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Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - JEE MCQ


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30 Questions MCQ Test - Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current

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Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 1

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.


Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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,

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 2

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

Q = Q0 cos ωt

∴ Instantaneous current,

∴ The magnitude of maximum current

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Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 3

Apply Kirchhoff 's law in the circuit

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 4


(a) is the correct option.

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 5

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.

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 6

= -πR2B

∴ E × 2πR = – πR2B

(b) is the correct option.

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 7

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.

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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.

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 8

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).

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 9

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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 9

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

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

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 10

The induced emf is

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 11

The inductance between A and D is

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 11

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

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 12

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

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 13

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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 13

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

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 14

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 15

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

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 16

The core of any transformer is laminated so as to 

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 16

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.

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 17

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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 17

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

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 18

Since the phase difference between L & C is π,

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

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 19

(∴ W2 &W1 are magnetic flux)

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 20


Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 21

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

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 22

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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   

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 23

Relative velocity = v + v = 2v

∴ emf. = B.l (2v)

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 24

For maximum power, X L =XC , which yields

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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?

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 25

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

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 26

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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 26

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 27

KEY CONCEPT : The charging of inductance given

Taking log on both the sides,

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 28

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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 28

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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 

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 29

Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - 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

Detailed Solution for Test: JEE Main 35 Year PYQs- Electromagnetic Induction & Alternating Current - Question 30

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