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Physics NEET MCQ
Quiz 6
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Q.1
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.2
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.3
Fuse wire should have [ BHU 1999]
0%
a)low resistance, high melting point
0%
b) low resistance low melting point
0%
c)high resistance, low melting point
0%
d)high resistance, high melting point
Explanation
Answer: (c)
Q.4
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
0%
c)25 minutes
0%
d)5 minutes
Explanation
for parallel combination of hater use formulaAnswer: (a)
Q.5
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
0%
c)200 watt, 150 watt
0%
d)50 watt, 200 watt
Explanation
parallel combination P=100+100=200 Wseries combination=(100×100) / ( 100+100)=50 W Answer:(d)
Q.6
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.7
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?
0%
a) 2
0%
b) 4
0%
c)6
0%
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.8
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
0%
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.9
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.10
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.11
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.12
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
0%
a) 1:3:5
0%
b) 5:3:1
0%
c)1:15:125
0%
d)125:15:1
Explanation
We know that H=I2R Answer:(d)
Q.13
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
0%
d) 1125 W
Explanation
Answer: (d)
Q.14
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
0%
a)14.7 A
0%
b) 5 A
0%
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.15
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
0%
b) a3R / 3b
0%
c)a3R / 2b
0%
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.16
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
0%
b) W1 > W2 > W3
0%
d)W1 < W2 < W3
0%
a) W₁ > W₂=W₃
0%
c)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.17
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
0%
a) 33.3%
0%
b) 66.6%
0%
c) 87.7%
0%
d) 93.3%
Explanation
efficiency η=Energy used / Energy supplied Answer: (d)
Q.18
Time taken by 836W water to heat one litre of water from 0°C to 40°C is
0%
a)50 sec
0%
b) 100 sec
0%
c)150 sec
0%
d)200 sec
Explanation
Answer: (c)
Q.19
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
0%
c)1:16
0%
d)4:9
Explanation
Answer:(a)
Q.20
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.21
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.22
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
0%
b) A will draw more power than B
0%
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)
Q.23
Ratio of the amount of heat developed in the four arms of a balanced wheat stone bridge, when the arms have resistance P=100Ω, Q=10Ω, R=300 Ω and S=30 Ω respectively is
0%
a) 3:30:1:10
0%
b) 30:3:10:1
0%
c)30:10:1:3
0%
d)30:1:3:10
Explanation
From diagram current through P and Q=V (110) Current through R and S - V/(330) Now heat=I2R t By taking ratio HP : HQ : HR : HS= 100/ (110)2 : 10 / (110)2 : 300/ (330)2 : 30 /(330)2 30:3:10:1 Answer:(b)
Q.24
Twenty one similar electric bulbs, each of resistance r are connected in series across 220 V supply. After one bulb is fused, the remaining bulbs are connected again in series cross the same supply. The percentage total change in illumination of bulbs is
0%
a) decrease by 5%
0%
b) increase by 5%
0%
c) decrease by 10%
0%
d) increase by 10%
Explanation
Answer: (b)
Q.25
The current capacity of a storage cell is 3 Ah. The maximum current it can supply for half hour is
0%
a)1.5 A
0%
b) 3 A
0%
c)4.5 A
0%
d)6.0 A
Explanation
Answer: (d)
Q.26
The current capacity of the charged secondary cell does not depend on
0%
a) rate of discharge
0%
b) temperature
0%
c)amount of active material
0%
d)rate of charging
Explanation
Answer: (d)
Q.27
The amount of ions liberated by 96500C charge passed through the electrolyte is called
0%
a) electrochemical equivalent
0%
b) chemical equivalent
0%
c)gram equivalent
0%
d)none of the above
Explanation
Answer:(c)
Q.28
Which of the following acts as depolariser in Leclanche cell?
0%
a) MnO2
0%
b) NH4Cl
0%
c) Mn2O3
0%
d) ZnCl2
Explanation
Answer: (a)
Q.29
The power of heater is 750W at 1000°C. What will be its power at 200°C if α=4 ×10⁻⁴C
0%
a) 500 W
0%
b) 990 W
0%
c)250 W
0%
d)1500 W
Explanation
Answer: (b)
Q.30
The power of heater is 500 W at 800° C. What will be its power t 200°C if α=4 ×10⁻⁴ per °C?
0%
a)484 W
0%
b) 672 W
0%
c)526 W
0%
d)611 W
Explanation
bResistance at 800° R'=V2 /P=220×220/ 500=484/5Let R be the resistance at 200°C and Ro be the resistance at 0°CFor the reaction of resistance and temperatureR=Ro [ 1 + αΔθ] R=Ro [ 1 + α200] R'=Ro [ 1 + α 800] Answer: (d)
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