Plus Two Physics Chapter 8 Electromagnetic Waves Chapter Wise Question and Answers PDF Download: Students of Standard 12 can now download Plus Two Physics Chapter 8 Electromagnetic Waves chapter wise question and answers pdf from the links provided below in this article. Plus Two Physics Chapter 8 Electromagnetic Waves Question and Answer pdf will help the students prepare thoroughly for the upcoming Plus Two Physics Chapter 8 Electromagnetic Waves exams.
Plus Two Physics Chapter 8 Electromagnetic Waves Chapter Wise Question and Answers
Plus Two Physics Chapter 8 Electromagnetic Waves question and answers consists of questions asked in the previous exams along with the solutions for each question. To help them get a grasp of chapters, frequent practice is vital. Practising these questions and answers regularly will help the reading and writing skills of students. Moreover, they will get an idea on how to answer the questions during examinations. So, let them solve Plus Two Physics Chapter 8 Electromagnetic Waves chapter wise questions and answers to help them secure good marks in class tests and exams.
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Board |
Kerala Board |
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Study Materials |
Chapter wise Question and Answers |
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For Year |
2021 |
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Class |
12 |
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Subject |
Physics |
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Chapters |
Physics Chapter 8 Electromagnetic Waves |
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Format |
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Provider |
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Plus Two Physics Chapter 8 Electromagnetic Waves Question and Answers PDF Download
We have provided below the question and answers of Plus Two Physics Chapter 8 Electromagnetic Waves Chapter wise study material which can be downloaded by you for free. These Plus Two Physics Chapter 8 Electromagnetic Waves Chapter Wise Question and answers will contain important questions and answers and have been designed based on the latest Plus Two Physics Chapter 8 Electromagnetic Waves, books and syllabus. You can click on the links below to download the Plus Two Physics Chapter 8 Electromagnetic Waves Chapter Wise Question and Answers PDF.
Question 1.
What physical quantity is the same for X-rays of wavelength 10-10, red light of wavelength 6800 Å and radio waves of wavelength 500m?
Answer:
The speed in vacuum is the same for all.
c = 3 × 108ms-1
Question 2.
A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
Answer:
\(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{B}}\) are in x-y plane and are mutually perpendicular. Given v = 30 MHz = 30 × 106 Hz, c = 3 × 108ms-1.
∴ \(\frac{c}{v}=\frac{3 \times 10^{8}}{30 \times 10^{6}}\) = 10m.
Question 3.
A radio can tune into any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Answer:
For v1 = 7.5 MHz = 7.5 × 106 Hz
For v2 = 12 MHz = 12 × 106 Hz
So the wavelength band: 40m – 25m.
Question 4.
Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad s-1t] \(\hat{i}\)
- What is the direction of propagation?
- What is the wavelength X?
- What is the frequency v?
- What is the amplitude of the magnetic field part of the wave?
- Write an expression for the magnetic field part of the wave.
Answer:
- \(-\hat{j}\)
- λ = \(\frac{2 \pi}{k}=\frac{2 \times 3.14}{1.8}\) = 3.48 = 3.5m
- v = C/λ = 3 × 108/ 3.5 = 85.7 × 106 = 86 MHz.
- B0 = \(\frac{E_{0}}{c}=\frac{31}{3 \times 10^{8}}\) = 10 × 10-8T = 100nT.
- 100nT) cos (1.8 rad/m) y + (5.4 × 106 rad/s)t \(\hat{k}\)
Plus Two Physics Electro Magnetic Waves One Mark Questions and Answers
Question 1.
The structure of solids is investigated by using
(a) cosmic rays
(b) X-rays
(c) y-rays
(d) infra-red radiations
Answer:
(b) X-rays
Question 2.
The wavelength of light of frequency 100Hz.
(a) 4 × 106m
(b) 3 × 106m
(c) 2 × 1o6m
(d) 5 × 10-5m
Answer:
(b) 3 × 106m
λ = \(\frac{3 \times 10^{8}}{100}\) = 3 × 106m.
Question 3.
What is the biological importance of ozone layer
(a) It stops ultraviolet rays
(b) ozone layer reduces green house effect
(c) ozone layer reflects radio waves
(d) ozone layer controls O2/H2 ratio in atmosphere
Answer:
(a) It stops ultraviolet rays
The ozone layer absorbs the harmful ultraviolet rays coming from sun.
Question 4.
Pick the odd one out from the following x-rays, visible light, matter waves, radio waves.
Answer:
Matter waves, (matter-wave is not a em wave).
Question 5.
What physical quantity is the same for X-rays of wavelength 10-10, red light of wavelength 6800 Å and radio waves of wavelength 500m?
Answer:
The speed in vacuum is the same for all.
c = 3 × 108ms-1.
Plus Two Physics Electro Magnetic Waves Two Mark Questions and Answers
Question 1.
State whether True or False.
- Electromagnetic waves propagate in the direction of electric field.
- For an electromagnetic wave the ratio of E to B is equal to speed of light.
- In an electromagnetic wave, electric field leads by π/2.
- Electromagnetic waves can be produced by accelerating electric charge.
Answer:
- False
- True
- False
- True
Plus Two Physics Electro Magnetic Waves Three Mark Questions and Answers
Question 1.
Match the following:
Answer:
- Radiowave – accelerated motion of charges in wires – cellular phones.
- Infrared waves – Hot bodies and molecule – Green house effect.
- X-rays – High energy electrons – diagnostic purpose
- Gama-rays – Radio active nuclei – destroy cancer cells.
Plus Two Physics Electro Magnetic Waves Four Mark Questions and Answers
Question 1.
In an electro magnetic wave electric and magnetic field vectors are given by E = 120 sin (ωt + kz) i, B = 40 × 10-8sin (ωt +kz)j
- What is the direction of propagation of electromagnetic wave?
- Determine the ratio of amplitude of electric field to magnetic field in the case of the above em wave?
- How can you relate the above ratio with m0 and e0.
Answer:
- A direction or perpendicular to the direction of variation of electric field and magnetic field, ie.z-direction.
- \(\frac{E_{0}}{B_{0}}=\frac{120}{40 \times 10^{-8}}\) = 3 × 108m/s.
- \(\frac{E_{0}}{B_{0}}=\sqrt{\frac{1}{\mu_{0} \varepsilon_{0}}}\) = C.
Question 2.
Match the following.
Answer:
- γ-ray – Radioactivity – Nucleus
- X-ray – Diagnosis- Photon emission by fast-moving e’s
- UV-ray- Sunburn – Electronicde excitation
- Microwave – Remote sensing – Oscillating current.
Question 3.
The magnetic field of a plane electromagnetic wave is given by B = 2 × 10-7 sin (0.5 × 103x + 1.5 × 1011t)
- What is the maximum amplitude of this magnetic field?
- What is wavelength and frequency?
- If this electromagnetic move in the z-direction. Write the equation of it’s electric field.
Answer:
1. 2 × 10-7.
2. Comparing with standard equation
B = Bx Sin (kx + ωt)
we get, kx = 0.5 × 103x.
m = 12.56 × 10-3m
ω = 1.5 × 1011, 2πf = 1.5 × 1011
f = 2.3 × 1011hz
3. Amplitude E0 = B0 C
= 2 × 10-7 × 3 × 108 = 60 N/C
If magnetic field vibrate in x direction,
then Ey = 60 sin (0.5 × 103z + 1.5 × 1011t).
Question 4.
The diagram shows the main regions of the electro-magnetic spectrum.
- Write the names of the blank regions.
- Which radiation shown in the table has the shortest wavelength?
- Which ray among these are responsible for the greenhouse effect? Explain the greenhouse effect.
Answer:
1. Ultraviolet, Infrared.
2. Gamma-ray
3. Infrared
Earth is heated by infrared radiations from the sun. At the same time earth radiates longer wavelength to the space. This radiation (outgoing) of longer wavelength is reflected back by CO2 and other gases.
Hence radiation (from earth) can’t escape from earth’s surface. Thus the temperature at earth’s surface increases. This phenomenon is called Green house effect.
Question 5.
- Sun is a very powerful source of ultraviolet rays. But we do not receive much ultraviolet rays on the surface of the earth. Why?
- The electric field of an ultraviolet ray is given by E = 60 sin (0.5 × 103 – 1.5 × 1011t) What is the amplitude of the magnetic field vector in the ultraviolet ray?
Answer:
1. Ozone prevents the UV ray.
2. \(B=\frac{E}{C}=\frac{60}{3 \times 10^{8}}\)
= 2 × 10-7wb/m2.
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