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NEET • PHYSICS • ELECTROMAGNETIC WAVES

Electromagnetic waves practice · Easy

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31 Easy

31 of 35Question permalink

Application +4 | -1

Wave length of light of frequency 100 Hz is

View answer and explanation

Correct option: B

3 \times {10^6}\,m

Option B (

3 \times {10^6}\,m

) is correct because the wavelength is calculated by dividing the speed of light by the frequency.

Solution approach

Correct Answer + Core Reason: $3 \times {10^6}\,m$ . The wavelength is calculated using the formula $\lambda = c/f$ .

Step-by-Step Solution:

Step 1: Identify the given information: Frequency $f = 100 \text{{ Hz}}$ and the speed of light $c = 3 \times 10^8 \text{{ m/s}}$ .

Step 2: Recall the formula relating wavelength, frequency, and speed of light: $\lambda = \frac{c}{f}$

Step 3: Substitute the given values into the formula: $\lambda = \frac{{3 \times 10^8 \text{{ m/s}}}}{{100 \text{{ Hz}}}}$

Step 4: Calculate the wavelength: $\lambda = 3 \times 10^6 \text{{ m}}$

Final Answer: $3 \times {10^6}\,m$

Why other options are wrong:
- Option A: $4 \times {10^6}\,m$ is incorrect. It might be the result of using a wrong value for the speed of light or incorrect calculation.
- Option C: $2 \times {10^6}\,m$ is incorrect. It might be the result of using a wrong value for the speed of light or incorrect calculation.
- Option D: $5 \times {10^{ - 5}}\,m$ is incorrect. This is a completely wrong order of magnitude and likely arises from dividing frequency by the speed of light instead of the other way around.

Key Concepts:
- Wave equation: $v = f \lambda$
- Speed of light
- Wavelength and frequency relationship

Memory Aids:
- Remember the formula: $\lambda = \frac{c}{f}$

Quick Tips:
- Always check the units to ensure they are consistent before performing calculations.

32 Easy

32 of 35Question permalink

Understanding +4 | -1

Which of the following are not electromagnetic waves

View answer and explanation

Correct option: C

\beta - rays

Option C is correct because beta rays are streams of particles (electrons or positrons), not electromagnetic waves.

Solution approach

Correct Answer + Core Reason: $\beta - rays$ . Beta rays are particles, not electromagnetic waves.

Step-by-Step Solution:

Step 1: Recall the definition of electromagnetic waves.

Step 2: Electromagnetic waves are disturbances in electric and magnetic fields that propagate through space.

Step 3: Cosmic rays, gamma rays, and X-rays are all forms of electromagnetic radiation.

Step 4: Beta rays are streams of electrons or positrons, which are particles, not electromagnetic waves.

Why other options are wrong:
- Option A: Cosmic rays. Cosmic rays are high-energy particles, but the term also refers to electromagnetic radiation from space.
- Option B: Gamma rays. Gamma rays are electromagnetic waves.
- Option D: $X - rays$ . X-rays are electromagnetic waves.

Key Concepts:
- Electromagnetic Waves
- Beta Rays
- Cosmic Rays
- Gamma Rays
- X-rays

Memory Aids:
- Remember that beta rays are particles, not waves.

Quick Tips:
- Know the difference between particles and electromagnetic waves.

33 Easy

33 of 35Question permalink

Application +4 | -1

The frequency of a plane electromagnetic wave traveling in vacuum in X-direction is
30 MHz, its wavelength is

View answer and explanation

Correct option: C

10 m

Option C (10 m) is correct because wavelength is the speed of light divided by the frequency.

Solution approach

Correct Answer + Core Reason: 10 m. Wavelength is the speed of light divided by the frequency.

Step-by-Step Solution:

Step 1: Recall the relationship between wavelength ( $\lambda$ ), frequency (f), and the speed of light (c): $\lambda = \frac{c}{f}$ .

Step 2: Given the frequency $f = 30 \text{{ MHz}} = 30 \times 10^6 \text{{ Hz}}$ , and the speed of light $c = 3 \times 10^8 \text{{ m/s}}$ .

Step 3: Substitute the values into the equation: $\lambda = \frac{3 \times 10^8}{30 \times 10^6} = 10 \text{{ m}}$ .

Why other options are wrong:
- Option A: 30 m is incorrect because it is not the correct calculation of wavelength.
- Option B: 15 m is incorrect because it is not the correct calculation of wavelength.
- Option D: 20 m is incorrect because it is not the correct calculation of wavelength.

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

Memory Aids:
- Remember the relationship $\lambda = \frac{c}{f}$ .

Quick Tips:
- Always use consistent units when calculating wavelength.

34 Easy

34 of 35Question permalink

Remembering +4 | -1

The range of frequency allotted for FM radio is

View answer and explanation

Correct option: B

88

108 \mathrm{~MHz}

Option B is correct because the frequency range allotted for FM radio broadcasting is 88 to 108 MHz.

Solution approach

Correct Answer + Core Reason: $88$ to $108 \mathrm{~MHz}$ . This is the standard frequency range for FM radio broadcasting.

Step-by-Step Solution:

Step 1: Recall the frequency ranges for different types of radio communication.

Step 2: FM radio typically operates in the Very High Frequency (VHF) band.

Step 3: The standard frequency range for FM radio is 88 MHz to 108 MHz.

Why other options are wrong:
- Option A: $88$ to $108 \mathrm{~kHz}$ . This is the range for AM radio, not FM.
- Option C: $88$ to $108 \mathrm{~GHz}$ . This is a much higher frequency range, typically used for microwave communication.
- Option D: $88$ to $109 \mathrm{~kHz}$ . This is incorrect because the FM range is in MHz, and the upper limit is 108 MHz.

Key Concepts:
- Electromagnetic Spectrum
- Frequency Bands
- FM Radio

Memory Aids:
- Remember that FM radio operates in the MHz range.

Quick Tips:
- Know the typical frequency ranges for different types of radio communication.

35 Easy

35 of 35Question permalink

Understanding +4 | -1

The property which is not of an electromagnetic wave travelling in free space is that:

View answer and explanation

Correct option: C

They originate from charges moving with uniform speed

The correct answer is that electromagnetic waves originate from accelerating charges, not charges moving with uniform speed. Option C is the correct answer.

Solution approach

Correct Answer + Core Reason: They originate from charges moving with uniform speed. Electromagnetic waves are produced by accelerating charges, not charges moving at a constant speed.

Step-by-Step Solution:

Step 1: Recall the properties of electromagnetic waves traveling in free space.

Step 2: Electromagnetic waves are transverse waves, meaning that the electric and magnetic fields are perpendicular to each other and to the direction of propagation.

Step 3: Electromagnetic waves travel with a speed equal to $\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ in free space.

Step 4: The energy density in the electric field is equal to the energy density in the magnetic field.

Step 5: Electromagnetic waves originate from accelerating charges, not charges moving with uniform speed.

Why other options are wrong:
- Option A: The energy density in the electric field is equal to the energy density in the magnetic field. This is a correct property of electromagnetic waves.
- Option B: They travel with a speed equal to $\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ . This is a correct property of electromagnetic waves.
- Option D: They are transverse in nature. This is a correct property of electromagnetic waves.

Key Concepts:
- Electromagnetic waves
- Transverse waves
- Speed of light in free space
- Energy density
- Accelerating charges

Memory Aids:
- Remember that electromagnetic waves are produced by accelerating charges.

Quick Tips:
- Understand the fundamental properties of electromagnetic waves.

Revision notes and key takeaways

Review ELECTROMAGNETIC WAVES in your textbook or class notes, then use this set to check your understanding.

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