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Physics NEET MCQ
Quiz 10
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Q.1
Amperes hour is the unit of
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a) quantity of charge
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b) strength of current
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c)power
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d)energy
Explanation
Answer: (a)
Q.2
In circuit shown, the cell E1 and E2 have emf of 4V and 8V. Internal resistance of 0.5Ω and 1Ω respectively. Then the potential difference across E1 and E2 will be
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a) 4.25 V; 4.25V
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b) 3.75V; 3.75V
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c)4.25V; 7.5V
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d)3.75V; 7.5V
Explanation
Effective emf=of the circuit=8-4=4VTotal resistance of the circuit=1+0.5+4.5 + (3×6/3+6)=8Ω∴ current in the circuit I=4/8=0.5ATerminal potential difference across E2 ( cell is discharging)=8 -0.5×1==7.5 VTerminal potential difference across E1 ( cell is charging)=4 +0.5×0.5=4.25 Answer:(c)
Q.3
In the circuit, when key K1 is closed, the ammeter reads Io whether K2 is open or closed. But when K1 is open, the ammeter reads Io/2 when K2 closed. Assuming that the ammeter resistance is much less than R2, find value of r and R1( in ohms)
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a) 100, 50
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b) 50, 100
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c) 0, 100
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d) 0, 50
Explanation
Case I) When K1 is closed, and K2 is open ,the resistance R1 is ineffective. I0=E / (100+r) CaseII) When K1 is closed, and K2 is closed ,the resistance R1 is ineffective.Total resistance in circuit=50+r and current through circuit I=E / (50+r) current through ammeter will be half of total thus Io=E / 2(50+r)from above equations E / (100+r)=E / 2(50+r) Therefore r=0 Case II) When k1 and K2 R1 becomes effective Total resistance in the circuit=R1 + r+ 50 But r=0 ∴ total resistance=R1 + 50 Current through circuit I=E/(R1 + 50) Current trough Ammeter will be half of total current=I/2 Current through ammeter=E/2(R1 + 50) Given that current through ammeter is Io/2 Thus Io/2=E/2(R1 + 50) Now from above I0=E / (100+r) but r=0 ∴ I0=E / (100) Thus E / 2(100)=E/2(R1 + 50) R1=50Ω Answer: (d)
Q.4
The terminal potential difference of a battery exceeds its emf when it is connected ( Pb.C.E.T. 1997)
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a)in parallel with battery of higher emf
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b) in series with a battery of higher emf
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c)in series with battery of lower emf
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d)in parallel with a battery of lower emf
Explanation
Answer: (a)
Q.5
The current I in a conductor varies with time t as I=2t + 3t2, where I is in ampere and t in seconds. Electric charge flowing through a section of the conductor during t=2s to t=3s is ( Orissa JEE 2003]
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a) 10 C
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b) 24C
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c)33C
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d)44C
Explanation
Answer: (b)
Q.6
According to Joule's law if potential difference across a conductor having a material of specific resistance ρ, remains constant, then heat produced in the conductor is directly proportional to
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a) 1 / √ρ
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b) 1/ρ
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c)ρ
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d)ρ2
Explanation
When potential is constant Heat ∝ 1/R But R ∝ρThus H ∝ 1/ρ Answer:(b)
Q.7
If in the circuit. The internal resistance of the battery is 1.5Ω and VP and VQ are potentials at P and Q respectively. What is the potential difference between the points P and Q?
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a) Zero
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b) 4 volts ( Vp > VQ)
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c) 4 volts (VQ > VP)
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d) 2.5 volts (VQ > VP)
Explanation
Total current in circuit=20/ (1.5+2.5)=5.0 A Current get equally divided in both arms=2.5 Total potential across each arm=5×2.5=12.5 V Potential across 3Ω resistance=3×2.5=7.5 Thus potential at P=12.5 - 7.5=5.0 V Potential across 2Ω resistance=2×2.5=5.0 V Potential at Q=12.5 - 5.0=7.5 VQ - VO=7.5-5.0=2.5 Answer: (d)
Q.8
In the circuit shown, the ammeter reads 2A. The resistance of ammeter is negligible. The value of R is
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a)2 Ω
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b) 4 Ω
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c)6 Ω
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d)8 Ω
Explanation
R, 2R and 3R are parallel and this combination equivalent resistance=6R/11 is is in series with 5R/11 resistance thus equivalent resistance=6R/11 + 5R/11=R Now V=IR 12=2 ×R ∴ R=6ΩAnswer: (c)
Q.9
All resistance shown un circuit are 2Ω each. The current in the resistance between D and E is
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a) 5 A
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b) 2.5 A
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c)1 A
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d)7.5 A
Explanation
Resistance in arm Ab has no significance because Resistance in arm Ab and Fg forms a balanced Whetstone bridgeResistance of Whetstone bridge=R switch is parallel to resistance in arm DGEquivalent resistance=R/2=1Ω∴ I=10/1=10AThus Current through arm DE using current division=10/4=2.5Answer: (b)
Q.10
In the circuit, each cell has an emf of 1.5 V and r=0.4Ω. The current I in the 36Ω resistor is
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a) 0.5 amp
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b) 0.1 amp
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c)0.083 amp
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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.11
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
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a) 3 Ω
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b) 2 Ω
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c) (1/3) Ω
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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.12
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
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a)2/5
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b) 3/1
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c)4/3
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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.13
Each resistance in the circuit is r. The equivalent resistance between A and B is
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a) r/4
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b) 4r
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c)2r/5
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d)0
Explanation
Circuit can be redrawn as From figure R=2r/5Answer: (c)
Q.14
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.
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a) both A and V will increase
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b) both A and V will decrease
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c)A will decrease, V will increase
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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.15
The equivalent resistance between A and B is
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a) 2R/3
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b) R/3
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c) R
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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.16
In a circuit shown, the galvanometer reads zero. If batteries have negligible internal resistance, the value of C will be
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a)10 Ω
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b) 100 Ω
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c)200 Ω
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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.17
The reading in the ammeter is
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a) 1 A
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b) 2 A
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c)0.67 A
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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.18
The potential difference between the points x and y in the adjoining figure will be
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a)zero
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b) 50 V
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c)10 V
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d)100V
Explanation
given diagram is of Whetstone balanced bridge. Potential between x and y is zero Answer:(a)
Q.19
In the adjoining diagram find the current through 10Ω resistance
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a) 0.4 A towards O
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b) 0.4 A away from O
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c) 0.6 A towards O
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d) 0.6 A away from O
Explanation
Answer: (b)
Q.20
In the circuit given below. The current in the arm AD will be , each resistance is 10Ω
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a)2i/5
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b) 3i/5
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c)4i/5
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d)i/5
Explanation
Answer: (a)
Q.21
A steady current is passing through a linear conductor of non-uniform cross-section. the current density in the conductor is
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a) independent of area of cross section
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b) directly proportional to area of cross section
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c)inversely proportional to area of cross-section
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d)inversely proportional to the square root of area of cess-section
Explanation
Answer: (c)
Q.22
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
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a) proportional to T
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b) proportional to √T
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c)zero
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d)finite but independent of temperature
Explanation
Answer:(c)
Q.23
Ohm's law is valid when the temperature of the conductor is
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a) constant
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b) very high
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c) very low
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d) varying
Explanation
Answer: (a)
Q.24
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
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a)40
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b) 40/3
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c)5/2
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d)100
Explanation
resistance ∝ (1/A) thus Resistance of thin wire=30 ΩAnswer: (a)
Q.25
Two heater wires of equal length are first connected in series and then in parallel The ratio of heatproduced in the two cases is....
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a) 2:1
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b) 1:2
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c) 4:1
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d) 1:4
Explanation
Answer:(d)
Q.26
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.
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a) 0.5 Ω
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b) (1/3) Ω
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c) (1/4) Ω
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d) 1 Ω
Explanation
E=I1( R1 + r )=I2 ( R1 + r ) r=(1/3) Ω Answer: (b)
Q.27
A wire of resistance 4 Ω is stretched to twice itsoriginal length. The resistance of stretched wirewould be
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a) 2 Ω
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b) 4 Ω
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c) 8 Ω
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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.28
The internal resistance of a 2.1 V cell which gives acurrent of 0.2 A through a resistance of 10 Omega; is
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a) 0.2 Ω
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b) 0.5 Ω
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c) 0.8 Ω
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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.29
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
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a) 1.0 A
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b) 0.2 A
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c) 0.1 A
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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.30
In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the resistance of ammeter will be
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a) G/500
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b) 500G/499
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c) G/499
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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)
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