JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev

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Q 34. An ideal capacitor of capacitance 0.2 µF is charged to a potential difference of 10 V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self-inductance 0.5 mH. The current at a time when the potential difference across the capacitor is 5 V, is:     [2018]
(1) 0.34 A
(2) 0.25 A
(3) 0.15 A
(4) 0.17 A
Ans:
(4)
Solution:

Qo = 0.2 × 10 µC = 2µC
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
Vo = 10V
Ei of capacitor = 1/2 × 0.2 µf × (10V)2
= 10µJ
Ef of capacitor = 1/2 × 0.2µf × (5V)2
= 2.5 µj
Einductor  = 7.5 µJ = 1/2 Li2
⇒ 7.5 × 10−6 = 1/2 × 0.5 × 10−3 × i2
⇒ 30 × 10−3 = i2
⇒ i = √3/10 = 0.17 A

Q 35.  A power transmission line feeds input power at 2300V to a step down transformer with its primary windings having 4000 turns giving the output power at 230V. If the current in the primary coil of the transformer is 5A and its efficiency is 90% the output current would be:    [2018]
(1) 45A
(2) 50A
(3) 20A
(4) 25A
Ans: 
A
Solution:

We know, formula
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
where, η is efficiency, Px denotes pawar of secondary transmission, Pη
denotes power of primary transmission
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
so, 0.9 = 230 x Ix / 2300 x 5
hence, Ix = 45A
therefore, output current would be 45A

Q 36.  A plane electromagnetic wave of wavelength λ has an intensity I. It is propagating along the positive Y-direction. The allowed expressions for the electric and magnetic fields are given by:    [2018]
(1)
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev

(2)
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev

(3)
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev

(4)
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
Ans:
A
Solution:
If E0 is magnitude of electric field then 1/2ε0E2 x C = I
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
direction of  JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev

Q 37. A charge q is spread uniformly over an insulated loop of radius r. If it is rotated with an angular velocity ω with respect to normal axis then magnetic moment of the loop is:    [2018]
(1) JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev

(2) JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
(3) qωr2
(4) JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
Ans:
(2)
Solution:
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev

Q 38. A coil of cross-sectional area A having N turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity ω, the maximum e.m.f. induced in the coil will be:    [2018]
(1) 3/2NBAω 
(2) 3NBAω
(3) NBAω
(4) 1/2NBAω
Ans: 
(3)
Solution:
Given, a coil of N turns having an area A, rotated with angular speed 'ω' in a uniform magnetic field 'B' connected to a resistor ' R'.
The flux linking to coil, φ = NBA sin (ωt)
Therefore,
Induced EMF,
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
When, sinωt = ±1, the induced emf value is maximum.
∴ the maximum EMF, E0 = NBAω

Q 39. In a coil of resistance 100 Ω, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is    [2017]
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
(1) 250 Wb
(2) 275 Wb
(3) 200 Wb
(4) 225 Wb
Ans
: (1)
Solution:
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
Magnitude of change in flux = R × area under current vs time graph
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev

Q 40. A small circular loop of wire of radius a is located at the centre of a much larger circular wire loop of radius b. The two loops are in the same plane. The outer loop of radius b carries an alternating current I = I0 cos (ωt). The emf induced in the smaller inner loop is nearly ?    [2017]
(1)JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
(2) JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
(3)JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
(4)JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
Ans:
(4)
Solution:
Magnetic field due to outer current loop 
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
Induced emf in inner loop 
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev

Q 41. Magnetic field in a plane electromagnetic wave is given by
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
Expression for corresponding electric field will be    [2017]
(1)JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
(2)JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
(3)JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
(4)JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
Ans: (3)
Solution:
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev

Q 42. A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R = 5 W, L = 25 μH and C = 1000 μF. The total impedance, and phase difference between the voltage across the source and the current will respectively be -    [2017]
(1) 10 Ω and tan-1(5/3)
(2) 7 Ω and 45°
(3) 7 Ω and tan-1(5/3)
(4) 10 Ω and tan-1(8/3)
Ans:
(2)
Solution:
e0 = 283 volt    ω = 320
xL = 320 × 25 × 10-3 = 8 Ω
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev

Q 43. The electric field component of a monochromatic radiation is given by JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
Its magnetic field JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
 is then given by:   [2017]
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
Ans:
(1)
Solution:
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev

Q 44. Hysteresis loops for two magnetic materials A and B are given below :
These materials are used to make magnets for electric generators, transformer core and electromagnet core.    [2016]
Then it is proper to use :
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
(1) A for electric generators and transformers
(2) A for electromagnets and B for electric generators
(3) A for transformers and B for electric generators
(4) B for electromagnets and transformers
Ans:
(4)
Conceptual (Requires low retentivity and low coercivity)

Q 45. An arc lamp requires a direct current of 10 A at 80 V to function. It is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to:    [2016]
(1) 80 H
(2) 0.08 H
(3) 0.044 H
(4) 0.065 H
Ans: 
(4)
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev

Q 46. Arrange the following electromagnetic radiations per quantum in the order of increasing energy :    [2016]
A: Blue light
B: Yellow light
C: X-ray
D: Radiowave
(1) D, B, A, C
(2) A, B, D, C
(3) C, A, B, D
(4) B, A, D, C
Ans:
(1)
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
radio wave<yellow light<blue light<X-rays
(Increasing order of energy)

Q 47. A series LR circuit is connected to a voltage source with V(t) = V0 sinΩt. After very large time current I(t) behaves as JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev    [2016]
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
Ans: (3)
Solution:
Current will be in the form of
I = I0 sin (Ωt - φ)
Graph will be sinusoidal

Q 48. The ratio (R) of output resistance r0, and the input resistance ri in measurements of input and output characteristics of a transistor is typically in the range:     [2016]
(1) R~0.1 –0.01 
(2) R~0.1 –1.0 
(3) R~102 –103 
(4) R~1 –10      
Ans: 
(3)
Solution:
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
The small signal model of a BJT is shown in the figure. From this figure, Rout = ro since the voltage across B and E will be 0 and hence current will only flow through r0.
Input resistance will be Rin = rπ since current will flow only through rπ
Hence
JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
Since r0 is very large and of the order of 100 kΩ and rπ  is of the order of 1- 10 kΩ, the ratio R will be of the order of 100 - 1000.

Q 49. Consider an electromagnetic wave propagating in vacuum. Choose the correct statement:
(1) For an electromagnetic wave propagating in +y direction the electric field is JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev and the magnetic field is JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev

(2) For an electromagnetic wave propagating in +y direction the electric field is JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev and he magnetic field is JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
(3) For an electromagnetic wave propagating in +x direction the electric field is JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev and the magnetic field is JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev
(4) For an electromagnetic wave propagating in +x direction the electric field is JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev and the magnetic field is JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev     [2016]
Ans: (4)
Solution:
If wave is propagating in x direction, JEE Main Previous year questions (2016-20): Electromagnetic Induction & Alternating Current- 2 JEE Notes | EduRev must be functions of (x, t) & must be in y- z plane.

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