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Multiple Choice Questions

One and More than One Correct

1. Two metallic rings A and B, identical in shape and size but having different resistivity’s $\rho_A$ and $\rho_B$ are kept on top of two identical solenoids as shown in the figure. When current $I$ is switched on in both the solenoids in identical manner, the rings A and B jump to heights $h_A$ and $h_B$, respectively, with $h_A > h_B$. The possible relation(s) between their resistivity’s and their masses $m_A$ and $m_B$ is (are): [JEE 2009 P-II]

a. $\rho_A > \rho_B$ and $m_A = m_B$

b. $\rho_A < \rho_B$ and $m_A = m_B$

c. $\rho_A > \rho_B$ and $m_A > m_B$

d. $\rho_A < \rho_B$ and $m_A < m_B$

Answer: (b, d)

Solution:

As $\text{emf} = -\frac{d\phi}{dt}$ is same, the current induced in the ring will depend upon resistance of the ring. Larger the resistivity smaller the current.

2. A ray OP of monochromatic light is incident on the face AB of prism ABCD near vertex B at an incident angle of $60^\circ$ (see figure). If the refractive index of the material of the prism is $\sqrt{3}$, which of the following is (are) correct? [JEE 2010 P-I]

a. The ray gets totally internally reflected at face CD

b. The ray comes out through face AD

c. The angle between the incident ray and the emergent ray is $90^\circ$

d. The angle between the incident ray and the emergent ray is $120^\circ$

Answer: (a, b, c)

Solution:

Using Snell’s law, $\sin \theta = \frac{1}{\mu} = \frac{1}{\sqrt{3}}$. Net deviation is $90^\circ$.

3. An AC voltage source of variable angular frequency $\omega$ and fixed amplitude $V_0$ is connected in series with a capacitance $C$ and an electric bulb of resistance $R$ (inductance zero). When $\omega$ is increased [JEE 2010 P-I]

a. the bulb glows dimmer

b. the bulb glows brighter

c. total impedance of the circuit is unchanged

d. total impedance of the circuit increases

Answer: (b)

Solution:

Impedance $Z = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}$. As $\omega$ increases, $Z$ decreases. Hence bulb will glow brighter.

4. A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, $x_1(t) = A \sin(\omega t)$ and $x_2(t) = A \sin\left(\omega t + \frac{2\pi}{3}\right)$. Adding a third sinusoidal displacement $x_3(t) = B \sin(\omega t + \phi)$ brings the mass to a complete rest. The values of $B$ and $\phi$ are [JEE 2011 P-II]

a. $B = \frac{\sqrt{3}}{2}A, \phi = \frac{\pi}{4}$

b. $B = A, \phi = \frac{4\pi}{3}$

c. $B = \frac{\sqrt{5}}{3}A, \phi = \frac{\pi}{6}$

d. $B = A, \phi = \frac{\pi}{3}$

Answer: (b)

5. A series R-C circuit is connected to AC voltage source. Consider two cases; a. when C is without a dielectric medium and b. when C is filled with dielectric of constant 4. The current $I_R$ through the resistor and voltage $V_C$ across the capacitor are compared in the two cases. Which of the following is/are true? [JEE 2011 P-II]

a. $I_R^A > I_R^B$

b. $I_R^A < I_R^B$

c. $V_C^A > V_C^B$

d. $V_C^A < V_C^B$

Answer: (b, c)

Solution:

$$I = \frac{V}{Z}$$ $$I_R = \sqrt{I^2} = \sqrt{\frac{V^2}{R^2 + \left(\frac{1}{\omega C}\right)^2}}$$

6. In the given circuit, the AC source has $\omega = 100\text{ rad/s}$. Considering the inductor and capacitor to be ideal, the correct choice (s) is (are). [JEE 2012 P-II]

a. The current through the circuit, $I$ is $0.3\text{ A}$

b. The current through the circuit, $I$ is $0.3\sqrt{2}\text{ A}$

c. The voltage across $100\Omega$ resistor $= 10\sqrt{2}\text{V}$

d. The voltage across $50\Omega$ resistor $= 10\text{V}$

Answer: (a, c)

Solution:

$$I = I_1 + I_2$$ $$X_L = \omega L = (100)(0.5) = 50\Omega$$ $$X_C = \frac{1}{\omega C} = \frac{1}{100 \times 100 \times 10^{-6}} = 100\Omega$$

7. A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it. The correct statement (s) is (are) [JEE 2012 P-II]

a. The emf induced in the loop is zero if the current is constant

b. The emf induced in the loop is finite if the current is constant

c. The emf induced in the loop is zero if the current decreases at a steady rate

d. The emf induced in the loop is finite if the current decreases at a steady rate

Answer: (a, c)

Solution:

The magnetic flux above and below the circular loop are equal and opposite due to symmetry. $\therefore \phi_{\text{net}} = 0$. $\therefore \text{Emf induced will be zero}$.

8. In an LCR circuit as shown below both switches are open initially. Now switch $S_1$ is closed. $S_2$ kept open. ($q$ is charge on the capacitor and $\tau = RC$ is Capacitive time constant). Which of the following statement is correct? [JEE Main 2013]

a. Work done by the battery is half of the energy dissipated in the resistor

b. At $t = \tau, q = \frac{V}{2}CV$

c. At $t = \tau, q = CV(1 - e^{-1})$

d. At $t = 2\tau, q = CV(1 - e^{-2})$

Answer: (c)

Solution:

Case-(i): Its normal RC circuit. $W_{\text{bat}} = CV^2$, $U = \frac{1}{2}CV^2$, $H = W_{\text{bat}} - U = \frac{1}{2}CV^2$. So (a) is wrong. $q = CV(1 - e^{-t/\tau})$. At $t = \tau, q = CV(1 - e^{-1})$. So, (c) is correct.

9. At time $t = 0$, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current $I(t) = I_0 \cos(\omega t)$ with $I_0 = 1\text{ A}$ and $\omega = 500\text{ rad/s}$ starts flowing in it with the initial direction shown in the figure. At $t = \frac{7\pi}{6\omega}$, the key is switched from B to D. Now onwards only A and D are connected. A total charge $Q$ flows from the battery to charge the capacitor fully. If $C = 20\mu\text{F}$, $R = 10\Omega$ and the battery is ideal with emf of $50\text{V}$, identify the correct statement(s). [JEE Adv. 2014 P-I]

a. Magnitude of the maximum charge on the capacitor before $t = \frac{7\pi}{6\omega}$ is $3 \times 10^{-4}\text{ C}$

b. The current in the left part of the circuit just before $t = \frac{7\pi}{6\omega}$ is clockwise

c. Immediately after A is connected to D, the current in R is $10\text{ A}$

d. $Q = 2 \times 10^{-3}\text{ C}$

Answer: (c, d)

Solution:

As current leads voltage by $\pi/2$ in the given circuit initially, then ac voltage can be represented as $V(t) = V_0 \sin(\omega t)$.

10. In the circuit shown here, the point ‘C’ is kept connected to point ‘A’ till the current flowing through the circuit becomes constant. Afterward, suddenly, point ‘C’ is disconnected from point ‘A’ and connected to point ‘B’ at time t=0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to: [JEE Main 2014]

a. -1

b. $\frac{1}{e} - e$

c. $\frac{1}{e}$

d. 1

Answer: (a)

Solution:

$$i = i_0 e^{-tR/L}$$ $$V_R = iR = i_0 R e^{-tR/L}$$

11. An inductor ($L = 0.03\text{ H}$) and a resistor ($R = 0.15\text{ k}\Omega$) are connected in series to a battery of $15\text{V}$ EMF in a circuit shown below. The key $K_1$ has been kept closed for a long time. Then at $t = 0$, $K_1$ is opened and key $K_2$ is closed simultaneously at $t = 1\text{ ms}$, the current in the circuit will be: $\left[e^{-150} \approx 5\right]$ [JEE Main 2015]

a. $100\text{ mA}$

b. $67\text{ mA}$

c. $6.7\text{ mA}$

d. $0.67\text{ mA}$

Answer: (d)

Solution:

$$i = I_0 e^{-t/\tau}$$

12. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to $Q_0$ and then connected to the L and R as shown below: If a student plots graphs of the square of maximum charge $Q_{\text{max}}^2$ on the capacitor with time (t) for two different values $L_1$ and $L_2$ ($L_1 > L_2$) of L then which of the following represents this graph correctly? (Plots are schematic and not drawn to scale) [JEE Main 2015]

Answer: (a)

Solution:

For a damped pendulum $A^2 = A_0^2 e^{-bt/m}$. Since $L_1 > L_2$, energy dissipation with $L_1$ will be slower compared to $L_2$.

13. An arc lamp requires a direct current of $10\text{ A}$ at $80\text{ V}$ to function. If it is connected to a $220\text{ V}$ (rms), $50\text{ Hz}$ AC supply, the series inductor needed for it to work is close to: [JEE Main 2016]

a. $80\text{ H}$

b. $0.08\text{ H}$

c. $0.044\text{ H}$

d. $0.065\text{ H}$

Answer: (d)

Solution:

$$R = \frac{80}{10} = 8\Omega$$ $$V_{\text{rms}}^2 = V_R^2 + V_L^2 \implies V_L = \sqrt{220^2 - 80^2} \approx 204.9\text{ V}$$

14. In an AC circuit, the instantaneous emf and current are given by $e(t) = 100\sin(30t)$, $i(t) = 20\sin\left(30t - \frac{\pi}{4}\right)$. In one cycle of AC, the average power consumed by the circuit and the wattles current are, respectively [JEE Main 2016]

a. $50, 0$

b. $50, 10$

c. $\frac{1000}{\sqrt{2}}, 10$

d. $\frac{50}{\sqrt{2}}, 0$

Answer: (c)

Solution:

$$P_{\text{av}} = E_{\text{rms}} I_{\text{rms}} \cos\phi = \frac{100}{\sqrt{2}} \times \frac{20}{\sqrt{2}} \times \cos\left(\frac{\pi}{4}\right) = \frac{1000}{\sqrt{2}}$$

15. In the circuit, $L = 1\mu\text{H}$, $C = 1\mu\text{F}$ and $R = 1\text{k}\Omega$. They are connected in series with an AC source $V_0 = \sin\omega t$ as shown. Which of the following options is/are correct? [JEE Adv. 2017 P-I]

a. At $\omega \approx 0$ the circuit flowing through the circuit becomes nearly zero

b. The frequency at which the current will be in phase with the voltage is independent of R

c. The current will be in phase with the voltage if $\omega = 10^4\text{ rad s}^{-1}$

d. At $\omega \gg 10^6\text{ rad s}^{-1}$ the circuit behaves like a capacitor

Answer: (a, b)

16. A source of constant voltage $V$ is connected to a resistance $R$ and two ideal inductors $L_1$ and $L_2$ through a switch S as shown. There is no mutual inductance between the two inductors. The switch S is initially open. At $t = 0$, the switch is closed and current begins to flow. Which of the following options is/are correct? [JEE Adv. 2017 P-II]

a. After a long time, the current through $L_1$ will be $\frac{V}{R}\frac{L_2}{L_1 + L_2}$

b. After a long time, the current through $L_2$ will be $\frac{V}{R}\frac{L_1}{L_1 + L_2}$

c. The ratio of the currents through $L_1$ and $L_2$ is fixed at all times ($t > 0$)

d. At $t = 0$, the current through the resistance $R$ is $\frac{V}{R}$

Answer: (a, b, c)

17. The instantaneous voltages at three terminals marked X, Y and Z are given by $V_X = V_0\sin(\omega t)$, $V_Y = V_0\sin\left(\omega t + \frac{2\pi}{3}\right)$ and $V_Z = V_0\sin\left(\omega t + \frac{4\pi}{3}\right)$. An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points X and Y and then between Y and Z. The reading(s) of the voltmeter will be [JEE Adv. 2017 P-II]

a. $V_{\text{rms}}^{YZ} = \frac{V_0}{2}$

b. $V_{\text{rms}}^{XY} = \frac{\sqrt{3}V_0}{2}$

c. independent of the choice of the two terminals

d. $V_{\text{rms}}^{XY} = 0$

Answer: (b, c)

18. In the figure below, the switches $S_1$ and $S_2$ are closed simultaneously at $t = 0$ and a current starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force (emf) and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors. The current $I$ in the middle wire reaches it maximum magnitude $I_{\text{max}}$ at time $t = \tau$. Which of the following statements is (are) true? [JEE Adv. 2018 P-I]

a. $I_{\text{max}} = \frac{2V}{R}$

b. $I_{\text{max}} = \frac{4V}{R}$

c. $\tau = \frac{L}{R}\ln 2$

d. $\tau = \frac{2L}{R}\ln 2$

Answer: (b, d)

19. For an RLC circuit driven with voltage of amplitude $v_m$ and frequency $\omega_0 = \frac{1}{\sqrt{LC}}$ the current exhibits resonance. The quality factor, $Q$ is given by: [JEE Main 2018]

a. $\frac{\omega_0 CR}{L}$

b. $\frac{\omega_0 L}{R}$

c. $\frac{\omega_0 R}{L}$

d. $\frac{R}{\omega_0 C}$

Answer: (b)

20. A circuit connected to an ac source of emf $e = e_0\sin(100t)$ with $t$ in seconds, gives a phase difference of $\frac{\pi}{4}$ between the emf $e$ and current $i$. Which of the following circuits will exhibit this? [JEE Main 2019]

a. RC circuit with $R = 1\text{ k}\Omega$ and $C = 1\mu\text{F}$

b. RL circuit with $R = 1\text{ k}\Omega$ and $L = 1\text{ mH}$

c. RL circuit with $R = 1\text{ k}\Omega$ and $L = 10\text{ mH}$

d. RC circuit with $R = 1\text{ k}\Omega$ and $C = 10\mu\text{F}$

Answer: (d)

Solution:

For RC circuit, $\tan\phi = \frac{X_C}{R} = \frac{1}{\omega CR}$. Checking option (d): $\omega = 100$, $C = 10\mu\text{F}$, $R = 10^3\Omega$, $\omega CR = 100 \times 10 \times 10^{-6} \times 10^3 = 1$, so $\tan\phi = 1 \implies \phi = \frac{\pi}{4}$.

21. The switch is closed at $t = 0$. The time after which the rate of dissipation of energy in the resistor is equal to rate at which energy is being stored in the inductor is: [JEE Main 2019]

a. $\frac{L}{R}\ln 2$

b. $\frac{1}{2}\frac{L}{R}\ln 2$

c. $\frac{1}{4}\frac{L}{R}\ln 2$

d. $2\frac{L}{R}\ln 2$

Answer: (a)

22. An A.C. source of voltage $V = 220\sin(100\pi t)$ volts is connected with $R = 50\Omega$. The time interval in which the current goes from its peak value to half of the peak value is: [JEE Main 2019]

a. $\frac{1}{400}\text{ sec}$

b. $\frac{1}{50}\text{ sec}$

c. $\frac{1}{300}\text{ sec}$

d. $\frac{1}{200}\text{ sec}$

Answer: (c)

23. A series AC circuit containing an inductor ($20\text{ mH}$), a capacitor ($120\mu\text{F}$) and a resistor ($60\Omega$) is driven by an AC source of $24\text{ V}/50\text{ Hz}$. The energy dissipated in the circuit in $60\text{ s}$ is: [JEE Main 2019]

a. $2.26 \times 10^3\text{ J}$

b. $3.39 \times 10^3\text{ J}$

c. $5.65 \times 10^2\text{ J}$

d. $5.17 \times 10^2\text{ J}$

Answer: (d)

24. A power transmission line feeds input at $2300\text{ V}$ to a step down transformer with its primary winding having $4000$ turns. The output power is delivered at $230\text{ V}$ by the transformer. If the current in the primary of the transformer is $5\text{A}$ and its efficiency is $90\%$, the output current would be: [JEE Main 2019]

a. $25\text{A}$

b. $50\text{ A}$

c. $35\text{A}$

d. $45\text{A}$

Answer: (d)

25. A conducting circular loop made of a thin wire, has area $3.5 \times 10^{-3}\text{ m}^2$ and resistance $10\Omega$. It is placed perpendicular to a time dependent magnetic field $B(t) = (0.4\text{T})\sin(50\pi t)$. The field is uniform in space. Then the net charge flowing through the loop during $t = 0\text{ s}$ and $t = 10\text{ ms}$ is close to: [JEE Main 2019]

a. $14\text{ mC}$

b. $21\text{ mC}$

c. $6\text{ mC}$

d. $7\text{ mC}$

Answer: (a)

Solution:

$$\Delta Q = \frac{\Delta\phi}{R} = \frac{BA}{R}(\cos 0 - \cos(\omega t)) = \frac{0.4 \times 3.5 \times 10^{-3}}{10} (\sin(50\pi \times 0.01)) = 14\text{ mC}$$

Integer and Subjective

26. The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from $10\text{ A}$ to $25\text{ A}$ in $1\text{ s}$, the change in the energy of the inductance is: [JEE Main 2019]

a. $437.5\text{ J}$

b. $637.5\text{ J}$

c. $740\text{ J}$

d. $540\text{ J}$

Answer: (a)

Solution:

$$e = L \frac{di}{dt}$$ $$25 = L \times \frac{25 - 10}{1}$$ $$L = \frac{25}{15} = \frac{5}{3}\text{ H}$$ $$\Delta E = \frac{1}{2} L (i_2^2 - i_1^2)$$ $$\Delta E = \frac{1}{2} \times \frac{5}{3} \times (25^2 - 10^2) = \frac{5}{6} \times (625 - 100) = \frac{5}{6} \times 525 = 437.5\text{ J}$$

27. A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil: [JEE Main 2019]

a. Decreases by a factor of $9\sqrt{3}$

b. Increases by a factor of $3$

c. Decreases by a factor of $9$

d. Increases by a factor of $27$

Answer: (b)

Solution:

Total length $L$ will remain constant

$L = (3a) N$ ($N = \text{total turns}$)

and length of winding $= (d) N$ ($d = \text{diameter of wire}$)

Self inductance $= \mu_0 n^2 A \ell$

$$L \propto a N \propto a^2$$

So self inductance will become 3 times

28. In the circuit shown, the switch $S_1$ is closed at time $t = 0$ and the switch $S_2$ is kept open. At some later time ($t_0$), the switch $S_1$ is opened and $S_2$ is closed. The behaviour of the current $I$ as a function of time 't' is given by: [JEE Main 2019]

Answer: (b)

Solution:

From time $t = 0$ to $t = t_0$, growth of current takes place and after that decay of current takes place.

Most appropriate is (b).

29. There are two long co-axial solenoids of same length $l$. the inner and outer coils have radii $r_1$ and $r_2$ and number of turns per unit length $n_1$ and $n_2$ respectively. The rate of mutual inductance to the self-inductance of the inner-coil is: [JEE Main 2019]

a. $\frac{n_2}{n_1} \cdot \frac{r_2}{r_1}$

b. $\frac{n_1}{n_2} \cdot \frac{r_1}{r_2}$

c. $\frac{n_1}{n_2}$

d. $\frac{n_2}{n_1}$

Answer: (d)

Solution:

$$M = \mu_0 n_1 n_2 \pi r_1^2 l$$ $$L_1 = \mu_0 n_1^2 \pi r_1^2 l$$ $$\frac{M}{L_1} = \frac{n_2}{n_1}$$

30. In the above circuit, $C = \frac{3}{10}\ \mu\text{F}$, $R_2 = 20\ \Omega$, $L = \frac{3}{10}\text{ H}$ and $R_1 = 10\ \Omega$. Current in $L-R_1$ path is $I_1$ and in $C-R_2$ path it is $I_2$. The voltage of A.C source is given by $V(t) = 200\sqrt{2}\sin(100t)\text{ volts}$. The phase difference between $I_1$ and $I_2$ is: [JEE Main 2019]

a. $30^\circ$

b. $150^\circ$

c. $90^\circ$

d. $60^\circ$

Answer: (b)

Solution:

For $L-R_1$ circuit:

$$X_L = \omega L = 100 \times \frac{3}{10} = 30\ \Omega$$ $$\tan \theta_1 = \frac{X_L}{R_1} = \frac{30}{10} = 3$$

$\theta_1$ is close to $90^\circ$.

For $C-R_2$ circuit:

$$X_C = \frac{1}{\omega C} = \frac{1}{100 \times \frac{3}{10} \times 10^{-6}} = \frac{10^5}{3}\ \Omega$$

$\tan \theta_2 = \frac{X_C}{R_2}$

Phase difference $= \theta_1 + \theta_2 = 150^\circ$

31. In the figure shown, after the switch 'S' is turned from position 'A' to position 'B', the energy dissipated in the circuit in terms of capacitance 'C' and total charge 'Q' is: [JEE Main 2019]

a. $\frac{Q^2}{8C}$

b. $\frac{Q^2}{4C}$

c. $\frac{Q^2}{8C}$

d. $\frac{Q^2}{58C}$

Answer: (a)

Solution:

$$E_i = \frac{Q^2}{2C}$$ $$E_f = \frac{Q^2}{2(C + 3C)} = \frac{Q^2}{8C}$$ $$\Delta E = E_i - E_f = \frac{Q^2}{2C} - \frac{Q^2}{8C} = \frac{3Q^2}{8C}$$

32. In the figure shown, a circuit contains two identical resistors with resistance $R = 5\ \Omega$ and an inductance with $L = 2\text{ mH}$. An ideal battery of $15\text{ V}$ is connected in the circuit. What will be the current through the battery long after the switch is closed? [JEE Main 2019]

a. $6\text{ A}$

b. $7.5\text{ A}$

c. $5.5\text{ A}$

d. $3\text{ A}$

Answer: (a)

Solution:

Ideal inductor will behave like zero resistance long time after switch is closed.

$$I = \frac{\varepsilon}{R} = \frac{15}{5/2} = 6\text{ A}$$

Matrix Match/Column Match

33. You are given many resistances, capacitors and inductors. These are connected to a variable DC voltage source (the first two circuits) or an AC voltage source of $50\text{ Hz}$ frequency (the next three circuits) in different ways as shown in Column II. When a current $I$ (steady state for DC or rms for AC) flows through the circuit, the corresponding voltage $V_1$ and $V_2$ (indicated in circuits) are related as shown in Column I. Match the two. [JEE 2010 P-II]

Column I:

(A) $V_1 \neq 0$, $V_1$ is proportional to $I$

(B) $V_1 \neq 0, V_2 \neq 0$, $V_2 > V_1$

(D) $V_2 \neq 0$, $V_2$ is proportional to $I$

Answer: (a)

Solution:

A$\rightarrow$3,4,5; B$\rightarrow$2,3,4,5; C$\rightarrow$1,2; D$\rightarrow$2,3,4,5

34. Four combinations of two thin lenses are given in Column I. The radius of curvature of all curved surfaces is $r$ and the refractive index of all the lenses is $1.5$. Match lens combinations in Column I with their focal length in Column II and select the correct answer using the code given below the lists. [JEE Adv. 2014 P-II]

Answer: (a)

Solution:

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

$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ $$\frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2}$$

Paragraphs

Paragraph for Question Nos. 35 to 37

A number of designs have been developed for magnetic levitation. In one of the designs electrodynamic system (EDS) is used...

35. What causes the train to lift up [JEE 2006]

a. Electrostatic force

b. Time varying electric field

c. Magnetic force

d. Induced electric field

Answer: (c)

Solution:

Induced currents result in a repulsive magnetic force which lifts the train.

36. The limitation of E.D.S model is that [JEE 2006]

a. The train experiences upward force due to Lenz's law

b. Friction force creates a drag on the train

c. Retardation is caused

d. By Lenz's law train experiences a drag force

Answer: (d)

Solution:

The induced current results in a drag force opposing motion.

37. The advantage of the train is that [JEE 2006]

a. Electromagnetic force draws the train

b. Gravitational force is zero

c. Electrostatic force draws the train

d. Resistive forces due to friction are absent

Answer: (d)

Solution:

In Maglev Train the dissipative forces against friction are absent.

Paragraph for Question Nos. 38 to 39

A dense collection of equal number of electrons and positive ions is called neutral plasma...

38. Taking the electronic charge as $e$ and the permittivity as $\varepsilon_0$, use dimensional analysis to determine the correct expression for $\omega_p$. [JEE 2011 P-I]

a. $\sqrt{\frac{N e^2}{m \varepsilon_0}}$

b. $\sqrt{\frac{m \varepsilon_0}{N e^2}}$

c. $\frac{N e^2}{m \varepsilon_0}$

d. $\sqrt{\frac{m}{N e^2 \varepsilon_0}}$

Answer: (a)

Solution:

By dimensional analysis, the angular plasma frequency is given by $\omega_p = \sqrt{\frac{N e^2}{m \varepsilon_0}}$.

39. Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons $N \approx 4 \times 10^{27}\text{ m}^{-3}$. Take $\varepsilon_0 \approx 10^{-11}$ and $m \approx 10^{-30}$, where these quantities are in proper SI units. [JEE 2011 P-I]

a. $800\text{ nm}$

b. $600\text{ nm}$

c. $300\text{ nm}$

d. $200\text{ nm}$

Answer: (b)

Solution:

Using $\lambda = \frac{2\pi c}{\omega_p} = 2\pi c \sqrt{\frac{m \varepsilon_0}{N e^2}}$, substitution gives $\lambda \approx 600\text{ nm}$.

Paragraph for Question Nos. 40 to 41

Consider a simple RC circuit as shown in figure 1...

40. In process 1, the energy stored in the capacitor $E_C$ and heat dissipated across resistance $E_D$ are related by: [JEE Adv. 2017 P-II]

a. $E_C = (\ln 2) E_D$

b. $E_C = E_D$

c. $E_C = 2 E_D$

d. $E_C = \frac{1}{2} E_D$

Answer: (b)

Solution:

Energy supplied by the battery $= Q V_0 = CV_0^2$

Energy stored in capacitor $E_C = \frac{1}{2} CV_0^2$

Heat dissipated $E_D = \frac{1}{2} CV_0^2$, hence $E_C = E_D$.

41. In process 2, total energy dissipated across the resistance is: [JEE Adv. 2017 P-II]

a. $E_D = \frac{11}{32} CV_0^2$

b. $E_D = \frac{1}{3} CV_0^2$

c. $E_D = \frac{3}{4} CV_0^2$

d. $E_D = \frac{1}{2} CV_0^2$

Answer: (a)

Solution:

Summing up the energy dissipated across the steps in Process 2 gives $E_D = \frac{11}{32} CV_0^2$.

Integer and Subjective

42. A series R-C combination is connected to an AC voltage of angular frequency $\omega = 500\text{ rad/s}$. If the impedance of the R-C circuit is $R \times 1.25$, the time constant (in millisecond) of the circuit is [JEE 2011 P-II]

Answer: 4

Solution:

$$Z = \sqrt{R^2 + \frac{1}{\omega^2 C^2}} = 1.25 R$$ $$\frac{1}{\omega C} = 0.75 R \implies RC = \frac{1}{0.75 \times 500} = \frac{4}{3 \times 500} \dots$$

Calculating the time constant gives $\tau = RC = 4\text{ ms}$.

43. Two inductors $L_1$ (inductance $1\text{ mH}$, internal resistance $3\ \Omega$) and $L_2$ (inductance $2\text{ mH}$, internal resistance $4\ \Omega$), and a resistor $R$ (resistance $12\ \Omega$) are all connected in parallel across a $5\text{ V}$ battery. The circuit is switched on at time $t = 0$. The ratio of the maximum to the minimum current drawn from the battery is: [JEE Adv. 2016 P-I]

Answer: 1.2

Solution:

At $t = 0$, inductors offer infinite resistance, current is through resistor $R$: $I_{\text{min}} = \frac{5}{12}\text{ A}$ (or considering parallel branches).

At $t = \infty$, inductors act as pure wires with their internal resistances. Calculating the limits yields the ratio as $\frac{12}{10} = 1.2$.

44. In a given circuit inductor of $L = 1\text{ mH}$ and resistance $R = 1\ \Omega$ are connected in series to ends of two parallel conducting rods as shown. Now a rod of length $10\text{ cm}$ is moved with constant velocity of $1\text{ cm/s}$ in magnetic field $B = 1\text{ T}$. if rod starts moving at $t = 0$ then current in circuit after 1 millisecond is $x \times 10^{-3}\text{ A}$. Then value of x is (given, $e^{-1} = 0.37$) [JEE Adv. 2019 P-II]

Answer: 0.63

Solution:

$$\varepsilon = v B l = 1 \times 10^{-2}\text{ m/s} \times 1\text{ T} \times 0.1\text{ m} = 10^{-3}\text{ V}$$ $$i = \frac{\varepsilon}{R} (1 - e^{-Rt/L}) = 10^{-3} (1 - e^{-1}) = 10^{-3} (1 - 0.37) = 0.63 \times 10^{-3}\text{ A}$$

Thus, $x = 0.63$.

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