NEET Practice Questions (MCQs) with Answers & Solutions

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If a source is transmitting an electromagnetic wave of frequency 8.2×106 Hz, then the wavelength of the electromagnetic wave transmitted from the source will be :

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Explanation

fλ=cλ=cf=3×1088.2×106=36.6 m

A plane electromagnetic wave

Ez=100 cos(6×108t+4x) V/m

propagates in a medium of dielectric constant

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Explanation

Ez=100 cos(6×108t+4x) V/m

Speed of wave in the medium, v=ωK=1.5×108 m/s

Refractive index, μ=cv=2

For non-magnetic material μ=K K=4

If μ0 be the permeability and K0 the dielectric constant of a medium, its refractive index is given by

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Explanation

μ=CV=μεμ0ε0ε=K0ε0 and μ=μ0μ=K0

If ε0 and μ0 represent the permittivity and permeability of vacuum and ε and  μ represent the permittivity and permeability of the medium, the refractive index of the medium is given by

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Explanation

C=1μ0ε0V=1μεμ=CV=μεμ0ε0

The magnetic field between the plates of radius 12 cm separated by a distance of 4 mm of a parallel plate capacitor of capacitance 100 pF along the axis of plates having conduction current of 0.15 A is

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A flood light is covered with a fitter that transmits red light. The electric field of the emerging beam is represented by a sinusoidal plane wave

Ex=36 sin (1.20×107 z-3.6×1015 t) V/m

The average intensity of beam in W/m2 will be

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Explanation

Intensity=12ε0E02C=12×8.85×10-12×(36)2×3×108=1.72 W/m2

The average energy - density of electromagnetic wave given by E = (50 N/C) sin(ωt-kx) will be nearly

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Explanation

E=50 sin (ωt-kx)Avg. energy density=12ε0E20=12×8.85×10-12×(50)2=1.1×10-8 J/m3

A larger parallel plate capacitor, whose plates have an area of 1 m2 are separated from each other by 1 mm, is being charged at a rate of 25.8 V/s. If the plates has dielectric constant 10, then the displacement current at this instant is

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Explanation

During charging of capacitor, 

Conduction current = Displacement current

dQdt=idQ=CVid=CdVdt  =KAε0d×dVdt  =10×1×8.85×10-1210-3×25.8  =2.28 μA

A parallel plate capacitor with plate area A and separation between the plates d, is charged by a constant current i.  Consider a plane surface of area A/2 parallel to the plates and drawn symmetrically between the plates. The displacement current through this area is

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Explanation

At any instant, let charge on capacitor be Q

Electric field between plates, E=QAε0

Flux through area A2, ϕE=EA2

Displacement current , 

id=ε0dϕEdt   =12dQdt   =i2 

The sun delivers 104 W/m2 of electromagnetic flux to the earth's surface. The total power that is incident on a roof of dimensions (10×10) m2 will be

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Explanation

P=Intensity×Area  =104×10×10  =106 W

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