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Physics
UTME / JAMB 2024
Heat Energy & Temperature
Solution
According to kinetic theory of gases, the pressure exerted by the gas on the wall is equal
A
change in momentum per unit volume
B
rate of change of momentum imparted by the walls per second per unit area
C
momentum imparted to the walls per unit area
D
change in momentum imparted to the walls per unit area
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Pressure is defined as force per unit area. In the context of gas molecules colliding with the walls of a container, the force exerted by the gas can be understood as the rate at which momentum is transferred to the walls.
As gas molecules collide with the walls, they change momentum, and this change occurs over time. Thus, the pressure can be described as the rate of change of momentum (change in momentum per collision multiplied by the number of collisions per second) divided by the area over which the collisions occur.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70467)
Physics
UTME / JAMB 2024
Light & Geometrical Optics
Solution
Rainbow is formed when sunlight undergoes
A
reflection
B
refraction
C
dispersion
D
diffraction
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
A rainbow is formed by the DISPERSION of sunlight by spherical raindrops floating in the air.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70355)
Physics
UTME / JAMB 2024
Light & Geometrical Optics
Solution
An object is placed 25cm in front of a convex mirror has its image formed 5cm behind the mirror. what is the focal length of the convex mirror
A
8.20cm
B
6.25cm
C
5.60cm
D
4.50cm
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Object distance (u) = -25 cm (negative because the object is in front of the mirror)
Image distance (v) = +5 cm (positive because the image is behind the convex mirror)
Using \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{1}{v}\)
\(\frac{1}{f}\) = \(\frac{1}{-25}\) + \(\frac{1}{5}\)
f = \(\frac{25}{4}\) = 6.250cm.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70497)
Physics
UTME / JAMB 2024
Light & Geometrical Optics
Solution
The power of a convex lens of focal length 20cm is
A
0.05D
B
0.50D
C
5.00D
D
50.00D
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The Power is given by: P = \(\frac{1}{f}\)
f = 20cm → 0.2m
P = \(\frac{1}{0.2}\) = 5 dioptres
Source: WAEC / JAMB / NECO Classroom Archive (Record #70506)
Physics
UTME / JAMB 2024
Light & Geometrical Optics
Solution
A monochromatic light is one that
A
can undergo dispersion
B
cannot be refracted
C
has one wavelength
D
has many wavelength
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
A monochromatic light is one that has a single wavelength or color. This means that it consists of light waves that all have the same frequency, resulting in a uniform appearance without any variation.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70669)
Physics
UTME / JAMB 2024
Light & Geometrical Optics
Solution
The diaphragm in the camera is similar to what part of the eyes?
A
Cornea
B
Lens
C
Ciliary muscle
D
Iris
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The diaphragm in a camera is similar to the D. Iris of the eyes.
The diaphragm controls the amount of light entering the camera, just as the iris controls the amount of light entering the eye by adjusting the pupil's size.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70686)
Physics
UTME / JAMB 2024
Mechanics
Solution
A wheelbarrow inclined at 60º to the horizontal is pushed with a force of 150N. What is the horizontal component of the applied force
A
130N
B
75N
C
60N
D
50N
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The horizontal component F\(_x\) can be calculated using the formula: F\(_x\) = F⋅ cos (\(\theta\))
F = 150N and \(\theta\) = 60º
F\(_x\) = 150 x 0.5 = 75N (cos 60º = 0.5)
Source: WAEC / JAMB / NECO Classroom Archive (Record #70368)
Physics
UTME / JAMB 2024
Satellite and Rocket
Solution
Which of the following is not a part of model rocket?
A
The body tube
B
The nose cones
C
The fin
D
Not recovery devices
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
D. Not recovery devices.
In contrast, the other options—A. The body tube, B. The nose cone, and C. The fin—are all essential components of a model rocket. Recovery devices (like parachutes or streamers) are also important for ensuring a safe landing, but "not recovery devices" are not a specific part of the rocket itself.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70405)
Physics
UTME / JAMB 2024
Motion Force Circular Motion
Solution
The friction due to air mass can be reduced by
A
streaming
B
lubricating
C
propeller
D
polishing
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Streaming refers to the streamlined shape of an object, which helps minimize air resistance (drag) as it moves through the air. This is often applied in aerodynamics to improve the efficiency of vehicles and objects moving through air.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70358)
Physics
UTME / JAMB 2024
Motion Force Circular Motion
Solution
The moon's acceleration due to gravity is \(\frac{1}{6}\) of the earth's value. The weight of a bowling ball on the moon would be
A
6 times its weight on the earth
B
\(\frac{1}{6}\) of its weight on the earth
C
36 times its weight on the earth
D
equal to its weight on the earth
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
W\(_{moon}\) = mass x g = mass x \(\frac{10}{6}\) = \(\frac{10m}{6}\)N
W\(_{earth}\) = mass x g = mass x 10 = 10m
mass = constant
g on earth = 10m/s\(^2\) and g on moon = \(\frac{1}{6}\) x that on earth
Therefore, The weight of a bowling ball on the moon would be \(\frac{1}{6}\), the weight on earth.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70372)
Physics
UTME / JAMB 2024
Motion Force Circular Motion
Solution
A body is whirled in a horizontal circle at the rate of 800 revolutions per minute. Determine the angular velocity
A
100\(\pi\)rads\(^{-1}\)
B
33.3\(\pi\)rads\(^{-1}\)
C
29.5\(\pi\)rads\(^{-1}\)
D
26.7\(\pi\)rads\(^{-1}\)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Angular velocity(\(\omega\)) = 2\(\pi\) x rev per sec
Rev. per sec = \(\frac{800}{60}\) \(\approxeq\) 13.33 rev/s
Angular velocity(\(\omega\)) = 2\(\pi\) x 13.33 \(\approxeq\) 26.7\(\pi\)rad/s
Source: WAEC / JAMB / NECO Classroom Archive (Record #70476)
Physics
UTME / JAMB 2024
Motion Force Circular Motion
Solution
The tangential force acting on an object that opposes it from sliding freely on the adjacent surface is called
A
Normal force
B
Friction force
C
Weight
D
Upthrust
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The tangential force acting on an object that prevents it from sliding freely on the adjacent surface is called friction. Friction plays a crucial role in preventing or resisting motion between surfaces in contact.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70488)
Physics
UTME / JAMB 2024
Motion Force Circular Motion
Solution
A body is pulled on a horizontal surface with a rope inclined at 30º to the vertical. If the effective force pulling the body along the horizontal surface is 15N, calculate the tension on the rope.
A
30N
B
60N
C
22.5N
D
75N
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Effective horizontal force = 15N
T = tension on the rope
F\(_x\) = T sin\(\theta\)
15 = T x sin 30º
T = \(\frac{15}{0.5}\) = 15 x 2 = 30N.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70536)
Physics
UTME / JAMB 2024
Work, Energy and Power
Solution
Under which conditions is work done
A
A man supports a heavy load on his head with hands.
B
A woman holds a pot of water
C
A boy climbs onto a table
D
A man pushes against a stationary petrol tanker
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Work is done when a force causes an object to move in the direction of that force. OPTION C. A boy climbs onto a table.
Work done (force is applied to lift himself).
when a boy climbs onto a table, he exerts a force to lift his body against gravity. As he moves upward, he is doing work because there is a displacement in the direction of the applied force. Thus, work is done in this case.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70352)
Physics
UTME / JAMB 2024
Work, Energy and Power
Solution
The energy in a moving car is an example of
A
Mechanical energy
B
Electrical energy
C
Potential energy
D
Kinetic energy
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The energy in a moving car is an example of kinetic energy which is the energy possessed by a body in motion.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70469)