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Physics
UTME / JAMB 1998
Reflection, Refraction & Dispersion of Light
Solution
The displacement d produced in a glass block of thickness t and refractive index n when an object is viewed through it is
A
t – n
B
t (1 + 1/n)
C
t (1-1/n)
D
t (1/n - 1)
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
n = \(\frac {\text{real depth}}{\text {apparent depth}{}}\)
n = \(\frac {t}{t-d}\)
\(\frac{t - d}{1}\) = \(\frac{t}{n}\)
-d = \(\frac{t}{n}\) - t
d = - \(\frac{t}{n}\) + t
d = t - \(\frac{t}{n}\)
d =t (1 - \(\frac{1}{n}\))
Source: Myschool Past Questions Archive (Record #27653)
Physics
UTME / JAMB 1998
Lenses & Optical Instruments
Solution
A man wears convex lens glasses of focal length 30cm in order to correct his eye defect. instead of the optimum 25cm, his least distance of distinct vision is
A
14cm
B
28cm
C
75cm
D
150cm
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
F = \(\frac{uv}{u + v}\)
F = +ve
v = -ve
30 = \(\frac {25 {\times} (-v)}{25 + (-v)}\)
-25v = 750 - 30v
5v = 750
v = 150cm
(Note that either negative or positive values go to show the positioning and hence, they are not a strong necessity in your final answer.)
Source: Myschool Past Questions Archive (Record #27655)
Physics
UTME / JAMB 1998
Gas Laws & Heat Transfer
Which of the following are true about the infra-red radiation? i. it is invisible ii. it is called heat ray iii. its frequency is higher than that of blue light iv. it travels as a transverse wave
A
i, ii, iii and iv
B
i, ii, iv only
C
i, iii, iv only
D
ii, iii, iv only
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 #27658)
Physics
UTME / JAMB 1998
Current Electricity & Electric Circuits
Solution
The force of repulsion between two point positive charges 5\(\mu\)C and 8\(\mu\)C separated at a distance of 0.02m apart is. [\(\frac{1}{4{\pi}{\varepsilon}_o}\) = 9 x 10\(^9\)Nm\(^2\)C\(^{-2}\)]
A
1.8 x 10-10N
B
9.0 x 10-8N
C
9.0 x 102N
D
4.5 x 103N
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
F =\(\frac{1}{4{\pi}{\varepsilon}_o}\) x \(\frac {q_1q_2}{r^2}\)
F =\(\frac {5 {\times} 10^{-6} {\times} 8 {\times} 10^{-6}}{0.002 {\times} 0.002}\) x 9 x 10\(^9\)
F = 900N
Source: Myschool Past Questions Archive (Record #27660)
Physics
UTME / JAMB 1998
Current Electricity & Electric Circuits
An ebonite rod rubbed with fur attracts a glass rod rubbed with silk because
A
ebonite has a negative charge while glass has a positive charge
B
ebonite has a positive charge while glass has a negative charge
C
both have negative charges
D
both have positive charges
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 #27661)
Physics
UTME / JAMB 1998
Current Electricity & Electric Circuits
Which of the following can be used to reduce local action in a Laclanche cell?
A
A carbon rod as the positive pole
B
Pure zinc as the negative pole
C
Potassium permanganate solution in contact with the positive pole
D
common salt solution
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 #27662)
Physics
UTME / JAMB 1998
Work, Energy, Power & Simple Machines
Solution
Find the work done in moving a 2 C charge between two points X and Y in an electric field if the potential difference is 100 volts
A
50J
B
100J
C
200J
D
400J
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
V = \(\frac{W}{Q}\)
W = VQ
W = 100 x 2
W = 200J
Source: Myschool Past Questions Archive (Record #27663)
Physics
UTME / JAMB 1998
Electrostatics & Capacitors
Solution
A parallel plate capacitor has a common plate area of 5 x 10\(^{-8}\)m2 and plate separation of 2 x 10\(^{-3}\)m, Assuming free space,
What is the capacitance? [\(\varepsilon\)o = 9.0 X 10-\(^{-12}\)C2N-1m-2]
A
2.25 x 10-17F
B
4.50 x 10-17F
C
2.25 x 10-16F
D
4.50 x 10-16F
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
C = \(\frac{{\varepsilon}_oA}{d}\)
= \(\frac{5{\times} 10^{-8} {\times} 10^{-12}{\times} 9}{2 {\times}10^{-3}}\)
= 2.25 x \(10^{-16}\)F
Source: Myschool Past Questions Archive (Record #27664)
Physics
UTME / JAMB 1998
Current Electricity & Electric Circuits
The iron core of an induction coil is made from bundles of wire so as to
A
minimize eddy-currents
B
generate eddy-current
C
prevent sparking at the contact breaker
D
get the greatest posible secondary voltage
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 #27665)
Physics
UTME / JAMB 1998
Current Electricity & Electric Circuits
Solution
Electricity is supplied to a school along a cable of total resistance 0.5\(\Omega\) with the maximum current drawn from the mains as 100A. The maximum energy dissipated as heat for 1 hr is
A
3.6 x 103J
B
5.0 x 103J
C
3.0 x 105J
D
1.8 x 107J
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Elect. energy = I2Rt
= 100 x 100 x 0.5 x 3600
= 1.8 x 107J
Source: Myschool Past Questions Archive (Record #27666)
Physics
UTME / JAMB 1998
Magnetism, Electromagnetism & AC Circuits
Solution
A transformer which can produce 8V from a 240v a.c. supply, has an efficiency of 80%. If the current in the secondary coil is 15A, calculate the current in the primary coil.
A
0.625A
B
1.600A
C
2.500A
D
6.250A
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Efficiency = \(\frac{I_sV_s}{I_pV_p}\) x 100
Source: Myschool Past Questions Archive (Record #27667)
Physics
UTME / JAMB 1998
Current Electricity & Electric Circuits
Solution
The electrochemical equivalent of a metal is 1.3 x 10-7kgC-1. The mass of the metal which 2.0 x 104C of electricity will deposit from a suitable electrolyte is
A
6.5 x 10-2kg
B
2.6 x 10-2kg
C
6.5 x 10-3kg
D
2.6 x 10-3kg
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
m = QZ
= 2 x 104 x 1.3 x 10-7
= 2.6 x 10-3kg
Source: Myschool Past Questions Archive (Record #27672)
Physics
UTME / JAMB 1998
Atomic, Quantum, Nuclear Physics & Electronics
Solution
In a nuclear fusion experiment, the loss of mass amounts to 1.0 x 10-6kg. The amount of energy obtained from the fusion is
A
3.0 x 10-4J
B
3.0 x 10-1J
C
9.0 x 104J
D
9.0 x 1010J
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Energy = mc2
= 1 x (3 x 108)2 x 106
= 9 x 1010J
Source: Myschool Past Questions Archive (Record #27675)
Physics
UTME / JAMB 1998
Current Electricity & Electric Circuits
Diagram
Solution
The total resistance measured at PQ in the diagram above is
Attached Figure / Diagram
A
18.0Ω
B
11.0Ω
C
4.0Ω
D
2.0Ω
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
The three 2 ohms are in series, therefore the total resistance is
\(R=R_1+R_2+R_3\)
\(R_ {total} =2+2+2= 6\) ohms.
resketching circuit;
6 ohms is used to replace the three 2 ohms which is therefore placed by the 3 ohms resistor.
The 3 ohms and 6 ohms resistors are in parallel, so the combined resistance is
\(R=\frac{R_1xR_2}{R_1+R_2}\)
\(R= \frac{3\times6}{3+6}\)
\(R=\frac{18}{9}=2\) ohms.
Resketching circuit:
The 2 ohms is used to replace the 3 ohms and 6 ohms laying side-by-side.
Therefore, the 3 resistors (1 ohm, 1 ohm, and 2 ohms) are in series.
Thus, the total resistance is
\(R= (R_1+R_2+R_3)\)
\(R_{total} =1+1+2=4\) ohms.
Therefore, the total resistance measured at PQ is 4 ohms.
Source: Myschool Past Questions Archive (Record #41744)
Physics
UTME / JAMB 1998
Magnetism, Electromagnetism & AC Circuits
Diagram
Solution
In the AC circuit diagram above, the resonance frequency is
Attached Figure / Diagram
A
5000π Hz
B
2500π Hz
C
5000/π Hz
D
2500/π Hz
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
resonance frequency \(f_r\) = \(\frac{ 1}{ 2\pi \sqrt{LC}}\)
\(f_r\) = \(\frac{1}{2\pi \sqrt{5\times10^{-6} \times 8\times 10^{-3}}}\)
\(f_r\) = \(\frac{1}{ \pi 0.0004}\)
\(f_r\) = \(\frac{ 2500}{\pi}\)
Source: Myschool Past Questions Archive (Record #41745)