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Physics NEET MCQ
Electro-Magentic Induction And Alternating Currents Mcq
Quiz 4
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Q.1
A step-up transformer operates on a 230 volt line and supplies to the load 2amp. The ratio of primary and secondary winding is 1:Determine the primary current [ JIMPER 1998]
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a) 12.5 amp
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b) 50 amp
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c) 8.8 amp
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d) 25.0 amp
Explanation
Answer:(b)
Q.2
Self inductance of solenoid is [ MPPMT 1993]
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a) Directly proportional to current flowing through coil
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b) Directly proportional to the length
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c) Directly proportional to its area of cross-section
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d) Inversely proportional to the area of cross section
Explanation
Answer: (c)
Q.3
The current passing through a chock coil of 5 henry is decreasing at the rate of 2 amp/sec. The emf developed across the coil is [ MPPMT 1990]
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a) -2.5 volt
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b) -10.0 volt
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c) +10 volt
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d) +2.5 volt
Explanation
Use formula E = -L(dI/dt) Here dI/dt = -2 amp/second as current is decreasing E = -5 × (-2) = 10V Answer: (c)
Q.4
An aeroplane in which the distance between the tips of the wings is 50 meters is flying horizontally with a speed of 360 km/hr over a place where the vertical component of the earth's magnetic field is 2.0×10-4 weber/metreThe potential difference between the tips of the wings would be [ CPMT 1990]
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a) 0.1 volt
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b) 1.0 Volt
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c) 0.2 Volt
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d) 0.01Volt
Explanation
From the formula E = Blv Here B = earths vertical magnetic component =2.0×10-4 weber/metre2 velocity v = 360 km/hr = 100 m/s l = length of the wing = 50 on substituting and solving we get E = 1.0 Volt Answer: (b)
Q.5
A coil of 40Ω resistance has 100 turns and radius 6mm is connected to ammeter of resistance of 160Ω. Coil is placed perpendicular to the magnetic field. When coil is taken out of the field, 32µC charge flows through it. The intensity of magnetic field will be [ Raj.PET 1997]
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a) 6.55 T
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b) 5.66T
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c) 0.655 T
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d) 0.566T
Explanation
Answer:(d)
Q.6
A transformer is employed to [ MPPMT 1988]
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a) obtain a suitable D.C voltage
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b) Convert D.D into A.C
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c) Obtain a suitable A.C. voltage
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d) convert A.C. into D.C
Explanation
Answer: (c)
Q.7
A coil of area 100cm2 has 500 turns. Magnetic field of 0.1 weber/m2 is perpendicular to the coil. The field is reduced to zero in 0.1 second. The induced emf in the coil is [ MPPMT 1991]
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a) 1 volt
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b) 5 volts
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c) 50 volts
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d) zero
Explanation
E = -N (dφ/dt) Here dφ/dt = AdB/dt = -10-2×1 E = -500×( -10-2) E = 5V Answer: (b)
Q.8
The armature current in a D.C. motor is maximum when [ PMT 1986]
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a) the motor has picked up maximum speed
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b) the motor has just started moving
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c) the motor has intermediate speed
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d) the motor has just been switched off
Explanation
Answer: (b)
Q.9
A 50 mH coil carries a current of 4 amp. The energy stored un joules is [ MPPMT 1999]
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a) 0.4 J
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b) 4.0 J
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c) 0.8 J
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d) 0.04J
Explanation
USe formula U = ½ LI2 Answer:(a)
Q.10
a magnet is moving with fast speed towards the coil at rest. Due to this, the induced electromotive force, the induced current and the induced charge in the coil are E, I and Q respectively. If the speed of the magnet is doubled, the incorrect statement is [ MPPMT 1995]
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a) E increases
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b) I increases
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c) Q remains constant
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d) Q increases
Explanation
Charge depends on the change in flux Answer: (d)
Q.11
The north pole of a magnet is brought near a metallic ring as shown in the figure. The direction of induced current in the ring will be [ MPPMT 1998]
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a) Anticlock wise
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b) clock wise
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c) First anti clock wise and then clockwise
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d) First clock wise and then anticlockwise
Explanation
Answer: (c)
Q.12
The current in a self inductance L = 40 mH is to be increased uniformly from 1A to 11A in 4 milliseconds. The emf induced in inductor during the process is [ CBSE 1990]
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a) 100volts
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b) 0.4 volts
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c) 40 volts
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d) 440 volts
Explanation
Answer: (a)
Q.13
power is transmitted from a power house on high voltage a.c. because [ Raj.PMT 1997]
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a) electric current travels faster at higher voltage
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b) it is more economical due to less power wastage
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c) it is difficult to generate power at low voltage
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d) chances of stealing transmission lines are minimized
Explanation
Answer:(b)
Q.14
Faraday's law of electromagnetic induction is [ CBSE 1993]
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a) related to law of conservation of charge
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b) related to law of conservation of energy
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c) related to third law of Newton
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d) related to law of conservation of angular momentum
Explanation
Answer: (b)
Q.15
The bob of simple pendulum is replaced by a magnet. The oscillations are set along the length of the magnet. A copper coil is added so that one pole of the magnet passes in and out of coil. The coil is short-circuited. Then which one of the following happens? [ CET 1994]
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a) Period decreases
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b) Period does not change
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c) Oscillations are damped
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d) Amplitude increases
Explanation
Answer: (c)
Q.16
Eddy current is produced in a material when it is [ CBSE 1993]
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a) heated
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b) Placed in time varying magnetic field
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c) placed iin an electric field
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d) placed in a uniform magnetic field
Explanation
Answer: (b)
Q.17
L, C, R represents the physical quantities inductance, capacitance and resistance respectively. The combination which have the dimensions of frequency are [ BIT 1992]
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a) 1 / RC
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b) RL / C
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c) R / LC
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d) C /L
Explanation
1/RC, R/L and 1/√(LC) have unit of frequency Can be proved using dimensional analysis Answer:(a)
Q.18
A wire coil carries the current i, The potential energy of the coil does not depend upon [ MPPMT 1993]
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a) the value of i
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b) the number of turns in the coil
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c) whether the coil has an iron core or not
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d) the resistance of coil
Explanation
Answer: (d)
Q.19
A long horizontal metallic rod with length along the east-west direction is falling under gravity. The potential difference between its ends will [ MPPMT 1997]
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a) be zero
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b) be constant
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c) increase with time
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d) decrease with time
Explanation
Answer: (c)
Q.20
The coil of a dynamo is rotating in a magnetic field. The developed induced emf changes and the number of magnetic lines of force also changes. Which of the following conditions is correct? [ MPPET 1993]
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a) lines of force minimum but induced emf is zero
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b) lines of force maximum but induced emf is zero
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c) lines of force maximum but induced emf is not zero
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d) lines of force maximum but induced emf is also maximum
Explanation
Answer: (b)
Q.21
Not to develop eddy currents in the core of the transformer [ MPPMT 1993]
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a) By increasing the number of turns in secondary coil
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b) By taking laminated core
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c) By making step-down transformer
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d) By using a weak-alternating current at hi high potential
Explanation
Answer: (c)
Q.22
A player with 3 meter long iron rod runs towards east with a speed of 30 km/hr. Horizontal component of earth's magnetic field is 4×10-5 weber/mIf he is running with rod horizontal and vertical positions, then the potential difference induced between the two ends of the rod in two cases will be
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a) zero in vertical position and 1×10-3 volt in horizontal position
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b) 1×10-3 volt in horizontal position and vertical position
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c) zero in both position
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d) 1×10-3 volts in both position
Explanation
When the rod is in vertical position, it cuts horizontal component of earth's field and an induced emf E = Blv E = 4×10-5 ×(25/3)×3 = 10-3 volt When the rod is horizontal it does not cut any of the field ( horizontal or vertical) as Horizontal component is parallel to length. So no induced emf is produced in this case Answer:(b)
Q.23
A coil has 2000 turns and area of 70 cmThe magnetic field perpendicular to the plane of the coil is 0.3 weber/m2 and takes 0.1 sec to rotate through 180°. The value of induced emf will be [ MPPET 1993]
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a) 8.4 volt
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b) 84 volt
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c) 42 volt
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d) 4.2 volt
Explanation
φ = BAcosθ φi = BAcosθ φf = BAcos(θ+π) = -BAcosθ dφ = φf - φi dφ = -BAcosθ -BAcosθ = -2BAcosθ let θ = 0 dφ = -2BA = -2×0.3 ×70×10-4 dφ = 42×10-4 dφ / dt = 42×10-3 E = -N(dφ / dt) = 2000×42×10-3 = 84V Answer: (b)
Q.24
In an induction coil, the secondary emf is [ CET 1994]
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a) zero during break of circuit
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b) very high during make of the circuit
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c) zero during make of the circuit
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d) very high during break of the circuit
Explanation
Answer: (d)
Q.25
An emf of 5 millivolt is induced in a coil when in a nearby placed another coil, the current changes by 5amp in 0.1 second. The coefficient of mutual induction between the two coils will be [ MPPMT 1993]
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a) 1 Henry
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b) 0.1 henry
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c) 0.1 millihenry
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d) 0.01 millihenry
Explanation
E = M ( dI/dt) 5×10-3 = M (5/.1) 5×10-3 = M 50 M = 1×10-4 Henry M = 0.1 milliHenry Answer: (c)
Q.26
When the current in a coil changes from 2 amp to 4 amp. in 0.05 sec. an emf of 8 volt is induced in the coil. The coefficient of self induction of the coil is [ MNR 1993]
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a) 0.1 henry
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b) 0.2 henry
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c) 0.4 henry
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d) 0.8 henry
Explanation
Answer:(b)
Q.27
In a step up transformer the ratio of turns in secondary and primary coil is 4:If the current in primary is 4A, the current in secondary will be [ Raj.PMT 1996]
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a) 8A
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b) 2A
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c) 1A
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d) 0.5A
Explanation
Answer: (c)
Q.28
In a transformer, the primary has 500 turns and secondary has 50 turns, 100 volts are applied to the primary coil, the voltage developed in the secondary will be [ MPPMT 1997]
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a) 1 V
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b) 10 V
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c) 1000 V
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d) 10,000 V
Explanation
Answer: (b)
Q.29
A two metre wire is moving with velocity of 1 m/ sec perpendicular to a magnetic field of 0.5 weber/mThe emf induced in it will be [ MPPMT 1998]
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a) 0.5 volt
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b) 0.1 volt
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c) 1 volt
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d) 2 volt
Explanation
Answer: (c)
Q.30
The inductance of coil is 60µH. A current in this coil increases from 1.0A to 1.5A in 0.1 second. The magnitude of the induced emf is [ PMT 1995]
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a) 60×10-6 V
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b) 300×10-4 V
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c) 30×10-4
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d) 3×10-4 V
Explanation
Answer:(d)
0 h : 0 m : 1 s
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