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Quiz 12
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Q.1
A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm all connected in series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be :-
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a) 1 A
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b) 0.5 A
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c) 0.25 A
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d) 2 A
Explanation
Resistance of coil of ammeter and shunt are parallel thus effective resistance Rp=480/25 Now Rp and 40.8 ohm are connected in series Current = V/R = 30/60 = 0.5 A Answer:(b)
Q.2
The charge flowing through a resistance R varies with time t as Q = at – bt2, where a and b are positive constants. The total heat produced in R is:
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a) (a3 R)/6b
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b) (a3 R)/3b
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c) (a3 R)/2b
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d) (a3 R)/b
Explanation
If I =0 then 0 =a-2bt t=a/2b Heat produced = I2Rt But current is time dependent thus we will integrate (a3 R)/6b Answer:(a)
Q.3
A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf is :- [ AIPMT 2015]
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a) 5 : 1
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b) 5 : 4
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c) 3 : 4
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d) 3 : 2
Explanation
10E1+10E2=50E1-50E2 E1+E2=5E1-5E2 4E1=6E2 E1/E2=3/2 Answer:(d)
Q.4
The potential difference (VA-VB) between the points A and B in the given figure is …[NEET II 2016]
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a) +6V
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b) +9V
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c) -3V
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d) +3V
Explanation
Applying Kirchhoff’s law VA-4 – 3-2=VB VB - VA = 9V Answer:(b)
Q.5
A filament bulb (500 W, 100 V) is be used in a 230V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500W. The value of R is … [NEET II 2016]
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a) 26 Ω
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b) 13 Ω
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c) 230 Ω
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d) 46 Ω
Explanation
Power = VI 500 = 100I I=5A Voltage across Resistance 130 = 5R R=26Ω Answer:(a)
Q.6
A potentiometer is an accurate and versatile deviceto make electrical measurements of E.M.F, becausethe method involves:
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a) Cells
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b) Potential gradients
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c) A condition of no current flow through thegalvanometer
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d) A combination
Explanation
When no current flows through the galvanometer, current from battery is zero, thus do not have effect of internal resistance of battery E= V- Ir as I = 0 E = V Answer:(c)
Q.7
The resistance of a wire is ‘R’ ohm. If it is meltedand stretched to ‘n’ times its original length, its newresistance will be
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a) nR
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b) R2
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c) n2R
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d) 2Rn
Explanation
Volume of the wire before melting = Al Volume of the wire of length = l1A1 Since volume is same Al = A1l1 Al = nA1l Thus A1 = A/n New resistance Answer:(c)
Q.8
This question has statement I and Statement II. Of the four choices given after the Statements,choose the one that best describes the two statements. Statement − I : Higher the range, greater is the resistance of ammeter. Statement − II : To increase the range of ammeter, additional shunt needs to be used across it.[ IIT Mains 2013]
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a) Statement - I is true, Statements - II is true, Statement - II is the correct explanation of Statements - I
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b) Statement - I is true, Statement - II is true, Statement - II is not the correct explanation ofStatement - I.
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c) Statement - I is true, Statement - II is false.
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d) Statement - I is false, Statement - II is true.
Explanation
I - False, II – True Answer:(d)
Q.9
than one correct answer Q341) Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by anew heater having two wires of the same material, each of length L anddiameter 2d. Theway these wires are connected is given in the options. How much time in minutes will ittake to raise the temperature of the same amount of water by 40 K?[ IIT Advance 2014]
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a) 4 if wires are in parallel
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b) 2 if wires are in series
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c) 1 if wires are in series
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d) 0.5 if wires are in parallel
Explanation
Amount of heat produced in given case H=0.5C40 Resistance of new wires Option a If connected in parallel resultant resistance = R/8 0.5×C×40=8×5C×t t= 0.5 minutes Option a is wrong but option d is correct Option b: Resistance in series = R/2 0.5C40=2×5C×t T= 2 minutes option b correct Option C is wrong Answer:(b, d)
Q.10
than one correct answer Q342) Two ideal batteries of emf V1 and V2 and three resistances R1, R2 and R3 are connected as shown in the figure. The current in resistance R2 would be zero if ..[IIT Advance 2014]
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a) V1 = V2 and R1 = R2 = R3
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b) V1 = V2 and R1 = 2R2 = R3
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c) V1 = 2V2 and 2R1 = 2R2 = R3
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d) 2V1= V2 and 2R1 = R2 = R3
Explanation
iR1 + i1R2 = V1 If i1 = 0 then i= V1/R1…(i) i1R2 – (i – i1)R3 = -V2 If i1 = 0 then i= V2/R3…(ii) From (i) and (ii) Option “a” correct Option “b” correct Option c Option d correct Answer:(a, b, d)
Q.11
During an experiment with a metre bridge, the galvanometer shows a null point when thejockey is pressed at 40.0 cm using a standard resistance of 90 Ω, as shown in the figure.The least count of the scale used in the metre bridge is 1 mm. The unknown resistance is…[ IIT Advance 2014]
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a) 60 ± 0.15 Ω
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b) 135 ±0.56 Ω
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c) 60 ± 0.25 Ω
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d) 135 ± 0.23 Ω
Explanation
R= 60Ω Now error in measurement is least count Δl= 1mm = 0.1cm We will assure R2 have no error R = 60±0.25Ω Answer:(c)
Q.12
In an aluminum (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivity’s of Al and Fe are 2.7 × 10-8Ωm and 1.0 × 10-7Ωm, respectively. The electrical resistance between the two faces P and Q of the composite bar is … [ IIT Advance 2015]
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a) (2475/64)µΩ
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b) (1875/64) µΩ
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c) (1875/49)µΩ
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d) (2475/132)µΩ
Explanation
Answer:(b)
Q.13
An infinite line charge of uniform electric charge density 位 lies along the axis of an electrically conducting infinite cylindrical shell of radius R. At time t = 0, the space inside the cylinder is filled with a material of permittivity ε and electrical conductivity σ. The electrical conduction in the material follows Ohm鈥檚 law. Which one of the following graphs best describes the subsequent variation of the magnitude of current density j(t) at any point in the material?
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a)
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b)
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c)
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d)
Explanation
It may be considered as charging of capacitor and graph given in option 'a' depects the same However detail calculations are as follows Let initial charge density be λ0 Consider inner shell of thickness dr and radius r and unit height Electric filed produced by infinitely long linear charge distribution |dv| = Edr Now resistance of cylindrical element of unit length and radius r and thickness dr which will be our length. Here current flowing radially out ward., also height of cylinder is unit. Thus Area = 2πr Now I = dq/dt, Charge for unit length= 位, as charge density is reducing On integrating Now J = I/A Option a follows the conditions Answer:(a)
Q.14
An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the bulb is powered at constant voltage, which of the following statement(s) is(are) true?
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a) The temperature distribution over the filament is uniform
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b) The resistance over small sections of the filament decreases with time
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c) The filament emits more light at higher band of frequencies before it breaks up
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d) The filament consumes less electrical power towards the end of the life of the bulb
Explanation
a) If we assume resistivity and cross-section is uniform then temperature will be uniform But it is given that wire break up at random locations thus option a wrong b) is wrong as cross section of wire gets reduced resistance increases c) is correct because temperature is not uniform, and frequency ∝ T d) correct as resistance increases with decreasing thickness Answer:(c, d)
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