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Physics
WAEC / WASSCE 2011
Mechanics and Properties of Matter
Solution
Explain why water in a narrow glass tube has a concave meniscus while mercury, in the same tube, has a convex meniscus.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
In the case of water in axarrow glass tube, the force of adhesion between the water and glass molecules is greater than the cohesive force among water molecule, hence, water clings to the glass tube forming a concave meniscus. In the case of mercury, the cohesive force among its molecules is greater than the adhesive force, between mercury molecules and water molecules hence mercury molecules cling to each other forming a convex meniscus in the capillary glass tube.
Source: Myschool Past Questions Archive (Record #53329)
Physics
WAEC / WASSCE 2011
Waves and Optics
Solution
State three methods of polarizing an unpolarized light
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
The methods of polarizing an unpolarized light are:
- reflection from plane, smooth surfaces.
- double refraction through crystals.
- selective absorption
- polaroid.
- Scattering by small particles, etc.
Source: Myschool Past Questions Archive (Record #53330)
Physics
WAEC / WASSCE 2011
Mechanics and Properties of Matter
Solution
An electron of charge 1.60 x 10\(^{-19}\) C is accelerated under a potential difference of 1.0 x 10\(^5\) V. Calculate the energy of the electron in joules.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
E = eV= 1.6 x 10\(^{-19}\) x 1.0 x 10\(^5\)
=1.6 x 10\(^{-14}\)J
Source: Myschool Past Questions Archive (Record #53331)
Physics
WAEC / WASSCE 2011
Mechanics and Properties of Matter
Solution
The uncertainty in determining the duration during which an electron remains in a particular energy level before returning to the ground state is 2.0 x 10\(^{-9}\)s. Calculate the uncertainty in determining its energy at that level. (\(\frac{h}{2 \pi}\) = h = 1.054 x 10\(^{-34}\)Js)
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
\(\Delta E \geq \frac{h}{2 \pi} \times \frac{1}{\Delta t}\)
\(\geq 1.054 \times 10^{-37} \times \frac{1}{2 \times 10^{-9}}\)
\(\geq 5.27 \times 10^{-28}\)J
Source: Myschool Past Questions Archive (Record #53333)
Physics
WAEC / WASSCE 2011
Mechanics and Properties of Matter
Solution
(a) What is a vector quantity?
(b) Three vectors 3 ms\(^{-1}\) N 45^\circ W, 12 ms\(^{-1}\)W and 5 ms\(^{-1}\)S act at a point.
(i) Sketch a vector diagram to illustrate the given information.
(ii) Calculate the resultant of the vectors.
(c) In a laboratory experiment to determine the force constant of a spiral spring, the mass or, the spring was varied and the corresponding extensions were measured and recorded as shown in the table below.
Mass M/g
Weight W/N
Extension e/cm
50
100
150
200
250
6.5
11.0
15.0
20.0
25.0
(i) Copy and complete the table. (Take g = 10 ms\(^{-2}\))
(ii) Plot a graph with weight, W, on the vertical axis and extension, e, on the horizontal axis.
(iii) Using the graph, determine the force constant of the spring.
(iv) Determine the natural length of the spring if its length was 38.0 cm when loaded with 250 g mass.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Vector quantity is any quantity that has both magnitude and direction, e.g. velocity.
(b)(i)
(ii)
x
y
-3 cos 45 = -2.121
-12.000
0.000
V_x = -14.121
+ 3 sin 45 = 2.121
0000
-5000
V_y = -2.879
R = \(\sqrt{V_x^2 + Vy^2}\)
= \(\sqrt{-14.121^2} + -2.879^2\) = 14.4m/s
Ta n \(\theta = \frac{2.879}{14.121} = 0.2039\)
\(\theta = tan^{-1} 0.2039 = 11.5^o\)
R = 14.4m/s S78.5\(^o\)W
(c)(i)
Mass/g
Weight W/N
Ext.e/cm
50
100
150
200
250
0.5
1.0
1.5
2.0
2.5
6,5
11.0
15.0
20.0
25.0
(iii) Force constant = slope
\(\frac{\Delta W}{\Delta e} = \frac{3.1 - 1.0}{30 - 11} = \frac{2.1}{9} = 0.1105\)
(iv) Nature length = 38.0 - 25 = 13cm
Source: Myschool Past Questions Archive (Record #53338)
Physics
WAEC / WASSCE 2011
Mechanics and Properties of Matter
Solution
(a)(i) State two advantages of alcohol over mercury as a thermometric liquid.
(ii) When the bulb of a thermometer is placed in a beaker of hot water, the level of the mercury first falls and then rises gradually. Explain this observation.
(b) List two factors, other than temperature, that affect the rate of evaporation of a liquid.
(b) A block of lead of mass 100 kg in a crucible and at a temperature of 40 ^\circC was placed in an electric furnace rated 10 kW. If the melting point of lead is 320 ^\circC, calculate the:
(i) quantity of heat required to heat the lead to its melting point;
(ii) additional heat energy required to melt the lead;
(iii) time taken to supply this additional energy. (Specific heat capacity of lead = 120 Jkg\(^{-1}\) K\(^{-1}\) (Specific Idtent heat of fusion lead = 2.5 x 10\(^{4} JK^{-1}\))
(d) State two precautions necessary in an experiment to determine the specific latent heat of vaporization of water.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
a)(i) Advantages of alcohol over mercury as a thermometric liquid are:
- Alcohol is cheaper
-Alcohol is easily available
-Alcohol has higher expansivity
- Alcohol has lower freezing point.
(ii) The glass bulb of the thermometer first got heated and expands, therefore the mercury level in it falls. Then the mercury expands and increases more than that of the glass and then rises.
(b) Factors that affect the rate of evaporation of a liquid, temperature exclusively are:
- Pressure, area of liquid surface exposed, humidity, nature of the liquid, wind and dryness of air, etc.
(c)(i) Q = mc
\(\Delta \theta = 100 \times 120 \times (320 - 40)\)
ii) Q = Ml = 100 x 0.25 x 10\(^5)\) = 2500000J
= 2.5MJ
ii) Pt = ml
t = \(\frac{ml}{p} = \frac{2500000}{10000} = 250s\)
(d) Precautions to be considered are:
- Calorimeter should be well lagged.
- Use only dry steam.
- Stir the mixture gently and continually.
- Shield the calorimeter from heat sources.
Source: Myschool Past Questions Archive (Record #53340)
Physics
WAEC / WASSCE 2011
Mechanics and Properties of Matter
Solution
(a)(i) What is an eclipse?
(ii) List the three types of eclipse.
(b) A student in a lecture theatre can read from the board clearly but requires a pair of spectacles to read from a book.
(i) What eye defect has this student?
(ii) What type of lens is needed to correct the eye defect?
(iii) The focal length of the lens used to correct this defect is 10cm. Calculate the power of the lens.
(c) A car B moves towards a stationary car A. If B produces an ultrasonic sound at a point and it takes 5.6 x 10\(^{-3}\)s for a beep to be heard in B, calculate the distance between the two cars at that instant. (Speed. sound in air = 340 ms\(^-\))
(d) The image of an object is located 9 cm behind a convex mirror. If magnification produced is 0.6, calculate the focal length of the mirror.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(i) An eclipse is a shadow that results from the obstruction of light from the sun by the moon or the earth.
(ii) Types of eclipse:
-Eclipse of the sun or solar eclipse.
-Annular eclipse
- Eclipse of the moon or lunar eclipse.
(b i) The eye defect is long sight or hypermetropia.
(ii) The type of lens to correct the defect is convex or converging lens.
(iii) Power of lens = \(\frac{1}{f} = \frac{1}{0.1} = dioptre(D)\)
(c) Velocity of an echo = \(\frac{2d}{t}\)
d = \(\frac{vt}{2} = \frac{340 \times 5 \times 10^{-3}}{2} = 0.952 m\)
(d) M = \(\frac{v}{u}\)
U = \(\frac{V}{M} = \frac{9}{0.6} = 15cm\)
\(\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{15} + \frac{1}{-9}\)
f = -22.5cm
Source: Myschool Past Questions Archive (Record #53343)
Physics
WAEC / WASSCE 2011
Mechanics and Properties of Matter
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
Mechanics and Properties of Matter
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
Mechanics and Properties of Matter
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
Electricity and Magnetism
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
Mechanics and Properties of Matter
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
Waves and Optics
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)
Physics
UTME / JAMB 2010
Heat Energy & Temperature
Solution
Which of the following has no effect on radiation?
A
nature of surface
B
density
C
surface area
D
temperature
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Radiation depends on nature of the surface area and temperature. It is independent of density.
Source: WAEC / JAMB / NECO Classroom Archive (Record #14314)
Physics
UTME / JAMB 2010
Motion Force Circular Motion
Solution
Two cars moving in the same direction have speeds of 100 kmh\(^{-1}\) and 130 kmh\(^{-1}\). What is the velocity of the faster car as measured by an observer in the slower car?
A
30 kmh-1
B
130 kmh-1
C
200 kmh
D
230 kmh-1
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Since the cars are moving in the same direction, the velocity of a body A, Va,relative to another body B, Vb is given by V = Va - Vb
∴ the velocity of faster car is measured by the observer in the slower car is V = 130kmh-1 - 100 kmh-1 = 30 kmh-1
Source: WAEC / JAMB / NECO Classroom Archive (Record #14283)