MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Games
NEET
Physics NEET MCQ
Thermal And Chemical Effects Of Current Mcq
Quiz 2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Q.1
A wire has a resistance of 3.1 Ω at 30°C and a resistance 4.5 Ω at 100°C. The temperature coefficient of resistance of the wire... [ CBSE-PMT 2001]
0%
a) 0.0064° /C
0%
b) 0.0034° /C
0%
c)0.0025° /C
0%
d)0.0012° /C
Explanation
Rt=Ro[ 1 + α(Δt)]R1=Ro [ 1 + α (30)]R2=Ro [ 1 + α (100)] Answer: (a)
Q.2
An electric kettle has two heating coils. When one of the coils is connected to an a.c. source, the water in the kettle boils in 10 minutes. When the other coil is used the water boils in 40 minutes. If both the coil are connected in parallel, the time taken by the same quantity of water to boil will be.. [ CBSE-PMT 2003]
0%
a) 15 min
0%
b) 8 min
0%
c)4 min
0%
d)25 min
Explanation
Water boils at same temperature thus amount of heat required is same for all cases. Voltage is same in all cases Now heat H=V2t/ R R=V2t/ H Now R1=(V2×10)/ H R2=(V2×40)/ H Requi=(V2×t)/ H Since R1 and R2 are connected in parallel equivalent resistance=1/R1 + 1/ R2 By substituting values of R1 and R2, Requi in above equation we get Answer:(b)
Q.3
The resistivity of a copper wire... [ CBSE-PMT 2002]
0%
a) increases with increase in its temperature
0%
b) decreases with increase in its cross-section
0%
c) increases with increase in its length
0%
d) increases with increase in its cross-section
Explanation
Answer: (a)
Q.4
Fuse wire is a wire of .. [ CBSE-PMT 2003]
0%
a)low resistance and high melting point
0%
b) high resistance and low melting point
0%
c)high resistance and low melting point
0%
d)low resistance and low melting point
Explanation
Fuse wire is used to control the maximum current flowing through the circuit. Thus it have maximum resistance and minimum melting pointAnswer: (c)
Q.5
When three identical bulbs of 60watt and 200 volts are connected in series to a 200 volt supply, the power drawn by them will be .. [ CBSE-PMT 2004]
0%
a) 20 watt
0%
b) 60 watt
0%
c)180 watt
0%
d)10 watt
Explanation
power p=V2 / R ∴ p ∝ (1/R) when bulbs connected in series effective resistance will be 3R ∴ power will be 1/3 of original=(1/3)×60=20 wattAnswer: (a)
Q.6
Two 220 volt, 100 watt bulbs are connected first in series and then parallel. Each time the combination is connected to 220 volt a.c. supply line. The power drawn by the combination in each case respectively will be .. [ CBSE-PMT 2003]
0%
a) 50 watt, 200 watt
0%
b) 50 watt, 100 watt
0%
c)100 watt, 50 watt
0%
d)200 watt, 150 watt
Explanation
In all the cases voltage is samePower P ∝ (1/R) When bulbs connected in series effective resistance will be 2RThus in series combination power=(1/2) ×100=50 wattWhen bulbs are connected parallel effective resistance will be=R/2 Thus in parallel combination power=2×(100)=200 watt Answer:(a)
Q.7
A battery is charged at a potential of 15V for 8 hours when the current flowing is 10A. The battery on discharge supplies a current of 5A for 15 hours. The mean terminal voltage during discharge is 14V. The "watt-hour" efficiency of the battery is ... [ CBSE-PMT 2004]
0%
a) 87.5%
0%
b) 82.5%
0%
c) 80%
0%
d) 90%
Explanation
Efficiency η=output / Input Output power=VTt=14×5×15=1050 watt Input power=VIt=15×10×8=1200 watt η=1050 /1200=0.875=87.5% Answer: (a)
Q.8
In India electricity is supplied for domestic use at 220V. It is supplied at 110V in U.S.A. If the resistance of a 60W bulb for use in India is R, the resistance of 60W bulb for use in USA will be ... [ CBSE-PMT 2004]
0%
a)R/2
0%
b) R
0%
c)2R
0%
d)R/4
Explanation
Power P=V2 /R for Indian voltage P=(220)2 / R --eq(1) for USA voltage p=1102 / R' --eq(2)from equation 1 and 2 we get(220)2 / R=1102 / R'R'=R/4Answer: (d)
Q.9
A 5 ampere fuse wire can withstand a maximum power of 1 watt in the circuit. The resistance of the fuse wire is .. [ CBSE-PMT 2005]
0%
a) 0.04 Ω
0%
b) 0.2 Ω
0%
c)5.0 Ω
0%
d)0.4 Ω
Explanation
power P=I2R 1=52 × R R=1/25=0.04ΩAnswer: (a)
Q.10
In producing chlorine through electrolysis, 100 watt power at 125V is being consumed. How much chlorine per minute is liberated? E.C.E of chlorine is 0.367×10-6 kg/coulomb [ CBSE-PMT 2006]
0%
a) 21.3 mg
0%
b) 24.3 mg
0%
c)13.6 mg
0%
d)17.6 mg
Explanation
power P=VI I=P/V=100 /125 Q=I×tQ=(100/125)×60=48 Coulombm=ZQm=0.367×10-6×48m=17.6 mg Answer:(d)
Q.11
The total power dissipated in watts in the circuit shown here is .. [ CBSE-PMT 2007]
0%
a) 40 W
0%
b) 54 W
0%
c) 4 W
0%
d) 16 W
Explanation
Total resistance in circuit is 6Ω Power P=V2 / R P=182 / 6 P=54 watt Answer: (b)
Q.12
A steady current of 1.5 amp flows through copper voltmeter for 10 minutes. If the electrochemical equivalent of copper is 30×10-5 g/coulomb, the mass of copper deposited on the electrode will be.. [ CBSE-PMT 2007]
0%
a)0.50 g
0%
b) 0.67 g
0%
c)0.27g
0%
d)0.40g
Explanation
m=ZItwhere Z is the electrochemical equivalent of copperm=30×10-5×1.5×10×60m=0.27 gmAnswer: (c)
Q.13
A current of 3 amp flows through the 2Ω resistor shown in the circuit. The power dissipated in the 5Ω resistor is ... [ CBSE-PMT 2008]
0%
a) 4 watt
0%
b) 2 watt
0%
c)1 watt
0%
d)5 watt
Explanation
1ω and 5Ω are connected in series hence effective resistance is 6ΩSince 2Ω ,4Ω and 6Ω resistance are parallel there fore potential difference across 6Ω resistance will be equal to potential difference across 2Ω resistance ∴ 6I=2× 3I=1 ampNow power across 5Ω resistance=I2RP=12×5P=5 wattAnswer: (d)
Q.14
In producing chlorine by electrolysis 100kW power at 125V is being consumed. How much chlorine per minute is liberated? (E.C.E. of chlorine is 0.367×10-6 kg/C) [ CBSE-PMT 2010]
0%
a) 1.76×10-3 kg
0%
b) 9.67×10-3 kg
0%
c)17.61×10-3 kg
0%
d)3.67×10-3 kg
Explanation
I=P/V=100×103 / 125 A=105 / 60 AE.C.E=0.367×10-6 kg/CCharge per minute=I×60 CCharge per minutes Q=105 / 125 × 60Q=6×106 / 125 C mass m=ZQ m=0.367×10-6 × (6×106 / 125)m=17.616×10-3kg Answer:(c)
Q.15
The thermo e.m.f E in volts of a certain thermocouple is found to vary with temperature difference θ in °C between the two junctions according to the relation E=30 - ( θ2 / 15 )The neutral temperature for the thermocouple will be
0%
a) 30°C
0%
b) 450°C
0%
c) 400 °C
0%
d) 225°C
Explanation
For neutral temperature dE /dθ=0 0=30 -(2/15)θ θ=225°C Answer: (d)
Q.16
If power dissipated in the 9Ω resistor in the circuit shown is 36 watt, the potential difference across the 2Ω resistor is .. [ CBSE-PMT 2011]
0%
a)4 volt
0%
b) 8 volt
0%
c)10 volt
0%
d)2 volt
Explanation
P=I2 RA Power dissipated in 9Ω resistor is 36 watt 36=I12(9) I1=2 ampNow potential across 9Ω resistance=Potential across 6Ω resistance9×2=6×I2 I2=3 amp Total current through 2Ω resistance I=I1 + I2 I=2+3=5 ampNow voltage across 2Ω resistance=IRV=5×2=10 VAnswer: (c)
Q.17
The rate of increase of thermo-emf with temperature at the neutral temperature of a thermo couple [ CBSE-PMT 2011]
0%
a) is positive
0%
b) is zero
0%
c)depends upon the choice of the two materials of the thermocouple
0%
d)is negative
Explanation
We know that E=αt + βt2 At neutral temperature rate of increase in thermo emf dE/dt=0Answer: (b)
Q.18
A thermocouple of negligible resistance produces an e.m.f. of 40µV / °C in the linear range of temperature. A galvanometer of resistance 10 Ω whose sensitivity is 1µA/div, is employed with the thermocouple. The smallest value of temperature difference that can be detected by the system will be.. [ CBSE-PMT 2011]
0%
a) 0.5°C
0%
b) 1.0°C
0%
c)0.1°C
0%
d)0.25°C
Explanation
1 division=1 µACurrent for 1°C=40µV/10Ω=4µA ∴ 1µ=(1/4)°C=0.25°C Answer:(d)
Q.19
What is the energy stored in the capacitor?
0%
a) 72 µJ
0%
b) 128µJ
0%
c) 96mJ
0%
d) 96MJ
Explanation
VD - VE=IR VD - VE=2 × 4=8 Volts Now energy in capacitor E=½ CV2 E=½ × 4× 10-6 ×(8) 2=128micro;J Answer: (b)
Q.20
A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. the heat developed is doubled if [ IIT 1980]
0%
a)both the length and the radius of the wire are halved
0%
b) both the length and the radius of the wire are doubled
0%
c)the radius of the wire is doubled
0%
d)the length of the wire is doubled
Explanation
A Heat produced H=V2 / R and R=ρl / πr2 Thus heat (H) is doubled if both length(l) and radius(r) are doubledAnswer: (b)
Q.21
In the circuit shown in figure the heat produced in the 5Ω resistor due to the current flowing through it is 10 calories per second. Heat generated in the 4Ω resistor is [ IIT 1981]
0%
a) 1 calorie/sec
0%
b) 2 calories/sec
0%
c)3 calories / sec
0%
d)4 calories / sec
Explanation
Potential across 5Ω=Potential across 4Ω + Potential across6 ΩI1×5=I2 ( 4+6)I2=I1 /2 Heat in 5Ω 10× J=I12 5 I12=2× JHeat in 4Ω=I22 4 H=(I1 /2 )2 4H=(2 /4)4=2 calories / sec Answer: (b)
Q.22
A piece of copper and another of germanium are cooled from room temperature to 80K. The resistance of [ IIT 1988]
0%
a) each of them increases
0%
b) each of them decreases
0%
c)copper increases and germanium decreases
0%
d)copper decreases and germanium increase
Explanation
temperature coefficient for Germanium is negative Answer:(d)
Q.23
A 100W bulb B1 and two 60W bulb B2 and B3 are connected to a 250V source, as shown in figure. Now W>1 , W2 and W3 are the output powers of the bulbs B1, B2 and B3 respectively. Then [ IIT 2002]
0%
a) W1 > W2=W3
0%
b) W1 > W2 > W3
0%
c) W1 < W2=W3
0%
d) W1 < W2 < W3
Explanation
R=V2 / P Here two bulbs are connected in series bulb 1 and bulb 2. Thus total resistance of combination is R1 + R2 Thus voltage across bulb 1 is V1 and bulb 2 is V2 which can be given as Thus new power of the bulbs W and W2 and W3 will be Now W3 : W2 : W1 will be From above it is clear that option 'd' is correct Answer: (d)
Q.24
A constant voltage is applied between the two ends of uniform metallic wire. Some heat is developed in it. The heat developed is doubled if
0%
a) the radius of the wire is doubled
0%
b) the length of the wire is doubled
0%
c) both the length and radius of the wire are halved
0%
d) both the length and radius of the wire are doubled
Explanation
Answer: (d)
Q.25
The negative Zn pole of a Daniell cell, sending a constant current through a circuit, decreases in mass by 0.13g in 30 minutes. If the electrochemical equivalent of Zn and Cu are 32.5 and 31.5 respectively, the increase in the mass of the positive Cu pole in this time is [ AIEEE 2003]
0%
a) 0.180 g
0%
b) 0.141g
0%
c)0.126g
0%
d)0.242g
Explanation
According to Faraday's law of electrolysis m=ZqFor same q m ∝Z Answer:(c)
Q.26
A 220 Volt, 1000 watt bulb is connected across 110 Volt mains supply. The power consumed will be [ AIEEE 2003]
0%
a) 750 watt
0%
b) 500 watt
0%
c) 250 watt
0%
d) 1000 watt
Explanation
Resistance of bulb=V/p R=(220)2 / 1000 Now connected to 110 V supply power p= Answer: (c)
Q.27
Time taken by a 836W heater to heat one litre of water from 10°C to 40°C is [ AIEEE 2004]
0%
a)150 s
0%
b) 100 s
0%
c)50 s
0%
d)200 s
Explanation
Heat produced by current=Heat absorbed by waterThus P×t=MCΔT specific heat of water 4180 J836×t=1×4180×(40-1)=4180×30 t=4180×30 / 836 t=150sAnswer: (a)
Q.28
A heater coil is cut into two equal parts and only one part is now used in the heater. the heat generated will now be [ AIEEE 2009]
0%
a) four times
0%
b) doubled
0%
c)halved
0%
d)one forth
Explanation
H=V2 t / R Resistance of half coil=R / 2 ∴ As R reduced to half, 'H' will be doubled Answer: (b)
Q.29
Two voltmeter, one copper and other of silver, are joined in parallel. When a total charge q flows through the voltmeters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are Z1 and Z2 respectively the charge which flows through the sliver voltmeter is [ AIEEE 2005]
0%
a) q /{1 +(Z2 / Z1)]
0%
b) q /{1 +(Z1 / Z2)]
0%
c)q ( Z2 / Z1)
0%
d)q(Z1 / Z2)
Explanation
MAss deposited=m=Zq Z ∝ (1 /q) Z1 / Z2=q2 / q1 ---eq(1) also q=q1 +q2 ---eq(2) dividing eq(2) by q2 ∴ q/q2=q1 / q2 + 1 q2=q / [ 1 +q1 / q20}--eq(3) From equation (i) and (iii) q2=q / [1 +(Z2 / Z1)] Answer:(a)
Q.30
A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is kept hot and the other is kept cold, then, an electric current will [ AIEEE 206]
0%
a) flow from Antimony to Bismuth at the hot junction
0%
b) flow from Bismuth to ANtimony at the cold junction
0%
c) now flow through the thermocouple
0%
d) flow from Antimony to Bismuth at the cold junction
Explanation
At cold junction, current flows from Antimony to Bismuth ( because current flows from metal occurring later in the series to the metal occurring earlier in the thermoelectric series ) Answer: (d)
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Support mcqgeeks.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page