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English Language
UTME / JAMB 2003
Lexis, Structure & Register (Cloze & Vocabulary)
In question number 16 above, choose the best option from letters A - D that best completes the gap.
A
advanced
B
analogus
C
discreet
D
polished
medium Difficulty
Verified Correct Answer: Option D
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #15125)
English Language
UTME / JAMB 2003
Lexis, Structure & Register (Cloze & Vocabulary)
In question number 17 above, choose the best option from letters A - D that best completes the gap.
A
vocal
B
bifocal
C
anticipatory
D
profuse
medium Difficulty
Verified Correct Answer: Option A
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #15126)
English Language
UTME / JAMB 2003
Lexis, Structure & Register (Cloze & Vocabulary)
In question number 18 above, choose the best option from letters A - D that best completes the gap.
A
decisive
B
positive
C
interactive
D
restrictive
medium Difficulty
Verified Correct Answer: Option B
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #15127)
English Language
UTME / JAMB 2003
Lexis, Structure & Register (Cloze & Vocabulary)
In question number 19 above, choose the best option from letters A - D that best completes the gap.
A
dialect
B
rhetoric
C
slang
D
language
medium Difficulty
Verified Correct Answer: Option D
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #15128)
English Language
UTME / JAMB 2003
Lexis, Structure & Register (Cloze & Vocabulary)
In question number 20 above, choose the best option from letters A - D that best completes the gap.
A
text
B
exchange
C
context
D
note
medium Difficulty
Verified Correct Answer: Option A
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #15129)
English Language
UTME / JAMB 2003
Reading Comprehension, Summary & Composition
Solution
Which of the following captures the writer's suggestion on how the Eagles could improve?
A
Good coaching, more strikers and more defenders
B
Better goalkeeping, a better attack and a stronger midfield
C
Fair officiating, good goalkeeping and fast players
D
Replacement of injured players, stronger attackers and a rugged defence
medium Difficulty
Verified Correct Answer: Option D
Grammar & Contextual Explanation
The writer clearly states that the defence needed to be strengthened to stop opponents from constantly threatening the goalkeeper. He also points out that the strikers performed poorly and needed to improve. In addition, the writer advises against using unfit players and recommends making better use of available reserves.
Source: Myschool Past Questions Archive (Record #15133)
English Language
UTME / JAMB 2003
Lexis, Structure & Register (Cloze & Vocabulary)
A suitable title for this passage is
A
The Eagles in World Cup 2002
B
The FIFA Korea/Japan 2002
C
The Nigeria and the Seneglese Teams
D
Africa Teams in the 2002 World Cup
medium Difficulty
Verified Correct Answer: Option B
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #15134)
English Language
UTME / JAMB 2003
Lexis, Structure & Register (Cloze & Vocabulary)
From the argument in the last paragraph, it can be concluded that the Eagles were
A
more timid and goal-shy than their opponents
B
not sure of getting to the next round of the tournament
C
not as strong as the Senegalese team
D
more experienced than the Senegalese team
medium Difficulty
Verified Correct Answer: Option D
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #15138)
English Language
UTME / JAMB 2003
Reading Comprehension, Summary & Composition
The word debutants, as used in the passage, means
A
first -timers
B
giant killers
C
hard-fighters
D
under-dogs
medium Difficulty
Verified Correct Answer: Option A
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #15140)
Physics
UTME / JAMB 2003
Newton's Laws, Momentum & Collisions
Solution
A bead traveling on a straight wire is brought to rest at 0.2m by friction. If the mass of the bead is 0.01kg and the coefficient of friction between the bead and the wire is 0.1 determine the work done by the friction
A
\( 2 \times 10^{-4}J \)
B
\( 2 \times 10^{-3}J \)
C
\( 2 \times 10^{1}J \)
D
\( 2 \times 10^{2}J \)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Work done by friction = Friction force x Displacement from the relation, f = uR
where u = coefficient of friction; R= Normal reaction
F = u mg (R =mg)
= 0.1 x 0.01 x 10
work done = F x Displacement
= 0.1 x 0.01 x 10 x 0.2
= 0.002 or \( 2 \times 10^{-3}J \)
Source: Myschool Past Questions Archive (Record #41452)
Physics
UTME / JAMB 2003
Reflection, Refraction & Dispersion of Light
Solution
A ray of light is incident on an equilateral triangular glass prism of refractive index 3/2, Calculate the angle through which the ray is minimally deviated in the prism
A
30.0\(^{\circ} \)
B
37.2\(^{\circ} \)
C
42.0\(^{\circ} \)
D
48.6\(^{\circ} \)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
For the equilateral glass prism, A = 60º
n = \(\frac{sin \frac{Dm + A}{2}}{sin \frac{A}{2}}\)
\(\frac{3}{2}\) = \(\frac{sin \frac{Dm + 60}{2}}{sin \frac{60}{2}}\)
= \(\frac{sin \frac{Dm + 60}{2}}{sin 30}\)
\(\frac{3}{2}\) x \(\frac{1}{2}\) = sin\(\frac{Dm + 60}{2}\)
\(\frac{3}{4}\) = 0.75 = sin\(\frac{Dm + 60}{2}\)
sin\(^{-1}\)(0.75) = \(\frac{Dm + 60}{2}\)
48.590 = \(\frac{Dm + 60}{2}\)
48.590 x 2 = Dm + 60
97.180 = Dm + 60
Dm = 97.180 - 60 = 37.18 ≈ 37 .2º
Source: Myschool Past Questions Archive (Record #41453)
Physics
UTME / JAMB 2003
Current Electricity & Electric Circuits
Solution
In the calibration of an ammeter using faraday's laws of electrolysis, the ammeter reading is kept constant at 1.2A. If 0.990g of copper is deposited in 40 minutes , the correction to be applied to the ammeter is
A
0.03A
B
0.04A
C
0.05A
D
0.06A
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
From faraday's law of electrolysis
M = I t Z
0.990 = I x (40 x 60) x 3.3 x 10-4
\( \implies \frac{0.990}{2400 \times 3.3 \times 10^{-4}} = 1.25A \\
\text{Thus the correction to be made } = 1.25 - 1.20 = 0.05A \)
Source: Myschool Past Questions Archive (Record #41454)
Physics
UTME / JAMB 2003
Atomic, Quantum, Nuclear Physics & Electronics
Solution
An electron makes a transition from a certain energy level Ek to the ground state E0. If the frequency of emission is 8.0 x 1014Hz The energy emitted is
A
8.25 x 10-19J
B
5.28 x 10-19J
C
5.28 x 1019J
D
8.25 x 1019J
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Atomic emission and absorption of energy is usually in packets referred to as photon or quantum of energy of magnitude E =hf, where f is the frequency and h is the plancks constant
Energy emitted = hf
= 6.6 x 10-34 x 8.0 x 10-14
= 5.28 x 10-19J
Source: Myschool Past Questions Archive (Record #41455)
Physics
UTME / JAMB 2003
Newton's Laws, Momentum & Collisions
Solution
A force of 100N was used to kick a football of mass 0.8kg. Find the velocity with which the ball moves if it takes 0.8s to be kicked
A
\(32ms^{-1} \)
B
\(50ms^{-1} \)
C
\(64ms^{-1} \)
D
\(100ms^{-1} \)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
From Newton's second law:
\( F = \frac{mV - mµ}{t} \)
Let the final velocity = V; and since the football is initially at rest, its initial vel µ = 0, M = 0.8kg
t = 0.8s
\(\implies 100 = \frac{0.8V - 0.8 \times 0}{0.8}\)
\(\text{Therefore } V = \frac{100 \times 0.8}{0.8} = 100ms^{-1}\)
Source: Myschool Past Questions Archive (Record #41456)
Physics
UTME / JAMB 2003
Elasticity, Density, Pressure & Fluids
Solution
Given that Young's modulus for aluminium is 7.0 x 10\(^{10}Nm^{-2}\) and density is 2.7 x 10\(^3\)kgm\(^{-3}\), find the speed of the sound produced if a solid bar is struck at one end with a hammer
A
5.1 x 103ms-1
B
4.2 x 103ms-1
C
3.6 x 103ms-1
D
2.8 x 103ms-1
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
The speed of sound in a material is given as:
V = \(\sqrt{\frac{E}{P}}\), Where
E = Young's modulus
P = Density
Therefore, V = \(\sqrt{\frac{7.0 \times 10^{10}}{2.7 \times 10^3}}\) = 5091.75
= 5.1 \(\times 10^3ms^{-1} \)
Source: Myschool Past Questions Archive (Record #41457)