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Physics
UTME / JAMB 2023
Simple Ac Circuit
Solution
In an AC circuit, resonance occurs when the impedance of the circuit is:
A
Zero
B
Maximum
C
Equal to the capacitive reactance of the circuit
D
Minimum
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Resonance in an AC circuit occurs when the inductive reactance (XL) is equal to the capacitive reactance (XC). This is because at this point, the two reactances cancel each other out, leaving only the resistance of the circuit. As a result, the impedance of the circuit is minimized.
Source: WAEC / JAMB / NECO Classroom Archive (Record #68586)
Physics
UTME / JAMB 2023
Structure of the Atom
Solution
When light of a certain frequency is incident on a metal surface, no photoelectrons are emitted. If the frequency of the light is increased, what happens to the stopping potential?
A
The stopping potential does not change.
B
The stopping potential decreases.
C
The stopping potential can either increase or decrease, depending on the intensity of the light.
D
The stopping potential increases.
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The stopping potential is the voltage that is required to stop the emitted photoelectrons from reaching the anode. When the frequency of the light is increased, the kinetic energy of the emitted photoelectrons also increases. This means that a higher stopping potential will be required to stop the photoelectrons from reaching the anode.
Source: WAEC / JAMB / NECO Classroom Archive (Record #68571)
Physics
UTME / JAMB 2023
Semi Conductor
Solution
The number of holes in an intrinsic semiconductor
A
is not equal to the number of free electrons
B
is greater than the number of free electrons
C
is equal to the number of free electrons
D
is less than the number of free electrons
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
An intrinsic semiconductor, also called an undoped semiconductor, is a pure semiconductor. In intrinsic semiconductors the number of excited electrons and the number of holes are equal: n = p
Source: WAEC / JAMB / NECO Classroom Archive (Record #67423)
Physics
UTME / JAMB 2023
Structure of the Nucleus
Solution
The half life of a radioactive material is 12 days. Calculate the decay constant.
A
0.8663 \(day^{-1}\)
B
0.04331 \(day^{-1}\)
C
0.17325 \(day^{-1}\)
D
0.05775 \(day^{-1}\)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
\(T_ {^1/_2}\) = 12 days ;V=?
\(T_ {^1/_2}\) = \(\frac{0.693}{γ}\)
⇒γ ⇒\(\frac{0.693}{t_{^1/_2}}\)
⇒γ =\(\frac{0.693}{12}\)
⇒γ= 0.05775 \(day^{-1}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #67376)
Physics
UTME / JAMB 2023
Structure of the Nucleus
Solution
Which process is responsible for production of energy in stars?
A
Nuclear reaction
B
Nuclear fission
C
Nuclear fusion
D
Radioactive decay
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Nuclear fusion is the process in which lighter nuclei are combined to form heavier ones. Matter lost in this process is converted into energy. The energy produced inside the stars is due to the process of nuclear fusion.
Source: WAEC / JAMB / NECO Classroom Archive (Record #67593)
Physics
UTME / JAMB 2023
Structure of the Nucleus
Diagram
Solution
The diagram above illustrates the penetrating power of some types of radiation. X, Y and Z are likely
Attached Figure / Diagram
A
X = α-particle; Y = γ-ray; Z = β-particle
B
X = α-particle; Y = β-particle; Z = γ-ray
C
X = γ-ray; Y = α-particle; Z = β-particle
D
X = β-particle; Y = γ-ray; Z = α-particle
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The penetrating power of alpha rays, beta rays, and gamma rays varies greatly. Alpha particles can be blocked by a few pieces of paper. Beta particles pass through paper but are stopped by aluminum foil. Gamma rays are the most difficult to stop and require concrete, lead, or other heavy shielding to block them.
Therefore, X = γ-ray; Y = \alpha-particle; Z = \beta-particle
Source: WAEC / JAMB / NECO Classroom Archive (Record #68121)
Physics
UTME / JAMB 2023
Structure of the Nucleus
Solution
What is the name of the model of the atom that describes electrons as orbiting the nucleus in specific energy levels?
A
Bohr model
B
Dalton model
C
Rutherford model
D
Thomson model
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
The Bohr model is a model of the atom that describes electrons as orbiting the nucleus in specific energy levels. This model was developed by Niels Bohr in 1913 and it is based on the work of Max Planck and Albert Einstein. The Rutherford model, the Thomson model, and the Dalton model are all older models of the atom that are not as accurate as the Bohr model.
Source: WAEC / JAMB / NECO Classroom Archive (Record #68574)
Physics
UTME / JAMB 2023
Structure of the Nucleus
Solution
Which of the following is NOT an example of elementary modern physics?
A
Classical mechanics
B
Quantum mechanics
C
Special relativity
D
Nuclear physics
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Classical mechanics is a branch of physics that deals with the motion of macroscopic objects. It is based on the principles of Newton's laws of motion and is not considered to be part of elementary modern physics.
The other three options, quantum mechanics, special relativity, and nuclear physics, are all considered to be part of elementary modern physics because they deal with the behavior of matter and energy at the atomic and subatomic levels.
Source: WAEC / JAMB / NECO Classroom Archive (Record #68578)
Physics
UTME / JAMB 2023
Reflection of Light
Solution
Rainbow formation is as a result of the combination of which of the following phenomena?
(i) Reflection
(ii) Dispersion
(iii) Total internal reflection
(iv) Refraction
A
(ii) and (iv) only
B
(i), (iii) and (iv) only
C
(ii), (iii) and (iv) only
D
(i), (ii) and (iv) only
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
As light ray enters a drop of water the light is refracted at the surface and at the end of the drop, it is totally internally reflected in which the reflected light returns to the front surface, where it again undergoes refraction as it moves from water to air. The result of this is a dispersed light of colours of different wavelengths.
Source: WAEC / JAMB / NECO Classroom Archive (Record #67437)
Physics
UTME / JAMB 2023
Reflection of Light
Solution
The pinhole camera works on
A
the principle of the reversibility of light
B
the principle of the rectilinear propagation of light
C
Huygen’s principle
D
the principle of total internal reflection of light
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The pinhole camera works on the principle of the rectilinear propagation of light. This principle states that light travels in straight lines. When light passes through the tiny hole in a pinhole camera, it forms an inverted image on the opposite side of the camera. The size of the image depends on the distance between the object and the pinhole.
Source: WAEC / JAMB / NECO Classroom Archive (Record #67986)
Physics
UTME / JAMB 2023
Reflection of Light
Solution
An object is placed 35 cm away from a convex mirror with a focal length of magnitude 15 cm. What is the location of the image?
A
26.25 cm behind the mirror
B
10.5 cm behind the mirror
C
26.25 cm in front of the mirror
D
10.5 cm in front of the mirror
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
f = 15cm (diverging mirror); u = -35cm(because object is in front of the mirror) v=?
⇒\(\frac{1}{f}\) = \(\frac{1}{v}\) + \(\frac{1}{u}\)
⇒ \(\frac{1}{15}\) = \(\frac{1}{v}\) - \(\frac{1}{35}\)
⇒ \(\frac{1}{15}\) + \(\frac{1}{35}\) = \(\frac{1}{v}\)
⇒ \(\frac{105}{10}\) = \(\frac{1}{v}\)
v = \(\frac{105}{10}\) = 10.5cm
∴The image is 10.5cm behind the mirror, since v is positive.
Source: WAEC / JAMB / NECO Classroom Archive (Record #68053)
Physics
UTME / JAMB 2023
Refraction of Light
Solution
Light of wavelength 589 nm in vacuum passes through a piece of fused quartz of index of refraction n = 1.458. What is the frequency of the light in fused quartz?
[Speed of light c = \(3.00×10^8 ms^{-1}\)]
A
\(5.09×10^{15}\)Hz
B
\(5.09×10^{14}\) Hz
C
\(1.77×10^{15}\) Hz
D
\(1.77×10^{14}\) Hz
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
n=1.458, c=\(3.00 \times10^8 ms^-1\) ,\lambdao = 589nm; f=?
Speed of light in a medium (v)=\(\frac{c}{n}\) where n is the refractive index of the medium
⇒\lambdan=\(\frac{589}{1.458}\) = 404nm
v=f\lambda
⇒f=\(\frac{v}{\lambda}\)
=\(\frac{2.06\times10^8}{404\times10^-9}\) \(1nano=10^{-9}\)
∴f=\(5.09\times10^{14}\) Hz
Source: WAEC / JAMB / NECO Classroom Archive (Record #67912)
Physics
UTME / JAMB 2023
Refraction of Light
Solution
A beam of light travelling in water is incident on a glass which is immersed in the water. The incident beam makes an angle of \(40^o\) with the normal. Calculate the angle of refraction in the glass.
[Refractive index of water = 1.33, Refractive index of glass = 1.5]
A
\(29.36^o\)
B
\(25.37^o\)
C
\(37.21^o\)
D
\(34.75^o\)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
\(n_1\)=1.33; \(n_2\)=1.5; i=40^\circ; r= ?
from snell's law:
\(n_1\) x sin i=\(n_2\)x sin r
⇒1.33 x sin 40^\circ=1.5 x sin r
⇒sin r = 0.5700
⇒r = \(sin^{-1}\) (0.5700)
⇒ r = 34.75
Source: WAEC / JAMB / NECO Classroom Archive (Record #68117)
Physics
UTME / JAMB 2023
Waves
Solution
A travelling wave of amplitude 0.80 m has a frequency of 16 Hz and a wave speed of 20 \(ms-1\)
Calculate the wave number of the wave.
A
3
B
4
C
5
D
2
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
A=0.8m; f=16Hz; v=20 \(ms^-1\); k=?
v = f\lambda
\lambda =1.25m
= k= \(\frac{2\pi}{\lambda}\) =\(\frac{2 \times 3.142}{1.25}\)
k= 5m ( to 1 s.f)
Source: WAEC / JAMB / NECO Classroom Archive (Record #67940)
Physics
UTME / JAMB 2023
Waves
Solution
Which of the following types of electromagnetic waves is used in night vision goggles?
A
Microwaves
B
Ultraviolet waves
C
Infrared waves
D
Radio waves
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Infrared waves are used in night vision goggles because they can penetrate the darkness and allow people to see objects that are not visible to the naked eye
Source: WAEC / JAMB / NECO Classroom Archive (Record #68584)