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# Test: 35 Year JEE Previous Year Questions: Electromagnetic Induction & Alternating Current

## 51 Questions MCQ Test Physics For JEE | Test: 35 Year JEE Previous Year Questions: Electromagnetic Induction & Alternating Current

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This mock test of Test: 35 Year JEE Previous Year Questions: Electromagnetic Induction & Alternating Current for JEE helps you for every JEE entrance exam. This contains 51 Multiple Choice Questions for JEE Test: 35 Year JEE Previous Year Questions: Electromagnetic Induction & Alternating Current (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: 35 Year JEE Previous Year Questions: Electromagnetic Induction & Alternating Current quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: 35 Year JEE Previous Year Questions: Electromagnetic Induction & Alternating Current exercise for a better result in the exam. You can find other Test: 35 Year JEE Previous Year Questions: Electromagnetic Induction & Alternating Current extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

### In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch S1. Also when S1 is opened and S2 is closed the capacitor is connected in series with inductor (L).       Q. At the start, the capacitor was uncharged. When switch S1 is closed and S2 is kept open, the time constant of this circuit is τ. Which of the following is correct

Solution:

For charging of  R – C circuit, Q = Q0 [1 – e–t/t] when the charging is complete, the potential difference between the capacitor plates will be V. The charge stored in this case will be maximum.
Therefore, Q0 = CV.

QUESTION: 2

### In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch S1. Also when S1 is opened and S2 is closed the capacitor is connected in series with inductor (L). Q. When the capacitor gets charged completely, S1 is opened and S2 is closed. Then,

Solution:

The instantaneous charge on plates at any time t during discharging is

Q = Q0 cos ωt

∴ Instantaneous current,

∴ The magnitude of maximum current

QUESTION: 3

### In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch S1. Also when S1 is opened and S2 is closed the capacitor is connected in series with inductor (L). Q. Given that the total charge stored in the LC circuit is Q0, for t > 0, the charge on the capacitor is

Solution:

Apply Kirchhoff 's law in the circuit

QUESTION: 4

A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumers' usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method , a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with power factor unity. All the currents and voltages mentioned are rms values.

Q. If the direct transmission method with a cable of resistance 0.4 Ω km–1 is used, the power dissipation| (in %) during transmission is

Solution:

(a) is the correct option.

QUESTION: 5

A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumers' usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method , a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with power factor unity. All the currents and voltages mentioned are rms values.

Q. In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1 : 10. If the power to the consumers has to be supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in the stepdown transformer is

Solution:

We know that P = V × I

∴ I = 150 A
Total resistance = 0.4 × 20 = 8 Ω

∴ Power dissipated as heat = I2R = (150)2 × 8

=  180,000W = 180 kW

(b) is the correct option.

QUESTION: 6

A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity w. This can be considered as equivalent to a loop carrying a steady current  A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant γ.

Q. The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is

Solution:

= -πR2B

∴ E × 2πR = – πR2B

(b) is the correct option.

QUESTION: 7

A point charge Q is moving in a circular orbit of radius R in the x-y plane with an angular velocity w. This can be considered as equivalent to a loop carrying a steady current  A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 0 to B in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant γ.

Q. The change in the magnetic dipole moment associated with the orbit, at the end of the time interval of the magnetic field change, is

Solution:

Given M = γ L
∴ M = γ mωR2
∴ M = γ m (Δω) R2 ...(1)

...(2)

The negative sign shows that change is opposite to the direction of B. (b) is the correct option.

QUESTION: 8

Statement-1 : A vertical iron rod has coil of wire wound over it at the bottom end. An alternating current flows in the coil. The rod goes through a conducting ring as shown in the figure. The ring can float at a certain height above the coil.
Statement-2 : In the above situation, a current is induced in the ring which interacts with the horizontal component of the magnetic field to produce an average force in the upward direction.

Solution:

As shown in the figure the horizontal component of the magnetic field interacts with the induced current produced in the conducting ring which produces an average force in the upward direction. (Fleming's left hand rule).

QUESTION: 9

The power factor of an AC circuit having resistance (R) and inductance (L) connected in series and an angular velocity ω is

Solution:

(b) The impedance triangle for resistance (R) and inductor

(L) connected in series is shown in the figure.

QUESTION: 10

A conducting square loop of side L and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A magnetic induction B constant in time and space, pointing perpendicular and into the plane at the loop exists everywhere with half the loop outside the field, as shown in figure. The induced emf is

Solution:

The induced emf is

QUESTION: 11

The inductance between A and D is

Solution:

These three inductors are connected in parallel. The equivalent inductance Lp is given by

QUESTION: 12

In a transformer, number of turns in the primary coil are 140 and that in the secondary coil are 280. If current in primary coil is 4 A, then that in the secondary coil is

Solution:

Np = 140, Ns = 280, Ip = 4A, Is = ?

QUESTION: 13

Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon

Solution:

Mutual conductance depends on the relative position and orientation of the two coils.

QUESTION: 14

When the current changes from +2 A to -2A in 0.05 second, an e.m.f. of 8 V  is induced in a coil. The coefficient of self -induction of the coil is

Solution:

QUESTION: 15

In an oscillating  LC circuit the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is

Solution:

When the capacitor is completely charged, the total energy in the L.C circuit is with the capacitor and that energy is

When half energy is with the capacitor in the form of electric field between the plates of the capacitor we get  where Q ' is the charge on one plate of the capacitor

QUESTION: 16

The core of any transformer is laminated so as to

Solution:

Laminated core provide less area of cross-section for the current to flow. Because of this, resistance of the core increases and current decreases thereby decreasing the eddy current losses.

QUESTION: 17

Alternating current can not be measured by D.C. ammeter because

Solution:

D.C. ammeter measure average current in AC current, average current is zero for complete cycle. Hence reading will be zero.

QUESTION: 18

In an LCR series a.c. circuit, the voltage across each of the components, L, C and R is 50V. The voltage across the LC combination will be

Solution:

Since the phase difference between L & C is π,

∴ net voltage difference across LC = 50 – 50 = 0

QUESTION: 19

A coil having n turns and resistance RΩ is connected with a galvanometer of resistance 4RΩ. This combination is moved in time t seconds from a magnetic field W1 weber to W2 weber. The induced current in the circuit is

Solution:

(∴ W2 &W1 are magnetic flux)

QUESTION: 20

In a uniform magnetic field of induction B a wire in the form of a semicircle of radius r rotates about the diameter of the circle with an angular frequency ω. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R, the mean power generated per period of rotation is

Solution:

QUESTION: 21

In a LCR circuit capacitance is changed from C to 2 C. For the resonant frequency to remain unchanged, the inductance should be changed from L to

Solution:

For resonant frequency to remain same LC should be const.  LC = const

QUESTION: 22

A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth’s magnetic field is 0.2×10–4T, then the e.m.f. developed between the two ends of the conductor is

Solution:

QUESTION: 23

One conducting U tube can slide inside another as shown in figure, maintaining electrical contacts between the tubes.
The magnetic field B is perpendicular to the plane of the figure . If each tube moves towards the  other at a constant speed v, then the emf induced in the circuit in terms of B, l and v where l is the width of each tube, will be

Solution:

Relative velocity = v + v = 2v

∴ emf. = B.l (2v)

QUESTION: 24

The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of

Solution:

For maximum power, X L =XC , which yields

QUESTION: 25

The phase difference between the alternating current and emf is π/2. Which of the following cannot be the constituent of the circuit?

Solution:

(a) Phase difference for R–L circuit lies between

QUESTION: 26

A circuit has a resistance of 12 ohm and an impedance of 15 ohm. The power factor of the circuit will be

Solution:

QUESTION: 27

A coil of inductance 300 mH and resistance 2 W is connected to a source of voltage 2 Ω The current reaches half of its steady state value in

Solution:

KEY CONCEPT : The charging of inductance given

Taking log on both the sides,

QUESTION: 28

Which of the following units denotes the dimension  where Q denotes the electric charge?

Solution:

QUESTION: 29

In a series resonant LCR circuit, the voltage across R is 100 volts and R = 1 kΩ with C = 2μF. The resonant frequency w is 200 rad/s. At resonance the voltage across L is

Solution:

QUESTION: 30

In an AC generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency ω in a magnetic field B. The maximum value of emf generated in the coil is

Solution:

QUESTION: 31

The flux linked with a coil at any instant 't' is given by

φ = 10t2 - 50t + 250

The induced emf at t = 3s is

Solution:

Growth in current in LR2 branch when switch is closed is given by

Hence, potential drop across

QUESTION: 32

An inductor (L = 100 mH), a resistor (R = 100 Ω) and a battery (E = 100 V) are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points A and B. The current in the circuit 1 ms after the short circuit is

Solution:

Initially, when steady state is achieved,

Let E is short circuited at t = 0. Then

Let during decay of current at any time the current flowing is

QUESTION: 33

In an a.c. circuit the voltage applied is E = E0 sin ωt. The resulting current in the circuit is  The power consumption in the circuit is given by

Solution:

KEY CONCEPT :  We know that power consumed in a.c. circuit is given by,

which implies that the phase difference,

QUESTION: 34

An ideal coil of 10H is connected in series with a resistance of 5Ω and a battery of 5V. 2 second after the connection is made, the current flowing in ampere in the circuit is

Solution:

(When current is in growth in LR circuit)

QUESTION: 35

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A = 10 cm2 and length = 20 cm. If one of the solenoid has 300 turns and the other 400 turns, their mutual inductance is

0 = 4π × 10 –7 Tm A–1)

Solution:

QUESTION: 36

An inductor of inductance L = 400 mH and resistors of resistance R1 = 2W and R2 = 2W are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is

Solution:

Growth in current in LR2 branch when switch is closed is given by

Hence, potential drop across

QUESTION: 37

A rectangular loop has a sliding connector PQ of length l and resistance RΩ and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I1, I2 and I are

Solution:

Due to the movement of resistor R, an emf equal to Blv will be induced in it as shown in figure clearly,

I = I1+I2
Also, I1 = I2
Solving the circuit, we get

QUESTION: 38

In the circuit shown below, the key K is closed at t = 0. The current through the battery is

Solution:

At t = 0, no current will flow through L and R1

∴ Current through battery

At t = ∞,

effective resistance,

QUESTION: 39

In a series LCR circuit R = 200Ω and the voltage and the frequency of the main supply is 220V and 50 Hz respectively.
On taking out the capacitance from the circuit the current lags behind the voltage by 30°. On taking out the inductor from the circuit the current leads the voltage by 30°. The power dissipated in the LCR circuit is

Solution:

When capacitance is taken out, the circuit is LR.

Again , when inductor is taken out, the circuit is CR.

Power dissipated = Vrms Irms cos φ

QUESTION: 40

A boat is moving due east in a region where the earth's magnetic field is 5.0 × 10–5 NA–1 m–1 due north and horizontal.
The boat carries a vertical aerial 2 m long. If the speed of the boat is 1.50 ms–1, the magnitude of the induced emf in the wire of aerial is:

Solution:

Induced emf = vBH l = 1.5 × 5 × 10–5 × 2 = 15 × 10–5
= 0.15 mV

QUESTION: 41

A fully charged capacitor C with initial charge q0 is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic fields is:

Solution:

Energy stored in magnetic field

Energy stored in electric field

QUESTION: 42

A resistor ‘R’ and 2µF capacitor in series is connected through a switch to 200 V direct supply. Across the capacitor is a neon bulb that lights up at 120 V. Calculate the value of R to make the bulb light up 5 s after the switch has been closed. (log10 2.5 = 0.4)

Solution:

We have, V = V0 (1 – e–t/RC)
⇒ 120 = 200 (1 - e-t / RC)
⇒ t = RC in (2.5)

⇒ R = 2.71 × 106 W

QUESTION: 43

A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; It is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to :

Solution:

Because of the Lenz's law of conservation of energy.

QUESTION: 44

A metallic rod of length ‘l’ is tied to a string of length 2l and made to rotate with angular speed w on a horizontal table with one end of the string fixed. If there is a vertical magnetic field ‘B’ in the region, the e.m.f. induced across the ends of the rod is

Solution:

Here, induced e.m.f.

QUESTION: 45

A circular loop of radius 0.3 cm lies parallel to amuch bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is

Solution:

As we know, Magnetic flux, φ = B.A

QUESTION: 46

In an LCR circuit as shown below both switches are open initially. Now switch S1 is closed, S2 kept open. (q is charge on the capacitor and τ = RC is Capacitive time constant).
Which of the following statement is correct ?

Solution:

Charge on he capacitor at any time t is given by

q = CV (1– et/τ)

at t = 2τ

q = CV (1 – e–2)

QUESTION: 47

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:

Solution:

Applying Kirchhoff's law of voltage in closed loop

QUESTION: 48

An inductor (L = 0.03 H) and a resistor (R = 0.15 kΩ) are connected in series to a battery of 15V EMF in a circuit shown below. The key K1 has been kept closed for a long time. Then at t = 0, K1 is opened and key K2 is closed simultaneously. At t = l ms, the current in the circuit will be :

Solution:

QUESTION: 49

An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q0 and then connected to the L and R as shown below :

If a student plots graphs of the square of maximum charge  on the capacitor with time(t) for two different values L1 and L2 (L1 > L,) of L then which of the following represents this graph correctly ? (plots are schematic and not drawn to scale)

Solution:

From KVL at any time t

From damped harmonic oscillator, the amplitude is given by A

Hence damping will be faster for lesser self inductance.

QUESTION: 50

Two coaxial solenoids of different radius carry current I in the same direction.  be the magnetic force on the inner solenoid due to the outer one and  be the magnetic force on the outer solenoid due to the inner one. Then :

Solution:

because of action and reaction pair

QUESTION: 51

An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to :

Solution: