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Physics NEET MCQ
Quiz 4
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Q.1
Two electric bulbs have tungsten filament of same length. If one of them gives 60watts and other 100watts, then
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a) 100 watt bulb has thicker filament
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b) 60 watt have thicker filament
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c)both filaments are of same thickness
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d)the thickness of filaments can not be compared
Explanation
R ∝ 1/P ∴ Resistance of 60 W bulb is more than that of 100 W bulb. Thus, the thickness of filament of 100W bulb is more. Answer:(a)
Q.2
A house served by a 220V supply line. In a circuit protected by a fuse marked 9A , the maximum number of 60W lamps in parallel that can be turned on is
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a) 44
0%
b) 22
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c) 20
0%
d) 33
Explanation
We know that in parallel connection power get added 220× 9=n×60 n=33 Answer: (d)
Q.3
A tap supply water at 22°C. A man takes 1 litre of water per minute at 37°C from the geyser. The power of geyser is
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a)2100 W
0%
b) 525W
0%
c)1050W
0%
d)1575W
Explanation
Heat produced by geyser=heat absorbed by water P t=mSΔθp×60=1000×4.2×(37-22) P=1050 WAnswer: (c)
Q.4
A uniform wire when connected directly across a 200V line produces heat H per sec. If the wire is divided into n parts and all parts are connected in parallel across a 200V line, the heat produced per second will be
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a) H
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b) nH
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c)n2H
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d)H/n²
Explanation
H=V2 / R resistance of each is R/n when connected in parallel to tal resistance of combination=R/n2 H'=V2 / [R/ n2]=n2 H Answer: (c)
Q.5
Two electric bulbs first of 2200volts, 100 watt and the second of 220 volt, 25 watt are connected in series across a 220 volt line. then the electric currents in the first and second bulb are respectively
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a) (1/11) amp and (1/11) amp
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b) (20/44) amp and (5/44) amp
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c)(25/88) amp and (25/88) amp
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d)(1/22) amp and (1/22) amp
Explanation
connected in series so current will be same calculate individual bulbs resistance from equation R=V2 /R then calculate total resistance and then V=IR to find current Answer:(a)
Q.6
A cell send a current through a resistance R1 for time t, next the same cell sends current through another resistance R2 for the same time t. If the same amount of heat is developed in both the resistance, then the internal resistance of the cell is
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c) √(R1 R2)
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a) (R₁ + R₂) /2
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b) (R₁ - R₂) /2
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d) √(R₁ R₂) / 2
Explanation
Emf of cell E=IR1 + Ir ∴ I=E / (R1 +r) Heat produced=I2R1t Answer: (c)
Q.7
The two head lamps of a car are in parallel. They to gather consume 48 watt with the help of a 6 volt battery. The resistance of each bulb is
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a) 0.67 ohms
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b) 3.0 ohms
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c)4.0 ohms
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d)1.5 ohms
Explanation
Answer: (d)
Q.8
A 200 watt and 100 watt bulbs, both meant for operation at 220 volt, are connected in series, When connected to a 220 volt supply, the power consumed by them will be
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a) 33 watt
0%
b) 66 watt
0%
c)100 watt
0%
d)300 watt
Explanation
Bulbs are connected in series1P2 / P1 + p2 Answer:(b)
Q.9
TWo identical heaters, each marked 1000 watt, 250 volt are placed in parallel with each other and connected to a 250 volt supply. Their combination rate of heat will be
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a) 250 W
0%
b) 500 W
0%
c) 1000 W
0%
d) 2000 W
Explanation
Answer: (d)
Q.10
An immersion heater is rated 830 watt, 220 volt. It is used on 220 volt line to heat 1 litre of water from 20°C to 30°C. Then the time taken is about
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a)100 sec
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b) 50 sec
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c)836 sec
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d)418 sec
Explanation
Heat supplied by heater=heat absorbed by water pt=msΔθs for water 4.2 joules/°gmAnswer: (b)
Q.11
Electric bulbs rated 50 watt and 100 volt and glowing at full power, are used in parallel with a battery of emf, 120Volts and internal resistance 10 ohm. The maximum number of bulbs that can be connected in the circuit when glowing at full power is
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a) 8
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b) 6
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c)4
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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 - Ir 100=120 - 10I I=2ACurrent capacity of each bulb i=50/100=0.5 AmpIf is the number of bulbs connected parallel then I=n(i) n=2/0.5=4Answer: (c)
Q.12
Two resistors having equal resistances are joined in series and current is passed through the combination. Neglect any variation in resistance as the temperature changes in a given time interval: [ MPPMT 1999]
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a) Equal amount of thermal energy must be produced in the resistors
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b) Unequal amount of thermal energy may be produced
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c)The temperature must rise equally in the resistors
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d)The temperature must rise unequally in the resistors
Explanation
Answer:(a)
Q.13
You are given resistance wire of length 50cm and a battery of legible resistance. In which case the maximum amount of heat is generated
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a)when the wire is connected across the battery directly
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b) when the wire is divided into two parts and both the parts are connected across the battery in parallel
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c)when the wire is divided into four parts and all the four parts are connected across the battery in parallel
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d)when only half the wire is connected across the battery
Explanation
Answer: (c)
Q.14
Resistance R1 and R2 are joined in parallel and a current is passed so that the amount of heat liberated in H1 and H2 respectively. The ratio H1 / H2 has the value: [ MPPMT 1994]
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c)R12 / R22
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d)R22 / R12
0%
a) R₂ / R₁
0%
b) R₁ / R₂
Explanation
Answer: (a)
Q.15
TWo bulbs of 500W and 300W are manufactured to operate on 220 V line. If their resistance are R1 and R2 respectively, the value of R1 and R2 is
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a) 5/3
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b) 3/5
0%
c)25/9
0%
d)9/25
Explanation
Answer:(b)
Q.16
An electric kettle work at 220 volt and 4 A current. To boil 1kg water at room temperature 20°C it will take time [ Boiling point of water 100°C]
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a) 6.0 minutes
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b) 6.3 minutes
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c) 12.6 minutes
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d) 12.8 minutes
Explanation
Use formula Pt=mSΔθ Answer: (b)
Q.17
Ten identical electric bulbs,each rated 220 volt, 50 watt are used in parallel on 220 line for 10 hours per day in a month of 30 days. Then the electrical energy consumed in kilowatt hours is [ CPMT 19991]
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a)1500
0%
b) 15000
0%
c)15
0%
d)150
Explanation
Answer: (d)
Q.18
Element in electric stove is made of
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a) copper
0%
b) nichrome
0%
c)platinum
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d)tungsten
Explanation
Answer: (b)
Q.19
The resistance of an electric bulb, marked 220V, 100W is [ MPPMT 1997]
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a)2.2 Ω
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b) 5/11 Ω
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c)500/11 Ω
0%
d)484 Ω
Explanation
Answer:(d)
Q.20
If the current in an electric bulb decreases by 0.5 %, then the power in the bulb decreases by approximately
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a) 0.5 %
0%
b) 1.0 %
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c) 2.0 %
0%
d) 0.25 %
Explanation
Answer: (b)
Q.21
The power rating of an electric motor which draws a current of 3.75 amp when operated at200volts is about
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a)1 H.P.
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b) 500 watts
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c)54 watts
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d)750 H.P.
Explanation
Answer: (a)
Q.22
A resistance 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 : [ CPMT 1985]
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a) decreases
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b) increases
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c)remains the same
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d)any one of the above depending upon the relative value of R1 and R2
Explanation
Answer: (a)
Q.23
The fuse wire is made of
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a) copper
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b) tungsten
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c)lead-tin alloy
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d)nichrome
Explanation
Answer:(c)
Q.24
Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per kilometer is 8 volt and the average resistance per kilometer is 0.5 Ω. The power loss in the wires is
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a) 19.2 watt
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b) 19.2 kilowatt
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c) 19.2 joule
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d) 19.2 kWh
Explanation
I=V/R=8/0.5=16A P=I2R P=(16)2 ×150×0.5=19.2 kW Answer: (b)
Q.25
A combination of two resistance of 2Ω and 2/3 Ω connected in parallel is joined across a battery of emf of 3 volt and of negligible internal resistance. The energy given out per sec will be
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a) (1/2) × 3 2 joules
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b) (1/2)2 × 3 2 joules
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c) 2 × 3 joules
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d) 32 × 2 joules
Explanation
Answer: (d)
Q.26
In the circuit shown , the heat produced in the 5Ω resistor due to current flowing it, is 10 calories per second. Then the heat generated in the 4 Ω resistor is
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a)1 calories per second
0%
b) 8 calories per sec
0%
c)3 calories per second
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d)4 calories per sec
Explanation
Let i1 be the current through resistor 5Ω and 4Ω Then (i12 ×5×t ) / J=10 cal/secHeat generated in 4Ω resistor (i12 ×4×t ) / J=8 cal/secAnswer: (b)
Q.27
The resistance of coil of a small motor of 10 watt, 20V is 12Ω. The rate of heat produced in the motor is
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a) 3 W
0%
b) 7 W
0%
c)10 W
0%
d)33.3 W
Explanation
I P/ V=10/20=1/2 H=I2R H=(1/4)×12=3 J/sec=3WAnswer: (a)
Q.28
Electric room radiator which operates at 225 volts has resistance of 50 Ω. Power of the radiator is approximately:
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a) 100 watt
0%
b) 450 Watt
0%
c)750 watt
0%
d)1000 watt
Explanation
Answer:(d)
Q.29
How much electric energy is consumed in using a 100 W lamp for 6 hours every day for 30 days?
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a) 0.5 kJ
0%
b) 0.5 kWh
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c) 18 kJ
0%
d) 18 kWh
Explanation
Answer: (d)
Q.30
A lamp is marked 60W, 220V. If it operates at 200V, the rate of consumption of energy will
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a)increase
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b) decrease
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c)remain unchanged
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d)cant say it may increase or decrease
Explanation
Answer: (b)
0 h : 0 m : 1 s
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