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Physics NEET MCQ
Quiz 5
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Q.1
A generator generates voltage at terminal voltage of 500 volt. This power is being transmitted to a distance to a 5 km by electric wires. The resistance of each wire is 0.01 ohm/metre. The potential difference across a load of 900Ω connected the terminals of these wires is
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a) 500 volts
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b) 300 volt
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c)450 volts
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d) cannot be calculated
Explanation
Resistance of wire=5 ×0.01 × 1000=50 ΩThe current carried by the wires=50/ (900+50+50)=(1/2)A , therefore the potential difference across the terminals of 900Œ. Resistor=IR=(1/2) × 900=450 VAnswer: (c)
Q.2
What are the essential requirements for the heater filament?
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a) High resistivity, low melting point
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b) Low resistivity, low melting point
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c)High resistivity, high melting point
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d)Low resistivity, high melting point
Explanation
Answer:(c)
Q.3
50 electric bulbs are connected in series across a 220V supply and the illumination produced is I2bulbs are fused . If the remaining 48 are again connected in series the illumination produced is IWhich of the following is true?
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b) I1 < I2
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c) I1 > I2
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d) It will depend on the resistance of each bulb
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a) I₁=I₂
Explanation
Answer: (b)
Q.4
If 2.2 kilowatt power is transmitted through a 10Ω line at 22000 volt, the power loss in the form of heat will be [ MPPMT 1998]
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a)0.1 watt
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b) 1 watt
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c)10 watt
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d)100 watt
Explanation
I=P/V=2200/22000 1/10 A H=(1/10) 2 × 10=0.1 wAnswer: (a)
Q.5
A heater of 500 watt is made to operate at 115 volt line. If the line voltage reduced to 110 volt, then the percentage reduction in heat produced by it is ( Assume resistance of heater remains constant) [ ISM Dhanbad 1994]
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a) 8.5 %
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b) 85%
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c)60%
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d)there is no change in heat produced
Explanation
R=V2 / P1=[(115)2 / 5000]Answer: (a)
Q.6
An electric heater operating at 220V boils 5 litres of water in 5 minutes. If it is used on 110V, it will boil the same amount of water in
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a) 10 min
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b) 20 min
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c)2.5 min
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d)the water will not boil
Explanation
By reducing the voltage by factor 1/2, the power of the heater becomes one-fourth and hence the time taken to boil water increases by four times is 20 min Answer:(b)
Q.7
Overloading of an electric circuit implies
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a) drawing of large current
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b) connecting higher resistance device
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c) connecting a number of electric devices in series with the power supply
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d) none of the above
Explanation
Answer: (a)
Q.8
Three bulbs A, B and C are connected as shown in the figure. B and C are identical. IF the bulb C is fused
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a)both A and B will glow more brightly
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b) both A and B will glow less brightly than before
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c)A will glow less brightly and B more brightly
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d)none of the bulbs will glow
Explanation
Answer: (c)
Q.9
Two 1000 W heaters when connected in parallel across 220 V supply produce heat Qp in times t. If they are connected in series across the same power supply, the heat produced in the same time is Qs ; then Qp/Qs is
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a) 4
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b) 2
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c)0.5
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d)0.25
Explanation
Qp=1000 +1000=2000 WQs=1000×1000 / ( 1000 + 1000 )=500 W Qp/Qs=2000 / 500=4Answer: (a)
Q.10
A 100 W lamp is connected in series with a 60 W lamp across a 220 V supply. Which of the following statement is wrong
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a) Potential drop across each lamp is same
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b) current through each lamp is same
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c)More heat is produced in 60W lamp
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d)Resistance of 60W lamp is more than that of 100Q lamp
Explanation
Answer:(a)
Q.11
Let H1 be the heat generated per second in the filament of 100W, 250 V lamp and H2 be that of 200W, 250V lamp. Which of the following is correct
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a) H1=2H2
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b) H1=4H2
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c) 2H1=H2
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d) 4H1=H2
Explanation
Answer: (c)
Q.12
An electric fan and a heater are marked as 100 watt , 220 volts and 1000 watt, 220 volts respectively. The resistance of heater is
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a) zero
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b) greater than that of the fan
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c)less than that of the fan
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d)equal to that of the fan
Explanation
Answer: (c)
Q.13
Three bulbs B1, B2 and B3 are connected as shown in figure. How will the incandescence of bulb B1 be affected if one of bulbs B2 or B3 is disconnected from the circuit
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a) no change in incandescence
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b) bulb B1 will become brighter
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c) bulb B1 will become less bright
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d)bulb B1 may become brighter or dimmer depending upon the cancel power of bulb which is disconnected
Explanation
Answer: (c)
Q.14
Two bulbs of equal wattage, one having carbon filament and the other having a tungsten filament, are connected in series to the mains
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a) both bulbs glow equally
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b) carbon filament bulb glows more
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c)tungsten filament bulb glows more
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d)carbon filament bulb glows less
Explanation
Both the bulbs are of same wattage, therefore same resistance, connected 1n series current through each bulb will be same. Note that The temperature coefficient for carbon is negative, the resistance is reduced with increasing temperature. where as temperature coefficient of tungsten is positive Now resistance of carbon filament will decrease with increase in temperature, and thus power consumed will be less. But tungsten resistance will remain more or the less same Answer:(c)
Q.15
A storage battery of 30 volts is being charged by a source of 120 volts. To limit the current at 15 amp, a resistance is connected in series with the charger. The rate of heat developed in the resistance and the time required to heat 1 kg water of 15°C to 100°C are
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a) 1350W, 4 minutes 24 seconds
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b) 1350 W, 6 minutes
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c) 1000 W, 4 minutes 24 seconds
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d) 1800 W, 4 minutes
Explanation
I=(120- 30) / R 15=90/R ∴ R=6 Ω Power consumed in resistance=I2R=15×15×6=1350 W Now Pt=J. mSΔθ t=[4.2×1000×1×(100 - 15)] / 1350=4 mon 24 sec Answer: (a)
Q.16
Of the two bulbs in a house, one glows brighter than other. Which of the two has a larger resistance
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a)the bright bulb
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b) the dim bulb
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c)both have the same resistance
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d)the brightness does not depend upon the resistance
Explanation
Answer: (b)
Q.17
You have following electric appliances, which one have highest resistancea) 1 kW 250 V electric heaterb) 1kW 250 V electric kettlec) 1kW 250 V electric bulb
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a) heater
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b) kettle
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c)bulb
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d)all have equal resistance's
Explanation
Answer: (d)
Q.18
Four equal resistors when connected in series dissipate 5 watt power. If they are connected in parallel, the power dissipated will be
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a) 20 W
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b) 40 W
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c)60 W
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d)80 W
Explanation
Answer:(d)
Q.19
Two identical heater coils are connected in parallel across the mains. If one of the coils breaks, the other coil
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a) will also not be heated
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b) will develop the same amount of temperature as before
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c) develop a higher temperature
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d) develop a lower temperature
Explanation
Identical Coils are connected in parallel, there is no difference between both the coil Voltage across each coil is same, Now if any one coil coils breaks, it will not affect the performance of other coil Answer: (b)
Q.20
The power dissipated in a conductor of resistivity ρ is proportional to
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a)ρ
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b) ρ2
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c)√ρ
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d)none of the above
Explanation
Power P ∝ (1/R) ∴ P ∝ (1/ρ)Answer: (d)
Q.21
A standard 40 watt tube light is in parallel with a room heater both connected to stable main A/C. Supply line. What will happen when the tube light is switched off
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a) the heater output will be larger
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b) it will remain smaller
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c)it will remain the same
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d)none of the above
Explanation
Answer: (c)
Q.22
A resistance coil is made by joining two resistance's of 2.5 ohm each. A steady potential difference of 1.0 volt is applied across the coil for 5 minutes. The heat produced in the wire is
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a) 14.3 cal
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b) 14.3 Joule
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c)0.2 Joule
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d)0.2 cal
Explanation
Answer:(a)
Q.23
Two heater wires of equal length are first connected in series and then connected in parallel ratio of heat produced in the two cases is [ MNR 1987]
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a) 2:1
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b) 1:2
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c) 4:1
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d) 1:4
Explanation
Answer: (d)
Q.24
If two bulbs of wattage 25W and 100W respectively each rated at 220V connected in series with the supply of 440 V, which bulb will fuse [ MNR 1988]
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a)25 W bulb
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b) 100 W bulb
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c)both of them
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d)none of them
Explanation
Answer: (a)
Q.25
How many calories of heat will approximately developed in a 210 watt electric bulb is 5 minutes
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a) 1050
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b) 15000
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c)63,000
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d)80,000
Explanation
Answer: (b)
Q.26
Two identical batteries, each of emf 2 volts and internal resistance 1Ω are available to produce heat in a resistance R=0.5Ω, by passing a current through it. Maximum joules power that can be developed across R using these batteries is [ CPMT 1990]
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a)(8/9) W
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b) 1.28 W
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c)2.0 W
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d)3.2 W
Explanation
Heat developed cross R will be maximum if the current through R is maximum. If batteries are connected in parallel internal resistance of battery combination will become 0.5Ω so total resistance in circuit=0.5+0.5=1.0Ω. Therefore current through circuit=2/1=2 A Joule power loss=i2R'=92)2×(0.5)=2 WAnswer: (c)
Q.27
A condenser having a capacity 2.0µF is charged to 200V and then the plates of the capacitor are connected to a resistance wire. The heat produced in joules will be [ Karnataka 1992]
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a) 2×10⁻² J
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c)4×104 J
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d)4×1010 J
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b) 4×10⁻² J
Explanation
Energy stored in capacitor=heat produced in resistor(1/2) CV2=Heat Substitute and solve Answer: (b)
Q.28
A current flowing through a conductor produces 80 joule of heat in ten seconds. The resistance of the conductor is
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a) 0.5 Ω
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b) 2Ω
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c)4 Ω
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d)20 Ω
Explanation
Answer:(b)
Q.29
An electric bulb rated 220V, 60 W. Its resistance is nearly [ CPMT 1994]
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a) 0.807 Ω
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b) 4 Ω
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c) 708Ω
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d) 807Ω
Explanation
Answer: (d)
Q.30
A bulb 100W, 200 V is attached to a voltage of 160V. The power dissipation is [ CBSE PMT 1997]
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a) 64 W
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b) 100 W
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c)32 W
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d)160 W
Explanation
Answer: (a)
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