Electromagnetism
Current Electricity
MCQ (Single Correct Answer)
Moving Charges and Magnetism
MCQ (Single Correct Answer)
Magnetism and Matter
MCQ (Single Correct Answer)
Electromagnetic Induction
MCQ (Single Correct Answer)
Alternating Current
MCQ (Single Correct Answer)
Electromagnetic Waves
MCQ (Single Correct Answer)
Modern Physics
Dual Nature of Radiation
MCQ (Single Correct Answer)
Semiconductor Devices and Logic Gates
MCQ (Single Correct Answer)
Communication Systems
MCQ (Single Correct Answer)
1
COMEDK 2024 Evening Shift
MCQ (Single Correct Answer)
+1
-0

A particle of mass $$2 \mathrm{mg}$$ has the same wavelength as a neutron moving with a velocity of $$3 \times 10^5 \mathrm{~ms}^{-1}$$. The velocity of the particle is (mass of neutron is $$1.67 \times 10^{-27} \mathrm{Kg}$$)

A
$$2.5 \times 10^{-16} \mathrm{~ms}^{-1}$$
B
$$1.5 \times 10^{-13} \mathrm{~ms}^{-1}$$
C
$$2.5 \times 10^{-13} \mathrm{~ms}^{-1}$$
D
$$1.5 \times 10^{-16} \mathrm{~ms}^{-1}$$
2
COMEDK 2024 Afternoon Shift
MCQ (Single Correct Answer)
+1
-0

The velocity of an electron so that its momentum is equal to that of a photon of wavelength $$660 \mathrm{~nm}$$ is

A
$$ 109.88 \mathrm{~cm} \mathrm{~s}^{-1} $$
B
$$ 1098 \mathrm{~ms} ^{-1} $$
C
$$ 1098 \mathrm{~cm} \mathrm{~s}^{-1} $$
D
$$ 109 \mathrm{~ms} ^{-1} $$
3
COMEDK 2024 Afternoon Shift
MCQ (Single Correct Answer)
+1
-0

$$\mathrm{K}_1$$ and $$\mathrm{K}_2$$ are maximum kinetic energies of photoelectrons emitted when lights of wavelength $$\lambda_1$$ and $$\lambda_2$$ respectively are incident on a metallic surface. If $$\lambda_1=3 \lambda_2$$, then

A
$$ \mathrm{K}_1=\left(\frac{1}{3}\right) \mathrm{K}_2 $$
B
$$ \mathrm{K}_1>3 \mathrm{~K}_2 $$
C
$$ \mathrm{K}_1>\left(\frac{1}{3}\right) \mathrm{K}_2 $$
D
$$ \mathrm{K}_1<\left(\frac{1}{3}\right) \mathrm{K}_2 $$
4
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

The threshold frequency for a metal surface is '$$n_0$$'. A photo electric current '$$I$$' is produced when it is exposed to a light of frequency $$\left(\frac{11}{6}\right) \mathrm{n}_{\mathrm{o}}$$ and intensity $$\mathrm{I}_{\mathrm{n}}$$. If both the frequency and intensity are halved, the new photoelectric current '$$\mathrm{I}^1$$' will become:

A
$$\mathrm{I}^1=\frac{1}{4} \mathrm{I}$$
B
$$\mathrm{I}^1=2 \mathrm{I}$$
C
$$\mathrm{I^1=0}$$
D
$$\mathrm{I}^1=\frac{1}{2} \mathrm{I}$$
COMEDK Subjects