Q. What is the main product of nuclear fusion in stars?
A.
Helium
B.
Hydrogen
C.
Carbon
D.
Iron
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Solution
The main product of nuclear fusion in stars is helium.
Correct Answer: A — Helium
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Q. What is the main reason for the appearance of a diffraction pattern when light passes through a narrow slit?
A.
Interference of light waves
B.
Refraction of light
C.
Absorption of light
D.
Reflection of light
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Solution
The diffraction pattern arises due to the interference of light waves that spread out after passing through the slit.
Correct Answer: A — Interference of light waves
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Q. What is the main reason for the stability of atomic nuclei?
A.
Electromagnetic force
B.
Gravitational force
C.
Strong nuclear force
D.
Weak nuclear force
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Solution
The stability of atomic nuclei is primarily due to the strong nuclear force, which binds protons and neutrons together.
Correct Answer: C — Strong nuclear force
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Q. What is the main reason for the stability of the electron orbits in an atom?
A.
Centripetal force
B.
Electrostatic force
C.
Quantum mechanics
D.
All of the above
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Solution
The stability of the electron orbits in an atom is primarily explained by quantum mechanics.
Correct Answer: C — Quantum mechanics
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Q. What is the main reason for the stability of the nucleus in an atom?
A.
Electromagnetic force
B.
Gravitational force
C.
Strong nuclear force
D.
Weak nuclear force
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Solution
The stability of the nucleus is primarily due to the strong nuclear force, which binds protons and neutrons together.
Correct Answer: C — Strong nuclear force
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Q. What is the main reason for the stability of the nucleus?
A.
Electromagnetic force
B.
Gravitational force
C.
Strong nuclear force
D.
Weak nuclear force
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Solution
The stability of the nucleus is primarily due to the strong nuclear force, which binds protons and neutrons together.
Correct Answer: C — Strong nuclear force
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Q. What is the majority charge carrier in a P-type semiconductor?
A.
Electrons
B.
Holes
C.
Neutrons
D.
Protons
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Solution
In a P-type semiconductor, holes are the majority charge carriers.
Correct Answer: B — Holes
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Q. What is the mass defect in a nucleus?
A.
The difference between the mass of the nucleus and the sum of the masses of its nucleons
B.
The mass of the nucleus itself
C.
The mass of the electrons surrounding the nucleus
D.
The mass of the binding energy
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Solution
The mass defect is the difference between the mass of the nucleus and the sum of the masses of its individual nucleons.
Correct Answer: A — The difference between the mass of the nucleus and the sum of the masses of its nucleons
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Q. What is the maximum intensity ratio in interference of two waves of equal amplitude?
A.
1:1
B.
2:1
C.
4:1
D.
3:1
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Solution
Maximum intensity (I_max) = 4I_0 for two waves of equal amplitude (I_0). Thus, the ratio is 4:1.
Correct Answer: C — 4:1
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Q. What is the maximum kinetic energy of photoelectrons emitted if the incident light has a frequency of 8 x 10^14 Hz and the work function is 3 eV?
A.
1 eV
B.
3 eV
C.
5 eV
D.
7 eV
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Solution
Maximum kinetic energy (K.E.) = hf - Φ = (6.626 x 10^-34 J·s)(8 x 10^14 Hz) - 3 eV = 5 eV.
Correct Answer: C — 5 eV
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Q. What is the maximum kinetic energy of photoelectrons emitted when light of frequency 8 x 10^14 Hz is incident on a metal with work function 3 eV?
A.
1 eV
B.
3 eV
C.
5 eV
D.
7 eV
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Solution
Maximum kinetic energy (K.E.) = hν - Φ = (4.14 x 10^-15 eV·s)(8 x 10^14 Hz) - 3 eV = 5 eV.
Correct Answer: C — 5 eV
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Q. What is the maximum kinetic energy of photoelectrons if the incident light has a frequency of 8 x 10^14 Hz and the work function is 3 eV?
A.
1 eV
B.
3 eV
C.
5 eV
D.
7 eV
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Solution
Maximum kinetic energy (K.E.) = hν - Φ = (4.14 x 10^-15 eV·s)(8 x 10^14 Hz) - 3 eV = 5 eV.
Correct Answer: C — 5 eV
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Q. What is the maximum number of diffraction maxima that can be observed for a double-slit experiment with a slit separation of 0.1 mm and light of wavelength 500 nm?
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Solution
The maximum number of observable maxima is given by d/λ, where d = 0.1 mm and λ = 500 nm. This gives approximately 10 maxima.
Correct Answer: B — 10
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Q. What is the maximum number of electrons in the d-subshell?
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Solution
The d-subshell can hold a maximum of 10 electrons.
Correct Answer: C — 10
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Q. What is the maximum number of electrons that can occupy a single p subshell?
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Solution
A p subshell can hold a maximum of 6 electrons, as it has three orbitals each capable of holding 2 electrons.
Correct Answer: B — 6
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Q. What is the maximum number of electrons that can occupy the d subshell?
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Solution
The d subshell can hold a maximum of 10 electrons.
Correct Answer: C — 10
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Q. What is the maximum number of electrons that can occupy the f subshell?
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Solution
The f subshell can hold a maximum of 14 electrons (7 orbitals, each can hold 2).
Correct Answer: D — 14
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Q. What is the maximum number of electrons that can occupy the n=3 energy level?
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Solution
The maximum number of electrons in an energy level is given by the formula 2n^2. For n=3, it is 2(3^2) = 18.
Correct Answer: C — 18
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Q. What is the maximum number of resistors that can be used in a Wheatstone bridge configuration?
A.
2
B.
3
C.
4
D.
Unlimited
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Solution
A standard Wheatstone bridge configuration uses four resistors.
Correct Answer: C — 4
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Q. What is the maximum percentage of light transmitted through two polarizers?
A.
0%
B.
25%
C.
50%
D.
100%
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Solution
The maximum percentage of light transmitted through two aligned polarizers is 100%.
Correct Answer: D — 100%
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Q. What is the maximum voltage that can be measured using a potentiometer with a 20 m wire and a potential gradient of 0.1 V/m?
A.
1 V
B.
2 V
C.
20 V
D.
10 V
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Solution
Maximum voltage = Potential gradient × Length = 0.1 V/m × 20 m = 2 V.
Correct Answer: C — 20 V
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Q. What is the maximum wavelength of light that can cause the photoelectric effect in a metal with a work function of 2.0 eV?
A.
620 nm
B.
400 nm
C.
500 nm
D.
300 nm
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Solution
The maximum wavelength λ_max can be calculated using λ_max = hc/E, where E = 2.0 eV = 3.2 x 10^-19 J. Thus, λ_max = (6.63 x 10^-34 * 3 x 10^8) / (3.2 x 10^-19) = 620 nm.
Correct Answer: A — 620 nm
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Q. What is the minimum energy required to remove an electron from the surface of a metal called?
A.
Work function
B.
Ionization energy
C.
Binding energy
D.
Threshold energy
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Solution
The minimum energy required to remove an electron from the surface of a metal is known as the work function.
Correct Answer: A — Work function
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Q. What is the minimum frequency of light required to eject electrons from a metal surface in the photoelectric effect?
A.
It depends on the intensity of light
B.
It is constant for all metals
C.
It depends on the work function of the metal
D.
It is equal to the energy of the incident photons
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Solution
The minimum frequency required to eject electrons is determined by the work function of the metal, given by the equation E = hf, where E is the work function, h is Planck's constant, and f is the frequency.
Correct Answer: C — It depends on the work function of the metal
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Q. What is the minimum speed required for a satellite to achieve a low Earth orbit?
A.
7.9 km/s
B.
11.2 km/s
C.
5.0 km/s
D.
9.8 km/s
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Solution
The minimum speed required for a satellite to achieve a low Earth orbit is approximately 7.9 km/s.
Correct Answer: A — 7.9 km/s
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Q. What is the minimum speed required for a satellite to maintain a low Earth orbit?
A.
7.9 km/s
B.
11.2 km/s
C.
5.0 km/s
D.
9.8 km/s
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Solution
The minimum speed required for a satellite in low Earth orbit is approximately 7.9 km/s.
Correct Answer: A — 7.9 km/s
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Q. What is the minimum thickness of a soap bubble for which the first order of bright fringe appears for light of wavelength 550 nm?
A.
275 nm
B.
550 nm
C.
1100 nm
D.
2200 nm
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Solution
For the first order bright fringe, thickness t = λ/2 = 550 nm / 2 = 275 nm.
Correct Answer: A — 275 nm
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Q. What is the minimum thickness of a soap bubble for which the first order of constructive interference occurs for light of wavelength 550 nm?
A.
275 nm
B.
550 nm
C.
1100 nm
D.
825 nm
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Solution
For constructive interference, the minimum thickness t = λ/2 = 550 nm / 2 = 275 nm.
Correct Answer: A — 275 nm
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Q. What is the minimum thickness of a soap bubble for which the first order of constructive interference occurs for light of wavelength 600 nm?
A.
100 nm
B.
200 nm
C.
300 nm
D.
400 nm
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Solution
For constructive interference in a soap bubble, the minimum thickness t = λ/2 = 600 nm / 2 = 300 nm.
Correct Answer: B — 200 nm
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Q. What is the minimum thickness of a soap bubble that appears black in reflected light?
A.
λ/4
B.
λ/2
C.
λ
D.
3λ/4
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Solution
A soap bubble appears black in reflected light when the thickness is λ/4, leading to destructive interference.
Correct Answer: A — λ/4
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