1
JEE Main 2025 (Online) 28th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Due to presence of an em-wave whose electric component is given by $E=100 \sin (\omega t-k x) \mathrm{NC}^{-1}$ a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

A
$50 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}$
B
$400 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}$
C
$200 \sin (\omega t-k x) \mathrm{NC}^{-1}$
D
$25 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}$
2
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Arrange the following in the ascending order of wavelength $(\lambda)$ :

(A) Microwaves $\left(\lambda_1\right)$

(B) Ultraviolet rays $\left(\lambda_2\right)$

(C) Infrared rays $\left(\lambda_3\right)$

(D) X-rays $\left(\lambda_4\right)$

Choose the most appropriate answer from the options given below :

A
$\lambda_4<\lambda_3<\lambda_2<\lambda_1$
B
$\lambda_4<\lambda_2<\lambda_3<\lambda_1$
C
$\lambda_3<\lambda_4<\lambda_2<\lambda_1$
D
$\lambda_4<\lambda_3<\lambda_1<\lambda_2$
3
JEE Main 2025 (Online) 23rd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A plane electromagnetic wave of frequency 20 MHz travels in free space along the $+x$ direction. At a particular point in space and time, the electric field vector of the wave is $\mathrm{E}_y=9.3 \mathrm{Vm}^{-1}$. Then, the magnetic field vector of the wave at that point is

A
$\mathrm{B}_z=1.55 \times 10^{-8} \mathrm{~T}$
B
$\mathrm{B}_z=6.2 \times 10^{-8} \mathrm{~T}$
C
$\mathrm{B}_z=3.1 \times 10^{-8} \mathrm{~T}$
D
$\mathrm{B}_z=9.3 \times 10^{-8} \mathrm{~T}$
4
JEE Main 2025 (Online) 23rd January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The electric field of an electromagnetic wave in free space is $\overrightarrow{\mathrm{E}}=57 \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](4 \hat{i}-3 \hat{j}) N / C$. The associated magnetic field in Tesla is

A
$\overrightarrow{\mathrm{B}}=\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](\hat{k})$
B
$\overrightarrow{\mathrm{B}}=\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})$
C
$\overrightarrow{\mathrm{B}}=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](\hat{k})$
D
$\overrightarrow{\mathrm{B}}=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 \mathrm{t}-5 \times 10^{-3}(3 x+4 y)\right](5 \hat{k})$
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