MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Games
NEET
Physics NEET MCQ
Current Electricity Mcq
Quiz 11
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
In the circuit, each cell has an emf of 1.5 V and r=0.4Ω. The current I in the 36Ω resistor is
0%
a) 0.5 amp
0%
b) 0.1 amp
0%
c)0.083 amp
0%
d)0.2 amp
Explanation
Potential between PQ=6V and internal resistance=0.8Ω 12, 18 and 36 ohm resistance are parallel equivalent resistance=6 Ω Thus circuit reduces to Resultant potential=4.5 V and total resistance in circuit=9Ω Current in circuit=4.5/9=0.5 amp Now we may consider equivalent of 12, 18 ohm resistance=36/5 ohm is connected parallel to 36 ohm By division of current current in 36Ω resistance Answer:(c)
Q.2
A galvanometer together with unknown resistance in series is connected to two identical batteries each of 1.5 V. When the batteries are connected in series the galvanometer registers a current of 1 ampere. When the batteries are in parallel the current is 0.6 ampere. The internal resistance of battery is
0%
a) 3 Ω
0%
b) 2 Ω
0%
c) (1/3) Ω
0%
d) (1/2) Ω
Explanation
Case I: When batteries connected in series together with unknown resistance total resistance=R'=2r + R + G G is resistance of galvanometer total potential=1.5 +1.5=3 V, given current=1A ∴ 1=3 / (2r + R + G) 2r + R + G=3 --eq(1) CaseII: When batteries are parallel resistance of circuit=[(r/2) + R +G] total potential=1.5 and current 0.6 given 0.6=1.5 / [(r/2) + R +G] --eq(2) from equation (1) and (2) r=(1/3) ΩAnswer: (c)
Q.3
Two cells with emf ε1=1.3 V and ε2=1.5 V are connected as shown. The voltmeter reads 1.15V. If r1 and r2 be the internal resistance of the cells, the ratio r1 / r2 is
0%
a)2/5
0%
b) 3/1
0%
c)4/3
0%
d)3/2
Explanation
Circuit may be redrawn as By applying Kirchoff's law I(r1 + r2 ) - ε1 + ε2=0 Thus I=- 0.2 /(r1 + r2 ) V>AB=1.5 - [0.2 /(r1 + r2 )] 1.45=1.5 - [0.2 /(r1 + r2 )] ∴ 0.2 /(r1 + r2 )=0.05 on solving r1 / r2=3/1Answer: (b)
Q.4
Each resistance in the circuit is r. The equivalent resistance between A and B is
0%
a) r/4
0%
b) 4r
0%
c)2r/5
0%
d)0
Explanation
Circuit can be redrawn as From figure R=2r/5Answer: (c)
Q.5
An ammeter and voltmeter are joined in series to a cell. There readings are A and V respectively. If a resistance is now joined in parallel with voltmeter.
0%
a) both A and V will increase
0%
b) both A and V will decrease
0%
c)A will decrease, V will increase
0%
d)A will increase, V will decrease
Explanation
Due to the new resistance effective resistance decreases hence current in Ammeter will increaseAs V1 + V 2=Constant V1 increases hence V2 decreases Answer:(d)
Q.6
The equivalent resistance between A and B is
0%
a) 2R/3
0%
b) R/3
0%
c) R
0%
d) 3R
Explanation
Vertical both resistance are ineffective as potential at upper end and lower end is same Given circuit is there 2R resistance connected in parallel Answer: (a)
Q.7
In a circuit shown, the galvanometer reads zero. If batteries have negligible internal resistance, the value of C will be
0%
a)10 Ω
0%
b) 100 Ω
0%
c)200 Ω
0%
d)500 Ω
Explanation
Potential at 'a' must be 2V and potential at 'b' must be 0 , so as current through G is zero and cell of 2V becomes ineffective Now potential across 500Ω must be 10V Thus current through 500 ohm resistance is 10/500=1/50 same current passes through X to produce potential drop of 2V This 2=(1/50)X X=100ΩAnswer: (b)
Q.8
The reading in the ammeter is
0%
a) 1 A
0%
b) 2 A
0%
c)0.67 A
0%
d)1.5 A
Explanation
Apply Kirchoff law for loop abcdea and loop cdfghc We get value of I1=I2=1/3Thus Current through ammeter=2/3=0.67AAnswer: (c)
Q.9
The potential difference between the points x and y in the adjoining figure will be
0%
a)zero
0%
b) 50 V
0%
c)10 V
0%
d)100V
Explanation
given diagram is of Whetstone balanced bridge. Potential between x and y is zero Answer:(a)
Q.10
In the adjoining diagram find the current through 10Ω resistance
0%
a) 0.4 A towards O
0%
b) 0.4 A away from O
0%
c) 0.6 A towards O
0%
d) 0.6 A away from O
Explanation
Answer: (b)
Q.11
In the circuit given below. The current in the arm AD will be , each resistance is 10Ω
0%
a)2i/5
0%
b) 3i/5
0%
c)4i/5
0%
d)i/5
Explanation
Answer: (a)
Q.12
A steady current is passing through a linear conductor of non-uniform cross-section. the current density in the conductor is
0%
a) independent of area of cross section
0%
b) directly proportional to area of cross section
0%
c)inversely proportional to area of cross-section
0%
d)inversely proportional to the square root of area of cess-section
Explanation
Answer: (c)
Q.13
A metallic block has no potential difference applied across it. Then the mean velocity of free electron in a conductor at absolute temperature T is
0%
a) proportional to T
0%
b) proportional to √T
0%
c)zero
0%
d)finite but independent of temperature
Explanation
Answer:(c)
Q.14
Ohm's law is valid when the temperature of the conductor is
0%
a) constant
0%
b) very high
0%
c) very low
0%
d) varying
Explanation
Answer: (a)
Q.15
Two wires of the metal have the same length but their cross-sections are in the ratio 3:1 They are joinedin series: The resistance of the thicker wire is 10Ω . The total resistance of the combination will be
0%
a)40
0%
b) 40/3
0%
c)5/2
0%
d)100
Explanation
resistance ∝ (1/A) thus Resistance of thin wire=30 ΩAnswer: (a)
Q.16
A potentiometer wire has length 10 m and resistance 20 Ω. A 2.5V battery of negligible internal resistanceis connected across the wire with an 80Ω series resistance. The potential gradient on the wire will be:
0%
a) 2.5 × 10 ×10-4 V/m
0%
b) 0.62 × 10 ×10-4 V/mm
0%
c)1 × 10-5 V/mm
0%
d)5 × 10-5 V/mm
Explanation
Voltage gradiant=V/l=0.5V / 10m=5 × 10-5 V/mmAnswer: (d)
Q.17
Two heater wires of equal length are first connected in series and then in parallel The ratio of heatproduced in the two cases is....
0%
a) 2:1
0%
b) 1:2
0%
c) 4:1
0%
d) 1:4
Explanation
Answer:(d)
Q.18
2 A current is obtained when a 2 Ω resistor is connected with battery having r Ω as internal resistance0.5A current is obtained if the above battery is connected to 9Ω resistor. Calculate the internal resistanceof the battery.
0%
a) 0.5 Ω
0%
b) (1/3) Ω
0%
c) (1/4) Ω
0%
d) 1 Ω
Explanation
E=I1( R1 + r )=I2 ( R1 + r ) r=(1/3) Ω Answer: (b)
Q.19
A wire of resistance 4 Ω is stretched to twice itsoriginal length. The resistance of stretched wirewould be
0%
a) 2 Ω
0%
b) 4 Ω
0%
c) 8 Ω
0%
d) 16Ω
Explanation
When wire is stretched twice it area becomes half as volume of wire remains same Thus New Resistance R’ = 4R, R’ = 4×4= 16Ω (R’ = n2R) , here n = 2 Answer:(d)
Q.20
The internal resistance of a 2.1 V cell which gives acurrent of 0.2 A through a resistance of 10 Omega; is
0%
a) 0.2 Ω
0%
b) 0.5 Ω
0%
c) 0.8 Ω
0%
d) 1.0 Ω
Explanation
E= I(R+r), r= internal resistance of cell, R =10Omega; 2.1 = 0.2(10+r) r= 0.5 Ω Answer:(b)
Q.21
The resistances of the four arms P, Q, R and S in a Wheatstone's bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be from the cell will be
0%
a) 1.0 A
0%
b) 0.2 A
0%
c) 0.1 A
0%
d) 2.0A
Explanation
Wheatstone’s bridge is balance hence no current lows through Galvanometer P and Q are in series = 40Ω. Resistance E and S are in series = 120Ω. 40 Ω and 120 Ω are in parallel have effective resistance = 30 Ω. Internal resistance of cell is in series with 30 Ω. Total resistance is 35V. Thus current drawn = 7/35 = 0.2A Answer:(b)
Q.22
In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be
0%
a) G/500
0%
b) 500G/499
0%
c) G/499
0%
d) 499G/500
Explanation
Potential across G = Potential across S 0.2IG = 99.8Ir r=G/499 G and r are parallel to each other thus Resistance of ammeter is G/500 Answer:(a)
Q.23
The resistance in the two arms of the meter-bridge are 5Ω and RΩ, respectively.When the resistance R is shunted with an equal resistance,the new balance point is at 1.6l1.The resistance 'R', is
0%
a) 20 Ω
0%
b) 25Ω
0%
c) 10Ω
0%
d) 15Ω
Explanation
According to Wheatstone bridge balance condition When another resistance of R is shunted by another R Multiply (i) by 1.6 Comparing (ii) and (iii) 160 -1.6l1 = 200 -3.2l1 1.6l1 = 40; l1=25 substituting l1 is (i) R = 15Ω Answer:(d)
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 km is 8 volt and the average resistance per km is0.5 立. The power loss in the wire is Answer : (d)
0%
a) 19.2J
0%
b) 12.2 kW
0%
c) 19.2W
0%
d) 19.2 kW
Explanation
Total resistance = 150 × 0.5 =75 Current in the wire = 8/0.5= 16 A Power loss = I2R = 162× 75 = 19200 w = 19.2kW Answer:(d)
Q.25
A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4m long.When the resistance R, connected across the given cell, has values of(i)infinity(ii) 9.5Ω, the 'balancing lengths', on the potentiometer wire are found to be 3m and 2.85 m, respectively. The value of internal resistance of the cell is
0%
a) 0.5Ω
0%
b) 0.75Ω
0%
c) 0.25Ω
0%
d) 0.95Ω
Explanation
Value of resistance is infinity means key is open length l1 = 3 Value of resistance is 9.5 Ω means key is closed length l2 =2.85 Answer:(a)
Q.26
A, B and C are voltmeters of resistance R, 1.5 R and 3R respectively as shown in the figure. When some potential difference is applied between X and Y, the voltmeter readings are VA, VB and VC respectively. Then :
0%
a) VA ≠ VB ≠ VC
0%
b) VA = VB = VC
0%
c) VA ≠ VB = VC
0%
d) VA = VB ≠ VC
Explanation
B and C are parallel thus VB = VC Effective resistance of parallel combination of B and C = 1R Thus VA = VB = VC Answer:(b)
Q.27
A potentiometer wire has length 4 m and resistance 8 Ω. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient 1 mV per cm on the wire is :
0%
a) 48 Ω
0%
b) 32 Ω
0%
c) 40 Ω
0%
d) 44 Ω
Explanation
Required potential across wire = 400×1mV=0.4V And current in wire =0.4/8= 0.05 A same current flow through resistance Current drop in resistance connected = 2-0.4=1.6 V Resistance =1.6/0.05=32 Ω Answer:(b)
Q.28
s a metallic conductor of non-uniform cross section a constant potential difference is applied. The quantity which remains constant along the conductor is :
0%
a) electric filed
0%
b) current density
0%
c) current
0%
d) drift velocity
Explanation
Answer:(c)
Q.29
A potentiometer wire of length L and a resistance are connected in series with a battery of e.m.f. E0 and a resistance rAn unknown e.m.f. E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by :
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Current in wire of length L and resistance ‘r’ and connected with resistance r1 because of E0 Now potential is balanced by length ‘l’, E= I ρl …(ii) Now ρ=r/L Substituting value of I from (i) and ρ in (ii) Answer:(c)
Q.30
Two metal wires of identical dimensions are connected in series. If σ1 and σ2 are the conductivities of the metal wires respectively, the effective conductivity of the combination is :-
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Let R1 and R2 are the resistance having conductivity σ1 and σ2 Since resistance connected in series, resistance of the combination R = R1+R2 If σ is the conductivity of combination, Length of resistance is now 2l as both resistance connected in series, but area A will be same thus From (i) and (ii) we get Answer:(b)
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