NEET Practice Questions (MCQs) with Answers & Solutions

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A plane E M wave of frequency 25 MHz travels in free space in the x-direction. At a particular point in space and time E=6.3 j^ V/m then B at the point is 

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Explanation

B=Ec=6.33×108=2.1×10-8  WbE×B should be in +x-dorection as the wave is propagating along +x-direction.So, B=2.1×10-8 k^

In an electromagnetic wave, the amplitude of electric field is 1 V/m. The frequency of wave is 5×1014 Hz. The wave is propagating along Z-axis. The average energy density of electric field in joule/m3, will be

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Explanation

Average energy density=120E22

12×8.85×10-12×122=2.21×10-12 J/m3

The infra-red spectrum lies between :

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Explanation

 The infra-red spectrum lies between the visible and ultraviolet regions.

Consider an electric charge oscillating with a frequency of 10 MHz. The radiation emitted will have a wavelength equal to

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Explanation

λ=cν=3×10810×106=30 m

Which of the following electromagnetic waves has minimum frequency?

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Explanation

Audible waves have the least frequency

A parallel plate capacitor consists of two circular plates each of radius 12 cm and separated by 5.0 mm. The capacitor is being charged by an external source. The charging current is constant and is equal to 0.15 A. The rate of change of the potential difference between the plates will be :

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Explanation

C=ε0Ad=8.85×10-12×π×0.1225×10-3=0.08×10-9 FdVdt=1Cdqdt=iC=0.150.08×10-9=1.873×109 V/sec

A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes 100 W of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of 5 m from the lamp will be

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Explanation

Power of Em waves = 3% of 100 W = 3 W

Intensity=PA=34π(5)2=0.0096 Wm2

Then, I=12 0CE02E0=2I0C=2.68 v/m

A parallel plate capacitor consists of two circular plates each of radius 2 cm, separated by a distance of 0.1 mm. If the voltage across the plates is varying at the rate of 5×1013 V/s, then the value of displacement current is :

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Explanation

ID=CdVdt=ε0AddVdtID=8.85×10-12×3.14×2×10-220.1×10-3×5×1013=5.56×103 A

In an electromagnetic wave, the direction of the magnetic field induction B is

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Explanation

Magnetic field and electric field are mutually perpendicular to each other as well as to the direction of wave

In an electromagnetic wave

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Explanation

power is transmitted in a direction perpendicular to both the fields.

 

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