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Physics NEET MCQ
Quiz 9
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Q.1
In an AC generator, a coil with N turns, all of the same area A and total resistance R, rotates with frequency ω in a magnetic field B. The maximum value of emf generated in the coil is [ AIEEE 2006]
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a) NABRω
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b) NAB
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c) NABR
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d) NABω
Explanation
Answer: (d)
Q.2
A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when
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a) Frequency of the AC source is decreased
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b) Number of turns in the coil is reduced
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c) A capacitance of reactance XC = XL is included in the same circuit
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d) An iron rod is inserted in the coil
Explanation
Brightness of bulb depends on the current. Increase in reactance decrease in brightness Reactant of coil is 2πfL thus when iron rod is inserted in the coil, value of inductance increases, thus option ‘d’ is correct. Answer:(d)
Q.3
A resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes 'Z', the power drawn will be
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a) p
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b)
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c)
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d)
Explanation
Power in A.C. is given p=I2Zcosθ For only resistance power θ=0 V2= pR…(i) For resistance and inductor in series combination impedance =Z Substituting value of V2 in above equation Answer:(b)
Q.4
A series R-C circuit is connected to an alternating voltage source. Consider two situations :- (a) When capacitor is air filled. (b) When capacitor is mica filled. Current through resistor is i and voltage acrosscapacitor is V then :-
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a) Va = Vb
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b) Va < Vb
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c) Va > Vb
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d) ia > ib
Explanation
When capacitor is filled with mica then capacitance C increases as Xc decreases so Va > Vb Answer:(c)
Q.5
An inductor 20 mH, a capacitor 50 µF and a resistor40Ωare connected inseries across a source of emf V = 10 sin 340 t. The power loss in A.C. circuit is :-
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a) 0.51 W
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b) 0.47 W
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c) 0.76 W
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d) 0.89 W
Explanation
Answer:(b)
Q.6
A small signal voltage V(t) = V0 sin ωt is applied across an ideal capacitor C :- ….[AIPMT 2015]
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a) Current I (t), lags voltage V(t) by 90°.
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b) Over a full cycle the capacitor C does not consume any energy from the voltage source.
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c) Current I (t) is in phase with voltage V(t).
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d) Current I (t) leads voltage V(t) by 180°.
Explanation
Power = Vrms .Irms cosθ as cosθ = 0 (Because θ = 90°) ∴ power consumed = 0 ( in one complete cycle) Answer:(b)
Q.7
Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication ?
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a) R = 15 Ω, L = 3.5 H, C = 30 µF
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b) R = 25 Ω, L = 1.5 H, C = 45 µF
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c) R = 20 Ω, L = 1.5 H, C = 35 µF
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d) R = 25 Ω, L = 2.5 H, C = 45 µF
Explanation
For better tuning, Q-factor must be high. R and C should be small and L should be high Answer:(a)
Q.8
A 100Ω resistance and a capacitor of 100Ω reactance are connected in series across a 220V source. When the capacitor is 50% charged, the peak value of the displacement current is …[ NEET II-2015]
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a) 4.4 A
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b) 11√2 A
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c) 2.2 A
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d) 11 A
Explanation
Asked about the peak value of the displacement current Peak voltage = 220√2 Answer:(c)
Q.9
Figure shows a circuit contains three identical resistors with resistance R = 9.0 Ω each, two identical inductors with inductance L = 2.0 mH each,and an ideal battery with emf ε = 18 V. The current'i' through the battery just after the switch closed is …. [ NEET 2017]
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a) 2 mA
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b) 0.2 A
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c) 2 A
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d) 0 ampere
Explanation
When switch is closed, inductors and capacitor will not conduct and no current flows through them, current will pass through middle resistance. Thus current will flow through middle resistance I = 18/9 = 2A Answer:(c)
Q.10
The supply voltage to a room is 120 V. The resistance of the lead wires is 6Ω. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb ? .. [ IIT Mains 2013]
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a) zero Volt
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b) 2.9 Volt
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c) 13.3 Volt
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d) 10.3 Volt
Explanation
Resistance of bulb Resistance of heater When only bulb is on Voltage across Bub Total resistance = 240 +6 = 246Ω Current in bulb = 120/246 Voltage across bulb when heater is on Total resistance of parallel combination R’ Current from supply when heater is switched on Resistance of circuit = 48+6 = 54 Current = 120/54 Potential across combination of bulb and heater Answer:(d)
Q.11
In an LCR circuit as shown below both switches are open initially.Now switch S1 is closed, S2 kept open. (q is charge on the capacitorand τ = RC is Capacitive time constant). Which of the followingstatement is correct?[ IIT Mains 2013]
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a) Work done by the battery is half of the energy dissipated in the resistor
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b) At t = τ, q = CV/2
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c) At t = 2τ, q = CV(1-e-2)
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d) At t = τ/2, q = CV(1-e-1)
Explanation
Option (a) V = VR + VC Work done by battery = qV , from formula for capacitor c = q/V Work done by battery = CV2 Energy of capacitor Loss of energy in resistance = Work done by battery – Energy stored in capacitance Since V ≠ VR Energy loss is less than ½CV2 which is half of the energy of battery For Charging of capacitor Option (b) At t = τ And Q0= CV Hence wrong Option c At t = 2τ, q = CV(1-e-2) Option c is correct Option d Option d wrong Answer:(c)
Q.12
raph Q272) A thermal power plant produces electric power of 600 kW at 4000 V, which is to betransported to a place 20 km away from the power plant for consumers usage. It can betransported either directly with a cable of large current carrying capacity or by using acombination of step-up and step-down transformers at the two ends. The drawback ofthedirect transmission is the large energy dissipation. In the method using transformers, thedissipation is much smaller. In this method, a step-up transformer is used at the plant side sothat the current is reduced to a smaller value. At theconsumers end, a step-down transformeris used to supply power to the consumers at the specified lower voltage. It is reasonable toassume that the power cable is purely resistive and the transformers are ideal with a powerfactor unity. All the currents and voltages mentioned are rms values. Q272A) If the direct transmission method with a cable of resistance 0.4Ω km−1 is used, the powerdissipation (in%) during transmission is
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a) 20
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b) 30
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c) 40
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d) 50
Explanation
P = 600 kW V = 4000 V Power Loss = I2R = (150)2 × 0.4 × 20 = 180 kW Answer:(a)
Q.13
In the circuit shown, L = 1 μH, C = 1 μF and R = 1 kΩ. They are connected in series with an a.c. source V = V0 sin ωt as shown. Which of the following options is/are correct ? [ IIT Advance 2016]
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a) a) The frequency at which the current will be in phase with the voltage is independent of R.
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b) At ω ˜ 0 the current flowing through the circuit becomes nearly zero
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c) At ω >> 106 rad.s-1, the circuit behaves like a capacitor.
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d) The current will be in phase with the voltage if ω = 104 rad.s-1.
Explanation
At resonance frequency current is in phase with voltage and circuit become purely resistive. Thus voltage is independent of resistance [ option a correct] When ω =0, after some time capacitor gets fully charged and do not conduct, current = 0 [ Option b is correct] At ω >> 106 rad.s-1 capacitive reactance =XC = 1/ωC tend to zero. And Inductive reactance increases thus ckt become inductive [ Option C wrong] At resonance frequency current will be in phase Thus option d wrong Answer:(a,b)
Q.14
More than one correct option. A source of constant voltage V is connected to a resistance R and two ideal inductors L1 and L2 through a switch S as shown. There is no mutual inductance between the two inductors. The switch S is initially open. At t = 0, the switch is closed and current begins to flow. Which of the following options is/ are correct? [ IIT Advance 2017]
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a) The ratio of the current through L1 and L2 is fixed at all time (t > 0)
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b) After a long time, the current through L1 will be
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c) After a long time, the current through L2 will be
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d) At t=0, the current trough the resistance R is V/R
Explanation
Final current from battery i= V/R Inductors are connected in parallel Thus current through L1 L1 i1= L2 i2 i1 + i2 = i And current through L2 Option b and c correct Option a is correct At t=0 current through source =0, option d wrong Answer:(a,b,c)
Q.15
The instantaneous voltages at three terminals marked X, Y and Z are given by Vx = V0 sinωt , Vy=V0 sin(ωt+2π/3) and Vz = V0 sin(ωt+4π/3) An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points X and Y and then between Y and Z. The reading(s) of the voltmeter will be … [ IIT Advance 2017]
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a)
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b)
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c) Independent of choice of terminal
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d)
Explanation
Integrating over a period 0 to T and taking average As average of cos is zero Taking square root Similarly it can be shown that Option c and option d correct Answer:(c, d)
Q.16
A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced e.m.f. is [NEET 2013]
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a) Once per revolution
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b) Twice per revolution
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c) Four times per revolution
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d) Six times per revolution
Explanation
When coil is rotated in magnetic field direction of current changes twice in one rotation Answer:(b)
Q.17
A thin semicircular conducting ring (PQR) ofradius 'r' is falling with its plane vertical in ahorizontal magnetic field B, as shown in figure.The potential difference developed across the ring when its speed is v, is … [ AIPMT 2014]
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a) πrBv and R is at higher potential
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b) 2rBv and R is at higher potential
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c) zero
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d) and P is at higher potential
Explanation
Induced emf = blv Distance between P and R is effective length l=2r Induced emf = 2rBv By using right hand palm rule we can find the direction of force on electrons. Figures in the direction of magnetic filed, thumb in the direction of velocity. Then perpendicular to palm will show direction of force on electrons. As electrons gets accumulated at R and P becomes positive or at higher potential Answer:(b)
Q.18
A transformer having efficiency of 90% is working on 200V and 3kW power supply. If the current in the secondary coil is 6A, the voltage across the secondary coil and the correct in the primary coil respectively are
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a) 450 V, 13.5 A
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b) 600V, 15A
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c) 300V, 15A
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d) 450V, 15A
Explanation
P1(0.9) = P2 P2 = 2.7 kW Now P = VI 2700 = 6V V= 450V Current in primary =3000/200 =15 A Answer:(d)
Q.19
A conducting square frame of side ‘a‘ and a long straight wire carrying current I are located in thesame plane as shown in the figure. The frame moves to the right with a constant velocity ‘V‘. The emf induced in the frame will be proportional to :
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a)
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b)
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c)
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d)
Explanation
Magnetic field produced by the current in wire By integrating from x-a/2 to x+a/2 we get the total magnetic filed Now velocity of frame is constant and area is not changing with time As ‘a’ is side which is constant Answer:(a)
Q.20
A uniform magnetic field is restricted within a region of radius r. The magnetic field changes with timeat a rate(dB)/dt. Loop 1 of radius R > r encloses theregion r and loop 2 of radius R is outside the region of magnetic field as shown in the figure below. Then the e.m.f. generated is …[NEET II -2016]
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a) in loop 1 and zero in loop 2
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b) in loop 1 and zero in loop 2
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c) Zero in loop 1 and zero in loop 2
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d) in loop 1 and zero in loop 2
Explanation
For Loop 1 For Loop 2 Εind = 0, as no flux linkage Answer:(b)
Q.21
A long solenoid of diameter 0.1 m has 2×104 turnsper meter. At the centre of the solenoid, a coil of 100turns and radius 0.01 m is placed with its axiscoinciding with the solenoid axis. The current in thesolenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10π2Ω, the total charge flowing through the coil during this time is
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a) 32πµC
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b) 16µC
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c) 32µC
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d) 16πµC
Explanation
Induced emf in coil Φ = BA Magnetic field is produced by solenoid = B =µ0nI dq = 32µC Answer:(c)
Q.22
A metallic rod of length ‘l’ is tied to a string of length 2l and made to rotate with angular speedω on a horizontal table with one end of the string fixed. If there is a vertical magnetic field ‘B’ in the region, the e.m.f. induced across the ends of the rod is … [ IIT mains 2013]
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a)
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b)
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c)
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d)
Explanation
Potential at A = VA = B2lv1and v = ω2l VA = 4Bωl2 Potential at B = VB = B 3l v2 and v2 = ω3l VA = 9Bωl2 Average potential difference Answer:(d)
Q.23
A rigid wire loop of square shape having side of length L and resistance R is moving along the x-axis with a constant velocity v0 in the plane of the paper. At t = 0, the right edge of the loop enters a region of length 3L where there is a uniform magnetic field B0 into the plane of the paper, as shown in the figure. For sufficiently large v0, the loop eventually crosses the region. Let x be the location of the right edge of the loop. Let v(x), I(x) and F(x) represent the velocity of the loop, current in the loop, and force on the loop, respectively, as a function of x. Counter-clockwise current is taken as positive. Which of the following schematic plot(s) is(are) correct? (Ignore gravity) [ IIT Advance 2016]
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a)
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b)
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c)
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d)
Explanation
When right edge of the loop enters the magnetic field, induced current is maximum. E/R =i= Blv/R …(i) As loop right edges passes through 3L direction of current will reverse be cause of decreasing magnetic filed (Option d correct and option a is wrong) The direction of current induced in loop is such that it opposes the cause of induction. As a result force act on loop left ward or negative ( Option c is correct) As loop right edges passes through 3L direction of force will reverse because of decreasing magnetic filed (Option d correct) If m is the weight of coil From (i) Above equation shows that velocity decreases and thus Current, and force also decreases and value is from (i) And force Consider the case when loop is completely inside the magnetic filed i.e., L < x < 3L. The magnetic flux through the loop is constant. Thus, the induced emf e = 0, induced current i = 0, and magnetic force F = 0 (as i = 0). The velocity of the loop remains constant (as F = 0) at its value at x = L V =0 ( Option b wrong) Answer:(c, d)
Q.24
More than one correct option. A circular insulated copper wire loop is twisted to form two loops of area A and 2A as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field B ⃗ points into the plane of the paper. At t = 0, the loop starts rotating about the common diameter as axis with a constant angular velocity ω in the magnetic field. Which of the following options is/are correct? [ IIT Advance 2017]
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a) The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper
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b) The net emf induced due to both the loops is proportional to cos ωt
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c) The emf induced in the loop is proportional to the sum of the areas of the two loops
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d) The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude of maximum emf induced in the smaller loop alone
Explanation
Flux Φ = BAcosω But θ = ωt If θ = 90 rate of change of flux is maximum. Or plane of loop is perpendicular to magnetic filed (Option a is correct) Now E ∝ sinωt (Option b is wrong) Option c Induced emf produced in both the loop is opposite to each other Thus net emf = e=B2Aωsinωt-BAωsinωt= BAωsinωt Option c is wrong Option d is correct Answer:(a,d)
Q.25
What is the coefficient of mutual inductance, when the magnetic flux changes by 2×10⁻² Wb, and change in current is 0.01A? [BHU 1998]
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a) 2 henry
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b) 3 henry
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c) 1/2 henry
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d) zero
Explanation
Caluclate M from formula E = -M(dI/dt) Answer: (a)
Q.26
When current in a coil changes to 2 amp from 8 amper in 3×10⁻³ second, the emf induced in the coil is 2 volt. The self-inductance of the coil in millihenry is [ MPPMT 1995]
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a) 1
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b) 5
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c) 20
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d) 10
Explanation
Answer: (a)
Q.27
If resistance of 100Ω, inductance of 0.5 henry and capacitance of 10×10⁻⁶ fd are connected in series through 50 Hz AC supply, then impedance is : [ BHU 1995]
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a) 1.876 Ω
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b) 18.76Ω
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c) 187.6Ω
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d) 101.3Ω
Explanation
Use formula Answer: (c)
Q.28
A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth's magnetic field is 0.2×10⁻⁴T, then the emf developed between the two ends of the conductor is [ AIEEE 2004]
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a) 5 mV
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b) 50 µV
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c) 5 µV
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d) 50 mV
Explanation
use the formula for induced emf E = Bvl / 2 Answer:(b)
Q.29
What is the maximum value of inductance L for which the current is maximum in a LCR circuit with C = 10µF and ω = 1000 s-1 [ CBSE-PMT 2007]
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a) 1 mH
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b) cannot be calculated unless R is known
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c) 10 mH
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d) 100 mH
Explanation
At resonance Impedance becomes only resistive at resonance ω = 1/ (LC) 1/2 L = 1/ ω2 C L = 1/ ( 106 × 10 ×10⁻⁶ ) L = 0.1 H L = 100 mH Answer: (d)
Q.30
In a series LCR circuit, resonance occurring at 105Hz. At that time, the potential difference across the 100Ω resistance is 40V while the potential difference across the pure inductor is 30v. The inductance L of the inductor is equal to [ AFMC 2011]
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a) 10-3 H
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b) 7.5× 10⁻⁴ H
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c) 5.0× 10⁻⁴ H
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d) 1.2×10⁻⁴ H
Explanation
At resonance imaginary part of current is zero Now V=IR Thus I = 40/100 = 0.4A XL = V L / I XL = 30/0.4 = 75Ω L = XL / ω L = 7.5× 10⁻⁴ Answer: (b)
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