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Physics
WAEC / WASSCE 2011
Current Electricity & Electric Circuits
Solution
(a) Explain mutual induction
(b) State four use of electromagnets
The diagram above illustrates two coils X an arranged so that their axes are collinear. X is connected to an a.c. supply and has an amme in series with it while Y is connected to a lamp Explain the following observations. The (i) lamp is lit when the a.c. supply is switched on;
(ii) brightness of the light from the lamp increases when distance between X. and Y is decreased;
(iii) filament of the lamp glows brighter whet bundle of insulatttriron wires is placed along ccmmon axis of the coils.
(d) Two cells, one have an emf of 2.0 V and an internal resistance of 0.4 and the other having an emf of 2.0 V and an internal resistance of 0.1 \(\Omega\), are connected in parallel. The combination is then connected in series with a resistor.
(i) Draw a circuit diagram of the arrangement.
(ii) Calculate the current through 5\(\Omega\) resistor.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Mutual Induction: Mutual Induction is a phenomenon which occurs when the current in a coil is changing, an e.m.f will be induced in a tear-by circuit due to some of the magnetic flux produced by the first circuit linking the second circuit.
(b) The uses of electromagnets are in:
- electric bells or telephone earpieces
- generators or electric motors
- industrial robots
- separation of iron scraps from debris
- electromagnetic switches or magnetic relays
- cranes fbr lifting or transporting heavy magnetic materials
- magnetic resonance scanners, i.e. in medical diagnosis.
(i) The changing magnetic flux in X induces changing magnetic flux in Y which induces the current that light up the bulb.
(ii) As the distance between X and Y decreases, the magnetic flux density linking coil Y increases, hence the induced e.m.f. increase hence the brightners of the lamp.
(iii) The insulated soft iron wires increases the magnetic flux through coil Y, hence increase in the induced current in Y, therefore, the filament of the lamp glows brighter.
(d)
Total e,m,f. E = 2V
Total internal resistance r
\(\frac{1}{r} + \frac{1}{r_2} = \frac{1}{0.4} + \frac{1}{0.1}\)
e = 0.08\(\Omega\)
E = I(R + r)
2 = I(5 + 0.08)
I = \(\frac{2}{5.08}\) = 0.394A
Source: Myschool Past Questions Archive (Record #53370)
Physics
WAEC / WASSCE 2011
Atomic, Quantum, Nuclear Physics & Electronics
Solution
(a) Explain the terms:
(i) transmutation as it relates to radioactivity; (ii) stopping potential.
(b) \(^{23}_{11}A + ^2_1B\) —> \(^p_qC\) + proton
\(^p_qC\) —> \(^r_sE\) + beta
A nucleus C, formed artificially from A and B radioactive and quickly decays to another nucli E as indicated in the nuclear equations abc Datermine the values of p, q, r and s.
(c) A certain metal of work function 1.6 eV is irradiated with ultra-violet light of wavelength 3.6 x 10\(^{-7}\) Calculate the maximum
(i) kinetic energy of ejected electron in joules;
(ii) speed of an emitted electron. (1eV = 1.6 x 10\(^{-18}\) J; C = 3.0 x 10\(^{8}\) ms\(^{-1}\); m, = 9.1 x 10\(^{-31}\) kg; h = 6.6 x 10\(^{-34}\) Js)
(d) If source of the ultra-violet light in (c) above is mo away from the surface of the metal, state the of on the maximum speed of the ejected electron
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) Transmutation is the change of one chemical element into another by radioactive disintegration or nuclear bombardment with particles. Example is the conversion of mercury to gold.
(ii) Stopping potential is the p.d of the anode of a photocell with respect to the cathode at which no electrons will reach the anode stopping potential Vs
= \(\frac{ht}{e} - \frac{Wo}{e}\)
(b) \(^{23}_{11}A + ^2_1B + ^1_1P\)
P + 1 = 23 + 2
P = 24
q + 1 = 11 + 1
q = 11
\(^{24}_{11}C \to ^r_sE + ^o_{-1}B\)
r + 0 = 24; r = 24
S - 1 = 11; S = 12
(c) (i) W = 1.6eV = 1.6 x 1.6 x 10\(^{-19}\)
= 2.56 x 10\(^{-19}\)J
E = \(\frac{hc}{\lambda} = \frac{6.6 \times 10{-34} \times 3 \times 10^{8}}{3.6 \times 10^{-7}}\)
E = W + K.E\(_{max}\)
ii) K.E = \(\frac{1}{2}mv^2\)
2.94 x 10\(^{-19}\) = \(\frac{1}{2} \times 9 \times 10^{-3} \times V^2_{max}\)
V\(_{max} = 8.04 \times 10^5 m/s\)
(d) No effect.
Source: Myschool Past Questions Archive (Record #53373)
Physics
WAEC / WASSCE 2011
Atomic, Quantum, Nuclear Physics & Electronics
Solution
State three properties of cathode rays which suggest the particle nature of matter
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Properties of cathode rays that show particle nature of matter:
- Casting of shadows of objects along their path.
- They have momentum
- Heat up objects that stop thorn, that is they obey energy conservation law.
- They are deflected in electric or magnetic fields, showing they are tiny particles of matter, etc.
Source: Myschool Past Questions Archive (Record #53378)
Physics
WAEC / WASSCE 2011
Current Electricity & Electric Circuits
Solution
You are provided with cells, a potentiometer, an ammeter, a voltmeter, a bulb, a key, a jockey, and other necessary materials.
Measure and record the emf E of the battery.
Set up a circuit as shown in the diagram above.
Close the key K and use the jockey to make a firm contact at J on the potentiometer wire such that PJ = x = 10cm.
Take the record of the voltmeter reading V and the corresponding ammeter reading I.
Evaluate log V and log I.
Repeat the procedure for other values of x =20, 30. 40, 50, and 60cm.
Tabulate your readings.
Plot a graph with the log I on the vertical axis and log V on the horizontal axis.
Determine the slope, s, of the graph.
Determine the intercept, c, on the vertical axis.
State two precautions taken to ensure accurate results.
(b)i. How is the brightness of the bulb affected as x increases? Give a reason for your answer.
ii. List two electrical devices whose actions do not obey ohm's law
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Observation/table of values
S/N
V(volts)
I(Amps)
Log V
Log I
1
0.790
0.030
-0.100
-0.520
2
0.690
0.240
-0.160
-0.620
3
0.600
0.180
-0.220
-0.740
4
0.600
0.120
-0.220
-0.920
5
0.500
0.060
-0.300
-0.220
6
0.300
0.060
-0.520
-0.220
Slope (s) = \(\frac{\bigtriangleup \text{log I}}{\bigtriangleup \text{log V}} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
(s) = \(\frac{-0.4-(-1.4)}{-0.04-(-0.62)} = \frac{-0.4+1.4}{-0.04+0.62}\)
= \(\frac{1}{0.58}\)
s = 1.72
Graph Log I against Log V
PRECAUTIONS; i
- Ensured clean/tight terminals
- Opened key when reading was not taken
- Avoided parallax error in ammeter/ voltmeter.
- Avoided sliding or jockey on the potentiometer wire M Zero error of voltmeter/ammeter were noted.
(b)i. The brightness of the bulb increases. Voltage/current through the bulb increases.
ii. The devices that do not obey ohm's law;
Diode
Transistor
Thyristor
Capacitor
Thermistor
Source: Myschool Past Questions Archive (Record #60015)
Physics
WAEC / WASSCE 2011
Units, Measurements & Dimensional Analysis
Solution
You are provided with a wooden block to which a hook is fixed, a set of masses, spring balance, and other necessary materials. Using the diagram above as a guide, carry out the following instructions.
Record the mass \(m_0\), indicated on the wooden block.
Place the block on the table.
Attach the spring balance to the hook.
Pull the spring balance horizontally with a gradual increase in force until the block just starts to move Record the spring balance reading F.
Repeat the procedure by placing in turn mass m=200, 400, 600, and 800g on top of the block. In each case, read and record the corresponding value of F.
Evaluate M = \(m_0\) + m and R = \(\frac{m}{100}\) in each case
Tabulate your readings.
Plot a graph with F on the vertical axis and R on the horizontal axis
Determine the slope, s, of the graph.
State two precautions taken to ensure accurate results.
(b)i. Define coefficient of static friction.
ii. A block of wood of mass 0.5 kg is pulled horizontally on a table by a force of 2.5 N. Calculate the coefficient of static friction between the two surfaces.(g = 10ms\(^{-2}\))
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Observation /table of values
Mass mo = 400g.
S/N
mo(g)
m(g)
(M=mo+m)g
R = \(\frac{M(g)}{100}\)
F(N)
1
400.0
0.00
400.0
4.00
1.70
2
400.0
200.0
600.0
6.00
2.40
3
400.0
400.0
800.0
8.00
3.60
4
400.0
600.0
1000.0
10.00
5.20
5
400.0
800.0
1200.0
12.00
6.30
Slope, s = \(\frac{\bigtriangleup{F}}{\bigtriangleup{R}} = \frac{F_{2}-F_{1}}{R_{2}-R_{1}}\)
Slope (S) = \(\frac{6.4-2.3}{12.0-6.0}\)
S = \(\frac{4.1}{6.0}\) = 0.68
PRECAUTION; i
- Avoided parallax error in reading spring balance.
- Tapped the wood/load slightly to avoid sticking to the table.
- Repeated readings are shown in the table.
- Zero error of the spring balance.
(b)i. Coefficient of static friction: This is the ratio of the frictional force F (when the body is just about to move) to the normal reaction R (between the two surfaces in contact).
ii. \(\mu\) = \(\frac{F}{R} = \frac{2}{5}\) = 0.5
Source: Myschool Past Questions Archive (Record #60006)
Physics
WAEC / WASSCE 2011
Reflection, Refraction & Dispersion of Light
Diagram
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.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
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: Myschool Past Questions Archive (Record #60007)
Chemistry
UTME / JAMB 2011
Stoichiometry, Mole Concept & Volumetric Analysis
What is the concentration of a solution containing 2g of NaOH in 100cm3 of solution?
[Na = 23, O = 16, H = 1]
A
0.40 mol dm-3
B
0.50 mol dm-3
C
0.05 mol dm-3
D
0.30 mol dm-3
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #4692)
Chemistry
UTME / JAMB 2011
Stoichiometry, Mole Concept & Volumetric Analysis
Which of the following properties is NOT peculiar to matter?
A
kinetic energy of particles increase from solid to gas
B
Random motion of particles increases from liquid to gas
C
Orderliness of particles increases from gas to liquid
D
Random motion of particles increases from gas to solid
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #4694)
Chemistry
UTME / JAMB 2011
Stoichiometry, Mole Concept & Volumetric Analysis
Diagram
From the diagram above, an ideal can be represented by
Attached Figure / Diagram
A
M
B
N
C
K
D
L
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #4696)
Chemistry
UTME / JAMB 2011
Periodic Table & Periodicity
Which of the following questions is correct about the periodic table?
A
The non-metallic properties of the elements tend to decrease across each period
B
The valence electrons of the elements increase progressively across the period
C
Elements in the same group have the same numbeer of electron shells
D
Elements in the same period have the number of valence electrons
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #4697)
Chemistry
UTME / JAMB 2011
Particulate Nature of Matter, Laws & Formulae
Solution
The relative atomic mass of a naturally occurring lithium consisting of 90% \(^7 _3\)Li and \(^6 _3\)10% Li is
A
6.9
B
7.1
C
6.2
D
6.8
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
R.A.M = m\(_1 \alpha_1\) + m\(_2 \alpha_2\)
= 7(90/100) + 6(10/100)
= 6.3 + 0.6
= 6.9
Source: Myschool Past Questions Archive (Record #4698)
Chemistry
UTME / JAMB 2011
Atomic Structure, Orbitals & Nuclear Chemistry
What type of bond exits between an element X with atomic number 12 and Y with atomic number 17?
A
Electrovalent
B
Metallic
C
Covalent
D
Dative
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #4770)
Chemistry
UTME / JAMB 2011
Water, Solutions & Solubility
Solution
Hardness of water is mainly due to the presence of
A
calcium hydroxide or magnesium hydroxide
B
calcium trioxocarbonate (IV) or calcium tetraoxosulphate (VI)
C
sodium hydroxide or magnesium hydroxide
D
calcium chloride or sodium chloride salts
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
Salts of Calcium and Magnesium are responsible for hardness of water. Bicarbonates, Carbonates, Chlorides and Sulfates of Calcium and Magnesium causes temporary and permanent hardness.
Source: Myschool Past Questions Archive (Record #4771)
Chemistry
UTME / JAMB 2011
Organic Chemistry I: Hydrocarbons & Crude Oil
Solution
A suitable solvent for iodine and naphthalene is
A
carbon (IV) sulphide
B
ethanol
C
water
D
benzene
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
Both Iodine and Naphthalene, being nonpolar compounds, dissolve best in nonpolar or slightly polar solvents like benzene, toluene, and carbon disulfide.
Source: Myschool Past Questions Archive (Record #4772)
Chemistry
UTME / JAMB 2011
Periodic Table & Periodicity
Solution
Which of the following noble gases is commonly found in the atmosphere?
A
Xenon
B
Neon
C
Helium
D
Argon
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
Argon, with the symbol Ar which is the third noble gas, in group 18 makes up about 1% of the Earth's atmosphere, making it the most common noble gas on Earth.
Source: Myschool Past Questions Archive (Record #4773)