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Physics MCQs for NEET — Practice Questions with Answers

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The ratio of $\frac{E_{0}}{H_{0}}$ for a plane electromagnetic wave has the dimension of

Impedance

Resistance

Both

None

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Explanation

$F o r v a c c u m, E_{o} = B_{o} c B_{o} = μ_{o} H_{o} S o, [E_{o}] = \frac{F o r c e}{C h a r g e} [H_{o}] = \frac{C u r r e n t}{L e n g t h} [\frac{E_{o}}{H_{o}}] = \frac{W o r k}{C h a r g e \times C u r r e n t} = \frac{P o t e n t i a l}{C u r r e n t} = R e s i s \tan c e$

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An electromagnetic wave going through a medium is given by $E = E_{0} \sin (k x - ω t) a n d B = B_{0} \sin (k x - ω t)$ then

$E_{0} k = B_{0} ω$

If Electric is in Z direction them magnetic field should be in -y direction

Both 'A' and 'B' are correct

Only A is correct

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Explanation

We know,

$E_{0} = v_{w} . B_{0} E_{0} = \frac{ω}{K} B_{0} E_{0} K = B_{0} ω$

As the wave is traveling along -x-direction, the magnetic field should be along +y-direction so that the direction of $\vec{E} \times \vec{B}$ is along the direction of the wave.

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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 $E = 6.3 \hat{j} V / m$ then B at the point is

$2.1 \times 10^{- 8} - \hat{k}$

$2.1 \times 10^{- 8} \hat{k}$

2.1 $\hat{k}$

$2.1 \times 10^{- 8} \hat{i}$

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Explanation

$B = \frac{E}{c} = \frac{6.3}{3 \times 10^{8}} = 2.1 \times 10^{- 8} W b \vec{E} \times \vec{B} s h o u l d b e i n + x - d o r e c t i o n a s t h e w a v e i s p r o p a g a t i n g a l o n g + x - d i r e c t i o n . S o, \vec{B} = 2.1 \times 10^{- 8} \hat{k}$

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In an electromagnetic wave, the amplitude of electric field is 1 V/m. The frequency of wave is $5 \times 10^{14} H z$ . The wave is propagating along Z-axis. The average energy density of electric field in joule/ $m^{3}$ , will be

$2.2 \times 10^{- 12}$

4.4 $\times 10^{- 12}$

6.6 $\times 10^{- 12}$

8.8 $\times 10^{- 12}$

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Explanation

Average energy density= $\frac{1}{2} \in_{0} {(\frac{E}{\sqrt{2}})}^{2}$

$\frac{1}{2} \times 8.85 \times 10^{- 12} \times {(\frac{1}{\sqrt{2}})}^{2} = 2.21 \times 10^{- 12} J / m^{3}$

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The infra-red spectrum lies between :

radio wave and microwave region

the micro-wave and visible region

the visible and ultraviolet region

the ultraviolet and the X-ray region

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Explanation

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

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

20 m

30 m

40 m

10 m

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Explanation

$λ = \frac{c}{ν} = \frac{3 \times 10^{8}}{10 \times 10^{6}} = 30 m$

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Which of the following electromagnetic waves has minimum frequency?

radio wave

ultrasonic wave

microwave

audible wave

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Explanation

Audible waves have the least frequency

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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 :

$1.873 \times 10^{7} V / s$

$1.873 \times 10^{8} V / s$

$1.873 \times 10^{9} V / s$

$1.873 \times 10^{10} V / s$

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Explanation

$C = \frac{ε_{0} A}{d} = 8.85 \times 10^{- 12} \times \frac{π \times {(0.12)}^{2}}{5 \times 10^{- 3}} = 0.08 \times 10^{- 9} F \frac{d V}{d t} = \frac{1}{C} \frac{d q}{d t} = \frac{i}{C} = \frac{0.15}{0.08 \times 10^{- 9}} = 1.873 \times 10^{9} V / s e c$

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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

1.34 V/m

2.68 V/m

4.02 V/m

5.36 V/m

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Explanation

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

Intensity= $\frac{P}{A} = \frac{3}{4 π {(5)}^{2}} = 0.0096 \frac{W}{m^{2}}$

$T h e n, I = \frac{1}{2} \in_{0} C E_{0}^{2} \Rightarrow E_{0} = \sqrt{\frac{2 I}{\in_{0} C}} = 2.68 v / m$

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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 $5 \times 10^{13}$ V/s, then the value of displacement current is :

5.50 A

$5.56 \times 10^{2} A$

5.56 $\times 10^{3}$ A

$2.28 \times 10^{4} A$

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Explanation

$I_{D} = C \frac{dV}{dt} = \frac{ε_{0} A}{d} \frac{dV}{dt} I_{D} = \frac{8.85 \times 10^{- 12} \times 3.14 \times {(2 \times 10^{- 2})}^{2}}{0.1 \times 10^{- 3}} \times 5 \times 10^{13} = 5.56 \times 10^{3} A$

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