Question Library
Browse, filter, and study verified past examination questions with step-by-step mathematical reasoning.
Searching 18,000+ Questions...
Sifting topics, KaTeX formulas & verified solutions
Showing 1951–1965 of 5315 verified questions
Sort by:
Physics
WAEC / WASSCE 2011
Heat Energy & Temperature
Uranium-234 disintegrated to form Thorium-230 by emitting
A
alpha particles
B
gamma rays
C
beta negative rays
D
beta positive rays
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 #35499)
Physics
WAEC / WASSCE 2011
Mechanics and Properties of Matter
Solution
The change in mass of fuel during a nuclear reaction is 2.0 x 10-27 kg. Calculate the amount of energy released. (Speed of light = 2.0 x 108ms-1)
A
1.8 x 10-10J
B
9.0 x 10-11J
C
6.0 x 10-19J
D
1.2 x 10-45J
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Q = \(\Delta MC^2\)
= 2 x 10-27(3 x 10)2
= 1.8 x 10-10J
Source: Myschool Past Questions Archive (Record #35500)
Physics
WAEC / WASSCE 2011
Heat Energy & Temperature
Nuclear fusion is not used as a source of enegy because
A
very high temperature is needed for the reaction
B
the fuel for the reaction is not easy to obtain
C
less energy is released
D
the reaction is too slow
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 #35501)
Physics
WAEC / WASSCE 2011
Electricity and Magnetism
Diagram
Solution
Using the circuit diagram above. Calculate the total charge in the circuit
Attached Figure / Diagram
A
80\(\mu C\)
B
35\(\mu C\)
C
20\(\mu C\)
D
16\(\mu C\)
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
C - \(C_1 + C_2 = 2 + 6 = \mu F\)
\(\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{8} + \frac{1}{2}\)
C = \(\frac{8}{5 \mu F}\)
Q = CV
= \(\frac{8}{5} \times 10 = 16\mu C\)
Source: Myschool Past Questions Archive (Record #35503)
Physics
WAEC / WASSCE 2011
Modern Physics and Nuclear Physics
Diagram
Solution
Using the circuit diagram above. Calculate the potential difference across the capacitors in parallel
Attached Figure / Diagram
A
10.0V
B
8.0V
C
6.7V
D
2.0V
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
V = \(\frac{Q}{V} = \frac{16}{8} = 2\Gamma\)
Source: Myschool Past Questions Archive (Record #35511)
Physics
UTME / JAMB 2011
Mechanics and Properties of Matter
Solution
If the volume of a gas increases steadily as the temperature decreases at constant pressure, the gas obeys
A
Graham's Law
B
Boyle's Law
C
Pressure law
D
Charles Law
E
No Correct Option
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
Rate of diffusion is inversely proportional to the square root of density - Graham's law
\( P \propto \frac{1}{v} \) at constant temperature - Boyle's law
\( P \propto T \) at constant volume - Pressure law
\( P \propto T \) at constant pressure - Charles law
therefore no correct option
Source: Myschool Past Questions Archive (Record #41404)
Physics
UTME / JAMB 2011
Electricity and Magnetism
Solution
In Homes, electrical appliances and lamps are connected in parallel because
A
Parallel connections do not heat up the wires
B
independence switches can be used
C
less current will be used
D
less voltage will be used
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
if appliances are connected in series, the effective voltage is the sum of individual voltages which will be high.
Source: Myschool Past Questions Archive (Record #41406)
Physics
UTME / JAMB 2011
Electricity and Magnetism
Solution
In an A.C circuit, the ratio of r.m.s value to peak value of current is
A
2
B
\( \frac{1}{2} \)
C
\( \frac{1}{\sqrt{2}} \)
D
\( \sqrt{2} \)
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Irms = \( \frac{I_0}{\sqrt{2}} \)
\( \text{Therefore } \sqrt{2Irms} = I_0 \\
\frac{Irms}{I_0} = \frac{1}{\sqrt{2}} \)
Source: Myschool Past Questions Archive (Record #41407)
Physics
UTME / JAMB 2011
Mechanics and Properties of Matter
Solution
Two resistors 5\Omega and 10\Omega are arranged first in series and later in parallel to a 24V source, the ratio of total power dissipated in the series and parallel arrangements respectively is
A
5:3
B
1:50
C
50:1
D
3:5
E
No Correct Option
medium Difficulty
Verified Correct Answer: Option E
Physical Principles & Numerical Solution
Power in Series Arrangement
Total Resistance (R_series): R\(_{series}\) = R\(_1\) + R\(_2\) = 5\Omega + 10\Omega = 15\Omega
Total Power (P_series): P = \(\frac{V^2}{\text{R}}\) = \(\frac{24^2}{15}\) = 38.4W
Power in parallel Arrangement
Total Resistance(R\(_{parallel}\)): R\(_{parallel}\) = \(\frac{R_1 \times R_2}{R_1 + R_2}\) = \(\frac{5 \times 10}{5 + 10}\) = 3.33
Total Power(P_{parallel}\): P = \(\frac{V^2}{\text{R}}\) = \(\frac{24^2}{3.33}\) = 172.8W
Ratio of Powers
Power ratio = \(\frac{\text{Pseries}}{\text{Pparallel}}\) = \(\frac{38.4}{172.8}\) = \(\frac{1}{4.5}\) = \(\frac{2}{9}\)
Therefore, there is no correct answer.
Source: Myschool Past Questions Archive (Record #41409)
Physics
WAEC / WASSCE 2011
Heat Energy & Temperature
Solution
Use the diagram above as a guide to carry out the following experiment.
Trace the outline ABCD of the rectangular glass prism on the drawing paper provided.
Remove the prism. Select a point N on AB such that AN is about one-quarter of AB.
Draw the normal LNM. Also, draw a line RN to make an angle \(\theta\) = 75^\circ with AB at N.
Fix two pins at \(P_1\) and \(P_2\) on line RN. Replace the prism on its outline.
Fix two other pins at \(P_3\) and \(P_4\) such that they appear to be in a straight line with the images of the pins at \(P_1\) and \(P_2\) when viewed through the prism from DC.
(vi) Remove the prism and the pins at \(P_3\) and \(P_4\). Draw a line to join \(P_3\) and \(P_4\)
Produce line \(P_4\) \(P_3\) to meet the line DC at O. Draw a line to join NO.
Measure and record the values of MO and NO.
Evaluate \(\theta\) = \(\frac{MO}{NO}\) and cos \(\theta\).
Repeat the procedure for four other values of \(\theta\) = 65^\circ 55^\circ, 45^\circ, and 35^\circ. In each case, evaluate \(\theta\) and cos \(\theta\).
Tabulate your readings.
Plot a graph with cos \(\theta\) on the vertical axis and \(\theta\) on the horizontal axis.
Determine the slope, s, of the graph.
State two precautions taken to ensure accurate results
(b)i. State Snell's law of refraction.
ii. Calculate the critical angle for the glass prism used in the experiment above if its refractive index is 1.5.
A
10.0V
B
8.0V
C
6.7V
D
2.0V
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Observation/table of values
S/N
\(\theta\)(0)
Mo(cm)
No(cm)
\(\theta\)=\(\frac{MO}{NO}\)
Cos \(\theta\)
1
75.0
1.60
6.60
0.100
0.260
2
65.0
3.00
7.00
0.290
0.420
3
55.0
3.10
7.30
0.390
0.570
4
45.0
3.10
7.60
0.480
0.700
5
35.0
4.50
8.00
0.520
0.820
Slope (s) = \(\frac{\bigtriangleup \text {cos} \theta}{\bigtriangleup \theta} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
\(\frac{MO}{NO} = \frac{0.77-0.20}{0.50-0.17}\)
\(\frac{0.57}{0.33}\)
Slope (s) = 1.72
Graph cos \(\theta\) against = \(\phi\) = \(\frac{MO}{NO}\)
see the graph above.
PRECAUTIONS; i
- Ensured pins are vertical/ erect.
- Ensured neat traces/sharp pencil (shown on trace)
- Ensured reasonable spacing of pins (about 4cm apart)
- Avoided parallax error in reading protractor/metre rule.
(b)i. Snell's law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media.
Or \(\frac {\text {Sin i}}{\text {Sin r}}\) = a constant for a given pair of media
Where i = angle of incidence;
r = angle of refraction
ii. n = \(\frac{i}{\text{Sin c}}\)
Sin c = \(\frac{1}{1.5}\) = 0.6667
C = sin\(^{-1}\) 0.6667
C = 41.80^\circ
Source: Official Exam / Past Question Archive
Physics
WAEC / WASSCE 2011
Heat Energy & Temperature
Solution
A stone projected horizontally from the top of a tower with a speed of 4 ms\(^{-1}\) lands on the level ground at a horizontal distance of 25 m from the foot of the tower. Calculate the height of the tower. = 10 ms\(^{-2}\)]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
R = U\(\sqrt{\frac{2h}{g}}\)
h = \(\frac{R^2g}{2u^2}\)
= \(\frac{25^2 \times 10}{2 \times 4^2}\)
= 195.3m
Source: Myschool Past Questions Archive (Record #53322)
Physics
WAEC / WASSCE 2011
Mechanics and Properties of Matter
Solution
(a) Explain diffusion.
(b) Give one reason why the rate of diffusion is higher in gases than in liquids at the same temperature.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Diffusion is a process whereby fluids mix intimately with one another due to random molecular motion.
(b) Diffusion rate in gases is higher than that in liquids at the same temperature because molecular motion in gases are faster, the intermolecular forces is less and the density is also less than those in liquids.
Source: Myschool Past Questions Archive (Record #53323)
Physics
WAEC / WASSCE 2011
Mechanics and Properties of Matter
Solution
State: (a) the difference between plane polarized light and ordinary light;
(b) two uses of polaroids.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) In plane polarized light the vibrations of light waves are perpendicular in one plane only while in ordinary light the vibrations of the particles are in all planes.
(b) Two uses of polaroids are:
- in sunglass to reduce sunlight intensity
- to eliminate reflected light or glare
- in the measurement of sugar solution concentration or saccharimetry; and
- in photo elasticity or stress analysis.
Source: Myschool Past Questions Archive (Record #53324)
Physics
WAEC / WASSCE 2011
Heat Energy & Temperature
Solution
(a). State: (a) two applications of electrolysis in an industry
(b) one application of electrolysis in a school laboratory.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
a) Industrial application of electrolysis are
- purification of some metals
- electroplating
- production of gases, e.g. Hydrogen from water
- extraction of some metals from their ores, e.g. Aluminium from bauxite.
(b) Applications of electrolysis in school laboratory
- electrolysis of acidified water
- calibration of ammeter.
Source: Myschool Past Questions Archive (Record #53325)
Physics
WAEC / WASSCE 2011
Mechanics and Properties of Matter
Solution
In fig. la and fig.1 b above, Ida and Wb represent the respective loads on a spring placedtiear a 30 cm rule, when in air and when in water:
(a) Identify the force causing a shrink in the spring in fig.(b).
(b) Given that the force constant of the spring is 2.0 x 10\(^{11}\) Nm\(^{-1}\), calculate the work done by the force in causing the shrink.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) The force causing a shrink is the upthrust of the water.
(b) Work done = \(\frac{1}{2}ke^2\)
= \(\frac{1}{2}(2 \times 10^{11}) \times (2 \times 10^{-2})^2\) = 4.0 x 10\(^7\)
Source: Myschool Past Questions Archive (Record #53328)