = V/d$, $E_1 / E_2 = 1$ (Both parts have common potential difference).

Assume $C_0$ be the capacitance without dielectric for whole capacitor.

$$C = \frac{2}{3} C_0 + \frac{1}{3} K C_0$$ $$\frac{C}{C_1} = \frac{2 + K}{K}$$ $$\frac{Q_1}{Q_2} = \frac{K}{2}$$

9. A parallel plate capacitor is made of two circular plates separated by a distance of $5\,\text{mm}$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3 \times 10^4\,\text{V/m}$, the charge density of the positive plate will be close to: [JEE Main 2014]

a. $3 \times 10^4\,\text{C/m}^2$

b. $6 \times 10^4\,\text{C/m}^2$

c. $6 \times 10^{-7}\,\text{C/m}^2$

d. $3 \times 10^{-7}\,\text{C/m}^2$

Answer: (c)

Solution:

$K = 2.2$, $d = 5\,\text{mm}$, $E = 3 \times 10^4\,\text{V/m}$

$$\frac{\sigma}{\varepsilon_0 K} = 3 \times 10^4$$ $$\sigma = K \varepsilon_0 \times 3 \times 10^4 = 2.2 \times 8.85 \times 10^{-12} \times 3 \times 10^4$$ $$\sigma = 58.41 \times 10^{-8} = 5.84 \times 10^{-7} \approx 6 \times 10^{-7}\,\text{C/m}^2$$

10. A parallel plate capacitor having plates of area S and plate separation d, has capacitance $C_1$ in air. When two dielectrics of different relative permittivities ($\varepsilon_1 = 2$ and $\varepsilon_2 = 4$) are introduced between the two plates as shown in the figure, the capacitance becomes $C_2$. The ratio $\frac{C_2}{C_1}$ is [JEE Adv. 2015 P-II]

a. $6/5$

b. $5/3$

c. $7/5$

d. $7/3$

Answer: (d)

Solution:

$$C_1 = \frac{\varepsilon_0 S}{d}$$ $$C_{10} = \frac{\varepsilon_0 (S/2)}{d/2} = \frac{\varepsilon_0 S}{d}, \quad C_{20} = \frac{2\varepsilon_0 S}{d}, \quad C_{30} = \frac{4\varepsilon_0 S}{d}$$ $$\frac{1}{C'} = \frac{1}{C_{10}} + \frac{1}{C_{20}} = \frac{d}{\varepsilon_0 S} \left[1 + \frac{1}{2}\right] \implies C' = \frac{2}{3} \frac{\varepsilon_0 S}{d}$$ $$C_2 = C' + C_{30} = \frac{2}{3} \frac{\varepsilon_0 S}{d} + \frac{4}{3} \frac{\varepsilon_0 S}{d} = \frac{7}{3} \frac{\varepsilon_0 S}{d}$$ $$\frac{C_2}{C_1} = \frac{7}{3}$$

11. In the given circuit, charge $Q_2$ on the $2\,\mu\text{F}$ capacitor changes as C is varied from $1\,\mu\text{F}$ to $3\,\mu\text{F}$. $Q_2$ as a function of 'C' is given properly by: (figures are drawn schematically and are not to scale) [JEE Main 2015]

Answer: (b)

Solution:

$$Q_2 = \frac{2 E}{C + 3}$$ $$Q_2 = \frac{E}{1 + 3/C}$$

As C increases from 1 to $3\,\mu\text{F}$, $Q_2$ increases as E is constant. Checking the concavity of the function using 2nd derivative we find option (b) is correct.

12. A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the $4\,\mu\text{F}$ and $9\,\mu\text{F}$ capacitors), at a point distant $30\,\text{m}$ from it, would equal: [JEE Main 2016]

a. $240\,\text{N/C}$

b. $360\,\text{N/C}$

c. $420\,\text{N/C}$

d. $480\,\text{N/C}$

Answer: (c)

Solution:

$$Q_1 = 24\,\mu\text{C}$$ $$Q_2 = 18\,\mu\text{C}$$ $$Q = Q_1 + Q_2 = 42\,\mu\text{C}$$ $$E = \frac{9 \times 10^9 \times 42 \times 10^{-6}}{30^2} = 420\,\text{N/C}$$

13. A capacitance of $2\,\mu\text{F}$ is required in an electrical circuit across a potential difference of $1.0\,\text{kV}$. A large number of $1\,\mu\text{F}$ capacitors are available which can withstand a potential difference of not more than $300\,\text{V}$. The minimum number of capacitors required to achieve this is: [JEE Main 2017]

a. $16$

b. $24$

c. $32$

d. $2$

Answer: (c)

Solution:

To hold $1\,\text{kV}$ potential difference, minimum 4 capacitors are required in series.

$C = \frac{1}{4}\,\mu\text{F}$ for one series branch. So, for $C_{\text{eq}}$ to be $2\,\mu\text{F}$, 8 parallel combinations are required.

Minimum number of capacitors $= 8 \times 4 = 32$.

14. In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be: [JEE Main 2017]

a. $C E \frac{r_1}{(r_1 + r_2)}$

b. $C E \frac{r_2}{(r_1 + r_2)}$

c. $C E \frac{r_1}{(r_1 + r)}$

d. $C E$

Answer: (b)

Solution:

In steady state, current through branch AB $= 0$.

$$V_{AB} = V_{CD}$$ $$V_{CD} = E \times \frac{r_2}{r_1 + r_2}$$ $$Q = C V_{AB} = C E \left(\frac{r_2}{r_1 + r_2}\right)$$

15. A parallel plate capacitor of capacitance $90\,\text{pF}$ is connected to a battery of emf $20\,\text{V}$. If a dielectric material of dielectric constant $K = \frac{5}{3}$ is inserted between the plates, the magnitude of the induced charge will be: [JEE Main 2018]

a. $0.9\,\text{nC}$

b. $1.2\,\text{nC}$

c. $0.3\,\text{nC}$

d. $2.4\,\text{nC}$

Answer: (b)

Solution:

$$C' = K C_0$$ $$Q = K C_0 V$$ $$Q_{\text{induced}} = Q \left(1 - \frac{1}{K}\right) = C_0 V (K - 1) = 90 \times 10^{-12} \times 20 \times \left(\frac{5}{3} - 1\right) = 1.2\,\text{nC}$$

16. A parallel plate capacitor has $1\,\mu\text{F}$ capacitance. One of its two plates is given $+2\,\mu\text{C}$ charge and the other plate, $+4\,\mu\text{C}$ charge. The potential difference developed across the capacitor is: [JEE Main 2019]

a. $5\,\text{V}$

b. $2\,\text{V}$

c. $3\,\text{V}$

d. $1\,\text{V}$

Answer: (d)

Solution:

Charges at inner plates are $1\,\mu\text{C}$ and $-1\,\mu\text{C}$.

Potential difference across capacitor:

$$V = \frac{q}{C} = \frac{1 \times 10^{-6}}{1 \times 10^{-6}} = 1\,\text{V}$$

17. A capacitor of capacitance $C = 15\,\text{pF}$ is charged with voltage $V = 500\,\text{V}$. The electric field inside the capacitor with dielectric is $10^6\,\text{V/m}$ and the area of the plate is $10^{-4}\,\text{m}^2$, then the dielectric constant of the medium is: ($\varepsilon_0 = 8.85 \times 10^{-12}$ in S.I units) [JEE Main 2019]

a. $12.47$

b. $8.47$

c. $10.85$

d. $14.85$

Answer: (b)

Solution:

$$E = \frac{V}{d} = \frac{V C}{A K \varepsilon_0}$$ $$K = \frac{V C}{A E \varepsilon_0} = \frac{500 \times 15 \times 10^{-12}}{10^{-4} \times 10^6 \times 8.85 \times 10^{-12}} = 8.47$$

18. A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants $K_1, K_2, K_3, K_4$ arranged as shown in the figure. The effective dielectric constant K will be: [JEE Main 2019]

a. $K = \frac{(K_1 + K_2)(K_3 + K_4)}{2(K_1 + K_2 + K_3 + K_4)}$

b. $K = \frac{(K_1 + K_2)(K_3 + K_4)}{(K_1 + K_2 + K_3 + K_4)}$

c. $K = \frac{(K_1 + K_4)(K_2 + K_3)}{2(K_1 + K_2 + K_3 + K_4)}$

d. $K = \frac{(K_1 + K_3)(K_2 + K_4)}{(K_1 + K_2 + K_3 + K_4)}$

Answer: (d)

Solution:

$$C_{13} = \frac{\varepsilon_0 (L/2)}{d/2} (K_1 + K_3) = \frac{\varepsilon_0 L}{d} (K_1 + K_3)$$ $$C_{24} = \frac{\varepsilon_0 (L/2)}{d/2} (K_2 + K_4) = \frac{\varepsilon_0 L}{d} (K_2 + K_4)$$ $$C_{\text{eq}} = \frac{(K_1 + K_3)(K_2 + K_4)}{K_1 + K_2 + K_3 + K_4} \frac{\varepsilon_0 L^2}{d}$$ $$K = \frac{(K_1 + K_3)(K_2 + K_4)}{K_1 + K_2 + K_3 + K_4}$$

19. A parallel plate capacitor is made of two square plates of side ‘a’, separated by a distance d ($d \ll a$). The lower triangular portion is filled with a dielectric of dielectric constant K, as shown in the figure. Capacitance of this capacitor is: [JEE Main 2019]

a. $\frac{K \varepsilon_0 a^2}{2d}$

b. $\frac{\varepsilon_0 a^2 \ln K}{d}$

c. $\frac{K \varepsilon_0 a^2 \ln K}{d(K - 1)}$

d. $\frac{\varepsilon_0 a^2}{2d(K + 1)}$

Answer: (c)

Solution:

Taking an elemental strip of width $dx$ at distance $x$:

$$dC = \frac{\varepsilon_0 a dx}{y + \frac{d-y}{K}}$$ $$C = \int_0^a \frac{K \varepsilon_0 a dx}{K y + (d - y)} = \frac{K \varepsilon_0 a^2 \ln K}{d(K - 1)}$$

20. A parallel plate capacitor having capacitance $12\,\text{pF}$ is charged by a battery to a potential difference of $10\,\text{V}$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant $6.5$ is slipped between the plates the work done by the capacitor on the slab is: [JEE Main 2019]

a. $692\,\text{pJ}$

b. $60\,\text{pJ}$

c. $508\,\text{pJ}$

d. $560\,\text{pJ}$

Answer: (c)

Solution:

Initial energy of capacitor:

$$U_i = \frac{1}{2} C V^2 = \frac{1}{2} \times 12 \times 10^2 = 600\,\text{pJ}$$

Final energy of capacitor:

$$U_f = \frac{U_i}{K} = \frac{600}{6.5} = 92\,\text{pJ}$$ $$W = U_i - U_f = 600 - 92 = 508\,\text{pJ}$$

21. A parallel plate capacitor is of area $6\,\text{cm}^2$ and a separation $3\,\text{mm}$. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants $K_1 = 10, K_2 = 12$ and $K_3 = 14$. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be: [JEE Main 2019]

a. $12$

b. $4$

c. $36$

d. $14$

Answer: (a)

Solution:

Let dielectric constant of material used be K.

$$\frac{\varepsilon_0 (A/3)}{d} \times 10 + \frac{\varepsilon_0 (A/3)}{d} \times 12 + \frac{\varepsilon_0 (A/3)}{d} \times 14 = \frac{K \varepsilon_0 A}{d}$$ $$\implies K = \frac{10 + 12 + 14}{3} = 12$$

22. Seven capacitors, each of capacitance $2\,\mu\text{F}$, are to be connected in a configuration to obtain an effective capacitance of $\left(\frac{6}{13}\right)\,\mu\text{F}$. Which of the combinations, shown in figures below, will achieve the desired value? [JEE Main 2019]

Answer: (d)

Solution:

$$C_{\text{eq}} = \frac{6}{13}\,\mu\text{F}$$

Therefore three capacitors must be in parallel to get $6\,\mu\text{F}$ in combination with others:

$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \frac{1}{C_4} + \frac{1}{C_5}$$ $$C_{\text{eq}} = \frac{6}{13}\,\mu\text{F}$$

23. In the figure shown below, the charge on the left plate of the $10\,\mu\text{F}$ capacitor is $-30\,\mu\text{C}$? The charge on the right plate of the $6\,\mu\text{F}$ capacitor is: [JEE Main 2019]

a. $-18\,\mu\text{C}$

b. $-12\,\mu\text{C}$

c. $+12\,\mu\text{C}$

d. $+18\,\mu\text{C}$

Answer: (d)

Solution:

$6\,\mu\text{F}$ and $4\,\mu\text{F}$ are in parallel and total charge on this combination is $30\,\mu\text{C}$.

Charge on $6\,\mu\text{F}$ capacitor $= \frac{6}{6 + 4} \times 30 = 18\,\mu\text{C}$.

Since, charge is asked on right plate therefore it is $+18\,\mu\text{C}$.

24. In the circuit shown, find C if the effective capacitance of the whole circuit is to be $0.5\,\mu\text{F}$. All values in the circuit are in $\mu\text{F}$. [JEE Main 2019]

a. $\frac{7}{10}\,\mu\text{F}$

b. $\frac{7}{11}\,\mu\text{F}$

c. $\frac{7}{6}\,\mu\text{F}$

d. $4\,\mu\text{F}$

Answer: (b)

Solution:

$$\frac{\frac{7}{3} C}{\frac{7}{3} + C} = \frac{1}{2}$$ $$14 C = 7 + 3C \implies 11 C = 7 \implies C = \frac{7}{11}\,\mu\text{F}$$

25. The charge on a capacitor plate in a circuit, as a function of time, is shown in the figure: What is the value of current at $t = 4\,\text{s}$? [JEE Main 2019]

a. $3\,\mu\text{A}$

b. $2\,\mu\text{A}$

c. $0$

d. $1.5\,\mu\text{A}$

Answer: (c)

Solution:

At $t = 4\,\text{s}$, charge q is constant (slope $\frac{dq}{dt} = 0$). Hence, current $I = 0$.

26. A parallel plate capacitor with plates of area $1\,\text{m}^2$ each, are at a separation of $0.1\,\text{m}$. If the electric field between the plates is $100\,\text{N/C}$, the magnitude of charge on each plate is: (Take $\varepsilon_0 = 8.85 \times 10^{-12}\,\text{C}^2/\text{N}\cdot\text{m}^2$) [JEE Main 2019]

a. $7.85 \times 10^{-10}\,\text{C}$

b. $6.85 \times 10^{-10}\,\text{C}$

c. $9.85 \times 10^{-10}\,\text{C}$

d. $8.85 \times 10^{-10}\,\text{C}$

Answer: (d)

Solution:

$$\sigma = \frac{Q}{A} = E \varepsilon_0$$ $$Q = A E \varepsilon_0 = (1)(100)(8.85 \times 10^{-12}) = 8.85 \times 10^{-10}\,\text{C}$$

27. Which statements is/are correct: [JEE Adv. 2019 P-I]

a. At time $t = 0$, this $S_1$ is closed instantaneous current in the closed circuit will be $25\,\text{mA}$

b. The key $S_1$ is kept closed for long time such that capacitors are fully charged. Now key $S_2$ is closed at this time the instantaneous current across $30\,\Omega$ resistor between P and Q will be $0.2\,\text{A}$.

c. If key $S_1$ is kept closed for long time such that capacitors are fully charged the voltage across $C_1$ will be $4\,\text{V}$.

d. If $S_1$ is kept closed for long time such that capacitors are fully charged the voltage difference between P and Q will be $10\,\text{V}$.

Answer: (a, c)

Solution:

Just after closing of switch, charge on C is zero. Replace all capacitors with wire.

$$i = \frac{5}{70 + 100 + 30} = \frac{5}{200} = 25\,\text{mA}$$

Now $S_1$ is kept closed for long time, circuit is in steady state.

$$q = 40\,\mu\text{C}$$ $$V_{\text{across } C_1} = \frac{40}{10} = 4\,\text{V}$$

Matrix Match/Column Match

28. Column I give certain situations in which a straight metallic wire of resistance R is used and Column II gives some resulting effects. Match the statements in Column I with the statements in Column II. [JEE 2007 P-I]

Column I

(A) A charged capacitor is connected to the ends of the wire

(B) The wire is moved perpendicular to its length with a constant velocity in a uniform magnetic field perpendicular to the plane of motion

(D) A battery of constant emf is connected to the ends of the wire

Column II

1. A constant current flows through the wire

2. Thermal energy is generated in the wire

3. A constant potential difference develops between the ends of the wire

4. Charges of constant magnitude appear at the ends of the wire

a. A$\rightarrow$2; B$\rightarrow$3,4; C$\rightarrow$3,4; D$\rightarrow$1,2,3

b. A$\rightarrow$1; B$\rightarrow$2,3; C$\rightarrow$1,2; D$\rightarrow$2,3,4

c. A$\rightarrow$3; B$\rightarrow$1,2; C$\rightarrow$2,4; D$\rightarrow$1,3,4

d. A$\rightarrow$4; B$\rightarrow$3,4; C$\rightarrow$1,3; D$\rightarrow$2,3,4

Answer: (a)

Solution:

A$\rightarrow$2; B$\rightarrow$3,4; C$\rightarrow$3,4; D$\rightarrow$1,2,3

Paragraphs

Paragraph for Question Nos. 29 to 31

The capacitor of capacitance C can be charged (with the help of a resistance R) by a voltage source V, by closing switch $S_1$ while keeping switch $S_2$ open. The capacitor can be connected in series with an inductor ‘L’ by closing switch $S_2$ and opening $S_1$. [JEE 2006]

29. Initially, the capacitor was uncharged. Now, switch $S_1$ is closed and $S_2$ is kept open. If time constant of this circuit is $\tau$, then

a. after time interval $\tau$, charge on the capacitor is $CV/2$

b. after time interval $2\tau$, charge on the capacitor is $CV(1 - e^{-2})$

c. the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged

d. after time interval $2\tau$, charge on the capacitor is $CV(1 - e^{-1})$

Answer: (b)

Solution:

$$Q = Q_0(1 - e^{-t/\tau}) = CV(1 - e^{-t/\tau})$$

After time interval $2\tau$, charge on the capacitor is $CV(1 - e^{-2})$.

30. After the capacitor gets fully charged, $S_1$ is opened and $S_2$ is closed so that the inductor is connected in series with the capacitor. Then

a. at $t = 0$, energy stored in the circuit is purely in the form of magnetic energy

b. at any time $t > 0$, current in the circuit is in the same direction

c. at $t > 0$, there is no exchange of energy between the inductor and capacitor

d. at any time $t > 0$, instantaneous current in the circuit may be $\frac{C V_0}{\sqrt{LC}}$

Answer: (d)

Solution:

In LC circuit, charge oscillates as $Q = Q_0 \cos(\omega t)$ and current $I = I_{\max} \sin(\omega t)$ where $I_{\max} = \omega Q_0 = \frac{1}{\sqrt{LC}} CV_0$.

31. If the total charge stored in the LC circuit is $Q_0$, then for $t \ge 0$

a. the charge on the capacitor is $Q = Q_0 \cos\left(\frac{\pi t}{2\sqrt{LC}}\right)$

b. the charge on the capacitor is $Q = -Q_0 \cos\left(\frac{\pi t}{2\sqrt{LC}}\right)$

c. $\frac{d^2Q}{dt^2} = -\frac{Q}{LC}$

d. $\frac{d^2Q}{dt^2} = -\frac{1}{LC} \frac{dQ}{dt}$

Answer: (c)

Solution:

For an LC circuit, the differential equation governing the charge is:

$$\frac{d^2Q}{dt^2} + \omega^2 Q = 0 \implies \frac{d^2Q}{dt^2} = -\frac{Q}{LC}$$

Paragraph for Question Nos. 32 to 33

Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius $r \ll h$. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at $+V_0$ and the top plate at $-V_0$. Due to their conducting surface the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity) [JEE Adv. 2016 P-II]

32. Which one of the following statements is correct?

a. The balls will execute simple harmonic motion between the two plates

b. The balls will bounce back to the bottom plate carrying the opposite charge they went up with

c. The balls will bounce back to the bottom plate carrying the same charge they went up with

d. The balls will stick to the top plate and remain there

Answer: (b)

Solution:

The balls will bounce back to the bottom plate carrying the opposite charge they went up with.

33. The average current in the steady state registered by the ammeter in the circuit will be:

a. proportional to $V_0^2$

b. proportional to $V_0^{1/2}$

c. proportional to the potential $V_0$

d. zero

Answer: (a)

Solution:

$$q \propto V_0 r$$ $$\frac{1}{2} m v^2 = q E h = q (2V_0)$$ $$t \propto \frac{h}{v} \propto \frac{1}{V_0}$$ $$I_{\text{avg}} = \frac{2q}{t} \propto V_0^2$$

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