1
JEE Main 2026 (Online) 6th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A point light source emits E.M. waves in free space. A detector, placed at a distance of $L \mathrm{~m}$, measures the intensity as $I_{\mathrm{o}}$. The detector is now shifted to another location on the same spherical surface ensuring the angle between original location and new location as $45^{\circ}$. The measured intensity at new location will be $\_\_\_\_$ .

A

$\frac{I_{\mathrm{o}}}{4}$

B

$I_{\mathrm{o}}$

C

$\frac{I_0}{\sqrt{2}}$

D

${\frac{I_{\mathrm{o}}}{2}}$

2
JEE Main 2026 (Online) 5th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : The electromagnetic wave exerts pressure on the surface on which they are allowed to fall.

Reason (R) : There is no mass associated with the electromagnetic waves.

In the light of the above statements, choose the correct answer from the options given below :

A

Both (A) and (R) are true and (R) is the correct explanation of (A)

B

Both (A) and (R) are true but (R) is not the correct explanation of (A)

C

(A) is true but (R) is false

D

(A) is false but (R) is true

3
JEE Main 2026 (Online) 5th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A displacement current of 4.0 A can be set up in the space between two parallel plates of $6 \mu \mathrm{~F}$ capacitor. The rate of change of potential difference across the plates of the capacitor is nearly $\alpha \times 10^6 \mathrm{~V} / \mathrm{s}$. The value of $\alpha$ is $\_\_\_\_$ .

A

0.58

B

0.67

C

0.82

D

0.75

4
JEE Main 2026 (Online) 4th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A magnetic field vector in an electromagnetic wave is represented by $\vec{B}=B_0 \sin \left(2 \pi v t-\frac{2 \pi x}{\lambda}\right) \hat{j}$. Its associated electric field vector is $\_\_\_\_$ .

A

$$ \vec{E}=-v \lambda B_0 \sin \left(2 \pi v t-\frac{2 \pi x}{\lambda}\right) \hat{k} $$

B

$$ \vec{E}=-v \lambda B_0 \sin \left(2 \pi v t-\frac{2 \pi x}{\lambda}\right) \hat{i} $$

C

$$ \vec{E}=v \lambda B_0 \sin \left(2 \pi v t-\frac{2 \pi x}{\lambda}\right) \hat{k} $$

D

$$ \vec{E}=v \lambda B_0 \sin \left(2 \pi v t-\frac{2 \pi x}{\lambda}\right) \hat{i} $$

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