2023 - JAMB Physics Past Questions and Answers - page 1
The branch of physics that deals with the motion of objects and the forces acting on them is called:
Electromagnetism
Thermodynamics
Mechanics
Quantum mechanics
Mechanics is the branch of physics that deals with the motion of objects and the forces acting on them. It is divided into two main branches: classical mechanics, which deals with the motion of objects that are much larger than atoms and move at speeds much slower than the speed of light, and quantum mechanics, which deals with the behaviour of particles on the atomic and subatomic scale.
How much net work is required to accelerate a 1200 kg car from 10\(ms^{-1}\) to 15\(ms^{-1}\)
1.95×\(10^5 j\)
1.35×\(10^4 j\)
7.5×\(10^4 j\)
6.0×\(10^4 j\)
The work done (\(W\)) in accelerating an object is given by the formula:
\[ W = \frac{1}{2} m (v_f^2 - v_i^2) \]
where:
\( m \) is the mass of the object,
\( v_f \) is the final velocity,
\( v_i \) is the initial velocity.
Given:
\( m = 1200 \, \text{kg} \),
\( v_i = 10 \, \text{m/s} \),
\( v_f = 15 \, \text{m/s} \).
\[ W = \frac{1}{2} \times 1200 \times (15^2 - 10^2) \]
\[ W = \frac{1}{2} \times 1200 \times (225 - 100) \]
\[ W = \frac{1}{2} \times 1200 \times 125 \]
\[ W = 75000 \, \text{Joules} \]
Therefore, the correct option is:
\(7.5 \times 10^4 \, \text{J}\)
A generator manufacturing company accidentally made an AC generator instead of a DC generator. To fix this error,
the magnetic field needs to be made stronger
the split rings should be replaced with slip rings
the number of turns of the armature coil needs to be decreased
the slip rings should be replaced with split rings
In an AC generator, slip rings are used to allow continuous rotation of the coil in the magnetic field. In a DC generator, split rings (also known as commutator) are used to convert the alternating current generated in the coil to direct current. Therefore, to convert an AC generator to a DC generator, the slip rings should be replaced with split rings.
The half life of a radioactive material is 12 days. Calculate the decay constant.
0.8663 \(day^{-1}\)
0.04331 \(day^{-1}\)
0.17325 \(day^{-1}\)
0.05775 \(day^{-1}\)
The decay constant (\(\lambda\)) can be calculated using the formula:
\[ \lambda = \frac{0.693}{\text{half-life}} \]
Given that the half-life is 12 days, we can substitute this into the formula:
\[ \lambda = \frac{0.693}{12} \]
Calculating this gives:
\[ \lambda \approx 0.05775 \, \text{day}^{-1} \]
Therefore, the correct answer is: \(0.05775 \, \text{day}^{-1}\).
Which of the following thermometers measures temperature from the thermal radiation emitted by objects?
Thermocouple thermometer
Platinum resistance thermometer
Pyrometer thermometer
Constant pressure gas thermometer
The thermometer that measures temperature from the thermal radiation emitted by objects is a Pyrometer thermometer.
The number of holes in an intrinsic semiconductor
is not equal to the number of free electrons
is greater than the number of free electrons
is equal to the number of free electrons
is less than the number of free electrons
The number of holes in an intrinsic semiconductor is equal to the number of free electrons. Therefore, the correct option is:
is equal to the number of free electrons
A lorry accelerates uniformly in a straight line with acceleration of \(4ms^-2\) and covers a distance of 250 m in a time interval of 10 s. How far will it travel in the next 10 s?
650
900
800
250
The lorry's initial velocity (u) is found using the formula \(s = ut + \frac{1}{2}at^2\). Given that \(a = 4 \, \text{m/s}^2\), \(s = 250 \, \text{m}\), and \(t = 10 \, \text{s}\):
\[
250 = 10u + \frac{1}{2} \times 4 \times (10^2) \\
250 = 10u + 200 \\
10u = 50 \\
u = 5 \, \text{m/s}
\]
For the next 10 seconds (\(t = 20 \, \text{s}\)), we use the formula \(s = ut + \frac{1}{2}at^2\) again:
\[
s = (5 \times 20) + \frac{1}{2} \times 4 \times (20^2) \\
s = 100 + 800 \\
s = 900 \, \text{m}
\]
The distance covered in the next 10 seconds is \(900 \, \text{m}\). Subtracting the initial distance (250 m), we get \(900 - 250 = 650 \, \text{m}\).
Therefore, the correct answer is: \(650 \, \text{m}\).
The terminals of a battery of emf 24.0 V and internal resistance of 1.0 Ω is connected to an external resistor 5.0 Ω. Find the terminal p.d.
18.0V
12.0V
16.0V
20.0V
The terminal potential difference (\(V\)) of a battery connected to an external resistor can be calculated using the formula:
\[V = \varepsilon - Ir\]
where:
- \(\varepsilon\) is the electromotive force (emf) of the battery,
- \(I\) is the current flowing through the circuit,
- \(r\) is the internal resistance of the battery.
Given that \(\varepsilon = 24.0 \, \text{V}\), \(r = 1.0 \, \Omega\), and the external resistor is \(5.0 \, \Omega\), we need to find the current (\(I\)) first using Ohm's Law:
\[I = \frac{\varepsilon}{R_T}\]
where \(R_T\) is the total resistance, which is the sum of the external resistor and the internal resistance:
\[R_T = R + r\]
\[R_T = 5.0 + 1.0 = 6.0 \, \Omega\]
Now, calculate \(I\):
\[I = \frac{24.0}{6.0} = 4.0 \, \text{A}\]
Now, substitute \(I\) and the given values into the first formula for \(V\):
\[V = 24.0 - (4.0 \times 1.0) = 20.0 \, \text{V}\]
Therefore, the terminal potential difference is \(20.0 \, \text{V}\).
So, the correct option is: 20.0V
An explosion occurs at an altitude of 312 m above the ground. If the air temperature is -10.00°C, how long does it take the sound to reach the ground?
[velocity of sound at \(0^oC\) = 331 ms-1]
0.94s
0.96s
0.93s
0.95s
The speed of sound (\(v\)) can be calculated using the formula:
\[v = v_0 + 0.6t\]
where:
- \(v_0\) is the velocity of sound at \(0^\circ C\),
- \(t\) is the air temperature in degrees Celsius.
Given that \(v_0 = 331 \, \text{m/s}\) and the air temperature (\(t\)) is \(-10.00^\circ C\), we can substitute these values to find \(v\):
\[v = 331 + 0.6 \times (-10) = 331 - 6 = 325 \, \text{m/s}\]
The time (\(t\)) it takes for the sound to reach the ground can be calculated using the formula:
\[v = \frac{d}{t}\]
where:
- \(d\) is the distance (altitude of the explosion),
- \(t\) is the time.
Solving for \(t\):
\[t = \frac{d}{v}\]
Given that \(d = 312 \, \text{m}\) and \(v = 325 \, \text{m/s}\), substitute these values:
\[t = \frac{312}{325} \approx 0.96 \, \text{s}\]
Therefore, the correct answer is: 0.96s
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?
The stopping potential does not change.
The stopping potential decreases.
The stopping potential can either increase or decrease, depending on the intensity of the light.
The stopping potential increases.
"The stopping potential increases."
In the photoelectric effect, when light of a certain frequency (equal to or greater than the threshold frequency) is incident on a metal surface, photoelectrons are emitted. The stopping potential is the minimum potential that must be applied to stop these emitted photoelectrons.
If the frequency of the incident light is increased beyond the threshold frequency:
- The kinetic energy of the emitted photoelectrons increases.
- To stop photoelectrons with higher kinetic energy, a higher stopping potential is required.
Therefore, as the frequency of the incident light increases, the stopping potential also increases to counteract the increased kinetic energy of the photoelectrons.