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Physics
UTME / JAMB 1987
Calorimetry, Latent Heat & Change of State
Solution
22,000J of heat is required to raise the temperature of 1.5kg of paraffin from 20oC to 30oC. Calculate the specific heat capacity of paraffin.
A
1466Jkg-1 oC-1
B
2933Jkg-1 oC-1
C
4400Jkg-1 oC-1
D
5866Jkg-1 oC-1
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
22,000 = 1.5 x c x (30 - 20)
c = 1466Jkg-1 oC-1
Source: Myschool Past Questions Archive (Record #28425)
Physics
UTME / JAMB 1987
Current Electricity & Electric Circuits
In which of the following are the substance arranged in descending order of their thermal conductivities?
A
copper, steel, glass
B
steel, copper, glass
C
steel, glass, copper
D
copper, glass, steel
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #28431)
Physics
UTME / JAMB 1987
Gas Laws & Heat Transfer
Solution
The vacuum in a thermos flask helps to reduce heat transfer by?
A
convection and radiation
B
convection and conduction
C
conduction and radiation
D
radiation only
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Conduction and convection rely on the presence of molecules to transfer heat. In a vacuum, however, there are no molecules available, making these processes impossible.
Source: Myschool Past Questions Archive (Record #28433)
Physics
UTME / JAMB 1987
Gas Laws & Heat Transfer
Which of the following phenomena explains the fact that a house whose roof is coated with white paint will be cooler in the hot season than one coated with black paint ?
A
conduction
B
convection
C
refraction
D
reflection
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #28434)
Physics
UTME / JAMB 1987
Current Electricity & Electric Circuits
Solution
A current of 0.5A flows through a resistor when connected to a 40-V battery. How much energy is dissipated in 2 minutes?
A
1200J
B
1500J
C
2400J
D
96000J
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Electrical energy = IVT
= 0.5 x 40 x 2 x 60
= 2400J
Source: Myschool Past Questions Archive (Record #28475)
Physics
UTME / JAMB 1987
Magnetism, Electromagnetism & AC Circuits
The direction of the magnetic field at a point in the vicinity of a bar magnet is
A
always towards the north pole of the magnet
B
always away from the south pole of the magnet
C
along the line joining the point to the neutral point
D
in the direction the north pole of a compass needle would point
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #28476)
Physics
UTME / JAMB 1987
Magnetism, Electromagnetism & AC Circuits
When two parallel wires carry currents in opposite directions, the force on either wire is
A
away from the other wire
B
zero, because the currents cancel each other
C
twice as much as when the currents are in the same direction
D
towards the other wire
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #28477)
Physics
UTME / JAMB 1987
Current Electricity & Electric Circuits
Solution
A galvanometer of resistance 20\(\Omega\) is to be provided with a shunt that \(\frac{1}{10}\) of the whole current in a circuit passes through the galvanometer. The resistance of the shunt is
A
2.00\(\Omega\)
B
2.22\(\Omega\)
C
18.00\(\Omega\)
D
18.22\(\Omega\)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Let the total current be \( I \). The galvanometer current is \( I_g = \frac{1}{10}I \), and the shunt current is \( I_s = I - I_g = \frac{9}{10}I \).
Since the galvanometer resistance \( R_g = 20 \Omega\) and shunt (resistance \( R_s \)) are in parallel, the voltage across them is equal:
\( I_g R_g = I_s R_s\)
\(\left( \frac{1}{10}I \right) \cdot 20 = \left( \frac{9}{10}I \right) \cdot R_s\)
Divide through by \( I \) (since \( I \neq 0 \)):
\(2 = \frac{9}{10} R_s\)
Solve for \( R_s \):
\(R_s = 2 \cdot \frac{10}{9} = \frac{20}{9}\) \(\approx\) 2.222\(\Omega\)
Source: Myschool Past Questions Archive (Record #28478)
Physics
UTME / JAMB 1987
Atomic, Quantum, Nuclear Physics & Electronics
What type of reaction is represented by the following scheme? \( ^2 _1X \) + \( ^{2} _{1}Y \) + \( ^{3} _{2}Z \) + \( ^{1} _{0}n \) + energy
A
Fusion reaction
B
Fission reaction
C
Chain reaction
D
Radioactive decay
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #28479)
Physics
UTME / JAMB 1987
Reflection, Refraction & Dispersion of Light
Solution
A ray of light is incident at an angle of 30o on one top surface of a parallel-sided glass block of refractive index 1.5. The ray finally emerges from the lower surface. What is the angular deviation of the emergent ray?
A
60o
B
39o
C
28o
D
0o
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
To find the angular deviation of a ray of light passing through a parallel-sided glass block, follow these steps:
Calculate the angle of refraction at the first surface: Using Snell's Law:
\(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\)
where:
\(n_1 = 1 \quad (\text{air}), \quad \theta_1 = 30^\circ, \quad n_2 = 1.5 \quad (\text{glass})\)
Substituting the values:
\(1 \cdot \sin(30^\circ) = 1.5 \cdot \sin(\theta_2) \Rightarrow 0.5 = 1.5 \cdot \sin(\theta_2) \Rightarrow \sin(\theta_2) = \frac{1}{3}\)
Thus,
\(\theta_2 \approx \arcsin\left(\frac{1}{3}\right) \approx 19.47^\circ\)
Angle of incidence at the second surface: Since the block is parallel-sided, the angle of incidence is: \(\theta_3 = \theta_2 \approx 19.47^\circ\)
Calculate the angle of refraction at the second surface: Applying Snell's Law again: \(n_2 \sin(\theta_3) = n_1 \sin(\theta_4)\)
Substituting values: \(1.5 \cdot \sin(19.47^\circ) = 1 \cdot \sin(\theta_4) \Rightarrow \sin(\theta_4) \approx 0.5 \Rightarrow \theta_4 = \arcsin(0.5) = 30^\circ\)
Calculate the angular deviation: The angular deviation \(D\) is given by: \(D = \theta_1 - \theta_4 = 30^\circ - 30^\circ = 0^\circ\)
The angular deviation of the emergent ray is \(0^\circ\).
Source: Myschool Past Questions Archive (Record #28463)
Physics
UTME / JAMB 1987
Lenses & Optical Instruments
Four lenses are being considered for use as a microscope objective. Which of the following focal lengths is most suitable?
A
-5mm
B
+5mm
C
-5cm
D
+5cm
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #28464)
Physics
UTME / JAMB 1987
Magnetism, Electromagnetism & AC Circuits
Solution
If the angle of declination in a place is 10oE, calculate the true geographic bearing if the compass reads N40oE
A
N50oE
B
N40oE
C
N30oE
D
N25oE
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
The true declination is calculated by adding the reading of compass bearing to the declination at that place.
= N40^\circE + 10^\circ = N50^\circE
Source: Myschool Past Questions Archive (Record #28466)
Physics
UTME / JAMB 1987
Electrostatics & Capacitors
Solution
Two capacitance of 6\(\mu\)F and 8\(\mu\)F are connected in series. What additional capacitance must be connected in series with this combination to give a total of 3\(\mu\)F
A
3\(\mu\)F
B
16\(\mu\)F
C
24\(\mu\)F
D
30\(\mu\)F
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
3 = \(\frac{6 \times 8 \times c}{(6 \times 8) + (6 \times c) + (8 \times c}\)
therefore, C = 24\(\mu\)F
Source: Myschool Past Questions Archive (Record #28472)
Physics
UTME / JAMB 1987
Current Electricity & Electric Circuits
Isotopes are nuclei which have
A
the same number of nuetrons and electrons
B
equal number of electrons and protons
C
the same atomic number but different number of neutrons
D
the same number of total particles
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #28481)
Physics
UTME / JAMB 1987
Atomic, Quantum, Nuclear Physics & Electronics
Solution
A radioactive sample initially contains N atoms. After three half-lives the number of atoms that have disintegrated is
A
\(\frac{N}{8}\)
B
\(\frac{3N}{8}\)
C
\(\frac{5N}{8}\)
D
\(\frac{7N}{8}\)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
N \(\to\) \(\frac{N}{2}\) \(\to\) \(\frac{N}{4}\) \(\to\) \(\frac{N}{8}\)
Number of atoms left = 1-1/8=\(\frac{7N}{8}\)
Source: Myschool Past Questions Archive (Record #28482)