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Physics NEET MCQ
Quiz 10
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Q.1
A condenser of capacity 1µF and resistance 0.5 MΩ are connected in series with D.C. supply of 2V. The time constant of the circuit is : [ AFMC 2010]
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a) 0.25 s
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b) 0.5s
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c) 1 s
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d) 2 s
Explanation
time constant τ = RC substituting R = 0.5×106 Ω C = 1 × 10⁻⁶F In above equation we get τ = 0.5sec Answer: (b)
Q.2
In an a.c circuit an alternating voltage e = 200√2 sin 100t volts is connected to a capacitor of capacity 1 µF. The r.m.s. value of the current in the circuit is ... [ CBSE-PMT 2011]
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a) 10 mA
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b) 100 mA
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c) 200 mA
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d) 20 mA
Explanation
Vr.m.s = Vmax / √2 Vr.m.s = 200√2 / √2 = 200 V Reactive capacitance XC =1/ ωC XC = 1/100× 10⁻⁶= 104 Ir.m.s = Vr.m.s / XC Irms = 200/104 = 20×10⁻³ = 20 mA Answer: (d)
Q.3
A solenoid has 2000 turns wound over a length of 0.3 m. The area of its cross-section is 1.2×10⁻³ mAround its central section a coil of 300 turns is wound. If an initial current of 2A is reversed in 0.25 sec, the emf induced in the coil is equal to [ JIMPER Pondy 1998]
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d) 48kV
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a) 6.0×10⁻⁴ V
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b) 4.8×10⁻²V
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c) 6.0×10⁻²V
Explanation
Given A =1.2×10⁻³ m2 N1 = 20000 and N2 = 300 length l = 0.3m dI = 2-(-2) = 4 dt = 0.25 sec From the formula for mutual inductance Emf E = M( dI/dt) E = 30.14×10⁻⁴ × (4/0.25) E = 4.82×10⁻² V Answer:(b)
Q.4
A square coil of 10-2 square metre area is placed perpendicular to a uniform magnetic field of intensity 103 Weber/metrethe magnetic flux through the coil is [ MPPMT 1990]
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a) 10 Weber
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b) 10-5 weber
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c) 105 weber
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d) 100 weber
Explanation
Flux Φ = BA = 103×10⁻² = 10 weber Answer: (a)
Q.5
An aeroplane in which the distance between the tips of the wings is 50 meters is flying horizontally with a speed of 360 km/hr over a place where the vertical component of the earth's magnetic field is 2.0×10⁻⁴ weber/metreThe potential difference between the tips of the wings would be [ CPMT 1990]
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a) 0.1 volt
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b) 1.0 Volt
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c) 0.2 Volt
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d) 0.01Volt
Explanation
From the formula E = Blv Here B = earths vertical magnetic component =2.0×10⁻⁴ weber/metre2 velocity v = 360 km/hr = 100 m/s l = length of the wing = 50 on substituting and solving we get E = 1.0 Volt Answer: (b)
Q.6
A player with 3 meter long iron rod runs towards east with a speed of 30 km/hr. Horizontal component of earth's magnetic field is 4×10⁻⁵ weber/mIf he is running with rod horizontal and vertical positions, then the potential difference induced between the two ends of the rod in two cases will be
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c) zero in both position
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a) zero in vertical position and 1×10⁻³ volt in horizontal position
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b) 1×10⁻³ volt in horizontal position and vertical position
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d) 1×10⁻³ volts in both position
Explanation
When the rod is in vertical position, it cuts horizontal component of earth's field and an induced emf E = Blv E = 4×10⁻⁵ ×(25/3)×3 = 10-3 volt When the rod is horizontal it does not cut any of the field ( horizontal or vertical) as Horizontal component is parallel to length. So no induced emf is produced in this case Answer:(b)
Q.7
A coil has 2000 turns and area of 70 cmThe magnetic field perpendicular to the plane of the coil is 0.3 weber/m2 and takes 0.1 sec to rotate through 180°. The value of induced emf will be [ MPPET 1993]
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a) 8.4 volt
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b) 84 volt
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c) 42 volt
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d) 4.2 volt
Explanation
φ = BAcosθ φi = BAcosθ φf = BAcos(θ+π) = -BAcosθ dφ = φf - φi dφ = -BAcosθ -BAcosθ = -2BAcosθ let θ = 0 dφ = -2BA = -2×0.3 ×70×10⁻⁴ dφ = 42×10⁻⁴ dφ / dt = 42×10⁻³ E = -N(dφ / dt) = 2000×42×10⁻³ = 84V Answer: (b)
Q.8
An emf of 5 millivolt is induced in a coil when in a nearby placed another coil, the current changes by 5amp in 0.1 second. The coefficient of mutual induction between the two coils will be [ MPPMT 1993]
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a) 1 Henry
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b) 0.1 henry
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c) 0.1 millihenry
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d) 0.01 millihenry
Explanation
E = M ( dI/dt) 5×10⁻³ = M (5/.1) 5×10⁻³ = M 50 M = 1×10⁻⁴ Henry M = 0.1 milliHenry Answer: (c)
Q.9
As shown in figure a metal rod of length 50cm makes contact and complete the circuit. The circuit is perpendicular to the magnetic field with B = 0.15 tesla. If the resistance is 3Ω, force needed to move the rod as indicated with a constant speed of 2m/s is : [ MPPMT 1994]
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c) 3.75×102 N
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a) 3.75×10⁻³ N
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b) 3.75×10⁻² N
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d) 3.75×10⁻⁴ N
Explanation
Induced current I = Blv/R Force on the wire = BIl F = B2l2V / R F = 3.75×10⁻³ Answer:(a)
Q.10
A solenoid is 1.5 m long and its inner diameter is 4.0 cm. It has three layers of windings of 1000 turns each and carries a current of 2.0 amperes. The magnetic flux for a cross-section of the solenoid is nearly [ AMU 1995]
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a) 2.5×10⁻⁷ weber
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b) 6.31 ×10⁻³ weber
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c) 5.2×10⁻⁵ weber
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d) 4.1×10⁻⁵ weber
Explanation
Magnetic field B = µonI n is number of turns per unit length = 3×1000 / 1.5 = 2000 B = 4π×10-7 ×2000 ×2 B = 1.6π×10-3 Fulx : φ = nBA φ = 1000(1.6π×10-3)(π4×10⁻⁴ φ = 63.10×10⁻⁴ = 6.31×10⁻³ wb Answer: (b)
Q.11
The root-mean-square value of an alternating current of 50hertz frequency is 10 ampere. The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be [ MPPMT 1993]
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a) 2×10⁻² sec and 14.14 amp
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b) 1×10⁻² sec and 7.07 amp
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c) 5×10⁻³ sec and 7.07 amp
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d) 5×10⁻³ sec and 14.14 amp
Explanation
Time for reading zero to maximum value = T/4 but T = 1/ f =1/50 = 0.02 Time for reading zero to maximum value = = 0.02/4 = 0.005= 5×10⁻³ io = irms√2 = 10×1.414 =14.14 A Answer:(d)
Q.12
An alternating voltage E( in volts) = 200√2sin(100t) is connected to a 1µf capacitor through an a.c. ammeter. The reading of the ammeter shall be [ MNR 1995]
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a) 10mA
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b) 20mA
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c) 40 mA
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d) 80mA
Explanation
Irms = Erms / Z But E0 = Eo / √2 = 200 Z = 1/ ωC 1/Z = ωC = 100×10⁻⁶ = 10-4 ∴ Irms = 200×10⁻⁴ = 2×10⁻² = 20mA Answer: (b)
Q.13
More than one correct option. At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key andan alternating current I(t) = I0cos (ωt), with I0 = 1A and ω = 500 rad s−1 starts flowing init with the initial direction shown in the figure. At =7π/6ω, the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery tocharge the capacitor fully. If C = 20µF, R = 10 Ω and the battery is ideal with emf of 50V,identify the correct statement (s). [ IIT Advance 2014]
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b) The current in the left part of the circuit just before t=7π/6ω is clockwise.
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c) Immediately after A is connected to D, the current in R is 10A.
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d) Q = 2×10⁻³ C.
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a) Magnitude of the maximum charge on the capacitor before t=7π/6ω is 1 × 10⁻³ C.
Explanation
Option a I(t) = I0cos (ωt) Integrating For maximum charge sinωt =1 Option a is wrong Option b Thus current is antilock wise Option C On disconnecting from B charge on capacitor And potential V=Q/C Capacitor act as battery with upper plate negative and is series with battery. total potential across R = 100 V thus current I=100/10 =10A Option c correct Option d charge on capacitor when capacitor is full charge is maximum cahrage on capacitor as calculated 2×10⁻³C Answer:(c, d)
Q.14
A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The center of the small loop is on the axis of the bigger loop. The distance between their centers is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is … [ IIT Mains 2017]
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a) 9.1 × 10⁻¹¹ weber
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b) 6 × 10⁻¹¹ weber
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c) 3.3 × 10⁻¹¹ weber
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d) 6.6 × 10⁻⁹ weber
Explanation
If we consider current is through big loop then we can calculate magnetic flux at smaller coil every easily as small loop is at axis of big loop and find mutual inductance, Then we can find magnetic flux at small loop as Φ2 = MI1 Current flowing through big loop produces a magnetic field given by Now Flux linked with small loop= Ba We know that Φ = M I Thus Now flux at bigger coil Φ = 9.1 × 10⁻¹¹ Answer:(a)
Q.15
An inductive circuit contains a resistance of 10Ω and an inductance of 2.0 henry. If an AC voltage of 120 Volts and frequency of 60Hz is applied to this circuit, the current in the circuit would be nearly [ CPMT 1990]
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a) 0.32 amp
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b) 0.16 amp
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c) 0.48 amp
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d) 0.80 amp
Explanation
Answer: (b)
Q.16
The potential difference across the resistance, capacitance and inductance are 80V, 40V and 100V respectively in an L-C-R circuit. The power factor of this circuit is …[NEET II 2016]
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a) 0.8
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b) 1.0
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c) 0.4
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d) 0.5
Explanation
Power factor =cosθ [since opposite side =3 and adjacent side =4, thus hypo = 5] Answer:(a)
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