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Q.1
The resistance of bulb filament is 100Ω at a temperature of 100°C. If its temperature coefficient of resistance be 0.005/ °C , its resistance will become 200Ω at a temperature of [ AIEEE 2006]
0%
a)300°C
0%
b) 400°C
0%
c)500°C
0%
d)200°C
Explanation
100=Ro [ 1 + α×100] --eq(1) 200=Ro [ 1 + α×T] --eq(2)On dividing eq(2) by eq(1)we get Answer: (b)
Q.2
An electric bulb is rated 220volt -100 watt. The power consumed by it when operated on 110 volt will be [ AIEEE 2006]
0%
a) 75 watt
0%
b) 40 watt
0%
c)25 watt
0%
d)5 watt
Explanation
Resistance of bulb is R=V2 / P=(220)2 / 100=(22)2 The power consumed when operated at 110 volts is P=(110)2 / (22)2=100/4=250 WAnswer: (c)
Q.3
In an ordinary heater if the length of the coil is halved, then a given quantity of water will boil is [ Andhra Medical]
0%
a) less time
0%
b) more time
0%
c)same time
0%
d)cannot be compared because specific resistance of material of wire is not given.
Explanation
Heat H=V2t / R Since heat required in both the cases is same and voltage is same Also resistance in second case is half the previous thus t1 / R=t2 / ( R/2)t1=2 (t2)t2=t1 / 2 Answer:(a)
Q.4
The supply voltage to a room is 120V. The resistance of the lead wire is 6Ω. A 60 watt bulb is already giving light. What is the decrease in voltage across the bulb when a 240 watt heater is switched on?
0%
a) zero
0%
b) 2.9 volt
0%
c) 13.3 volt
0%
d) 10.4 volt
Explanation
Resistance of bulb=Rb=V2 / P Rb=(120)2 / 60=240Ω potential across bulb in absence of heater Total resistance in circuit=240 + 6=246 Ω Current in circuit i=120 /246 Potential across bulb=i× resistance of bulb Potential=(120/246)×240=117.7 volts potential across bulb in presence of heater Resistance of heater=Rh=(120)2 / 240=6Ω Total resistance in circuit Current through circuit i'=120/54 Potential cross bulb=(120/54)×48=106.66 volt ∴ change in voltage across the bulb, on switching the heater=( 117.7 - 106.66)=10.4 volt Answer: (d)
Q.5
A 5°C rise in temperature is observed in conductor by passing a current. When the current is doubled, the rise in temperature will be approximately [ CBSE 1998]
0%
a)20°C
0%
b) 16°C
0%
c)12°C
0%
d)10°C
Explanation
Heat generated H ∝ (current ) 2 when current is doubled heat produced is 4times of earlier heat Thus increase in temperature by 20°CAnswer: (a)
Q.6
TWo identical batteries each of emf E=2V and internal resistance r=1Ω are available to produce heat in an external resistance by passing a current through it. The maximum Joulean power that can be developed across R using these batteries is [ CBSE 1990]
0%
a) 1.28 W
0%
b) 2W
0%
c)8/9 W
0%
d)3.2 W
Explanation
If batteries are connected in parallel internal resistance of batteries in parallel combination=r/2=1/2 For maximum power output external resistance should be equal to (1/2)ΩTotal resistance in circuit will be 1Ω Current through the circuit=2 / 1=2 Amp Maximum power=I2× resistance R=(2)2 × (1/2)=2 Watt If batteries are connected in series internal resistance of batteries in parallel combination=r+r=2r=2Ω For maximum power output external resistance should be equal to 2Ω Total resistance in circuit will be=2+ 2=4Ω Current through the circuit=(2+2) / 4=1 Amp Maximum power=I2× resistance R=(1)2 × 2=2 Watt Answer: (b)
Q.7
If two bulbs of wattage 25 and 100 respectively each rated at 220 volts,are connected in series across 440V which bulb will fuse? [ MNR 1988]
0%
a) 100 watt bulb
0%
b) 25 watt bulb
0%
c)none of them
0%
d)both of them
Explanation
Resistance of bulb of 25W=R=V2 / P R=(220)2 / 25 Resistance of bulb of 100W=R=V2 / P R=(220)2 / 100 When both the bulbs are connected in series total resistance R"=R +R'=(220)2 / 25 + (220)2 / 100 R"=[5×(220)2] / 100When connected to 440 V supply current=V/ R"=440 / [(5×(220)2) / 100] I=(440)(100) / (5×(220)2)=40/220 ACurrent capacity of 25 watt bulb P=VI I'=P/V=25/220 A Current capacity of 100 watt bulb P=VI I"=P/V=100/220 A As I" > I' , 25watt bulb will fuse Answer:(b)
Q.8
A 40 W-220 V bulb , 60W-220V bulb and a 100W-220V bulb are joined in series and connected to the mains. Which bulb will glow brighter [ MPT 1995]
0%
a) 40 W bulb
0%
b) 100 watt bulb
0%
c) first 40 W bulb and then 100 W bulb
0%
d) both will glow with same brightness
Explanation
resistance of the bulb 40 W is maximum as R=V2/P) for same voltage When bulbs are connected in series current will remain same for all the bulbs Now heat produced=I2Rt Here I and t is same for all bulb H ∝ R Thus 40 w bulb will glow brighterAnswer: (a)
Q.9
An electric kettle has two coils, when one of these is switched on, the water in the kettle boils in 6 minutes. When the other coil is switched on, the water boils in 3 minutes. If the two coils are connected in series, the time taken to boil water in the kettle is :
0%
a)3 minutes
0%
b) 6 minutes
0%
c)2 minutes
0%
d)9 minutes
Explanation
Same amount of heat is required in all the cases to boil the water First coil JH=V2t1 / R1R1=V2t1 / JH Second Coil JH=V2t2 / R2R2=V2t2 / JHWhen coils connected in series total resistance=Thus t=t1 + t2 t=6 + 3=9 minutesAnswer: (d)
Q.10
In above example if coils are connected in parallel the time taken to boil the water in the kettle is
0%
a) 3 minutes
0%
b) 6 minutes
0%
c) 2 minutes
0%
d)9 minutes
Explanation
When connected in parallel effective resistance R Answer: (c)
Q.11
The speed of rotation of a fan is reduced. For convenience let us say the supply is D.C. it will consume
0%
a) more power than a full speed
0%
b) same power than at full speed
0%
c)less power than at full speed but less efficiently
0%
d)less power than at full speed but more efficiently.
Explanation
Answer: (a)
Q.12
Three equal resistors connected in series across a source of e.m.f together dissipates 10 watts of power. What will be the power dissipated if the same resistors are connected in parallel across the same source of e.m.f? [ CBSE 1998]
0%
a) 10 watts
0%
b) 30 watts
0%
c)90 watts
0%
d)3.33 watts
Explanation
In series H1=V2 / 3RHere H=10 W given∴ 30=V2 / HIn parallel H2=3V2 / RHere resistance=R/3∴ H2=3×30W=90 W Answer:(c)
Q.13
The resistance of a hater coil is 110 Ω. A resistance R is connected in parallel with it and the resistance R is connected in parallel with it and the combination is joined in series with a resistance of 11Ω to 220 Volts main line. The heater operates with a power of 110 watt. The value of R in ohm is [ ISM Dhanbad 1994]
0%
a) 12.22
0%
b) 24.42
0%
c) negative
0%
d) that the given values are not correct
Explanation
Power of heater P=I2RheaterI2=110 / 110=1 A is the current passing through heater Now potential across Resistance R=potential across heater potential across resistance R=(110)×1 Now Supply potential=potential across R and potential across 11Ω resistance 220=110 + Potential across 11Ω resistance ∴ Potential across resistance=220 - 110=110 V Potential across resistance 11Ω is V=IR ∴ 110=I (11) ⇒ I=10 A Thus 9 A current is passing through resistance R and have potential of 110 V Thus 110=9× R R=110/9=12.22ΩAnswer: (a)
Q.14
The electric bulbs whose resistance's are R1 and R2 are connected in parallel to a constant voltage source. The power dissipated in them have ratio [ CPMT 1999]
0%
a)R1 / R2
0%
b) R2 / R1
0%
c)(R1 / R2) 2
0%
d)(R2 / R1) 2
Explanation
Power P=V2 / R For constant voltage sourceP1 / P2=R2 / R1 Answer: (b)
Q.15
The time required for 1KW heater to raise the temperature of 10 lit of water through 10°C is [ EAMCET 1987]
0%
a) 210 sec
0%
b) 420 sec
0%
c)42 sec
0%
d)840 sec
Explanation
Heat produced by heater=HEat absorbed by water Heat produced by heater=P×t in joules=P×t / J Answer: (b)
Q.16
If potential difference across a conductor having a material of specific resistance ρ, remains constant, then according to Joules law, heat developed in the conductor is directly proportional to [ MP 1986]
0%
a) ρ
0%
b) ρ2
0%
c)1 / √ρ
0%
d)1 / ρ
Explanation
When voltage is constant H ∝ 1/R ∴ H ∝ 1 /ρ Answer:(d)
Q.17
A torch bulb rated as 4.5 W, 1.5V is connected as shown in figure. The e.m.f of the cell, needed to make the bulb glow at full intensity is [
0%
a) 4.5 V
0%
b) 1.5 V
0%
c) 2.67 V
0%
d) 13.5V
Explanation
Resistance of bulb R=V2 / P R=(1.5)2 / 4.5=0.5 Ω Current through bulb=V / R=1.5/05=3Amp Current through 1 Ω resistance V=IR 1.5=I(1) I=1.5 Amp Total current from battery=3 + 1.5=4.5 Amp Total resistance in circuit=0.5×1/1.5=0.333 E=I(R + r) E=4.5( 0.33 + 2.67)=13.5V Answer: (d)
Q.18
A 25 W, 220V bulb and a 100W, 220V bulb are joined in series and connected to the mains. Which bulb glow brighter? [ MPPMT 1999]
0%
a)25 W bulb
0%
b) 100 W bulb
0%
c)First 25 W bulb and then 100 W bulb
0%
d)Both will low with same brightness
Explanation
Resistance of bulb of 25 W will be more than 100 W bulb since voltage is sameThus when connected in series potential across 25 W bulb will be more than 100W bulb thus 25 W bulb will glow brighterAnswer: (a)
Q.19
Electric bulbs rated 50 Watt and 100 volt and glowing at full power, are used in parallel with a battery of e.m.f. 120 volts and internal resistance 10Ω. The maximum number of bulbs that can be connected in the circuit when glowing at full power is
0%
a) 8
0%
b) 6
0%
c)4
0%
d)2
Explanation
If the bulbs are glowing at full power, then the potential difference across the terminals should be 100 volts. Thus V=E - Ir100=120 -10I or I=2A If n bulbs are used each carrying a current i=P/V=50/100=0.5A then n ×(1/2)=2 or n=4Answer: (c)
Q.20
The safe current for the fuse wire of radius r is I then
0%
a) I ∝ √r
0%
b) I ∝ r
0%
c)I ∝ r3/2
0%
d)I ∝ r2
Explanation
Amount of heat produced due to current=Amount of heat absorbed by fuse wireI2Rt=mCΔT here m is mass of wire, Δ is change in temperature, C is latent heat of meltingbut m=Area ×length × density m=πr2 ×l×d Answer:(d)
Q.21
A resistor R1 dissipates the power P when connected to a certain generator, if a resistance R2 is put in series with R1, the power dissipated by R1 is [ CPMT 1985]
0%
a) decreases
0%
b) increases
0%
c) remains same
0%
d) any one of the above depending upon the relative values of R1 and R2
Explanation
Answer: (a)
Q.22
A generator generates voltage at terminal voltage of 500 Volts. This power is being transmitted to a distance of 5km. by electric wires. The resistance of each wire is 0.01 ohm/ metre. The potential difference across a load of 900Ω connected between the terminals of these wires is
0%
a)500 volt
0%
b) 300 volt
0%
c)450 volt
0%
d)cannot be calculated
Explanation
Resistance of wire=5×0.01×1000=50ΩThe current carried by the wires=500 / ( 900 + 50 + 50)=0.5A.Therefore potential difference across the terminals of 900Ω.Resistance isV=IRV=0.5×900=450VAnswer: (c)
Q.23
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 that the resistance of heater remains same) [ ISM Dhanbad 1994]
0%
a) 8.5%
0%
b) 85%
0%
c)60%
0%
d)there is no change in heat produced
Explanation
resistance of heater=V2 / P R=(115)2 / 500 Power on connecting to 110 volt P2 % loss=[(P2 - P1) / P1)]×100Answer: (a)
Q.24
A heater coil is cut into two equal length and only one of them is used in the heater. The ratio of the heat produced by this half coil to the original coil is
0%
a) 2 :1
0%
b) 1:2
0%
c)1:4
0%
d)4:1
Explanation
H=V2 R half part H'=V2 / (R//2)=2H Answer:(a)
Q.25
How much heat is produced by 1500W heater in 7 minute
0%
a) 1.5 kcal
0%
b) 15 kcal
0%
c) 150 kcal
0%
d) 1500kcal
Explanation
Heat=Power × t / 4.2 in Cal HEat=1500×(7×60) / 4.2=150,000 cal=150 kcal Answer: (c)
Q.26
Two resistance R1 and R2 when connected across a 120V line consume power at the rate of 25W and 100 W respectively when connected in series and parallel across the same 120V line. Then the ratio of power consumed by R1 to that consumed by R2 will be
0%
a)1:1
0%
b) 1:2
0%
c)2:1
0%
d)1:4
Explanation
CaseI :Series connection25=(120)2 / (R1 + R2) R1 + R2=(120)2 / 25 Case: Parallel connection 100=(120)2 (R1 + R2 /(R1 × R2) Now Answer: (a)
Q.27
An electric bulb is rated 220 V and 100 Watts. Power consumed by it when operated on 110 V is [ MPPMT 1986]
0%
a) 50 watts
0%
b) 75 watts
0%
c)90 watts
0%
d)25 watts
Explanation
Use formula R=V2 / P to determine resistance and use same formula for 110 voltsAnswer: (d)
Q.28
If R1 and R2 are respectively the filament resistance of 200 watt and 100 watt bulb respectively, the bulb, designed to operate on the same voltage then [ CPMT 1991]
0%
a) R1=2R2
0%
b) R2=2R1
0%
c)R2=4R1
0%
d)R1=4R2
Explanation
Use formula R=V2 / P Then use P1× R1= P2 × R2 Answer:(b)
Q.29
A heater coil is cut into two parts of equal length and only one of them is used in the heater. The ratio of the heat produced by this half-coil to that by the original coil is
0%
a) 2:1
0%
b) 1:2
0%
c) 1:4
0%
d) 4:1
Explanation
Use P=V2 / R Answer: (a)
Q.30
A 60 watt incandescent lamp operates at 120V. The number of electrons passing through the filament per second
0%
a)1.6 × 1012
0%
b) 3.125 × 1018
0%
c)7200
0%
d)12.5 × 1018
Explanation
Use P=VI Q=I × tt=1 sec ∴ Q=I Q=ne ne=I Answer: (b)
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