# Match List - I with List - II. List -I (Electromagnetic waves) List - II (Wavelength) (a) AM radio waves (i) $$10^{-10}~\text{m}$$ (b) Microwaves (ii) $$10^{2} ~\text{m}$$ (c) Infrared radiation (iii) $$10^{-2} ~\text{m}$$ (d) $$X$$-rays (iv) $$10^{-4} ~\text{m}$$ Choose the correct answer from the options given below: (a) (b) (c) (d) 1. (ii) (iii) (iv) (i) 2. (iv) (iii) (ii) (i) 3. (iii) (ii) (i) (iv) 4. (iii) (iv) (ii) (i)

Subtopic:  Electromagnetic Spectrum |
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NEET - 2022
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When light propagates through a material medium of relative permittivity, $$\varepsilon_{r}$$ and relative permeability, $$\mu_{r}$$ the velocity of light, $$v$$ is given by:
($$c$$ = velocity of light in vacuum)
1. $$v=\dfrac{{c}}{\sqrt{\varepsilon_{r} \mu_{{r}}}}$$
2. $$v={c}$$
3. $$v=\sqrt{\dfrac{\mu_{{r}}}{\varepsilon_{{r}}}}$$
4. $$v=\sqrt{\dfrac{\varepsilon_{{r}}}{\mu_{{r}}}}$$
Subtopic:  Properties of EM Waves |
85%
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NEET - 2022
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If $$\lambda_X,\lambda_I,\lambda_M$$ and $$\lambda_\gamma$$ are the wavelengths of $$X$$-rays, infrared rays, microwaves and $$\gamma$$-rays respectively, then:
 1 $$\lambda_\gamma<\lambda_X<\lambda_I<\lambda_M$$ 2 $$\lambda_M<\lambda_I<\lambda_X<\lambda_\gamma$$ 3 $$\lambda_X<\lambda_\gamma<\lambda_M<\lambda_I$$ 4 $$\lambda_X<\lambda_I<\lambda_\gamma<\lambda_M$$
Subtopic:  Electromagnetic Spectrum |
82%
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NEET - 2022
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An electromagnetic wave is moving along negative $$z (-z)$$ direction and at any instant of time, at a point, its electric field vector is $$3\hat j~\text{V/m}$$. The corresponding magnetic field at that point and instant will be: (Take $$c=3\times10^{8}~\text{ms}^{-1}$$)
 1 $$10\hat i~\text{nT}$$ 2 $$-10\hat i~\text{nT}$$ 3 $$\hat i~\text{nT}$$ 4 $$-\hat i~\text{nT}$$
Subtopic:  Properties of EM Waves |
53%
From NCERT
NEET - 2022
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The magnetic field of a plane electromagnetic wave
is given by,
$$\vec B=3\times10^{-8}\text{cos}(1.6\times10^3x+48\times10^{10}t)\hat j~\text{T}.$$
The associated electric field will be:
 1 $$3 \times 10^{-8} \text{cos}\left(1.6 \times 10^3 x+48 \times 10^{10} t\right) \hat{i}~\text{ V/m}$$ 2 $$3 \times 10^{-8} \text{sin} \left(1.6 \times 10^3 {x}+48 \times 10^{10} {t}\right) \hat{{i}}~ \text{V} / \text{m}$$ 3 $$9 \text{sin} \left(1.6 \times 10^3 {x}-48 \times 10^{10} {t}\right) \hat{{k}} ~~\text{V} / \text{m}$$ 4 $$9 \text{cos} \left(1.6 \times 10^3 {x}+48 \times 10^{10} {t}\right) \hat{{k}}~~\text{V} / \text{m}$$
Subtopic:  Properties of EM Waves |
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From NCERT
NEET - 2022
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The ratio of the magnitude of the magnetic field and electric field intensity of a plane electromagnetic wave in free space of permeability $$\mu_0$$ and permittivity $$\varepsilon_0$$ is:
(Given that $$c=$$ velocity of light in free space)
1.  $$c$$
2.  $$\dfrac1c$$
3.  $$\dfrac{c}{\sqrt{\mu_0\varepsilon_0}}$$
4.  $$\dfrac{\sqrt{\mu_0\varepsilon_0}}{c}$$
Subtopic:  Properties of EM Waves |
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NEET - 2022
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In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of $$2.0\times 10^{10}~ \text{Hz}$$ and amplitude $$48~\text{Vm}^{-1}$$. Then the amplitude of the oscillating magnetic field is: (Speed of light in free space $$3\times 10^{8}~ \text{ms}^{-1}$$)
1. $$1.6 \times 10^{-6} ~\text{T}$$
2. $$1.6 \times 10^{-9} ~\text{T}$$
3. $$1.6 \times 10^{-8} ~\text{T}$$
4. $$1.6 \times 10^{-7} ~\text{T}$$
Subtopic:  Properties of EM Waves |
58%
From NCERT
NEET - 2023
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$$\varepsilon_0$$ and $$\mu_0$$ are the electric permittivity and magnetic permeability of free space respectively. If the corresponding quantities of a medium are $$2\varepsilon_0$$ and $$1.5\mu_0$$ respectively, the refractive index of the medium will nearly be:
1. $$\sqrt2$$
2. $$\sqrt3$$
3. $$3$$
4. $$2$$
Subtopic:  Properties of EM Waves |
78%
From NCERT
NEET - 2023
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To produce an instantaneous displacement current of $$2~\text{mA}$$ in the space between the parallel plates of a capacitor of capacitance $$4~\mu\text{F}$$, the rate of change of applied variable potential difference $$\left(\frac{dV}{dt}\right)$$ must be:
1. $$800~ \text{V} / \text{s}$$
2. $$500~ \text{V} / \text{s}$$
3. $$200~ \text{V} / \text{s}$$
4. $$400 ~\text{V} / \text{s}$$
Subtopic:  Displacement Current |
77%
From NCERT
NEET - 2023
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A capacitor of capacitance $$C$$ is connected across an AC source of voltage $$V$$, given by;
$$V=V_0 \sin \omega t$$$\mathrm{}$
The displacement current between the plates of the capacitor would then be given by:
1. $$I_d=\frac{V_0}{\omega C} \sin \omega t$$
2. $$I_d=V_0 \omega C \sin \omega t$$
3. $$I_d=V_0 \omega C \cos \omega t$$
4. $$I_d=\frac{V_0}{\omega C} \cos \omega t$$

Subtopic:  Displacement Current |
56%
From NCERT
NEET - 2021
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