1
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
2
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
3
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
$\frac{\lambda_0}{\sqrt{1+\frac{\mathrm{e}^2 \mathrm{~B}_0^2 \mathrm{t}^2}{\mathrm{~m}^2}}}$
4
JEE Main 2025 (Online) 2nd April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A monochromatic light is incident on a metallic plate having work function $\phi$. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point $B$. The distance between $A$ and $B$ is : (Given : The magnitude of charge of an electron is e and mass is $\mathrm{m}, \mathrm{h}$ is Planck's constant and c is velocity of light. Take the magnetic field exists throughout the path of electron)

A
$\sqrt{2 m\left(\frac{h c}{\lambda}-\phi\right)} / \mathrm{eB}$
B
$\sqrt{8 \mathrm{~m}\left(\frac{\mathrm{hc}}{\lambda}-\phi\right)} / \mathrm{eB}$
C
$\sqrt{\mathrm{m}(\mathrm{hc} / \lambda-\phi)} / \mathrm{eB}$
D
$2 \sqrt{\mathrm{~m}(\mathrm{hc} / \lambda-\phi)} / \mathrm{eB}$
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