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A copper rod of length l is rotated about one end perpendicular to the magnetic field B with constant angular velocity ω. The induced e.m.f. between the two ends is
1. $\frac{1}{2} B ω l^{2}$
2. $\frac{3}{4} B ω l^{2}$
3. $B ω l^{2}$
4. $2 B ω l^{2}$

A uniform but time-varying magnetic field B(t) exists in a circular region of radius a and is directed into the plane of the paper, as shown. The magnitude of the induced electric field at point P at a distance r from the centre of the circular region

1. Is zero
2. Decreases as $\frac{1}{r}$
3. Increases as r
4. Decreases as $\frac{1}{r^{2}}$

As shown in the figure, P and Q are two coaxial conducting loops separated by some distance. When the switch S is closed, a clockwise current I_P flows in P (as seen by E) and an induced current $I_{Q_{1}}$ flows in Q. The switch remains closed for a long time. When S is opened, a current $I_{Q_{2}}$ flows in Q. Then the directions of $I_{Q_{1}}$ and $I_{Q_{2}}$ (as seen by E) are

1. Respectively clockwise and anticlockwise
2. Both clockwise
3. Both anticlockwise
4. Respectively anticlockwise and clockwise

A square metallic wire loop of side $0.1~\text m$ and resistance of $1~\Omega$ is moved with a constant velocity in a magnetic field of $2~\text{wb/m}^2$ as shown in the figure. The magnetic field is perpendicular to the plane of the loop and the loop is connected to a network of resistances. What should be the velocity of the loop so as to have a steady current of $1~\text{mA}$ in the loop?

1. $1~\text{cm/s}$
2. $2~\text{cm/s}$
3. $3~\text{cm/s}$
4. $4~\text{cm/s}$

Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction B = B₀e^–t is established inside the coil. If the key (K) is closed, the electrical power developed right after closing the switch, at t=0, is equal to

1. $\frac{B_{0}^{2} π r^{2}}{R}$
2. $\frac{B_{0} 10 r^{3}}{R}$
3. $\frac{B_{0}^{2} π^{2} r^{4} R}{5}$
4. $\frac{B_{0}^{2} π^{2} r^{4}}{R}$

A rectangular loop with a sliding connector of length $l= 1.0$ m is situated in a uniform magnetic field $B = 2T$ perpendicular to the plane of the loop. Resistance of connector is $r=2~\Omega$. Two resistances of $6~\Omega$ and $3~\Omega$ are connected as shown in the figure. The external force required to keep the connector moving with a constant velocity $v = 2$ m/s is:

1. $6~\text{N}$
2. $4~\text{N}$
3. $2~\text{N}$
4. $1~\text{N}$

A wire cd of length l and mass m is sliding without friction on conducting rails ax and by as shown. The vertical rails are connected to each other with a resistance R between a and b. A uniform magnetic field B is applied perpendicular to the plane abcd such that cd moves with a constant velocity of

1. $\frac{m g R}{B l}$
2. $\frac{m g R}{B^{2} l^{2}}$
3. $\frac{m g R}{B^{3} l^{3}}$
4. $\frac{m g R}{B^{2} l}$

A conducting rod AC of length 4l is rotated about a point O in a uniform magnetic field $\vec{B}$ directed into the paper. AO = l and OC = 3l. Then

1. $V_{A} - V_{O} = \frac{B ω l^{2}}{2}$
2. $V_{O} - V_{C} = \frac{7}{2} B ω l^{2}$
3. $V_{A} - V_{C} = 4 B ω l^{2}$
4. $V_{C} - V_{O} = \frac{9}{2} B ω l^{2}$

The network shown in the figure is a part of a complete circuit. If at a certain instant the current i is 5 A and is decreasing at the rate of 10³ A/s then V_B – V_A is

1. 5 V
2. 10 V
3. 15 V
4. 20 V

10.

A simple pendulum with bob of mass m and conducting wire of length L swings under gravity through an angle 2θ. The earth’s magnetic field component in the direction perpendicular to swing is B. Maximum potential difference induced across the pendulum is

1. $2 B L \sin (\frac{θ}{2}) {(g L)}^{1 / 2}$
2. $B L \sin (\frac{θ}{2}) {(g L)}^{\frac{1}{2}}$
3. $B L \sin (\frac{θ}{2}) {(g L)}^{3 / 2}$
4. $B L \sin (\frac{θ}{2}) {(g L)}^{2}$

11.

An alternating current is given as $i = i_1 \cos\omega t+i_2\sin\omega t$. The RMS current is given by;
1. $\dfrac{i_1+i_2}{\sqrt{2}}$
2. $\dfrac{(i_1+i_2)^2}{\sqrt{2}}$
3. $\sqrt{\dfrac{i_1^2+i^2_2}{2}}$
4. $\dfrac{\sqrt{i_1^2+i^2_2}}{2}$

12.

In a step-up transformer, the turn ratio is 1:2. A Leclanche cell (e.m.f. 1.5V) is connected across the primary coil. The voltage developed in the secondary coil would be-
1. 3.0 V
2. 0.75 V
3. 1.5 V
4. Zero

13.

The peak value of an alternating e.m.f. E is given by $E = E_{0} \cos ω t$ is 10 volts and its frequency is 50 Hz. At time $t = \frac{1}{600} s e c$ , the instantaneous e.m.f. is
1. 10 V
2. $5 \sqrt{3} V$
3. 5 V
4. 1 V

14.

If a current I given by $I_{0} \sin (ω t - \frac{π}{2})$ flows in an ac circuit across which an ac potential of $E = E_{0} \sin ω t$ has been applied, then the power consumption P in the circuit will be
1. $P = \frac{E_{0} I_{0}}{\sqrt{2}}$
2. $P = \sqrt{2} E_{0} I_{0}$
3. $P = \frac{E_{0} I_{0}}{2}$
4. P = 0

15. A resistance of $20~ \Omega$ is connected to a source of an alternating potential, $V=220\sin(100 \pi t).$ The time taken by the current to change from its peak value to its rms value will be:

1.	$ 0.2~\text{sec}$	2.	$ 0.25~\text{sec}$
3.	$25 \times10^{-3}~\text{sec}$	4.	$2.5 \times10^{-3}~\text{sec}$

16.

In a LCR circuit having L = 8.0 henry, C = 0.5 μF and R = 100 ohm in series. The resonance frequency in radian per second is
1. 600 radian/second
2. 600 Hz
3. 500 radian/second
4. 500 Hz

17.

In a series LCR circuit, resistance R = 10Ω and the impedance Z = 20Ω. The phase difference between the current and the voltage is
1. 30°
2. 45°
3. 60°
4. 90°

18.

In the circuit shown below, the AC source has voltage $V = 20\cos(\omega t)$ volts with $\omega =2000$ rad/sec. The amplitude of the current is closest to:

1. $2$ A
2. $3.3$ A
3. $\frac{2}{\sqrt{5}}$ A
4. $\sqrt{5}~\text{A}$

19.

When an AC source of emf $e = E_0 \sin (100t)$ is connected across a circuit, the phase difference between the emf $e$ and the current $i$ in the circuit is observed to be $\frac{\pi}{4}$ as shown in the diagram. If the circuit consists only of $RC$ or $LC$ in series, then what is the relationship between the two elements?

1.	$R=1~\text{k} \Omega, C=10 ~\mu \text{F}$
2.	$R=1~\text{k}\Omega, C=1~\mu \text{F}$
3.	$R=1 ~\text{k}\Omega, L=10 ~\text{H}$
4.	$R=1 ~\text{k}\Omega, L=1~\text{H}$

20.

In an electrical circuit R, L, C, and an AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and the current in the circuit is $π / 3 .$ If instead, C is removed from the circuit, the phase difference is again $π / 3 .$ The power factor of the circuit is
1. 1/2
2. 1/ $\sqrt{2}$
3. 1
4. $\sqrt{3} / 2$

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1.	\( 0.2~\text{sec}\)	2.	\( 0.25~\text{sec}\)
3.	\(25 \times10^{-3}~\text{sec}\)	4.	\(2.5 \times10^{-3}~\text{sec}\)

1.	\(R=1~\text{k} \Omega, C=10 ~\mu \text{F}\)
2.	\(R=1~\text{k}\Omega, C=1~\mu \text{F}\)
3.	\(R=1 ~\text{k}\Omega, L=10 ~\text{H}\)
4.	\(R=1 ~\text{k}\Omega, L=1~\text{H}\)

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