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Quiz 1
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Q.1
The power of heater is 750W at 1000°C. What will be its power at 200°C if α=4 ×10-4C
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a) 500 W
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b) 990 W
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c)250 W
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d)1500 W
Explanation
Answer: (b)
Q.2
Two electric bulbs, one of 200 volt - 60 watt and the other of 200 volt - 200 watt are connected in a house wiring circuits
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a)they have equal currents through them
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b) the resistance of the filaments in both the bulbs is same
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c)the resistance of the filament in 60 watt bulb is more than the resistance in 200 watt bulb
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d)the resistance of the filament in 200 watt bulb is more than the resistance in 60 watt bulb
Explanation
Answer: (c)
Q.3
The electric bulbs have tungsten filaments of same length. If one of them gives 60 watts and the other 100 watts then
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a) 100 watts bulbs has thicker filament
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b) 60 watts bulb has thicker filament
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c)both filaments are of same thickness
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d)it is possible to get different wattage unless the lengths are different
Explanation
Answer: (a)
Q.4
Find the power wasted in the transmission cable of resistance 0.05Ω when 10 kW is transmitted at 200 volts
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a) 0.0125 kW
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b) 0.125 kW
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c)25 kW
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d)37.5 kW
Explanation
Answer:(b)
Q.5
A current of 0.5 amp flows in a 60 W lamp, then the total charge passing through it in one hour will be
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a) 1800 coulombs
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b) 2400 coulombs
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c) 3000 coulombs
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d) 36000 coulombs
Explanation
Answer: (a)
Q.6
A resistance coil of 60Ω is immersed in 42 kg of water. A current of 7A is passed through it. The rise in temperature of water per minutes is
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a)4°C
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b) 8°C
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c)1°C
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d)12°C
Explanation
Answer: (c)
Q.7
It is well known that resistance of the filament of bulb charges with temperature. If an electric bulb rated 220 volt, 100 watt is connected to a source of voltage 220(0.5) volt then the actual power will be
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a) 100(0.8) watt
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b) (100) ( 0.8)2 watt
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c)more than 100(0.8) watt but less than 100 watt
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d)more than 100(0.8) 2 watt but less than 100(0.8) watt
Explanation
Answer:(d)
Q.8
A wire when connected to 220V means supply has power dissipation PNow the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is PThen P2 : P1 is [ AIEEE 2002]
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a)1
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b) 4
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c)2
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d)3
Explanation
P1=V2 ROn connect two pieces in parallel effective resistance will be R/4Thus P2=V2 / (R/4)=4P1 P2 : P1=4 : 1Answer: (b)
Q.9
If 2.5kW power is transmitted through a 10Ω line at 22000Volt, the power loss in the form of heat will be [ AFMC 1998]
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a)100 watt
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b) 10 watt
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c)0.1 watt
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d)1 watt
Explanation
Power=VI 2200=22000 ×II=0.1 A Power loss as heat=I 2RPower loss's as heat=(0.1) 2 ×10=0.1 WAnswer: (c)
Q.10
A relation between Faraday's constant F, chemical equivalent E and Electrochemical equivalent Z [ AFMC 2000]
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a) F=EZ2
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b) F=E / Z2
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c)F=E2 / Z
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d)F=E / Z
Explanation
According to Second law of electrolysisAnswer: (d)
Q.11
Three equal resistors are connected as shown in figure. The maximum power consumed by each resistor is 18 watt. Then maximum power consumed by the combination is [ AFMC 2001]
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a) 36 watt
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b) 18 watt
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c)27 watt
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d)54 watt
Explanation
Given that all the resistance are equal let its value be 'R'Maximum power consumed is 18watt thus I2R=18 Now total resistance in circuit will br R/2 +R=3R/2Power consumed by combination=I2(3R/2)P=I2R(3/2)P=18(3/2)P=27 watt Answer:(c)
Q.12
The filament of bulb is made of :.. [ AFMC 2003]
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a) mercury
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b) copper
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c) tungsten
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d) none of these
Explanation
Answer: (c)
Q.13
Current provided by battery is maximum when ... [ AFMC 2004]
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a)internal resistance is equal to external resistance
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b) internal resistance is greater than external resistance
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c)internal resistance is less than external resistance
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d)none of these
Explanation
Answer: (a)
Q.14
The mass of product liberated on anode in an electrochemical cell depends upon [ AIEEE 2002]
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a) ( It)2
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b) It
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c)I/t
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d)I2t
Explanation
According to Faraday's law of electrolysis m=ZItAnswer: (b)
Q.15
A heater coil connected to a supply of a 220V is dissipating some power PThe coil is cut into two halves and the two halves are connected in parallel. The heater now dissipates a power PThe ratio of P1: P2 is . [ AFMC 2004]
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a) 2:1
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b) 1:2
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c)1:4
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d)4:1
Explanation
If R is the resistance of the coil, power of the coil P1=v2R Coil cut in two halves thus resistance of each halve is R/2 Such two halves are connected in parallel thus equivalent resistance is R/4 Power of the coil P2=(V2×4 )/ R P2=4P1 P1: v2R=1 :4Answer: (c)
Q.16
Which of the following is secondary cell [ AFMC 2004]
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a)Voltaic cell
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b) Daniell cell
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c)Leclanche cell
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d)Edison cell
Explanation
Answer:(d)
Q.17
Five equal resistors when in series dissipated 5 watt power. If they are connected in parallel, the power dissipated will be [ AFMC 2009]
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a) 25 watt
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b) 50 watt
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c) 100 watt
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d) 125 watt
Explanation
Let R be the resistance . When five connected in series total resistance is 5R Power dissipated 5=V2 / 5R 25=V2 / R When connected in parallel equivalent resistance is R/5 Power dissipated P=V2 ×5) / R P=25×5=125 watt Answer: (d)
Q.18
Two identical galvanometers are taken, one is to be converted into an ammeter and other into a milliammeter. Shunt of milliammeter compared to ammeter is [AIIMS 1999]
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a) less
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b) more
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c)zero
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d)none of these
Explanation
The shunt of milliammeter will be of larger resistance, because a large part of the main current will pass through the coil of the gravimeter which will work as a smaller range ammeter.Answer: (b)
Q.19
A current of 2A passing through a conductor produces 80J of heat in 10 sec's. The resistance of conductor in ohm is ..[ CBSE-PMT 1989]
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a) 0.5
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b) 2
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c)4
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d)20
Explanation
H=I2Rt 80=(2)(2)R(10) R=2ΩAnswer: (b)
Q.20
Two identical batteries each of e.m.f 2V and internal resistance 1Ω are available to produce heat in an external resistance by passing a current through it. The maximum power that can be developed across R using these batteries is [1990]
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a) 3.2W
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b) 2.0 W
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c) 1.28W
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d) (8/9) W
Explanation
maximum power will develop across resistance R if maximum current flows through R and R is equal to the internal resistance of batteryFor maximum current two batteries should be connected in series. Also R=1+1=2ΩHence power developed across resistance R will be P=I2R --eq(1)Now form the formula for terminal voltage V=E - IrIR=E - IrFor two batteriesIR=2E-I(2r)I=2E/ (R +2r)Substituting E=2V, R=2Ω r=1ΩI=4/(2+2)=1 amp substituting value of I in equation 1 we getP=(1)(1) (2)=2 watt Answer:(b)
Q.21
Faraday's laws are consequence of conservation of ..{CBSE PMT 1991]
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a) Energy
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b) energy and magnetic field
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c) charge
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d) magnetic field
Explanation
Faradays laws are based on the conservation of electrical energy into mechanical energy. Which is according to law of conservation of energy Answer: (a)
Q.22
In electrolysis, the amount of mass deposited or liberated at an electrode is directly proportional to [ CBSE-PMT 2000]
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a)square of electric charge
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b) amount of charge
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c)square of current
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d)concentration of electrolyte
Explanation
By Faraday's Ist LawAmount deposited m=ZIt=ZqAmount deposited is directly proportional to chargeAnswer: (b)
Q.23
A heating coil is labeled 100W, 220V. the coil is cut in half and the two pieces are joined un parallel to the same source. the energy now liberated per second is ... [ CBSE-PMT 1995]
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a) 25J
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b) 50J
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c)200J
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d)400J
Explanation
Let resistance of Coil be 'R' is cut in two parts and connected in parallel thus equivalent resistance will be R/4Thus power will be 4 times of initial=4×100=400JAnswer: (d)
Q.24
Forty 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.. [ CBSE-PMT1989]
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a) more with 40 bulbs than with 39
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b) more with 39 bulbs than with 40
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c)equal in both the cases
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d)in the ratio 402 : 392
Explanation
voltage is same and resistance is reducedHeat ∝(1/R) . Hence combination of 39 bulbs will glow more Answer:(b)
Q.25
Direct current is passed through a copper sulphate solution using platinum electrodes. The elements liberated at the electrodes are ... [ CBSE-PMT 1993]
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a) copper at anode and sulphur at cathode
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b) sulphur at anode an copper at cathode
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c) oxygen at anode and copper at cathode
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d) copper at anode and oxygen at cathode
Explanation
in the electrolysis of copper sulphate solution, oxygen is liberated at anode and copper is deposited at cathode. Answer: (c)
Q.26
If the resistance of a conductor is 5 Ω at 50°C and 7 Ω at 100°C, then the mean temperature coefficient of resistance ( of material ) is .. [ CBSE-PMT 1996]
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a)0.001 / °C
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b) 0.004 / °C
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c)0.006 / °C
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d)0.008 / °C
Explanation
Resistance varies with temperature as , let reference temperature be 0°C R=Ro [ 1 + α(Δt)]Ist case : 5=Ro [ 1 + α(50)] --eq(1)IIst case : 7=Ro [ 1 + α(100)] --eq(1)Divide I by II we getAnswer: (a)
Q.27
A 100 W and 200 V bulb is connected to a 160V power supply. The power consumption would be .. [ CBSE-PMT 1997]
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a) 125 W
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b) 100 W
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c)80 W
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d)64 W
Explanation
Resistance of the bulb R=V2 / PHere P=100 W and voltage=200 VThus R=2002 / 100R=400 Ω When bulb is connected to 160 V Current in bulb=V/R=160 / 400 A Power Consumed=VI=160 × ( 160 /400 )=64 WAnswer: (d)
Q.28
Three equal resistors connected across a source of e.m.f. together dissipate 10 watts if same resistors are connected in parallel across the same source of e.m.f what will be the power dissipated? [ CBSE-PMT 1998]
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a) 10
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b) 10/3
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c)30
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d)90
Explanation
Let each resistance be R thus total resistance is 3RWhen connected in parallel new resistance R/3Thus resistance in parallel is 1/9 times the series connections Power dissipated H ∝ 1/RPower dissipated=9 ×10=90 W Answer:(d)
Q.29
If nearly 105 coulombs liberate 1 gm equivalent of aluminum, then the amount of aluminum ( equivalent weight 9), deposited through electrolysis in 20 minutes by a current of 50amp will be... [ CBSE PMT 1998]
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a) 0.6 gm
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b) 0.09 gm
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c) 5.4 gm
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d) 10.8 gm
Explanation
mass liberated m=Z(It)=ZQ given m=9 and Q=9×10-5 ∴ 9=Z×10-5 Z=9×10-5×50×20×60=5.4gm Answer: (c)
Q.30
Si and Cu are cooled to a temperature of 300K then resistivity? [ CBSE-PMT 2001]
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a) For Si increases and for Cu decreases
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b) For Cu increases and for Si decreases
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c)Decreases fro both Si and Cu
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d)Increases for both Si and Cu
Explanation
Conductivity of semiconductor increases with increase in temperature while conductivity of metal decreases with increase in temperatureAnswer: (b)
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