Test: Electromagnetic Induction & Alternating Current (February 9) - NEET MCQ

# Test: Electromagnetic Induction & Alternating Current (February 9) - NEET MCQ

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## 15 Questions MCQ Test Daily Test for NEET Preparation - Test: Electromagnetic Induction & Alternating Current (February 9)

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Test: Electromagnetic Induction & Alternating Current (February 9) - Question 1

### There are two stationary coils near each other. If current in coil-1 is changing with time, relationship of current in coil-2 to it's emf is described by

Detailed Solution for Test: Electromagnetic Induction & Alternating Current (February 9) - Question 1

M is mutual inductance

Test: Electromagnetic Induction & Alternating Current (February 9) - Question 2

### The domestic power supply is at 220 volt. The amplitude of emf will be

Detailed Solution for Test: Electromagnetic Induction & Alternating Current (February 9) - Question 2

We know: Vrms​=​Vo​​/√2

Substituting values (Here RMS Voltage is 220 V)
We get: Maximum V (or amplitude) =311V

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Test: Electromagnetic Induction & Alternating Current (February 9) - Question 3

### The instantaneous value of current in an ac circuit is I = 2 sin (100pt + p/3) A. The current at the beginning (t = 0) will be

Detailed Solution for Test: Electromagnetic Induction & Alternating Current (February 9) - Question 3

Instantaneous current i=2sin(100πt+π/3)A
At (t=0),  i(0)=2sin(π/3)=2×√3/2​​=√3​A
The current at the beginning will be √3​ A

Test: Electromagnetic Induction & Alternating Current (February 9) - Question 4

A resistance of 50W, an inductance of 20/p henry and a capacitor of 5/p mF are connected in series with an A.C. source of 230 volt and 50Hz. The impedance of circuit is

Detailed Solution for Test: Electromagnetic Induction & Alternating Current (February 9) - Question 4

Given, R=50W, L=(20/π)H and C=(5/π)mF
Voltage supply, V=230V.
v=50Hz
Since, XL= ω=2πvL=2πx50x(20/π)=XL=2000 Ω
And, XC= 1/ωC=1/2πvL=1/[2πx50x(5/π)x10-6]=(1000/500) x103
So, impedance, Z=√[(XL-XC)2+R2]
Z=√[(2000-2000)2+502]
Z=50Ω
Hence, Option B is correct.

Test: Electromagnetic Induction & Alternating Current (February 9) - Question 5

The inductive reactance of a coil is 1000W. If its self inductance and frequency both are increased two times then inductive reactance will be

Detailed Solution for Test: Electromagnetic Induction & Alternating Current (February 9) - Question 5

We know that,
XL= ωL
Or, XL =2πfL   [ ω can be written as, 2πf]
Initially,
XL1=2πf1L1=1000Ω
For the second case,
F  and L both are doubled, so,
XL2 =2π2f12L1 =4x2πf1L1
=4000 Ω

Test: Electromagnetic Induction & Alternating Current (February 9) - Question 6

Predict the polarity of the capacitor in the situation described by fig.

Detailed Solution for Test: Electromagnetic Induction & Alternating Current (February 9) - Question 6

Explanation:A will become positive with respect to B because current  indused is in clockwise direction

Test: Electromagnetic Induction & Alternating Current (February 9) - Question 7

An ac generator consists of 8 turns of wire, each of area A=0.0900m2 , and the total resistance of the wire is 12.0Ω. The loop rotates in a 0.500-T magnetic field at a constant frequency of 60.0 Hz. Maximum induced emf is

Detailed Solution for Test: Electromagnetic Induction & Alternating Current (February 9) - Question 7

e = NBAω = NBA2πn = 8 x 0.5 x 0.09 x 2 x 3.14 x 60 = 136V

Test: Electromagnetic Induction & Alternating Current (February 9) - Question 8

Coils C1 and C2 are stationary. C1 connected in series with one very small diwali bulb while C2 is connected to a battery .If the connection is suddenly snapped

Detailed Solution for Test: Electromagnetic Induction & Alternating Current (February 9) - Question 8

Explanation:Flux will be changed due to this current is induced momentarily

Test: Electromagnetic Induction & Alternating Current (February 9) - Question 9

A coil of inductance 0.1 H is connected to an alternating voltage generator of voltage E = 100 sin (100t) volt. The current flowing through the coil will be

Detailed Solution for Test: Electromagnetic Induction & Alternating Current (February 9) - Question 9

L =0.14
E=100sin(100t)
⇒XL​=wL​=100×0.1
=10Ω
⇒io​=Eo/XL​=100/10​=10A
⇒i=io​sin(100t− π/2​)
⇒i=−io​sin[(π/2)​−100]
So we get
⇒i=−10cos(100t)A
therefore, option D is correct.

Test: Electromagnetic Induction & Alternating Current (February 9) - Question 10

The emf and the current in a circuit are E = 12 sin (100pt) ; I = 4 sin (100pt + p / 3) then

Test: Electromagnetic Induction & Alternating Current (February 9) - Question 11

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, estimate of the self-inductance of the circuit is

Detailed Solution for Test: Electromagnetic Induction & Alternating Current (February 9) - Question 11

Test: Electromagnetic Induction & Alternating Current (February 9) - Question 12

If AC generator coil has area A, rotates in a magnetic field of B with an angular speed ω , the voltage generated is

Detailed Solution for Test: Electromagnetic Induction & Alternating Current (February 9) - Question 12

Explanation:e=NBAωSinωt

Test: Electromagnetic Induction & Alternating Current (February 9) - Question 13

If instantaneous value of current is I = 10 sin (314 t) A, then the average current for the half cycle will be

Detailed Solution for Test: Electromagnetic Induction & Alternating Current (February 9) - Question 13

I=10sin(314t)
dt      [ω=314, T=2π/​ω=2π/​314, T=π/​157]

Iavg=(2/T)x10 [-cos314/314]T/2
=-20/(Tx314)[cos{314x(π/2x157)}-cos0]
=20/3.14
=6.37A

Test: Electromagnetic Induction & Alternating Current (February 9) - Question 14

A metallic rod of 1 m length is rotated with a frequency of 50 rev/s about an axis passing through the centre point O. Other end of the metallic rod slides on a Metallic ring . A constant and uniform magnetic field of 2 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?

Detailed Solution for Test: Electromagnetic Induction & Alternating Current (February 9) - Question 14

l = 1 m

= 314V

Test: Electromagnetic Induction & Alternating Current (February 9) - Question 15

RMS value of ac i = i1 cos wt + i2 sin wt will be

Detailed Solution for Test: Electromagnetic Induction & Alternating Current (February 9) - Question 15

I=(I1)cosωt+(I2)sinωt
(I2)mean=I12cos2ωt+I22sin2ωt+I1I2cosωtsinωt
I12×(1/2)×i22×(1/2)+2I1I2×0
Irms=√(i2)mean=√(I12+I22/2)=(1/√2)( I12+I22)1/2

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