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Physics
WAEC / WASSCE 2007
Current Electricity & Electric Circuits
Diagram
Solution
The diagram above illustrates three forces acting on an object at point O. If the object is in equilibrium, determine the magnitude of the force P.
Attached Figure / Diagram
A
10.5N
B
11.0N
C
17.3N
D
20.0N
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
10 - P sin 30 = 0
10 - P x 0.5 = 0
P = 20N
Source: Myschool Past Questions Archive (Record #34975)
Physics
WAEC / WASSCE 2007
Current Electricity & Electric Circuits
Diagram
Solution
Calculate the effective resistance between P and Q in the diagram above
Attached Figure / Diagram
A
0.75\(\Omega\)
B
4.00\(\Omega\)
C
5.00\(\Omega\)
D
7.00\(\Omega\)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
R = \(\frac{1}{2} + \frac{1}{2} + 3 = 1 + 3 = 4\Omega\)
Source: Myschool Past Questions Archive (Record #34977)
Physics
WAEC / WASSCE 2007
Current Electricity & Electric Circuits
Solution
You have been provided with an accumulator E, a standard resistor Rx, two resistance boxes R\(B_1\) and R\(B_2\), two keys \(K_1\) and \(K_2\) and other necessary apparatus.
Measure and record the e.m.f of the accumulator.
Connect a circuit as shown above.
Set the resistance R, in the resistance box such that R in R\(B_1\) = R in R\(B_2\) = 1\(\Omega\).
With \(K_1\) open and \(K_2\) closed, measure and record the potential difference V across the standard resistor Rx. (v) Close K, and K,. Read and record the potential difference \(V_o\) across Rx.
Evaluate V\(_{1}^{-1}\)
Repeat procedure (v) for four other values of R = 2, 3, 4 and 5\(\Omega\) respectively. In each case, ensure that the value of R in R\(B_1\) is equal to the value of R in R\(B_2\).
Evaluate V\(_{1}^{-1}\) in each case. Tabulate your readings.
Plot a graph of V\(_{1}^{-1}\) on the vertical axis against R on the horizontal axis starting both axes from the origin (0,0).
Determine the slope, s, of the graph and the intercept on the vertical axis.
Evaluate y = \(\frac{1}{s}\)
State two precautions taken to ensure accurate results
(b)i. Explain what is meant by the potential difference between two points in an electric circuit
ii. A cell has an e.m.f. of 3 V. When it is connected across a resistor of resistance 4\(\Omega\), a current 0.5A passes through the circuit. Calculate the internal resistance of the cell.
A
0.75\(\Omega\)
B
4.00\(\Omega\)
C
5.00\(\Omega\)
D
7.00\(\Omega\)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Table of values (Observation)
R (\(\Omega\)
\(V_1\)(v)
V\(_{1}^{-1}\)
1
1.50
0.667
2
1.30
0.769
3
1.15
0.870
4
1.10
0.909
5
0.90
1.111
Slope (s) = \(\frac{\bigtriangleup {V}_{1}^{-1}}{DR} = \frac{1.22 - 0.6}{6.5 - 0.3}\)
= \(\frac{0.62}{6.20}\) = 0.1
Intercept (c) = 0.57
Evaluation: y = \(\frac{1}{S} = \frac{1}{0.1}\) = 10
Precautions:
I removed the key when not taking readings
I ensured tight connections/clean terminals
I avoided parallax error in reading the voltmeter
I corrected for zero error on the voltmeter
(b)i. The potential difference (pd) (in volts) between two points in a circuit is the work done (in joules) when one unit of positive charge (coulomb) moves from one point to the other
ii. E = IR + Ir, 3 =0.5 x 4 + 0.5r, 3 = 2 +0.5r
r = \(\frac{3 - 2}{0.5}\) = 2\(\Omega\)
Source: Official Exam / Past Question Archive
Physics
UTME / JAMB 2007
Work, Energy, Power & Simple Machines
Solution
A particle of weight 120N is placed on a plane inclined at 300 to the horizontal. If the plane has an efficiency of 60% to the floor what is the force required to push the weight uniformly up the plane?
A
50N
B
100N
C
175N
D
210N
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Efficiency = \( \frac{W(120N)INPUT}{WORK OUTPUT} \times \frac{100}{1}\)
And for incline plane
V.R =\( \frac{1}{Sin\theta}\)
= V.R = \(\frac{1}{Sin30} =\frac{1}{0.5}\) = 2
\(\text{therefore } \frac{60}{120} = \frac{M.A}{2}\)
M.A = \(\frac{120}{100}\) = 1.2
\(\text{but M.A } = \frac{load}{effort} = \frac{120}{\text{E}}\)
\(\text{therefore} \frac{120}{\text{E}}\) = 1.2
E = \(\frac{120}{1.2 }\) = 100N
Therefore, Effort up the plane = 100N
Source: Myschool Past Questions Archive (Record #41373)
Physics
UTME / JAMB 2007
Elasticity, Density, Pressure & Fluids
Solution
A string of length 4m is extended by 0.02m when a load of 0.4kg is suspended at its end. What will be the length of the string when the applied force is 15N?
A
4.05m
B
4.08m
C
5.05m
D
6.08m
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
from hooke's law F=Ke
\(
\frac{F_1}{e_1} - \frac{F_2}{e_2} \text{therefore}
\frac{4}{0.02} = \frac{15}{e_2} \)
e1 = \( \frac{0.02 x 15}{4} = 0.075M \)
The new length of the string is 4 + 0.075
= 4.075
= 4.08
Source: Myschool Past Questions Archive (Record #41376)
Physics
UTME / JAMB 2007
Gas Laws & Heat Transfer
Solution
Which of the following gas laws is equivalent to the work done?
A
Boyle' law
B
Charles law
C
Pressure law
D
van der Waal's law
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Mathematically Boyle's law = PV = K
Where P = Pressure
and V = Volume
By definition Pressure = \( \frac{F}{A} = \frac{F}{M \times M} \)
Again Workdone = force x displacement
therefore \( PV = \frac{F}{A} \times V = \frac{F}{M^2} \times \frac{M^3}{1} \\
PV = \frac{F}{x\times x} \times \frac{x \times x \times x}{1} = F \times x \)
= force x displacement
Source: Myschool Past Questions Archive (Record #41379)
Physics
UTME / JAMB 2007
Current Electricity & Electric Circuits
Solution
The difference observed in solids, liquids and gases can be accounted for by
A
The different molecules in each of them
B
The spaces and forces acting between the molecules
C
Thier melting points
D
Their relative massess
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Spaces and Forces: The differences observed in solids, liquids, and gases can primarily be attributed to the arrangement, spacing, and intermolecular forces acting between the molecules.
Solids: Tightly packed molecules with strong intermolecular forces, giving a definite shape and volume.
Liquids: Close molecules that can move past each other, resulting in a definite volume but no definite shape.
Gases: Widely spaced molecules with weak intermolecular forces, filling the entire container.
Source: Myschool Past Questions Archive (Record #41383)
Physics
UTME / JAMB 2007
Work, Energy, Power & Simple Machines
Solution
If two charged plates are maintained at a potential difference of 3 kv, the work done in taking a charge of 600µc across the field is
A
0.8j
B
1.8j
C
9.0j
D
18.0j
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Workdone = product of the charge Q, and the p.d V.
i.e Work done = QV
= 600 * 10-6 *2.1 * 103
=1.8j
Source: Myschool Past Questions Archive (Record #41386)
Physics
UTME / JAMB 2007
Current Electricity & Electric Circuits
Solution
A conductor has a diameter of 1.00mm and length of 2.00m. If the resistance of the material is 0.1\Omega, the resistivity is
A
2.55 * 105 Ωm
B
2.55 * 103 Ωm
C
3.93 * 10-6 Ωm
D
3.93 * 10-8 Ωm
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
\( R = \frac{\rho L}{A}\) \(\implies \rho\) = \(\frac{R.A}{L}\) = \(\frac{0.1 \times 3.14}{2} \left ( \frac{10^{-3}}{2} \right)^2\)
\(\therefore\) \(\rho\) = \(\frac{0.1 \times 3.14 \times 10^{-6}}{2 \times 4}\)
= 0.3925 x 10\(^{-7}\) OR 3.930 x 10 \(^{-8}\)\Omegam
Source: Myschool Past Questions Archive (Record #41387)
Physics
UTME / JAMB 2007
Current Electricity & Electric Circuits
Solution
A 40KW electric cable was used to transmit electricity through a resistor of resistance 2.00\Omega at 800V. The power loss as internal energy is
A
4.0 x 102W
B
5.0 x 102W
C
4.0 x 103W
D
5.0 x 103W
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
In general, Power = IV; \( \implies 40KW = IV \\
\text{Therefore } 40000 = 1 \times 800\\
\implies I = \frac{40000}{800} = 50A,\text{tune the current through} \)
Resistor = 50A
power loss= I2R = 502 x 2
= 2500 x 2 = 5.0 x 103W
Source: Myschool Past Questions Archive (Record #41388)
Physics
UTME / JAMB 2007
Magnetism, Electromagnetism & AC Circuits
Solution
The instrument that measures both ac and dc is called
A
A current balance
B
a moving coil ammenter
C
a moving iron ammeter
D
an inverter
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
A moving iron ammeter can measure both alternating current (AC) and direct current (DC) due to its design, which allows it to respond to the magnetic field created by both types of current.
Source: Myschool Past Questions Archive (Record #41389)
Physics
UTME / JAMB 2007
Current Electricity & Electric Circuits
Solution
Two long parallel wires X and Y carry currents 3A and 5A each. if the force experienced by unit length is 5 x 10-5N. the force per unit length experienced by wire Y is
A
3 x 10-5
B
3 x 10-6
C
5 x 105
D
5 x 10-5
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
since the wires are parallel to each other, the forces which they exert on each other are equal and opposite, since they must agree with newton's law of action an reaction. Hence the force exerted by the wire is equally
5 x 10-5
Source: Myschool Past Questions Archive (Record #41392)
Physics
UTME / JAMB 2007
Magnetism, Electromagnetism & AC Circuits
Solution
Lenz's law is a law of conservation of
A
electric charge
B
electric current
C
Energy
D
Momentum
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Lenz's lew of electrical magnetic induction is usefully referred to as the law of conservation of energy
Source: Myschool Past Questions Archive (Record #41393)
Physics
UTME / JAMB 2007
Current Electricity & Electric Circuits
Solution
A 120V, 60W lamp is to be operated on 220V ac supply mains. calculate the value of non inductive resistance that would be required to ensure that the lamp is run on correct value
A
100Ω
B
200Ω
C
300Ω
D
500Ω
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Power of the lamp = 60W;
\(I = {\frac{P}{V}}\)
\(I = {\frac{60}{120}}\) = \(0.5A\)
\(R = {\frac{V}{I}}\)
\(R_mains\) = \(\frac{220}{0.5}\) = \(440\Omega\)
\(R_lamp\) = \(\frac{120}{0.5}\) = \(240\Omega\)
\(\therefore\) The non-inductive resistance to keep the lamp = \((440 - 240)\Omega\)
= \(200\Omega\)
Source: Myschool Past Questions Archive (Record #41395)
Physics
UTME / JAMB 2007
Atomic, Quantum, Nuclear Physics & Electronics
Solution
A radioactive substance has a half life of 20 days. What fraction of the original radioactive nuclei will remain after 80 days?
A
\( \frac{1}{32} \)
B
\( \frac{1}{16} \)
C
\( \frac{1}{8} \)
D
\( \frac{1}{4} \)
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
80 days = 4 half lives (i.e 4 x 20dqs = 80dq)
But the fraction always left by any radioactive material after 4 half lives is
\( \left ( \frac{1}{2} \right)^4 = \frac{1}{16} \)
Source: Myschool Past Questions Archive (Record #41396)