Physics · Electromagnetic Waves
JEE Main 2024 — 5 April, Shift 2 — Question 43
Match List-I with List-II :-
| List-I EM-Wave | List-II Wavelength Range | ||
|---|---|---|---|
| (A) | Infra-red | (I) | |
| (B) | Ultraviolet | (II) | 400 nm to 1 nm |
| (C) | X-rays | (III) | 1 mm to 700 nm |
| (D) | Gamma rays | (IV) | 1 nm to |
Choose the correct answer from the options given below :
- Option A:
(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
- Option B:Correct
(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
- Option C:
(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
- Option D:
(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
Answer: B
Step-by-step solution
Infrared is the least energetic thus having biggest wavelength & gamma rays are most energetic thus having smallest
wavelength .
Answer key and solution verified before publishing.
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- Exam
- JEE Main 2024
- Paper
- 5 April, Shift 2
- Subject
- Physics
- Chapter
- Electromagnetic Waves
- Topic
- Properties of EM Waves and Electromagnetic Spectrum