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Physics NEET MCQ
Quiz 3
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Q.1
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.2
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.3
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.4
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.5
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.6
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.7
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.8
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.9
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.10
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.11
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.12
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.13
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.14
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.15
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.16
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.17
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.18
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.19
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.20
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.21
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.22
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.23
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.24
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)
Q.25
Two bulb of equal wattage, one having carbon filament, are connected in parallel to the mains
0%
a) both bulbs glow equally
0%
b) carbon filament bulb glows more
0%
c)tungsten filament bulb glows more
0%
d)none of these
Explanation
Resistance of carbon decreases on heating. and P ∝ 1/rAnswer: (b)
Q.26
A constant potential difference is applied across the ends of a wire. Which one of the following operations will reduce the rate of heat generation to half
0%
a) Both length and diameter are halved
0%
b) Both length and diameter are doubled
0%
c)Diameter is doubled and length is halved
0%
d)Diameter is halved and length is doubled
Explanation
Answer:(a)
Q.27
Fourty electric bulbs are connected in series across a 220V supply. After one bulb is fused the remaining 39 are connected again in series across the same supply. The illumination will be
0%
a) more with 40 bulb than with 39
0%
b) more with 39 bulbs than with 40
0%
c)equal in both the cases
0%
d)in the ratio 402 : 392
Explanation
given that voltage is same but one bulb is less thus resistance of combination decreases and heat H ∝ (1/r) Hence combination of 39 bulb will glow more Answer:(b)
Q.28
A 3°C rise of temperature is observed in a conductor by passing a certain current. When the current is doubled, the rise of temperature will be
0%
a) 15°C
0%
b) 12°C
0%
c) 9°C
0%
d) 3°C
Explanation
H ∝ I2 Answer: (b)
Q.29
Two electric bulb rated P1 watt, V volt and P2 V volt are connected in parallel across V volts mains, the the total power is
0%
b) √(P1P2)
0%
c)P1P2 / (P1 + P2)
0%
d)(P1 + P2) / P1P2
0%
a)P₁ + P₂
Explanation
Answer: (a)
Q.30
Find the power wasted in the transmission cables of resistance 0.05Ω when 10kW is transmitted at 200 volts
0%
a) 0.0125 kW
0%
b) 0.125 kW
0%
c)25 kW
0%
d)37.5 kW
Explanation
I=P/V=10,000/200=50 AH=I2R=2500×0.05=0.125kWAnswer: (b)
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