1
NEET 2025
MCQ (Single Correct Answer)
+4
-1

A photon and an electron (mass $m$ ) have the same energy $E$. The ratio ( $\lambda_{\text {photon }} / \lambda_{\text {electron }}$ ) of their de Broglie wavelengths is: ( $c$ is the speed of light)

A
$c \sqrt{\frac{2 m}{E}}$
B
$\frac{1}{c} \sqrt{E / 2 m}$
C
$\sqrt{E / 2 m}$
D
$c \sqrt{2 m E}$
2
NEET 2025
MCQ (Single Correct Answer)
+4
-1

A sphere of radius $R$ is cut from a larger solid sphere of radius $2 R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the $Y$-axis is:

NEET 2025 Physics - Rotational Motion Question 1 English

A
$\frac{7}{57}$
B
$\frac{7}{64}$
C
$\frac{7}{8}$
D
$\frac{7}{40}$
3
NEET 2025
MCQ (Single Correct Answer)
+4
-1

An electron (mass $9 \times 10^{-31} \mathrm{~kg}$ and charge $1.6 \times 10^{-19} \mathrm{C}$ ) moving with speed $c / 100(c=$ speed of light) is injected into a magnetic field $\vec{B}$ of magnitude $9 \times 10^{-4} \mathrm{~T}$ perpendicular to its direction of motion. We wish to apply an uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $c=3$ $\times 10^3 \mathrm{~ms}^{-1}$)

A
$\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 \times 10^2 \mathrm{~V} \mathrm{~m}^{-1}$
B
$\vec{E}$ is parallel to $\vec{B}$ and its magnitude is $27 \times 10^4 \mathrm{~V} \mathrm{~m}^{-1}$
C
$\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^4 \mathrm{~V} \mathrm{~m}^{-1}$
D
$\vec{E}$ is perpendicular to $\vec{B}$ and its magnitude is $27 \times 10^2 \mathrm{~V} \mathrm{~m}^{-1}$
4
NEET 2025
MCQ (Single Correct Answer)
+4
-1

The electric field in a plane electromagnetic wave is given by $$ E_z=60 \cos \left(5 x+1.5 \times 10^9 t\right) \mathrm{V} / \mathrm{m}$$ Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field) :

A
$B z=60 \cos \left(5 x+1.5 \times 10^9 t\right) T$
B
$B_y=60 \sin \left(5 x+1.5 \times 10^9 t\right) T$
C
$B_y=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) T$
D
$B_x=2 \times 10^{-7} \cos \left(5 x+1.5 \times 10^9 t\right) T$
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