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Physics
UTME / JAMB 2024
Deformation of Solids
Solution
A force of 10N extends a spring of natural length 1m by 0.02m, calculate the length of the spring when the applied force is 40N.
A
0.08m
B
0.09m
C
1.08m
D
1.80m
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
F = ke
k = \(\frac{F_1}{e_1}\)
F\(_1\) = 10N, e\(_1\) = 0.02m
F\(_2\) = 40N, e\(_2\) = ?
\(\frac{F_1}{e_1}\) = \(\frac{F_2}{e_2}\)
\(\frac{10}{0.02}\) = \(\frac{40}{e_2}\)
e\(_2\) = \(\frac{40 \times 0.02}{10}\) = 0.08m
The new length = 1 + 0.08 = 1.08m
Source: WAEC / JAMB / NECO Classroom Archive (Record #70671)
Physics
WAEC / WASSCE 2024
Scalars and Vectors
Solution
Which of the following terms has different dimension from the others?
A
Stress x Strain
B
Pressure
C
\(\frac{\text{stress}}{\text{strain}}\)
D
Torque
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The term with a different dimension from the others is D. Torque, as its dimensions ML\(^2\)T\(^{−2}\) which differ from the dimensions of stress and pressure, which are ML\(^{−1}\)T\(^{−2}\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #72063)
Physics
WAEC / WASSCE 2024
Scalars and Vectors
Solution
(a) Derive the dimension of surface tension.
(b) Name the instrument used to measure the force of gravity at a place
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) SURFACE TENSION = \(\frac{\text{force}}{\text{length}}\)
FORCE = mass x accel. due to gravity
Dimension of force = M.LT\(^{-2}\)
Dimension of length = L
Dimension of surface area = \(\frac{MLT^{-2}}{L}\) = MT\(^{-2}\)
(b) The instruments used for measuring the force of gravity at a place are:
(1) Gravimeter, (2) Spring balance, (3) Gravitational accelerometer.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72084)
Physics
WAEC / WASSCE 2024
Scalars and Vectors
Solution
(a) State the S.I base unit of the following quantities: (i) a stress, (ii) force constant (iii) Plank's constant
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(i) Stress: The base unit of stress is the Pascal (Pa), which is equivalent to kg⋅m\(^{−1}\)⋅s\(^{−2}\)
(ii) Force Constant (Spring Constant,k): The base unit of the spring constant is also in Newtons per meter (N/m), which is equivalent to kg⋅s\(^{−2}\)
(iii) Planck's Constant (h): The base unit of Planck's constant is Joule seconds (J·s), which is equivalent to kg⋅m\(^2\)s\(^{−1}\)
.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72086)
Physics
UTME / JAMB 2024
Heat and Temperature
Solution
Pilots uses aneroid barometer to know the height above sea level because
A
Very accurate
B
Small and convenient
C
Heavy and huge
D
Costly
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Aneroid barometers are compact and lightweight, making them suitable for use in aircraft where space and weight are critical considerations. They provide a reliable measurement of altitude based on changes in atmospheric pressure.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70361)
Physics
UTME / JAMB 2024
Heat and Temperature
Solution
A thick glass tumbler cracks when boiling water is poured into it because
A
The inside expands more rapidly than the outside
B
The inside expands less rapidly than the outside
C
The expansion is even
D
It does not expand
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
When boiling water is poured into the tumbler, the inner surface heats up and expands quickly, while the outer surface remains cooler and expands more slowly. This difference in expansion rates creates stress in the glass, leading to cracks.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70406)
Physics
UTME / JAMB 2024
Heat and Temperature
Solution
A mass of gas at 40mmHg is heated from 298k to 348k at constant volume. Cal the pressure exerted by the gas.
A
50.21mmHg
B
46.71mmHg
C
42.35mmHg
D
35.12mmHg
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
From pressure law
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)
P\(_2\) = \(\frac{P_1 \times T_2}{T_1}\)
P\(_1\) = 40mmHg, P\(_2\) = ?
T\(_1\) = 298k, T\(_2\) = 348k
\(\frac{40}{298} = \frac{P_2}{348}\)
P\(_2\) = \(\frac{40\times348}{298}\) = 46.71mmHg
Source: WAEC / JAMB / NECO Classroom Archive (Record #70409)
Physics
UTME / JAMB 2024
Heat and Temperature
Solution
The thermometer whose thermometric property is change in volume with temperature is
A
Resistance thermometer
B
Thermocouple
C
Bimetallic thermometer
D
Liquid-in-glass thermometer
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
D. Liquid-in-glass thermometer.
In a liquid-in-glass thermometer, the liquid (usually mercury or colored alcohol) expands and contracts with temperature changes, allowing for the measurement of temperature based on the volume change of the liquid.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70411)
Physics
UTME / JAMB 2024
Heat and Temperature
Solution
Which of these gas laws is equivalent to workdone
A
Charle's law
B
Pressure law
C
Boyle's law
D
Gay Lussac's law
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Boyle's law states that at a constant temperature, the pressure of a gas is inversely proportional to its volume (P ∝\(\frac{1}{V}\) This relationship is fundamental when calculating the work done during the isothermal expansion or compression of a gas, as it directly involves changes in volume and pressure
Source: WAEC / JAMB / NECO Classroom Archive (Record #70415)
Physics
UTME / JAMB 2024
Heat and Temperature
Solution
The average translational kinetic energy of gas molecules depends on
A
Temperature only
B
Number of moles and temperature
C
Pressure and temperature
D
Pressure only
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Temperature only.
This means that the average kinetic energy is directly related to the gas's absolute temperature. The relationship can be defined by the equation:
K.E = \(\frac{3}{2}\)kT
where K.E = average kinetic energy, k = Boltzmann's constant, and T = absolute temperature.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70424)
Physics
UTME / JAMB 2024
Heat and Temperature
Solution
Convert 60ºC to degree Fahrenheit
A
111.15\(^0\)F
B
115.56\(^0\)F
C
132.23\(^0\)F
D
140.00\(^0\)F
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Using \(^0\)F = ( \(^0\) C x \(\frac{9}{5}\)) + 32 → ( ºc x 1.8 ) + 32.
\(^0\) C = 60º
Then, Using \(^0\)F = ( 60 x \(\frac{9}{5}\)) + 32
= 108 + 32 = 140ºF
Source: WAEC / JAMB / NECO Classroom Archive (Record #70471)
Physics
UTME / JAMB 2024
Heat and Temperature
Solution
At absolute zero temperature, the average velocity of the molecules
A
Decreases
B
Increases
C
Is zero
D
Is uncertain
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Absolute zero is the lowest possible temperature where molecular motion is minimal. At this temperature, the kinetic energy of the molecules is effectively zero, meaning they do not have enough energy to move.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70473)
Physics
UTME / JAMB 2024
Heat and Temperature
Solution
I It wets glass
II It needs to be coloured
III It has a low density
Water is not suitable for use as a thermometric liquid because
A
I only
B
I and II only
C
I, II, and III only
D
II and III only
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Water is not suitable for use as a thermometric liquid because:
a) It wets glass: This can cause issues with reading the level of the liquid.
b) It needs to be coloured: Water is typically clear, making it difficult to see the level without coloring.
c) It has a low density: This can affect the sensitivity and accuracy of the thermometer.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70504)
Physics
UTME / JAMB 2024
Heat and Temperature
Solution
At a pressure of 10\(^5\)Nm\(^{-2}\), a gas has a volume of 20m\(^3\). Calculate the volume at 4 x 10\(^5\)Nm\(^{-2}\) at constant temperature.
A
2m\(^3\)
B
3m\(^3\)
C
4m\(^3\)
D
5m\(^3\)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Using \(P_1\) \(V_1\) = \(P_2\) \(V_2\)
\(V_2\) = \(\frac{P_1\times V_2}{P_2}\)
\(V_2\) = \(\frac{10^5\times 20}{4 \times 10^5}\)
\(V_2\) = 5m\(^3\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #70530)
Physics
UTME / JAMB 2024
Heat and Temperature
Solution
If the pressure on an enclosed gas is doubled and its temperature rises from 27ºC to 127ºC, then its volume will
A
increase by \(\frac{2}{3^{rd}}\)
B
increase by \(\frac{1}{3^{rd}}\)
C
decrease by \(\frac{2}{3^{rd}}\)
D
decrease by \(\frac{1}{3^{rd}}\)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Using the general gas equation;
\(\frac{P_1 \times V_1}{T_1}\) = \(\frac{P_2 \times V_2}{T_2}\)
P\(_1\) = P, P\(_2\) = 2P, T\(_1\) = 27ºC → 300K, T\(_2\) = 127ºC → 400k,
V\(_2\) = \(\frac{P_1 \times V_1 \times T_2}{T_1 \times P_2}\)
V\(_2\) = \(\frac{P \times V_1 \times 400}{300 \times 2P}\) = \(\frac{2V_1}{3}\)
\DeltaV = V\(_1\) - \(\frac{2V_1}{3}\) = \(\frac{V_1}{3}\)
Therefore, The volume decreases by \(\frac{1}{3}\) of the original volume.
Source: WAEC / JAMB / NECO Classroom Archive (Record #70667)