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Physics NEET MCQ
Quiz 5
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Q.1
A transformer connected to 220 volt line shows an out put of 2A at 11000 volt. The efficiency is 100%. The current drawn from the line is [ PMT 1995]
0%
a) 100A
0%
b) 200A
0%
c) 22A
0%
d) 11A
Explanation
Answer:(a)
Q.2
In a circuit with a coil of resistance 2Ω the magnetic flux changes from 2Wb to 10.0 Wb in 0.2 second. The charge that flows in the coil during this time is [ MPPMT 1997]
0%
a) 5.0 coulomb
0%
b) 4.0 coulomb
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c) 1.0 coulomb
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d) 0.8 coulomb
Explanation
E = -dφ/dt RI= -dφ/dt 2×I = 8/0.2 I = 20 Amp In 1 second charge flown = 20 coulomb ∴ in 0.2 second = 4 coulomb Answer: (b)
Q.3
The primary winding of a transformer has 500 turns where as its secondary has 500 turns. The primary is connected to an a.c supply of 20V, 50Hz. The secondary will have an output of [ CBSE 1997]
0%
a) 200v, 50Hz
0%
b) 20V, 500Hz
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c) 2V, 50Hz
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d) 2V, 5Hz
Explanation
Answer: (a)
Q.4
Two identical coaxial circular loops carry current I each circulating in the clockwise direction. If the loops are approaching each other, then [ PMT 1995]
0%
a) current in each loop increases
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b) current in each loop remains the same
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c) current in each loop decreases
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d) current in one loop increases and in the other it decreases
Explanation
Answer: (c)
Q.5
A magnet is brought near a ring a) quickly b) slowly, induced emf will be [ Raj.PMT 1997]
0%
a) More in case(a)
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b) Less in case (a)
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c) same in both the cases
0%
d) Depends on the radius of ring
Explanation
Answer:(a)
Q.6
Two loops of different wires are placed concentrically in a plane. If the current in the outer loop is made to pass clockwise and current increases with time, the induced current in the inner loop will be [ Rj.PET 1996]
0%
a) clockwise
0%
b) anticlockwise
0%
c) zero
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d) will depend upon the radius of loop
Explanation
Answer: (b)
Q.7
Mutual inductance of two coils can be increased by [ MPPMT 1994]
0%
a) Decreasing the number of turns in the coils
0%
b) Increasing the number of turns in the coils
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c) Winding the coils on wooden core
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d) None of above
Explanation
Answer: (b)
Q.8
The number of turns of primary and secondary coils of a transformer are 5 and 10, respectively and the mutual inductance of the transformer is 25henry. Now the number of turns in the primary and secondary of the transformer are made 10 and 5 respectively. the mutual inductance of the transformer in Henry will be [ MPPET 1995]
0%
a) 6.25
0%
b) 12.5
0%
c) 25
0%
d) 50
Explanation
Answer: (c)
Q.9
The mutual inductance between a primary and secondary circuit is 0.5H. The resistance of the primary and the secondary circuit are 20Ω and 5Ω respectively. To generate a current of 0.4A in the secondary, the current in the primary must be changed at the rate of [ MPPMT 1997]
0%
a) 4.0A/s
0%
b) 16.A/s
0%
c) 1.6A/s
0%
d) 8.0A/s
Explanation
E = M(dI/dt) dI/dt = E /M dI/dt = IR /M dI/dt = (0.4)(5) /0.5 = 4 A/s Answer: (a)
Q.10
The secondary of transformer gives 200 volts when 2.0 kilowatt power is suppled to its 500 turn primary at 0.5 ampere. The number of turns in the secondary is [ AMU 1995]
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a) 25
0%
b) 30
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c) 35
0%
d) 40
Explanation
Primary voltage I1V1 = 2000 watt V1 = 2000/0.5 = 4000V from formula V1 / V2 = N1 / N2 4000 / 200 = 500 / N2 20 = 500 / N2 = 500 /20 = 25 N2 Answer:(a)
Q.11
Tick out the wrong statement [ AMU 1995]
0%
a) An emf can be inducted between the ends of a straight conductor by moving the ends of a straight conductor by moving it through a uniform magnetic field
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b) The self induced emf produced by changing current in a coil always tends to decrease the current
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c) Inserting an iron core in a coil increases the coefficient of self-inductance
0%
d) According to lenz's law, direction of the induced current is such that it opposes the flux change that causes it
Explanation
Answer: (b)
Q.12
The self-inductance of a coil is L. Keeping the length and area same, the number of turns in the coil is increased to four times. the self inductance of the coil will now be [ MPPMT 1997]
0%
a) L/4
0%
b) L
0%
c) 4L
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d) 16L
Explanation
Answer: (d)
Q.13
A galvanometer is connected to the secondary coil. The galvanometer shows an instantaneous deflection of 7 division when current is started in primary coil of solenoid. Now if primary coil is suddenly rotated through 180°, then the new instantaneous deflection will be [ CPMT 1991]
0%
a) 7 units
0%
b) 14 units
0%
c) zero units
0%
d) 21 units
Explanation
Answer:(b)
Q.14
An emf E = 4cos(1000 t) volt is applied to an LR-circuit of inductance 3mH and resistance 4Ω. The amplitude of current in the circuit is [ MPPMT 1997]
0%
a) 4/√7 A
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b) 1.0A
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c) 4/7 A
0%
d) 0.8 A
Explanation
I = E /Z Z = √ ( R2 + (ωL)2 Z = √(42 + ( 3mH×1000)2 Z = 5 ω I = 4/5 = 0.8 amp. Answer: (d)
Q.15
The average power dissipation in pure inductance is [ MPPET 1999]
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a) ½ LI2
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b) 2LI2
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c) ¼ LI2
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d) zero
Explanation
Answer: (d)
Q.16
A choke coil has [ Raj.PMT 1997]
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a) high inductance an high resistance
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b) low inductance and low resistance
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c) high inductance and low resistance
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d) low inductance and high resistance
Explanation
Answer: (c)
Q.17
In an A.C. circuit, the resonance is obtained when [ Raj/.PET 1996]
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a) Z = R
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b) Z = ωL - (1/ ωC)
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c) L and C are in same phase
0%
d) The phase in C varies with source voltage
Explanation
Answer:(a)
Q.18
An L.C.R circuit is connected to a source of alternating current. At resonance, the applied voltage and current flowing through the circuit will have a phase difference of [ CBSE 1994]
0%
a) π/4
0%
b) zero
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c) π
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d) π/2
Explanation
Answer: (b)
Q.19
In an LR-circuit, the inductive reactance is equal to the resistance R of the circuit. An emf E = Eocosωt is applied to the circuit. The power consumed in the circuit is [ MPPMT 1997]
0%
a) Eo2 / R
0%
b) Eo2 / 2R
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c) Eo2 /4R
0%
d) Eo2 / 8R
Explanation
given XL = R Z = √ [ XL2 +R2] = R√2 power P = EoIocosφ / 2 but Io = Eo / Z and cosφ = R/Z = R/ R√2 = 1 / √2 Answer: (c)
Q.20
The current in resistance R at resonance is [ CPMT 1993]
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a) zero
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b) Minimum but finite
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c) Maximum but finite
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d) Infinite
Explanation
L and C are connected in parallel, therefore impedance will be low at other frequencies At resonance circuit will have only resistance Answer: (c)
Q.21
The resistance of coil at frequency 104 Hz is 104 Ω, the reactance at 2×104 Hz frequency will become [ Raj.PMT 1996]
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a) 104 Ω
0%
b) 2×104 Ω
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c) 3×107 Ω
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d) 4×104 Ω
Explanation
XL = ωL XL = 2πfL XL ∝f If frequency is double, then reactance will be double. Answer:(b)
Q.22
In pure resistance A.C. circuit the phase difference between current and voltage is [ Rj.PMT 1996]
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a) zero
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b) π/2
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c) -π/2
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d) π/4
Explanation
Answer: (a)
Q.23
An alternating current of frequency f is flowing in a circuit containing a resistance R and a choke L in series. The impedance of the circuit is equal to [ MPPET 1999]
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a) R
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b) R +πfL
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c) √[ R2 +4π2f2L2]
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d) R/2πfL
Explanation
Answer: (c)
Q.24
A 10Ω resistance, 5mH coil and 10µF capacitor are joined in series. When a suitable frequency alternating current source is joined to this combination, the circuit resonates. If the resistance is halved, the resonance frequency [ MPPET 1995]
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a) is halved
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b) is doubled
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c) remain unchanged
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d) is quadrupled
Explanation
resonance frequency depends on L and C. Answer:(c)
Q.25
The frequency for which a 5.0µF capacitor has a reactance of 1000Ω is given by [ MPPMT 1993]
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a) 1000/π cycles/sec
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b) 100/π cycles/sec
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c) 200 cycles/sec
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d) 5000 cycles/sec
Explanation
Answer: (b)
Q.26
In an AC circuit, a resistance of RΩ is connected in series with an inductance L. If phase angle between voltage and current be 45°, the value of inductive reactance will be [ MPPMT 1998]
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a) R/4
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b) R/2
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c) R
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d) can not found with the given data
Explanation
Answer: (c)
Q.27
In an A.C circuit. V and I are given by V = 100sin100t volts and I = 100sin(100t + π/3) mA. The power dissipated in the circuit is [ MPPMT 1999]
0%
a) 104 watts
0%
b) 10 watts
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c) 2.5 watts
0%
d) 50 watts
Explanation
From the equation Io = 100mA and Vo = 100 phase is π/3 = 60° Power Answer: (c)
Q.28
In a series resonant circuit, the a.c. voltage across resistance R, inductance L and capacitance C are 5V, 10V and 10V respectively. The a.c. voltage applied to the circuit will be [ CET 1994]
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a) 20 V
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b) 10 V
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c) 5 V
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d) 25 V
Explanation
We know that V2 = (Vl -Vl)2 + VR2 V2 = (10-10)2 +52 V = 5 V Answer: (c)
Q.29
The impedance of a circuit consists of 3Ω resistance and 4Ω reactance. The power factor of the circuit is [ MPPMT 1994]
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a) 0.4
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b) 0.6
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c) 0.8
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d) 1.0
Explanation
Power factor = R/Z Z = √( 42 + 32) = 5 Power factor = 2 / 5 = 0.6 Answer: (b)
Q.30
Same current is flowing in two alternating circuits. The first circuit contains only inductance and the other contains only a capacitor. If the frequency of the emf is increased the effect on the value of the current will be [ MPPMT 1993]
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a) Increase in first circuit and decrease in the other
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b) Increase in both circuits
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c) Decrease in both circuits
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d) Decrease in first circuit and increase in the other
Explanation
Imedane of Inductor increases with increase in frequency. While impedance of capaciotor decreases with increase in frequency Answer: (d)
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