1
JEE Main 2025 (Online) 4th April Morning 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 $\mathbf{R}$

Assertion A : In photoelectric effect, on increasing the intensity of incident light the stopping potential increases.

Reason R : Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than threshold frequency.

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

A
Both $\mathbf{A}$ and $\mathbf{R}$ are true and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$
B
$\mathbf{A}$ is false but $\mathbf{R}$ is true
C
Both $\mathbf{A}$ and $\mathbf{R}$ are true but $\mathbf{R}$ is NOT the correct explanation of $\mathbf{A}$
D
$\mathbf{A}$ is true but $\mathbf{R}$ is false
2
JEE Main 2025 (Online) 3rd April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
The work function of a metal is 3 eV . The color of the visible light that is required to cause emission of photoelectrons is
A
Red
B
Green
C
Blue
D
Yellow
3
JEE Main 2025 (Online) 3rd April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2 m away from it, is
A
$3 \times 10^{-8}$ Pascals
B
0
C
$1.5 \times 10^{-8}$ Pascals
D
$6 \times 10^{-8}$ Pascals
4
JEE Main 2025 (Online) 2nd April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

An electron with mass ' m ' with an initial velocity $(\mathrm{t}=0) \overrightarrow{\mathrm{v}}=\mathrm{v}_0 \hat{i}\left(\mathrm{v}_0>0\right)$ enters a magnetic field $\overrightarrow{\mathrm{B}}=\mathrm{B}_0 \hat{j}$. If the initial de-Broglie wavelength at $\mathrm{t}=0$ is $\lambda_0$ then its value after time ' t ' would be :

A
$\frac{\lambda_0}{\sqrt{1-\frac{\mathrm{e}^2 \mathrm{~B}_0^2 \mathrm{t}^2}{\mathrm{~m}^2}}}$
B
$\lambda_0$
C
$\lambda_0 \sqrt{1+\frac{\mathrm{e}^2 \mathrm{~B}_0^2 \mathrm{t}^2}{\mathrm{~m}^2}}$
D
$\frac{\lambda_0}{\sqrt{1+\frac{\mathrm{e}^2 \mathrm{~B}_0^2 \mathrm{t}^2}{\mathrm{~m}^2}}}$
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