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Thermal And Chemical Effects Of Current Mcq
Quiz 6
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Q.1
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.2
How many calories of heat will approximately developed in a 210 watt electric bulb is 5 minutes
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a) 1050
0%
b) 15000
0%
c)63,000
0%
d)80,000
Explanation
Answer: (b)
Q.3
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]
0%
a)(8/9) W
0%
b) 1.28 W
0%
c)2.0 W
0%
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.4
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]
0%
a) 2×10-2 J
0%
b) 4×10-2 J
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c)4×104 J
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d)4×1010 J
Explanation
Energy stored in capacitor=heat produced in resistor(1/2) CV2=Heat Substitute and solve Answer: (b)
Q.5
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.6
An electric bulb rated 220V, 60 W. Its resistance is nearly [ CPMT 1994]
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a) 0.807 Ω
0%
b) 4 Ω
0%
c) 708Ω
0%
d) 807Ω
Explanation
Answer: (d)
Q.7
A bulb 100W, 200 V is attached to a voltage of 160V. The power dissipation is [ CBSE PMT 1997]
0%
a) 64 W
0%
b) 100 W
0%
c)32 W
0%
d)160 W
Explanation
Answer: (a)
Q.8
If 2.2 kilo watt power is transmitted through a 10Ω line at 22000 volt, the power loss in the form of heat will be [ MPPMT 1998]
0%
a) 0.1 watt
0%
b) 1.0 watt
0%
c)10 watt
0%
d)100 watt
Explanation
Answer:(a)
Q.9
If a high power heater is connected to electric mains, then the bulb in the house become dim because [ BHU 1999]
0%
a) current drop
0%
b) potential drop
0%
c) no current drop
0%
d) no potential drop
Explanation
Answer: (a)
Q.10
Fuse wire should have [ BHU 1999]
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a)low resistance, high melting point
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b) low resistance low melting point
0%
c)high resistance, low melting point
0%
d)high resistance, high melting point
Explanation
Answer: (c)
Q.11
An electric kettle has two heating elements. One brings it to boil in 10 minutes and the other in fifteen minutes. If two heating filaments are connected in parallel, the water in kettle will boil is [ KCET 2000]
0%
a) 6 minutes
0%
b) 8 minutes
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c)25 minutes
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d)5 minutes
Explanation
for parallel combination of hater use formulaAnswer: (a)
Q.12
Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220 volt a.c. supply line. The power drawn by the combination in each case respectively will be [ CBSE 2003]
0%
a) 50 watt, 100 watt
0%
b) 100 watt, 50 watt
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c)200 watt, 150 watt
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d)50 watt, 200 watt
Explanation
parallel combination P=100+100=200 Wseries combination=(100×100) / ( 100+100)=50 W Answer:(d)
Q.13
A 40µF capacitor in a defibrillator is charged to 3000V. The energy stored in the capacitor is sent through the patient during a pulse of duration 2ms. The power delivered to the patient is [ AIIMS 2004]
0%
a) 45 kW
0%
b) 90 kW
0%
c) 180 kW
0%
d) 360 kW
Explanation
Energy in capacitor=(1/2) CV2 Energy in capacitor U=(1/2) ×40×10-6 ×(3000)2 U=180 J Power=U/times=180/(2×10-3=90kW Answer: (b)
Q.14
A d.c. voltage with appreciable ripple expressed as V=V1 + V2 cos ωt is applied to a resistor R. The amount of heat generated per second is given by [ IAPT 1999]
0%
a)
0%
b)
0%
c)
0%
d)None of the above
Explanation
Heat produced H=V2 / R average value of cos2ωt over a cycle=1/2 Average value of cosωt over a cycle=0 ∴ heat generated in one second Answer: (b)
Q.15
How many lamps each of 50 W and 100 V can be connected in parallel across a 120 V battery of internal resistance 10Ω, so that each flows to full power?
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a) 2
0%
b) 4
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c)6
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d)8
Explanation
Resistance of each bulb=V2 /P=200 Ω Current rating of the bulb=V/R=100/200=1/2 A Let n bulbs are connected so current through battery=n/2 Amp Total resistance in circuit=10 + 200/ n Current through battery=120/ (10 +200/n) thus Answer: (b)
Q.16
A heater boils 1 kg water in time t1 and other heater in time t2 if the heaters are connected in series, the combination will boil the same water in times
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a)
0%
b)
0%
c)
0%
d)
Explanation
We know that when heaters are connected in series time taken is t1 + t2on simplifying option 'd' we get same answer option d is correct Answer:(d)
Q.17
Fifty electric bulbs are connected in series across a 220 V supply and the illumination produced is I1 . Five bulbs are fused and remaining forty five bulbs are connected in series, the illumination produced is I2, by what percentage illumination will change
0%
a) increase by 10%,
0%
b) decrease by 11%,
0%
c) increase by 11%,
0%
d) decrease by 11%,
Explanation
Let R be the resistance of bulb Total resistance in circuit=50R Power P=2202 / 50R After removing 5 bulbs total resistance in circuit=45R Power P'=2202 / 45R Thus increase in power=P' -P= Answer: (c)
Q.18
A uniform wire connected across a supply produces heat H per second. If the wire is cut into three equal parts and all the parts are connected in parallel across the same supply, the heat produced per second will be
0%
a) H/9
0%
b) 9H
0%
c)3H
0%
d)H/3
Explanation
New resistance=R/9 Thus Heat produced=pH Answer: (b)
Q.19
Find heat produced per minute in the resistance R2 shown in circuit
0%
a) 640 J
0%
b) 1280 J
0%
c)960 J
0%
d)320 J
Explanation
Total resistance in circuit=1 + 6×3 / ( 3+6)=3Ω Current in circuit=12/3=4 ANow Current through R2=4×6 / ( 6+3)=24/9=8/3 A Power=(8/3) 2×3=64/3 Heat=Pt Heat=(64/3)×60=1280 J Answer: (b)
Q.20
Masses of three wires are in the ratio of 1:3:Their lengths are in ratio of 5:3:When connected in series with battery the ratio of heat produced in them will be
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a) 1:3:5
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b) 5:3:1
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c)1:15:125
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d)125:15:1
Explanation
We know that H=I2R Answer:(d)
Q.21
An electric heating element consumes 500W when connected to a 100V line. If the line voltage becomes 150V, the power consumed will be [ Roorkee 1990]
0%
a) 500W
0%
b) 750 W
0%
c) 1000 W
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d) 1125 W
Explanation
Answer: (d)
Q.22
A fuse wire with circular cross-sectional radius of 0.02 mm blows with a current of 5 amp. For what current, another fuse wire made from the same material with cross sectional radius of 0.4mm will blow
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a)14.7 A
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b) 5 A
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c)3 A
0%
d)1.5 A
Explanation
Heat lost per second per unit surface area of fuse wire is given by Answer: (a)
Q.23
The charge Q flowing through a resistance R varies with time t as Q=at - btThe total heat produced in R is
0%
a) a3R / 6b
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b) a3R / 3b
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c)a3R / 2b
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d)a3R / b
Explanation
We know that I=dQ/dt=a - 2bt for time t=t0 Current I=0 thus 0=a - bt0 ∴ t=a/2bThe current flows from time t=0 to t=t0. The heat producedAnswer: (a)
Q.24
A 100 W bulb B1 and 60W bulb B2 and B3 are connected to a 250 V source as shown in figure . Now W1, W2, W3 are the output power of the bulb B1, B2 and B3 respectively. Then
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a) W1 > W2=W3
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b) W1 > W2 > W3
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c)W1 < W2=W3
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d)W1 < W2 < W3
Explanation
Voltage across B3 is greatest, hence B3 will show maximum brightness. In series combination of bulb, the bulb of lesser wattage will glow more bright. Hence W2 > W1. So option 'd' is correct Answer:(d)
Q.25
An electric kettle taking 3A at 200V brings one litre of water from 20°C to the boiling point in 10 minutes. Its efficiency is
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a) 33.3%
0%
b) 66.6%
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c) 87.7%
0%
d) 93.3%
Explanation
efficiency η=Energy used / Energy supplied Answer: (d)
Q.26
Time taken by 836W water to heat one litre of water from 0°C to 40°C is
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a)50 sec
0%
b) 100 sec
0%
c)150 sec
0%
d)200 sec
Explanation
Answer: (c)
Q.27
The same mass of copper is drawn into two wires of thickness 1mm and 2mm. If two wires are connected in series and current is passed, then heat produced in the wire is in the ratio of
0%
a) 16:1
0%
b) 9:4
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c)1:16
0%
d)4:9
Explanation
Answer:(a)
Q.28
The water in an electric kettle begins to boil in 15 minutes after being switched on. Using the same mains supply, should the length of the wire used as heating element, be increased or decreased if the water is to boil in 10 minutes
0%
a) increased by 10%
0%
b) decreased
0%
c) increased by 50%
0%
d) unchanged
Explanation
Q=(V2/R)×t Thus for first heater Q=(V2/R)×15 second heater Q=(V2/R')×1 Since Q is same 15/R=10/R' R'=(10/15)R Bur R ∝ l as area of cross section is same l'=(10/15)l l'=(2/3)l (l-l')/ l=(1/3) Thus decrease in length should be 33.33% Answer: (b)
Q.29
A 100 W immersion heater is placed in pot containing 1 litre of water at 20°C. How long it will take to heat the water to boiling temperature, if 20% of the available energy is lost to surroundings?
0%
a)18 minutes
0%
b) 16 minutes
0%
c)14 minutes
0%
d)20 minutes
Explanation
Answer: (c)
Q.30
Two electric bulbs A and B are designed for the same voltage. Their power ratings are PA and PB respectively, with PB > PA. If they are joined in series across a V volt supply
0%
a) B will draw more power than A
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b) A will draw more power than B
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c)the ratio of powers drawn by them will depend on V
0%
d)A and B will draw the same power
Explanation
Answer: (b)
0 h : 0 m : 1 s
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