ELECTROMAGNETIC WAVES MCQs for NEET UG PHYSICS — Easy

1 Easy

Understanding +4 | -1

An electromagnetic wave is propagating along the

+\mathrm{z}

-axis. Direction of electric and the magnetic field of this wave can be represented by

View answer and explanation

Correct option: A

\vec{E}=E_0 \hat{i}, \vec{B}=B_0 \hat{j}

Option A (

\vec{E}=E_0 \hat{i}, \vec{B}=B_0 \hat{j}

) is correct because the electric and magnetic fields are perpendicular to each other and to the direction of propagation, which is the z-axis.

Solution approach

Correct Answer + Core Reason: $\vec{E}=E_0 \hat{i}, \vec{B}=B_0 \hat{j}$ . E and B fields are perpendicular to each other and the direction of propagation.

Step-by-Step Solution:

Step 1: Recall the properties of electromagnetic waves.

Step 2: In an electromagnetic wave, the electric field ( $\vec{E}$ ), magnetic field ( $\vec{B}$ ), and the direction of propagation ( $\vec{k}$ ) are mutually perpendicular. Since the wave is propagating along the +z-axis, $\vec{k} = \hat{k}$ .

Step 3: If $\vec{E}$ is along the x-axis ( $\hat{i}$ ), then $\vec{B}$ must be along the y-axis ( $\hat{j}$ ) to satisfy the condition $\vec{E} \times \vec{B} \propto \vec{k}$ .

Step 4: Final Answer: $\vec{E}=E_0 \hat{i}, \vec{B}=B_0 \hat{j}$

Why other options are wrong:
- Option B: $\vec{E}=E_0 \hat{k}, \vec{B}=B_0 \hat{i}$ is incorrect because the electric field is along the z-axis, which is the direction of propagation, violating the perpendicularity condition.
- Option C: $\vec{E}=E_0 \hat{j}, \vec{B}=B_0 \hat{i}$ is incorrect because $\vec{E} \times \vec{B}$ would be in the -z direction, not the +z direction.
- Option D: $\vec{E}=E_0 \hat{j}, \vec{B}=B_0 \hat{k}$ is incorrect because the magnetic field is along the z-axis, which is the direction of propagation, violating the perpendicularity condition.

Key Concepts:
- Electromagnetic waves
- Direction of propagation
- Perpendicularity of E, B, and k

Memory Aids:
- Remember that E, B, and the direction of propagation are mutually perpendicular.

Quick Tips:
- Use the right-hand rule to determine the direction of propagation: $\vec{E} \times \vec{B}$ gives the direction of propagation.

2 Easy

2 of 35Question permalink

Application +4 | -1

Calculate the momentum transferred to a surface when a radiation of energy E falls normally on it. Assume that the reflectivity of the surface is unity.

View answer and explanation

Correct option: B

\frac{2 E}{c}

Option B (

\frac{2 E}{c}

) is correct because the momentum transferred to a surface with reflectivity of unity is twice the energy divided by the speed of light.

Solution approach

Correct Answer + Core Reason: $\frac{2 E}{c}$ . Momentum transferred for total reflection is twice the energy divided by c.

Step-by-Step Solution:

Step 1: Recall the relationship between energy and momentum of radiation.

Step 2: The momentum of radiation with energy E is given by $p = \frac{E}{c}$ .

Step 3: For a surface with reflectivity of unity (total reflection), the momentum transferred is twice the initial momentum.

Step 4: Therefore, the momentum transferred is $2p = \frac{2E}{c}$ .

Why other options are wrong:
- Option A: $\frac{E}{c}$ is incorrect because it represents the momentum transferred for total absorption, not reflection.
- Option C: $E c$ is incorrect because it multiplies energy by the speed of light, which is not the correct relationship for momentum.
- Option D: $\frac{E}{c^2}$ is incorrect because it divides energy by the square of the speed of light, which is not the correct relationship for momentum.

Key Concepts:
- Momentum of radiation
- Reflection of radiation

Memory Aids:
- Remember that reflection doubles the momentum transfer.

Quick Tips:
- For total reflection, the momentum transferred is $2E/c$ .

3 Easy

3 of 35Question permalink

Application +4 | -1

Radiations of intensity 0.5

W{m^{ - 2}}

are striking a metal plate. The pressure on the plate is

View answer and explanation

Correct option: A

0.166 \times {10^{ - 8}}\;\mathrm{N}{\mathrm{m}^{ - 2}}

Option A (

0.166 \times {10^{ - 8}}\;\mathrm{N}{\mathrm{m}^{ - 2}}

) is correct because the pressure is calculated as intensity divided by the speed of light when the radiation is fully absorbed.

Solution approach

Correct Answer + Core Reason: $0.166 \times {10^{ - 8}}\;\mathrm{N}{\mathrm{m}^{ - 2}}$ . The pressure is calculated by dividing the intensity by the speed of light.

Step-by-Step Solution:

Step 1: Identify the given information: Intensity, $I = 0.5 \;W/m^2$ .

Step 2: Recall the formula for radiation pressure when the radiation is fully absorbed: $P = \frac{I}{c}$ , where $P$ is the pressure, $I$ is the intensity, and $c$ is the speed of light.

Step 3: Substitute the given values: $P = \frac{0.5 \;W/m^2}{3 \times 10^8 \;m/s}$

Step 4: Calculate the pressure: $P = 0.166 \times 10^{-8} \;N/m^2$

Step 5: Final Answer: $P = 0.166 \times {10^{ - 8}}\;\mathrm{N}{\mathrm{m}^{ - 2}}$

Why other options are wrong:
- Option B: This value is incorrect. It might be obtained by multiplying the correct answer by 2, possibly confusing the case of full reflection with full absorption.
- Option C: This value is incorrect. It might be obtained by using an incorrect formula or making a calculation error.
- Option D: This value is incorrect. It might be obtained by dividing the intensity by a value other than the speed of light or making a calculation error.

Key Concepts:
- Radiation pressure
- Intensity of electromagnetic waves
- Speed of light

Memory Aids:
- Pressure = Intensity / Speed of Light (P = I/c) for full absorption

Quick Tips:
- Remember the formula for radiation pressure. Ensure you use the correct formula based on whether the radiation is absorbed or reflected.

4 Easy

4 of 35Question permalink

Remembering +4 | -1

Part of electromagnetic spectrum that has largest penetrating power is

View answer and explanation

Correct option: A

\gamma

-rays

Option A is correct because gamma rays have the largest penetrating power due to their high energy and short wavelength in the electromagnetic spectrum.

Solution approach

Correct Answer + Core Reason: $\gamma$ -rays. Gamma rays have the highest penetrating power due to their high energy.

Step-by-Step Solution:

Step 1: Recall the relationship between energy, frequency, and wavelength of electromagnetic radiation: $E = hf = \frac{hc}{\lambda}$ , where E is energy, f is frequency, λ is wavelength, h is Planck's constant, and c is the speed of light.

Step 2: Higher energy electromagnetic radiation has greater penetrating power.

Step 3: Among the given options, gamma rays have the highest energy and shortest wavelength, therefore the largest penetrating power.

Why other options are wrong:
- Option B: X-rays. X-rays have less penetrating power than gamma rays.
- Option C: radio-waves. Radio waves have much lower energy and penetrating power than gamma rays.
- Option D: Infrared rays. Infrared rays have much lower energy and penetrating power than gamma rays.

Key Concepts:
- Electromagnetic Spectrum
- Energy of Electromagnetic Radiation
- Penetrating Power

Memory Aids:
- Remember that higher energy radiation has greater penetrating power.

Quick Tips:
- Know the relationship between energy, frequency, and wavelength.

5 Easy

5 of 35Question permalink

Application +4 | -1

If frequency of electromagnetic wave is

60 \mathrm{~MHz}

and it travels in air along

z

direction then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other and the wavelength of the wave (in

\mathrm{m}

) is

View answer and explanation

Correct option: D

5

Option D (5) is correct because the wavelength is calculated using the formula

\lambda = c/f

, where

c

is the speed of light and

f

is the frequency.

Solution approach

Correct Answer + Core Reason: 5. The wavelength is calculated using $\lambda = c/f$ .

Step-by-Step Solution:

Step 1: Formula: The wavelength of an electromagnetic wave is given by $\lambda = \frac{c}{f}$ , where $c$ is the speed of light in air (approximately $3 \times 10^8 \text{ m/s}$ ) and $f$ is the frequency.

Step 2: Substitution: Given $f = 60 \text{ MHz} = 60 \times 10^6 \text{ Hz}$ , we have $\lambda = \frac{3 \times 10^8}{60 \times 10^6}$ .

Step 3: Calculation: $\lambda = \frac{300}{60} = 5 \text{ m}$ .

Step 4: Final Answer: 5

Why other options are wrong:
- Option A: This is incorrect. The correct calculation yields 5 m.
- Option B: This is incorrect. The correct calculation yields 5 m.
- Option C: This is incorrect. The correct calculation yields 5 m.

Key Concepts:
- Wavelength of electromagnetic waves
- Frequency
- Speed of light

Memory Aids:
- $\lambda = c/f$ (Lambda equals c over f).

Quick Tips:
- Always convert frequency to Hz before calculating wavelength.

6 Easy

6 of 35Question permalink

Application +4 | -1

n

photons of wavelength

\lambda

are absorbed by a black body of mass '

m

'. The momentum gained by the body is

View answer and explanation

Correct option: A

\frac{nh}{\lambda}

Option A (

\frac{nh}{\lambda}

) is correct because the momentum gained by the body is the sum of the momenta of the absorbed photons.

Solution approach

Correct Answer + Core Reason: $\frac{nh}{\lambda}$ . The momentum gained is the sum of the momenta of the absorbed photons.

Step-by-Step Solution:

Step 1: Recall the momentum of a photon.

Step 2: The momentum of a single photon is given by $p = \frac{h}{\lambda}$ , where h is Planck's constant and $\lambda$ is the wavelength.

Step 3: Calculate the total momentum gained by the black body.

Step 4: Since n photons are absorbed, the total momentum gained is $P = n \times p = n \frac{h}{\lambda}$ .

Why other options are wrong:
- Option B: $\frac{nh}{m \lambda}$ is incorrect because it divides the momentum by the mass of the black body, which is not relevant to the momentum gained.
- Option C: $\frac{mnh}{\lambda}$ is incorrect because it multiplies the momentum by the mass of the black body, which is not relevant to the momentum gained.
- Option D: $\frac{h}{m \lambda}$ is incorrect because it considers only one photon and divides by the mass of the black body, which is not relevant to the momentum gained.

Key Concepts:
- Momentum of a photon
- Absorption of photons

Memory Aids:
- Remember that the momentum of a photon is inversely proportional to its wavelength.

Quick Tips:
- The total momentum gained is the sum of individual photon momenta.

7 Easy

7 of 35Question permalink

Understanding +4 | -1

During the porpagation of electromagnetic waves in a medium,

View answer and explanation

Correct option: C

electric energy density is equal to the magnetic energy density

Option C (electric energy density is equal to the magnetic energy density) is correct because in an electromagnetic wave, the electric and magnetic energy densities are equal on average.

Solution approach

Correct Answer + Core Reason: electric energy density is equal to the magnetic energy density. In an EM wave, energy is equally distributed between electric and magnetic fields.

Step-by-Step Solution:

Step 1: Recall the energy densities of electric and magnetic fields.

Step 2: The energy density of the electric field is given by $u_E = \frac{1}{2} \varepsilon_0 E^2$ , and the energy density of the magnetic field is given by $u_B = \frac{1}{2} \frac{B^2}{\mu_0}$ .

Step 3: In an electromagnetic wave, $E = cB$ , and $c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}$ . Substituting these relationships, we find that $u_E = u_B$ .

Step 4: Final Answer: $u_E = u_B$

Why other options are wrong:
- Option A: electric energy density is double of the magnetic energy density is incorrect because the energy densities are equal, not doubled.
- Option B: electric energy density is half of the magnetic energy density is incorrect because the energy densities are equal, not halved.
- Option D: Both electric and magnetic energy densities are zero is incorrect because electromagnetic waves carry energy, so the energy densities are non-zero.

Key Concepts:
- Electromagnetic waves
- Electric and magnetic energy densities

Memory Aids:
- Remember that energy is equally distributed in EM waves.

Quick Tips:
- In EM waves, the electric and magnetic fields contribute equally to the energy density.

8 Easy

8 of 35Question permalink

Application +4 | -1

In an electromagnetic wave, at any instant the electric field was found to oscillate with an amplitude of

18\, \mathrm{v} / \mathrm{m}

. The magnitude of oscillating magnetic field will be -

View answer and explanation

Correct option: B

6 \times 10^{-8} \mathrm{~T}

Option B (

6 \times 10^{-8} \mathrm{~T}

) is correct because the magnitude of the oscillating magnetic field is calculated using the formula

B_0 = \frac{E_0}{c}

, where c is the speed of light.

Solution approach

Correct Answer + Core Reason: $6 \times 10^{-8} \mathrm{~T}$ . The magnitude of the oscillating magnetic field is calculated using the formula $B_0 = \frac{E_0}{c}$ .

Step-by-Step Solution:

Step 1: Identify the given value: Amplitude of the electric field $E_0 = 18 \text{ V/m}$ .

Step 2: Recall the relationship between the amplitudes of the electric and magnetic fields in an electromagnetic wave: $E_0 = cB_0$ , where $c$ is the speed of light in vacuum ( $c = 3 \times 10^8 \text{ m/s}$ ).

Step 3: Solve for the amplitude of the magnetic field: $B_0 = \frac{E_0}{c} = \frac{18 \text{ V/m}}{3 \times 10^8 \text{ m/s}}$ .

Step 4: Calculate the amplitude of the magnetic field: $B_0 = 6 \times 10^{-8} \text{ T}$ .

Why other options are wrong:
- Option A: $4 \times 10^{-6} \mathrm{~T}$ is incorrect because it does not follow from the correct calculation.
- Option C: $9 \times 10^{-9} \mathrm{~T}$ is incorrect because it does not follow from the correct calculation.
- Option D: $11 \times 10^{-11} \mathrm{~T}$ is incorrect because it does not follow from the correct calculation.

Key Concepts:
- Electromagnetic waves
- Relationship between electric and magnetic field amplitudes
- Speed of light

Memory Aids:
- Remember the relationship $E_0 = cB_0$ .

Quick Tips:
- Always remember the relationship between E and B fields in EM waves: E = cB.

9 Easy

9 of 35Question permalink

Application +4 | -1

The sun delivers

{10^3}\,W/{m^2}

of electromagnetic flux to the earth’s surface. Find the total power that is incident on a roof of dimensions

8\,m \times 20\,m

and the radiation force on the roof is given by

5.33 \times {10^{ - 4}}N

View answer and explanation

Correct option: C

1.6 \times {10^5}\mathrm{ watt}

Option C (

1.6 \times {10^5}\mathrm{ watt}

) is correct because the total power is the product of the intensity and the area of the roof.

Solution approach

Correct Answer + Core Reason: $1.6 \times {10^5}\mathrm{ watt}$ . Total power is intensity times area.

Step-by-Step Solution:

Step 1: Calculate the area of the roof.

Step 2: Area = length x width = 8 m x 20 m = 160 m $^2$ .

Step 3: Calculate the total power incident on the roof.

Step 4: Power = Intensity x Area = 10 $^3$ W/m $^2$ x 160 m $^2$ = 1.6 x 10 $^5$ W.

Why other options are wrong:
- Option A: 20 watt is incorrect. This value is much smaller than the actual power.
- Option B: 40 watt is incorrect. This value is much smaller than the actual power.
- Option D: Zero is incorrect because the sun delivers electromagnetic flux to the earth's surface, hence there is power incident on the roof.

Key Concepts:
- Intensity of electromagnetic waves
- Power calculation

Memory Aids:
- Remember that Power = Intensity x Area.

Quick Tips:
- Always ensure consistent units when calculating power.

10 Easy

10 of 35Question permalink

Application +4 | -1

If equation of wave is given by

y=0.05 \sin (2 x-4 t)

. Find velocity of wave?

View answer and explanation

Correct option: A

2

Option A (2) is correct because the velocity of the wave is the coefficient of t divided by the coefficient of x.

Solution approach

Correct Answer + Core Reason: 2. The velocity of the wave is calculated as the ratio of the coefficient of t to the coefficient of x.

Step-by-Step Solution:

Step 1: Recall the general form of a wave equation: $y = A \sin(kx - \omega t)$ , where A is the amplitude, k is the wave number, and $\omega$ is the angular frequency.

Step 2: Given the wave equation $y = 0.05 \sin(2x - 4t)$ .

Step 3: Identify the values of k and $\omega$ : $k = 2$ and $\omega = 4$ .

Step 4: Calculate the velocity of the wave (v) using the formula $v = \frac{\omega}{k}$ .

Step 5: Substitute the values of $\omega$ and k into the equation: $v = \frac{4}{2} = 2$ .

Why other options are wrong:
- Option B: 0 is incorrect because it is not the correct calculation of the velocity of the wave.
- Option C: 0 is incorrect because it is not the correct calculation of the velocity of the wave.
- Option D: 0 is incorrect because it is not the correct calculation of the velocity of the wave.

Key Concepts:
- Wave equation
- Wave number
- Angular frequency
- Velocity of wave

Memory Aids:
- Remember the relationship $v = \frac{\omega}{k}$ .

Quick Tips:
- Understand the general form of a wave equation and how to extract the values of k and $\omega$ .

11 Easy

11 of 35Question permalink

Understanding +4 | -1

A plane electromagnetic wave propagating along z-direction can be generated by

View answer and explanation

Correct option: A

E_x, B_y

Option A (

E_x, B_y

) is correct because in an electromagnetic wave propagating along the z-direction, the electric field and magnetic field are perpendicular to each other and to the direction of propagation.

Solution approach

Correct Answer + Core Reason: $E_x, B_y$ . The electric and magnetic fields must be perpendicular to each other and the direction of propagation.

Step-by-Step Solution:

Step 1: Recall that electromagnetic waves are transverse waves, meaning that the electric field (E) and magnetic field (B) are perpendicular to each other and to the direction of propagation.

Step 2: Given that the wave is propagating along the z-direction, the electric and magnetic fields must be in the x-y plane.

Step 3: Therefore, if the electric field is in the x-direction ( $E_x$ ), the magnetic field must be in the y-direction ( $B_y$ ), and vice versa.

Why other options are wrong:
- Option B: $E_x, B_x$ is incorrect because the electric and magnetic fields must be perpendicular to each other.
- Option C: $E_x, B_z$ is incorrect because the magnetic field must be perpendicular to the direction of propagation (z-direction).
- Option D: $E_y, B_z$ is incorrect because the magnetic field must be perpendicular to the direction of propagation (z-direction).

Key Concepts:
- Electromagnetic waves
- Transverse waves
- Direction of propagation

Memory Aids:
- Remember that the electric field, magnetic field, and direction of propagation are mutually perpendicular.

Quick Tips:
- Use the right-hand rule to determine the direction of the electric and magnetic fields.

12 Easy

12 of 35Question permalink

Application +4 | -1

A plane electromagnetic wave of frequency

35 {MHz}

travels in free space along the

X

-direction. At a particular point (in space and time)

\vec{E}=9.6 \hat{j} {~V} / {m}

. The value of magnetic field at this point is :

View answer and explanation

Correct option: D

3.2 \times 10^{-8} \hat{k}\, T

Option D (

3.2 \times 10^{-8} \hat{k}\, T

) is correct because the magnetic field is calculated using

B = \frac{E}{c}

and the direction is determined by the right-hand rule, resulting in the k direction.

Solution approach

Correct Answer + Core Reason: $3.2 \times 10^{-8} \hat{k}\, T$ . The magnetic field is calculated using $B = \frac{E}{c}$ and the direction is determined by the right-hand rule.

Step-by-Step Solution:

Step 1: Identify the given values: Frequency $f = 35 \text{ MHz} = 35 \times 10^6 \text{ Hz}$ and electric field $\vec{E} = 9.6 \hat{j} \text{ V/m}$ . The wave travels along the X-direction.

Step 2: Recall the relationship between the amplitudes of the electric and magnetic fields in an electromagnetic wave: $E = cB$ , where $c$ is the speed of light in vacuum ( $c = 3 \times 10^8 \text{ m/s}$ ).

Step 3: Solve for the magnitude of the magnetic field: $B = \frac{E}{c} = \frac{9.6 \text{ V/m}}{3 \times 10^8 \text{ m/s}} = 3.2 \times 10^{-8} \text{ T}$ .

Step 4: Determine the direction of the magnetic field. Since the wave travels along the X-direction and the electric field is along the Y-direction ( $\hat{j}$ ), the magnetic field must be along the Z-direction ( $\hat{k}$ ) to satisfy the condition that the direction of propagation is given by $\vec{E} \times \vec{B}$ .

Step 5: Therefore, the magnetic field is $\vec{B} = 3.2 \times 10^{-8} \hat{k} \text{ T}$ .

Why other options are wrong:
- Option A: $9.6 \hat{j} T$ is incorrect because it has the wrong magnitude and direction.
- Option B: $9.6 \times 10^{-8} \hat{k}\, T$ is incorrect because it has the wrong magnitude.
- Option C: $3.2 \times 10^{-8} \hat{i} {~T}$ is incorrect because it has the wrong direction.

Key Concepts:
- Electromagnetic waves
- Relationship between electric and magnetic field amplitudes
- Speed of light
- Direction of propagation

Memory Aids:
- Remember the relationship $E = cB$ and the right-hand rule for the direction of propagation.

Quick Tips:
- Always remember to consider the direction of the fields in addition to their magnitudes.

13 Easy

13 of 35Question permalink

Understanding +4 | -1

A plane electromagnetic wave propagating along z-direction can be generated by

View answer and explanation

Correct option: B

E_y, B_x

Option B,

E_y, B_x

, is correct because for a wave propagating along the z-direction, the electric and magnetic fields must be perpendicular to each other and to the direction of propagation.

Solution approach

Correct Answer + Core Reason: $E_y, B_x$ . For a wave propagating in the z-direction, the electric and magnetic fields must be perpendicular to each other and to the direction of propagation.

Step-by-Step Solution:

Step 1: Recall the properties of electromagnetic waves.

Step 2: Electromagnetic waves are transverse waves, meaning that the electric field ( $\overrightarrow E$ ) and the magnetic field ( $\overrightarrow B$ ) are perpendicular to each other and to the direction of propagation ( $\overrightarrow k$ ).

Step 3: If the wave is propagating along the z-direction, then $\overrightarrow k$ is along the z-axis. Therefore, $\overrightarrow E$ and $\overrightarrow B$ must be in the x-y plane. If $\overrightarrow E$ is along the y-axis ( $E_y$ ), then $\overrightarrow B$ must be along the x-axis ( $B_x$ ) to satisfy the condition that $\overrightarrow E$ , $\overrightarrow B$ , and $\overrightarrow k$ are mutually perpendicular and $\overrightarrow E \times \overrightarrow B$ is in the direction of $\overrightarrow k$ .

Why other options are wrong:
- Option A: $E_y, B_y$ is incorrect because the electric and magnetic fields are parallel to each other, which is not possible for an electromagnetic wave.
- Option C: $E_x, B_z$ is incorrect because the magnetic field is along the direction of propagation, which is not possible for an electromagnetic wave.
- Option D: $E_y, B_z$ is incorrect because the magnetic field is along the direction of propagation, which is not possible for an electromagnetic wave.

Key Concepts:
- Properties of electromagnetic waves
- Transverse waves
- Direction of propagation

Memory Aids:
- Remember that the electric and magnetic fields are perpendicular to each other and to the direction of propagation.

Quick Tips:
- Always remember the properties of electromagnetic waves.

14 Easy

14 of 35Question permalink

Remembering +4 | -1

Part of electromagnetic spectrum that corresponds to wavelength of

10^{-10}

m is

View answer and explanation

Correct option: D

X-rays

Option D is correct because X-rays correspond to a wavelength of approximately 10^-10 m in the electromagnetic spectrum.

Solution approach

Correct Answer + Core Reason: X-rays. X-rays have wavelengths around $10^{-10}$ m.

Step-by-Step Solution:

Step 1: Recall the approximate wavelengths of different parts of the electromagnetic spectrum:
- Radio waves: > 1 m
- Microwaves: $10^{-3}$ m to 1 m
- Infrared: $10^{-6}$ m to $10^{-3}$ m
- Visible light: $4 \times 10^{-7}$ m to $7 \times 10^{-7}$ m
- Ultraviolet: $10^{-8}$ m to $4 \times 10^{-7}$ m
- X-rays: $10^{-8}$ m to $10^{-12}$ m
- Gamma rays: < $10^{-12}$ m

Step 2: Compare the given wavelength ( $10^{-10}$ m) with the approximate wavelengths of the options.

Step 3: $10^{-10}$ m falls within the range of X-rays.

Why other options are wrong:
- Option A: Microwaves. Microwaves have much longer wavelengths than $10^{-10}$ m.
- Option B: Ultraviolet rays. Ultraviolet rays have longer wavelengths than $10^{-10}$ m.
- Option C: Infrared rays. Infrared rays have much longer wavelengths than $10^{-10}$ m.

Key Concepts:
- Electromagnetic Spectrum
- Wavelength

Memory Aids:
- Remember the approximate wavelengths of different parts of the electromagnetic spectrum.

Quick Tips:
- Know the typical wavelengths for different types of electromagnetic radiation.

15 Easy

15 of 35Question permalink

Application +4 | -1

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of

5 \times 10^{10} \mathrm{~Hz}

and an amplitude of

50 \mathrm{Vm}^{-1}

. The total average energy density of the electromagnetic field of the wave is : [Use

\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2

]

View answer and explanation

Correct option: C

1.106 \times 10^{-8} \mathrm{Jm}^{-3}

Option C (

1.106 \times 10^{-8} \mathrm{Jm}^{-3}

) is correct because it is the total average energy density calculated using the formula

u = \varepsilon_0 E_0^2 / 2

.

Solution approach

Correct Answer + Core Reason: $1.106 \times 10^{-8} \mathrm{Jm}^{-3}$ . The total average energy density is calculated using the electric field amplitude and permittivity of free space.

Step-by-Step Solution:

Step 1: Identify the given information:\nFrequency, $f = 5 \times 10^{10} \mathrm{Hz}$ \nAmplitude of electric field, $E_0 = 50 \mathrm{Vm}^{-1}$ \nPermittivity of free space, $\varepsilon_0 = 8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2$

Step 2: Recall the formula for the total average energy density of an electromagnetic wave:
$u = \frac{1}{2} \varepsilon_0 E_0^2$

Step 3: Substitute the given values into the formula:
$u = \frac{1}{2} \times (8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2) \times (50 \mathrm{Vm}^{-1})^2$

Step 4: Calculate the energy density:
$u = \frac{1}{2} \times 8.85 \times 10^{-12} \times 2500 \mathrm{Jm}^{-3}$
$u = 4.425 \times 10^{-12} \times 2500 \mathrm{Jm}^{-3}$
$u = 11.0625 \times 10^{-9} \mathrm{Jm}^{-3}$
$u = 1.10625 \times 10^{-8} \mathrm{Jm}^{-3}$

Step 5: Round to three significant figures:
$u \approx 1.106 \times 10^{-8} \mathrm{Jm}^{-3}$

Final Answer: $1.106 \times 10^{-8} \mathrm{Jm}^{-3}$

Why other options are wrong:
- Option A: $4.425 \times 10^{-8} \mathrm{Jm}^{-3}$ : This is incorrect because it seems to have missed the factor of 1/2 in the energy density formula.
- Option B: $2.212 \times 10^{-8} \mathrm{Jm}^{-3}$ : This is incorrect, likely from a calculation error or using the wrong formula.
- Option D: $2.212 \times 10^{-10} \mathrm{Jm}^{-3}$ : This is incorrect because it seems to have used the wrong power of 10, possibly due to a unit conversion error.

Key Concepts:
- Electromagnetic waves
- Energy density of electromagnetic waves
- Permittivity of free space

Memory Aids:
- Remember the formula for energy density: $u = \frac{1}{2} \varepsilon_0 E_0^2$

Quick Tips:
- Ensure correct units are used in the calculation.
- Double-check the formula before substituting values.

16 Easy

16 of 35Question permalink

Application +4 | -1

In an electromagnetic wave the amplitude of magnetic field is

3 \times 10^{-8} \mathrm{~T}

. If the frequency of wave is

10^{12} \mathrm{~Hz}

, then find the amplitude of associated electric field in S.I. unit.

View answer and explanation

Correct option: A

9

Option A (9) is correct because the amplitude of the electric field is calculated using the relation

E_0 = cB_0

, where c is the speed of light.

Solution approach

Correct Answer + Core Reason: 9. The amplitude of the electric field is calculated using the relation $E_0 = cB_0$ .

Step-by-Step Solution:

Step 1: Identify the given values: Amplitude of magnetic field $B_0 = 3 \times 10^{-8} \text{ T}$ , and frequency $f = 10^{12} \text{ Hz}$ .

Step 2: Recall the relationship between the amplitudes of the electric and magnetic fields in an electromagnetic wave: $E_0 = cB_0$ , where $c$ is the speed of light in vacuum ( $c = 3 \times 10^8 \text{ m/s}$ ).

Step 3: Substitute the given values into the equation: $E_0 = (3 \times 10^8 \text{ m/s}) \times (3 \times 10^{-8} \text{ T})$ .

Step 4: Calculate the amplitude of the electric field: $E_0 = 9 \text{ V/m}$ .

Why other options are wrong:
- Option B: 0 is incorrect because it does not follow from the correct calculation.
- Option C: 0 is incorrect because it does not follow from the correct calculation.
- Option D: 0 is incorrect because it does not follow from the correct calculation.

Key Concepts:
- Electromagnetic waves
- Relationship between electric and magnetic field amplitudes
- Speed of light

Memory Aids:
- Remember the relationship $E_0 = cB_0$ .

Quick Tips:
- Always remember the relationship between E and B fields in EM waves: E = cB.

17 Easy

17 of 35Question permalink

Understanding +4 | -1

Given below are two statements:
Statement I: Electromagnetic waves carry energy
as they travel through space and this energy is
equally shared by the electric and magnetic fields
Statement II: When electromagnetic waves strike
a surface, a pressure is exerted on the surface
In the light of the above statements, choose the
most appropriate answer from the options given
below

View answer and explanation

Correct option: A

Both Statement I and Statement II are correct

Option A (Both Statement I and Statement II are correct) is correct because electromagnetic waves carry energy and exert pressure when they strike a surface.

Solution approach

Correct Answer + Core Reason: Both Statement I and Statement II are correct. Electromagnetic waves carry energy and exert pressure.

Step-by-Step Solution:

Step 1: Statement I: Electromagnetic waves carry energy. This is a fundamental property of electromagnetic waves. The energy is indeed shared by the electric and magnetic fields.

Step 2: Statement II: Electromagnetic waves exert pressure. This is also true. When electromagnetic waves strike a surface, they transfer momentum, resulting in radiation pressure.

Step 3: Final Answer: Both statements are correct.

Why other options are wrong:
- Option B: This is incorrect because Statement II is also correct.
- Option C: This is incorrect because Statement I is also correct.
- Option D: This is incorrect because both statements are correct.

Key Concepts:
- Energy of electromagnetic waves
- Radiation pressure

Memory Aids:
- Electromagnetic waves are like tiny bullets carrying energy and momentum.

Quick Tips:
- Remember that electromagnetic waves can transfer both energy and momentum.

18 Easy

18 of 35Question permalink

Application +4 | -1

An electromagnetic wave has electric field given by

\vec{E}=(9.6 \hat{j}) \sin \left[2 \pi\left\{30 \times 10^6 t-\frac{1}{10} x\right\}\right], x

and

t

are in SI units. The maximum magnetic field is

View answer and explanation

Correct option: A

3.2 \times 10^{-9}

Option A (

3.2 \times 10^{-9}

) is correct because the maximum magnetic field is calculated using the relation

B_0 = \frac{E_0}{c}

, where c is the speed of light.

Solution approach

Correct Answer + Core Reason: $3.2 \times 10^{-9}$ . The maximum magnetic field is calculated using the relation $B_0 = \frac{E_0}{c}$ .

Step-by-Step Solution:

Step 1: Identify the given value: Amplitude of the electric field $E_0 = 9.6 \text{ V/m}$ .

Step 2: Recall the relationship between the amplitudes of the electric and magnetic fields in an electromagnetic wave: $E_0 = cB_0$ , where $c$ is the speed of light in vacuum ( $c = 3 \times 10^8 \text{ m/s}$ ).

Step 3: Solve for the amplitude of the magnetic field: $B_0 = \frac{E_0}{c} = \frac{9.6 \text{ V/m}}{3 \times 10^8 \text{ m/s}}$ .

Step 4: Calculate the amplitude of the magnetic field: $B_0 = 3.2 \times 10^{-8} \text{ T}$ .

Why other options are wrong:
- Option B: $9.6 \times 10^{-8}$ is incorrect because it does not follow from the correct calculation.
- Option C: $1.7 \times 10^{-9}$ is incorrect because it does not follow from the correct calculation.
- Option D: $10^{-7}$ is incorrect because it does not follow from the correct calculation.

Key Concepts:
- Electromagnetic waves
- Relationship between electric and magnetic field amplitudes
- Speed of light

Memory Aids:
- Remember the relationship $E_0 = cB_0$ .

Quick Tips:
- Always remember the relationship between E and B fields in EM waves: E = cB.

19 Easy

19 of 35Question permalink

Application +4 | -1

In a place e.m. wave, the electric field oscillates sinusoidally at a frequency of

2.5 \times 10^{10} {~Hz}

and amplitude

480 {~V} / {m}

. The amplitude of oscillating magnetic field will be :-

View answer and explanation

Correct option: C

1.6 \times 10^{-6} {~Wb} / {m}^2

Option C (

1.6 \times 10^{-6} {~Wb} / {m}^2

) is correct because the amplitude of the oscillating magnetic field is calculated using the formula

B_0 = \frac{E_0}{c}

, where c is the speed of light.

Solution approach

Correct Answer + Core Reason: $1.6 \times 10^{-6} {~Wb} / {m}^2$ . The amplitude of the oscillating magnetic field is calculated using the formula $B_0 = \frac{E_0}{c}$ .

Step-by-Step Solution:

Step 1: Identify the given values: Frequency $f = 2.5 \times 10^{10} \text{ Hz}$ and amplitude of the electric field $E_0 = 480 \text{ V/m}$ .

Step 2: Recall the relationship between the amplitudes of the electric and magnetic fields in an electromagnetic wave: $E_0 = cB_0$ , where $c$ is the speed of light in vacuum ( $c = 3 \times 10^8 \text{ m/s}$ ).

Step 3: Solve for the amplitude of the magnetic field: $B_0 = \frac{E_0}{c} = \frac{480 \text{ V/m}}{3 \times 10^8 \text{ m/s}}$ .

Step 4: Calculate the amplitude of the magnetic field: $B_0 = 1.6 \times 10^{-6} \text{ T} = 1.6 \times 10^{-6} \text{ Wb/m}^2$ .

Why other options are wrong:
- Option A: $1.52 \times 10^{-8} {~Wb} / {m}^2$ is incorrect because it does not follow from the correct calculation.
- Option B: $1.52 \times 10^{-7} {~Wb} / {m}^2$ is incorrect because it does not follow from the correct calculation.
- Option D: $1.6 \times 10^{-7} {~Wb} / {m}^2$ is incorrect because it does not follow from the correct calculation.

Key Concepts:
- Electromagnetic waves
- Relationship between electric and magnetic field amplitudes
- Speed of light

Memory Aids:
- Remember the relationship $E_0 = cB_0$ .

Quick Tips:
- Always remember the relationship between E and B fields in EM waves: E = cB.

20 Easy

20 of 35Question permalink

Remembering +4 | -1

Rays in electromagnetic spectrum that has shortest wavelength are

View answer and explanation

Correct option: D

X-rays

Option D is correct because X-rays have the shortest wavelength among the given options in the electromagnetic spectrum.

Solution approach

Correct Answer + Core Reason: X-rays. X-rays have the shortest wavelength among the given options.

Step-by-Step Solution:

Step 1: Recall the order of the electromagnetic spectrum from longest to shortest wavelength: Radio waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, Gamma rays.

Step 2: Compare the wavelengths of the given options:
- Microwaves have longer wavelengths than infrared, visible light, ultraviolet, X-rays, and gamma rays.
- Ultraviolet rays have shorter wavelengths than microwaves, infrared, and visible light, but longer than X-rays and gamma rays.
- Infrared rays have longer wavelengths than visible light, ultraviolet, X-rays, and gamma rays.
- X-rays have shorter wavelengths than microwaves, infrared, visible light, and ultraviolet rays, but longer than gamma rays.

Step 3: From the given options, X-rays have the shortest wavelength.

Why other options are wrong:
- Option A: Microwaves. Microwaves have longer wavelengths than X-rays.
- Option B: Ultraviolet rays. Ultraviolet rays have longer wavelengths than X-rays.
- Option C: Infrared rays. Infrared rays have longer wavelengths than X-rays.

Key Concepts:
- Electromagnetic Spectrum
- Wavelength

Memory Aids:
- Remember the order of the electromagnetic spectrum.

Quick Tips:
- Know the relative positions of different types of electromagnetic radiation in the spectrum.

21 Easy

21 of 35Question permalink

Remembering +4 | -1

Displacement Current is:

View answer and explanation

Correct option: D

Due to time Varying Electric field

Option D is correct because displacement current arises from a time-varying electric field, as described by Maxwell's equations.

Solution approach

Correct Answer + Core Reason: Due to time Varying Electric field. Displacement current is caused by a changing electric field.

Step-by-Step Solution:

Step 1: Recall Maxwell's modification to Ampere's Law: Maxwell added the displacement current term to Ampere's Law to account for changing electric fields. The displacement current is given by $I_D = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 \frac{d(EA)}{dt} = \epsilon_0 A \frac{dE}{dt}$ , where $\Phi_E$ is the electric flux.

Step 2: Analyze the options: The displacement current is directly proportional to the rate of change of the electric field.

Why other options are wrong:
- Option A: Flow of free electrons constitutes conduction current, not displacement current.
- Option B: Flow of positive ions constitutes conduction current, not displacement current.
- Option C: Flow of both positive and negative ions constitutes conduction current, not displacement current.

Key Concepts:
- Displacement current
- Maxwell's equations
- Time-varying electric fields

Memory Aids:
- Remember that displacement current is related to the *change* in electric field.

Quick Tips:
- Displacement current exists even in the absence of free charges.

22 Easy

22 of 35Question permalink

Remembering +4 | -1

The S.I. unit of displacement current is:

View answer and explanation

Correct option: C

Ampere

Option C (Ampere) is correct because displacement current, like regular current, is measured in Amperes.

Solution approach

Correct Answer + Core Reason: Ampere. Displacement current is a current, so it has the same unit as current.

Step-by-Step Solution:

Step 1: Recall the definition of displacement current: $I_D = \epsilon_0 \frac{d\Phi_E}{dt}$ .

Step 2: Analyze the units: The units of displacement current are the same as the units of regular current, which is Ampere.

Why other options are wrong:
- Option A: Henry is the unit of inductance.
- Option B: Coulomb is the unit of electric charge.
- Option D: Faraday is a (non-SI) unit of electric charge, and also a unit of capacitance.

Key Concepts:
- Displacement current
- Units of current

Memory Aids:
- Displacement current is a *current*, so its unit must be Ampere.

Quick Tips:
- Always remember the units of physical quantities.

23 Easy

23 of 35Question permalink

Understanding +4 | -1

Let the velocities of light in vacuum and glass are represented by

v

and

v_{\mathrm{g}}

. When electrons move through glass with a velocity

v_e

, there emitted a visible radiation in glass if

View answer and explanation

Correct option: A

v_e > v_g

Option A (

v_e > v_g

) is correct because Cherenkov radiation is emitted when a charged particle moves through a dielectric medium faster than the speed of light in that medium.

Solution approach

Correct Answer + Core Reason: $v_e > v_g$ . Cherenkov radiation is emitted when a charged particle moves faster than the speed of light in that medium.

Step-by-Step Solution:

Step 1: Understand Cherenkov Radiation: This phenomenon occurs when a charged particle travels through a dielectric medium at a speed greater than the phase velocity of light in that medium.

Step 2: Define variables: $v_e$ is the velocity of the electron in the glass, and $v_g$ is the velocity of light in the glass.

Step 3: Condition for Cherenkov Radiation: For visible radiation to be emitted, the electron's velocity must be greater than the speed of light in the glass: $v_e > v_g$ .

Step 4: Final Answer: $v_e > v_g$

Why other options are wrong:
- Option B: This is incorrect because if $v_e < v_g$ , the electron is moving slower than light in the glass, and Cherenkov radiation will not occur.
- Option C: This is incorrect because the speed of light in vacuum ( $v$ ) is always greater than the speed of light in a medium ( $v_g$ ). The condition depends on the speed of light in the glass, not in vacuum.
- Option D: This is incorrect because it introduces an unnecessary ratio of speeds and doesn't represent the condition for Cherenkov radiation.

Key Concepts:
- Cherenkov Radiation
- Speed of light in a medium

Memory Aids:
- Remember that Cherenkov radiation is the optical equivalent of a sonic boom.

Quick Tips:
- Cherenkov radiation occurs when a particle's speed exceeds the phase velocity of light in the medium.

24 Easy

24 of 35Question permalink

Analyzing +4 | -1

A proton moving with a constant velocity passes through a region of space without any change in its velocity. If

\vec{E}

and

\vec{B}

represent the electric and magnetic fields respectively, then the region of space may have:

(A)

E=0, B=0

(B)

E=0, B \neq 0

(C)

E \neq 0, B=0

(D)

E \neq 0, B \neq 0

Choose the most appropriate answer from the options given below:

View answer and explanation

Correct option: C

(A), (B) and (D) only

Option C ((A), (B) and (D) only) is correct because a proton can move with constant velocity if there are no fields, only a magnetic field (velocity parallel to B), or both electric and magnetic fields such that their forces cancel out.

Solution approach

Correct Answer + Core Reason: (A), (B) and (D) only. A proton can move with constant velocity if the net force on it is zero.

Step-by-Step Solution:

Step 1: Analyze each case:
- (A) $E=0, B=0$ : If both electric and magnetic fields are zero, the net force on the proton is zero, and it will move with constant velocity. This is possible.
- (B) $E=0, B \neq 0$ : If the electric field is zero and there is only a magnetic field, the force on the proton is given by the Lorentz force $\vec{F} = q(\vec{v} \times \vec{B})$ . If the velocity $\vec{v}$ is parallel to the magnetic field $\vec{B}$ , then $\vec{v} \times \vec{B} = 0$ , and the force is zero. This is possible.
- (C) $E \neq 0, B=0$ : If there is an electric field but no magnetic field, the force on the proton is given by $\vec{F} = q\vec{E}$ . Since $E \neq 0$ , the force is non-zero, and the proton will accelerate. This is not possible.
- (D) $E \neq 0, B \neq 0$ : If both electric and magnetic fields are present, the net force on the proton is given by $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$ . It is possible for the electric and magnetic forces to cancel each other out, resulting in zero net force and constant velocity. This is possible.

Step 2: Based on the analysis, cases (A), (B), and (D) are possible.

Final Answer: (A), (B) and (D) only

Why other options are wrong:
- Option A ((A), (C) and (D) only): Option (C) is incorrect because a non-zero electric field without a magnetic field will cause the proton to accelerate.
- Option B ((A), (B) and (C) only): Option (C) is incorrect because a non-zero electric field without a magnetic field will cause the proton to accelerate.
- Option D ((B), (C) and (D) only): Option (C) is incorrect because a non-zero electric field without a magnetic field will cause the proton to accelerate, and option (A) is possible.

Key Concepts:
- Lorentz force
- Electric force
- Magnetic force
- Constant velocity

Memory Aids:
- $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$ (Lorentz force equation)
- If $\vec{v}$ is parallel to $\vec{B}$ , then $\vec{v} \times \vec{B} = 0$ .

Quick Tips:
- Remember that the net force on the particle must be zero for it to move with constant velocity.
- Consider the direction of the velocity relative to the magnetic field.

25 Easy

25 of 35Question permalink

Understanding +4 | -1

If

\overrightarrow E \,\,\mathrm{and}\,\,\overrightarrow B

are the electric and magnetic field vectors of E.M waves then the direction of propagation of E.M wave is along the direction of

View answer and explanation

Correct option: C

\overrightarrow E \times \overrightarrow B

Option C,

\overrightarrow E \times \overrightarrow B

, is correct because the direction of propagation of an electromagnetic wave is given by the Poynting vector, which is proportional to the cross product of the electric and magnetic field vectors.

Solution approach

Correct Answer + Core Reason: $\overrightarrow E \times \overrightarrow B$ . The direction of propagation is given by the Poynting vector, which is proportional to the cross product of the electric and magnetic fields.

Step-by-Step Solution:

Step 1: Recall the definition of the Poynting vector.

Step 2: The Poynting vector ( $\overrightarrow S$ ) is given by: $\overrightarrow S = \frac{1}{\mu_0} (\overrightarrow E \times \overrightarrow B)$

Step 3: The Poynting vector represents the direction of energy flow, which is also the direction of propagation of the electromagnetic wave. Therefore, the direction of propagation is along the direction of $\overrightarrow E \times \overrightarrow B$ .

Why other options are wrong:
- Option A: $\overrightarrow E$ is incorrect because the direction of propagation is not solely determined by the electric field vector.
- Option B: $\overrightarrow B$ is incorrect because the direction of propagation is not solely determined by the magnetic field vector.
- Option D: None of these is incorrect because the direction of propagation is given by $\overrightarrow E \times \overrightarrow B$ .

Key Concepts:
- Poynting vector
- Direction of propagation of electromagnetic waves

Memory Aids:
- Remember that the direction of propagation is given by the cross product of the electric and magnetic fields.

Quick Tips:
- Always remember the definition of the Poynting vector.

26 Easy

26 of 35Question permalink

Application +4 | -1

A plane electromagnetic wave travels in a medium of relative permittivity 2 and relative permeability

8.0.

If the magnitude of magnetic intensity is

\frac{6}{\pi} \times 10^{-2} \mathrm{~A} / \mathrm{m}

at a point, then the electric field at that point in ___

\mathrm{V} / \mathrm{m}

.

View answer and explanation

Correct option: A

1.80

Option A (1.80) is correct because the electric field is calculated using the formula

E = vB

, where

v = \frac{c}{\sqrt{\mu_r \epsilon_r}}

is the speed of light in the medium.

Solution approach

Correct Answer + Core Reason: $1.80$ . The electric field is calculated using the formula $E = vB$ , where $v$ is the speed of light in the medium.

Step-by-Step Solution:

Step 1: Identify the given values: Relative permittivity $\epsilon_r = 2$ , relative permeability $\mu_r = 8.0$ , and magnetic intensity $H = \frac{6}{\pi} \times 10^{-2} \text{ A/m}$ .

Step 2: Calculate the magnetic field $B$ using the relation $B = \mu H = \mu_0 \mu_r H$ , where $\mu_0 = 4\pi \times 10^{-7} \text{ H/m}$ .
$B = (4\pi \times 10^{-7} \text{ H/m}) \times 8.0 \times (\frac{6}{\pi} \times 10^{-2} \text{ A/m}) = 192 \times 10^{-9} \text{ T}$

Step 3: Calculate the speed of light in the medium: $v = \frac{c}{\sqrt{\epsilon_r \mu_r}} = \frac{3 \times 10^8 \text{ m/s}}{\sqrt{2 \times 8}} = \frac{3 \times 10^8}{4} = 0.75 \times 10^8 \text{ m/s}$ .

Step 4: Calculate the electric field $E$ using the relation $E = vB = (0.75 \times 10^8 \text{ m/s}) \times (192 \times 10^{-9} \text{ T}) = 1.8 \text{ V/m}$ .

Why other options are wrong:
- Option B: $1.79$ is close but incorrect due to rounding errors.
- Option C: $1.5$ is incorrect because it doesn't follow from the correct calculation.
- Option D: 1 is incorrect because it doesn't follow from the correct calculation.

Key Concepts:
- Electromagnetic waves
- Relative permittivity and permeability
- Speed of light in a medium
- Relationship between electric and magnetic fields

Memory Aids:
- Remember the relation $E = vB$ and $v = \frac{c}{\sqrt{\epsilon_r \mu_r}}$ .

Quick Tips:
- Always remember to account for the relative permittivity and permeability when calculating the speed of light in a medium.

27 Easy

27 of 35Question permalink

Application +4 | -1

The wave function [in S.I unit] for an electromagnetic wave is given as

\psi \left( {x,t} \right) = {10^3}\sin \pi \left[ {3 \times {{10}^6}x - 9 \times {{10}^{14}}t} \right]

The speed of the wave is [in m/s]

View answer and explanation

Correct option: B

3 \times {10^8}

Option B (

3 \times {10^8}

) is correct because the speed of the wave can be found by dividing the coefficient of t by the coefficient of x.

Solution approach

Correct Answer + Core Reason: $3 \times {10^8}$ . The speed of the wave is found by dividing the coefficient of t by the coefficient of x.

Step-by-Step Solution:

Step 1: Identify the general form of a wave function: $\psi(x, t) = A \sin(kx - \omega t)$ , where A is the amplitude, k is the wave number, and $\omega$ is the angular frequency.

Step 2: Given the wave function $\psi(x, t) = {10^3}\sin \pi [3 \times {10^6}x - 9 \times {10^{14}}t]$ .

Step 3: Identify the values of k and $\omega$ : $k = 3 \times {10^6}\pi$ and $\omega = 9 \times {10^{14}}\pi$ .

Step 4: Calculate the speed of the wave (v) using the formula $v = \frac{\omega}{k}$ .

Step 5: Substitute the values of $\omega$ and k into the equation: $v = \frac{9 \times {10^{14}}\pi}{3 \times {10^6}\pi} = 3 \times {10^8} \text{{ m/s}}$ .

Why other options are wrong:
- Option A: $9 \times {10^{14}}$ is incorrect because it is the angular frequency, not the speed of the wave.
- Option C: $3 \times {10^6}$ is incorrect because it is related to the wave number, not the speed of the wave.
- Option D: $3 \times {10^7}$ is incorrect because it is not the correct calculation of the speed of the wave.

Key Concepts:
- Wave function
- Wave number
- Angular frequency
- Speed of wave

Memory Aids:
- Remember the relationship $v = \frac{\omega}{k}$ .

Quick Tips:
- Understand the general form of a wave function and how to extract the values of k and $\omega$ .

28 Easy

28 of 35Question permalink

Remembering +4 | -1

Rays in electromagnetic spectrum that has lowest frequency are

View answer and explanation

Correct option: A

Microwaves

Option A is correct because Microwaves have the lowest frequency among the given options in the electromagnetic spectrum.

Solution approach

Correct Answer + Core Reason: Microwaves. Microwaves have the lowest frequency among the given options.

Step-by-Step Solution:

Step 1: Recall the order of the electromagnetic spectrum from lowest to highest frequency: Radio waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, Gamma rays.

Step 2: Compare the frequencies of the given options:
- Microwaves have lower frequencies than infrared, visible light, ultraviolet, and X-rays.
- Ultraviolet rays have higher frequencies than microwaves, infrared, and visible light.
- Infrared rays have higher frequencies than microwaves.
- X-rays have higher frequencies than microwaves, infrared, and ultraviolet rays.

Step 3: From the given options, microwaves have the lowest frequency.

Why other options are wrong:
- Option B: Ultraviolet rays. Ultraviolet rays have higher frequencies than microwaves.
- Option C: Infrared rays. Infrared rays have higher frequencies than microwaves.
- Option D: X-rays. X-rays have higher frequencies than microwaves.

Key Concepts:
- Electromagnetic Spectrum
- Frequency

Memory Aids:
- Remember the order of the electromagnetic spectrum.

Quick Tips:
- Know the relative positions of different types of electromagnetic radiation in the spectrum.

29 Easy

29 of 35Question permalink

Understanding +4 | -1

In the case of a photon :-

View answer and explanation

Correct option: D

Its speed can change but cannot become equal to zero.

Option D (Its speed can change but cannot become equal to zero) is correct because a photon always travels at the speed of light in a given medium, and it cannot be stopped.

Solution approach

Correct Answer + Core Reason: Its speed can change but cannot become equal to zero. A photon always travels at the speed of light in a given medium.

Step-by-Step Solution:

Step 1: Recall properties of a photon: A photon is a massless particle that always travels at the speed of light in vacuum ( $c$ ).

Step 2: Photon speed in a medium: When a photon enters a medium, its speed changes to $v = c/n$ , where $n$ is the refractive index of the medium. However, it never becomes zero.

Step 3: Final Answer: Its speed can change but cannot become equal to zero.

Why other options are wrong:
- Option A: This is incorrect because the energy of a photon is proportional to its frequency ( $E = hf$ ) and not the square of its velocity (which is constant in a given medium).
- Option B: This is incorrect because a photon's energy remains constant unless it interacts with matter. It doesn't lose energy simply by entering a denser medium.
- Option C: This is incorrect because photons can interact, although photon-photon scattering is a rare event.

Key Concepts:
- Properties of a photon
- Speed of light in a medium
- Energy of a photon

Memory Aids:
- Photons are always moving!

Quick Tips:
- Remember that photons are massless and always travel at the speed of light (in a vacuum).

30 Easy

30 of 35Question permalink

Application +4 | -1

The velocity of electromagnetic waves in a dielectric medium of constant 4 is

View answer and explanation

Correct option: A

1.5 \times {10^8}\mathrm{m}{\mathrm{s}^{ - 1}}

Option A (

1.5 \times {10^8}\mathrm{\;m}{\mathrm{s}^{ - 1}}

) is correct because the velocity of electromagnetic waves in a dielectric medium is given by

v = c / \sqrt{\epsilon_r}

, where

\epsilon_r

is the dielectric constant.

Solution approach

Correct Answer + Core Reason: $1.5 \times {10^8}\mathrm{\;m}{\mathrm{s}^{ - 1}}$ . The velocity of EM waves in a dielectric is reduced by a factor of $\sqrt{\epsilon_r}$ .

Step-by-Step Solution:

Step 1: Formula: The velocity of electromagnetic waves in a dielectric medium is given by $v = \frac{c}{\sqrt{\epsilon_r}}$ , where $c$ is the speed of light in vacuum and $\epsilon_r$ is the dielectric constant.

Step 2: Substitution: Given $\epsilon_r = 4$ and $c = 3 \times 10^8 \text{ m/s}$ , we have $v = \frac{3 \times 10^8}{\sqrt{4}}$ .

Step 3: Calculation: $v = \frac{3 \times 10^8}{2} = 1.5 \times 10^8 \text{ m/s}$ .

Step 4: Final Answer: $1.5 \times {10^8}\mathrm{\;m}{\mathrm{s}^{ - 1}}$

Why other options are wrong:
- Option B: This is the speed of light in vacuum, not in a dielectric medium with a dielectric constant of 4.
- Option C: This is incorrect. The velocity is reduced, not doubled.
- Option D: This is incorrect. The velocity is reduced, not quadrupled.

Key Concepts:
- Velocity of electromagnetic waves in a dielectric medium
- Dielectric constant

Memory Aids:
- Dielectric slows down EM waves.

Quick Tips:
- Remember the formula $v = \frac{c}{\sqrt{\epsilon_r}}$ .

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