JEE Main 2024 (Online) 6th April Morning Shift | Electromagnetic Waves Question 12 | Physics

Electromagnetic waves travel in a medium with speed of $$1.5 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$$. The relative permeability of the medium is 2.0. The relative permittivity will be:

JEE Main 2024 (Online) 5th April Evening Shift

MCQ (Single Correct Answer)

-1

Match List I with List II :

	LIST I EM-Wave		LIST II Wavelength Range
A.	Infra-red	I.	$$<10^{-3}$$ nm
B.	Ultraviolet	II.	400 nm to 1 nm
C.	X-rays	III.	1 mm to 700 nm
D.	Gamma rays	IV.	1 nm to $$10^{-3}$$ nm

Choose the correct answer from the options given below :

(A)-(I), (B)-(III), (C)-(II), (D)-(IV)

(A)-(III), (B)-(II), (C)-(IV), (D)-(I)

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

JEE Main 2024 (Online) 4th April Evening Shift

MCQ (Single Correct Answer)

-1

Arrange the following in the ascending order of wavelength:

A. Gamma rays $$\left(\lambda_1\right)$$

B. $$x$$ - rays $$\left(\lambda_2\right)$$

C. Infrared waves $$\left(\lambda_3\right)$$

D. Microwaves $$\left(\lambda_4\right)$$

Choose the most appropriate answer from the options given below

$$\lambda_1<\lambda_2<\lambda_3<\lambda_4$$

$$\lambda_2<\lambda_1<\lambda_4<\lambda_3$$

$$\lambda_4<\lambda_3<\lambda_2<\lambda_1$$

$$\lambda_4<\lambda_3<\lambda_1<\lambda_2$$

JEE Main 2024 (Online) 4th April Morning Shift

MCQ (Single Correct Answer)

-1

The electric field in an electromagnetic wave is given by $$\overrightarrow{\mathrm{E}}=\hat{i} 40 \cos \omega(\mathrm{t}-z / \mathrm{c}) \mathrm{NC}^{-1}$$. The magnetic field induction of this wave is (in SI unit) :

$$\overrightarrow{\mathrm{B}}=\hat{j} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})$$

$$\overrightarrow{\mathrm{B}}=\hat{i} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})$$

$$\vec{B}=\hat{j} 40 \cos \omega(t-z / c)$$

$$\overrightarrow{\mathrm{B}}=\hat{k} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})$$

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