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

A proton of mass ' $m_P$ ' has same energy as that of a photon of wavelength ' $\lambda$ '. If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.

A
$\frac{1}{c} \sqrt{\frac{E}{m_p}}$
B
$\frac{1}{\mathrm{c}} \sqrt{\frac{2 \mathrm{E}}{\mathrm{m}_{\mathrm{p}}}}$
C
$\frac{1}{\mathrm{2c}} \sqrt{\frac{ \mathrm{E}}{\mathrm{m}_{\mathrm{p}}}}$
D
$\frac{1}{\mathrm{c}} \sqrt{\frac{\mathrm{E}}{2 \mathrm{~m}_{\mathrm{p}}}}$
2
JEE Main 2025 (Online) 28th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A wire of resistance R is bent into an equilateral triangle and an identical wire is bent into $a$ square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is

A
$8 / 9$
B
$9 / 8$
C
$32 / 27$
D
$27 / 32$
3
JEE Main 2025 (Online) 28th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Consider a long thin conducting wire carrying a uniform current I. A particle having mass "M" and charge " $q$ " is released at a distance " $a$ " from the wire with a speed $v_0$ along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance $x$ from the wire. The value of $x$ is [ $\mu_0$ is vacuum permeability]

A
$a\left[1-\frac{m v_o}{2 q \mu_0 \mathrm{I}}\right]$
B
$a e^{-\frac{4 \pi \mathrm{mv}_o}{q \mu_o \mathrm{I}}}$
C
$a\left[1-\frac{\mathrm{mv}}{\mathrm{q}} \mu_{\mathrm{o}} \mathrm{I}\right]$
D
$\frac{a}{2}$
4
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
$25 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{NC}^{-1}$
JEE Main Papers
2023
2021
EXAM MAP