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Physics
WAEC / WASSCE 2015
Mechanics and Properties of Matter
Solution
(a)(i) what is resonance?
(ii) State two eXwilples of resonance.
(iii) Differentiate between loudness and intensity of sound.
(b) When a ray is refracted through a rectangular glass prism, which of the following properties of the ray will change? Wavelength, frequency and speed.
(c)
(i) Copy the diagram above in your answer booklet.
(iii) On the copied diagram, sketch the pattern of the waves immediately after passing through the opening.
(d) A diverging lens of focal length 18.0 in is used to view a shark that is 90.0 in away from the lens. If the image formed is 1.0 m long, calculate the:
(i) image distance;
(ii) length of the shark.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) Resonance is a phenomenon which occurs whenever a body is set in oscillation at its own natural frequency as a result of impulses received from another vibrating body, vibrating with the same frequency.
(ii) Examples of resonance are:
(1) Loud sound from vibrating prongs of a tuning fork.
(2) Bridge vibrating under uniform matching of soldiers.
(3) Repeated jumping of a diver at the free end of a diving board causes the board to resonate.
(4) Rattling of car engine at high speed causing vibrating of the body of the car. (5) \(X_c\) equals X\(_L\) in an L-C-R series circuit for tuning a radio set.
(iii)
Loudness
Intensity
- The magnitude of the sensation resulting from sound reaching the ear.
-Sensation that depends on the individual hearing ability.
- The time rate of flow of energy per unit area perpendicular to the direction of propagation of the sound.
- Depends on physical quantities of energy, time and area
(b) The following properties will change in speed and wavelength.
(c)
(d) \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\)
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{90}\)
= \(\frac{-5 -1}{90}\)
v = -15m
M = \(\frac{u}{v} = \frac{h_i}{h_o}\)
\(h_o = \frac{hi \times u}{v} = \frac{1 \times 90}{1.5}\)
= 6.0 m
Source: Myschool Past Questions Archive (Record #52990)
Physics
WAEC / WASSCE 2015
Mechanics and Properties of Matter
Solution
(a) Define:(i) the coulomb; (ii) resistance.
(b) State the three effects of an electric current.
(c) State the standard international colour convention for the insulating material covering the following electrical wires in a three-pin plug: (i) live; (ii) neutral; (iii) earth.
(d)
A water melon of mass 5.0 kg is suspended on a uniform rod of mass 4.0 kg and 4.0 in long as illustrated in the diagram above. If the rod is in equilibrium by the action of the force between the charges +q and -q, calculate the:
(i) anti-clockwise moment;
(ii) value of q. [ g = 10 ms\(^{-2}\)]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) The coulomb is defined as the quantity of electric charge passing through a point in a conductor when a steady current of one ampere passes for one second.
(ii) Resistance is the opposition offered to the flow or passage of an electric current in a conductor.
(b) The effects of electric current are:
(i) Magnetic (ii) Chemical
(iii) Heating or thermal.
(c) The standard international colour for insulating cover for a three in plug are:
(i) Live is brown or red colour.
(ii) Neutral is blue or black colour.
(iii) Earth is green/yellow/green and yellow colour.
(d) Clock wise moment = 4gl\(_2\) + \(\frac{kq^2l_3}{12}\)
\(\frac{4 \times 10 \times 1 + 8.99 \times 10^9 \times q^2 \times 3}{(2.5)^2}\) = 40 + 4.315 x 10\(^{9}\) x \(q^{2}\)
Anti-clockwise moment = 5gl\(_1\) = 5 x 10 x 1 = 50 Nm
But sum of clockwise moment = Anti-clockwise moment 40 + 4.315 x 10\(^9\) q\(^2\) = 50
q = \(\sqrt{\frac{10}{4.315 \times 10^9}}\)
= 4.814 x 10\(^{-5}\)C
Source: Myschool Past Questions Archive (Record #52991)
Physics
WAEC / WASSCE 2015
Mechanics and Properties of Matter
Solution
(a) Define nucleon number
(b) A radioactive isotope of Americium (Am —241) decays into a nucleus of Neptunium (Np — 237) and an alpha (\(\alpha\)) particle as shown in the nuclear equation below. \(^{241}_{95}Am\) \(\to\) \(^{237}_{c} + ^b_a \alpha\)
(i) State the number of neutrons in the nucleus of Americium — 241.
(ii) Determine the values of a,b and c.
(c)(i) Why are y-rays not deflected by electromagnetic field?
(ii) State two properties of gamma rays that make them suitable for sterilizing medical equipment.
(d) A sample of radioactive substance was found to be left with h of its initial count rate after 110 years. Calculate its decay constant.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Nucleon number is the total number of protons and neutrons in the nucleus of an atom.
(b)(i) Number of neutrons in the nucleus of Americium = 241 - 90 =146.
(ii) a = 2
b = 4
c = 95 - 2
= 93.
(c)(i) Y-ray are electrically neutral or carry no charge.
(ii) properties of gamma rays that makes them suitable for medical sterilization of equipment are:
(1) Possession of strong penetrating power.
(2) Posession of low ionizing power.
(d) \(\frac{N}{N_o} = (\frac{1}{2})^a = \frac{1}{32} = (\frac{1}{2})^a\) = (\(\frac{1}{2})^5\)
But T\(_\frac{1}{2}\) = \(\frac{t}{n} = \frac{110}{5}\) = 22 years
Hence \(\lambda\) = \(\frac{0.693}{T_{\frac{1}{2}}} = \frac{0,693}{22}\)
Decay constant \(\lambda\) 3.2 x 10\(^{-2}\) per year
Source: Myschool Past Questions Archive (Record #52993)
Physics
WAEC / WASSCE 2015
Heat Energy & Temperature
Solution
You have been provided with a resistance box, a voltmeter, a key, a battery, and other necessary materials.
Connect the circuit as shown in the diagram above.
With the key K closed, read and record the voltmeter reading \(V_0\).
Set the resistance R in the resistance box equal to I\(\Omega\).
Close the key, read and record the potential difference V on the voltmeter.
EvaluateR\(^{-1}\) and V\(^{-1}\) .
Repeat the procedure for five other values or R = 2\(\Omega\), 3\(\Omega\), 4\(\Omega\), 5\(\Omega\) and 6\(\Omega\).
Tabulate your readings.
Plot a graph with V\(^{-1}\) on the vertical axis and R on the horizontal axis.
Determine the slope, s, of the graph and the intercept, C, on the vertical axis.
Evaluate c\(^{-1}\)
State two precautions taken to obtain accurate results.
(b)i. Define the potential difference between two points in an electric circuit.
ii. Explain why the emf of a cell is greater than the p.d. across the call when it is supplying Current through an external resistance.
A
72o
B
100o
C
120o
D
144o
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
S/N
R/W
V/vat
R\(^{-1}\)/W\(^{-1}\)
V\(^{-1}\)/vat
1
1.0
1.80
1.00
0.56
2
2.0
2.10
0.50
0.48
3
3.0
2.30
0.33
0.43
4
4.0
2.40
0.25
0.42
5
5.0
2.50
0.20
0.40
6
6.0
2.60
0.17
0.30
Slope = \(\frac{0.65 - 0.40}{1.10 - 0.20} = \frac{0.25}{0.9}\)
= 0.3
X = c = 0.35 = C\(^{-1}\) = 2.86
Precautions; I
ensured that the key was open after each reading so that the battery will not run down quickly.
ensured there was no Zero error when taking the voltmeter reading.
ensured the electrical wire was tightly connected to the terminals in order to avoid current fluctuations.
(b) i. Potential difference is the amount of work done in moving a unit positive charge from a point to another, mostly from a point of lower potential to the point of higher potential. Its density p.d. measured in Volts.
ii. The e.m.f. of a cell is the p.d. across the terminals when it is in an open circuit i.e. hot supplying current to an external circuit. While terminal p.d. 5 is defined as the p.d, between the terminals of a cell when it is delivering current to an external circuit. Therefore, emf is greater because it has not yet been connected to an external circuit.
Source: Official Exam / Past Question Archive
Physics
WAEC / WASSCE 2015
Mechanics and Properties of Matter
Solution
You are provided with a retort stand, boss head, clamp, stopwatch, slotted weights, hanger, grooved pulley, thread, measuring tape, and other necessary materials.
i. Measure and record the radius R of the pulley.
ii. Setup the apparatus as illustrated in the diagram above, such that the clamp is 1.5 m above the floor.
iii. Tie one end of the thread to the pulley.
iv. Tie the other end of the thread to the hanger.
v. Slot a mass m= 50 g on the hanger.
vi. Wind the thread around the groove of the pulley until the base of the hanger is at a height h = 1.4 m above the floor. Maintain this height h for every other value of m through out the experiment.
vii. Release the mass to unwind the thread.
viii. Determine and record the time t taken by the mass m to reach the floor.
ix. Evaluate t\(^{2}\)
x. Also evaluate
a =\(\frac{2h}{t^{2}}\),T = \(\frac{m}{1000}\) (10 - a) and \(\propto\) = \(\frac{a}{R}\)
xi. Repeat the procedure for four other values of m = 70 g, 90 g, 110 g and 130 g
xii. Tabulate your readings.
xiii. Plot a graph with \(\propto\) on the vertical axis and T on the horizontal axis.
xiv. Determine the slope s, of the graph.
xv. Evaluate I = \(\frac{R}{s}\).
xvi. State two precautions taken to obtain accurate results.
(b)i. Define centripetal force
ii. An object drops to the ground from a height of 2.0 m. Calculate the speed with which it strikes the ground. [g=10 ms\(^{-2}\)]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
S/N
m/g
\(t_1\)/s
\(t_2\)/s
\(t_1\)+\(t_2\)/s
t\(^{2}\)/5\(^{2}\)
a=\(\frac{24}{t^{2}}\)
T=\(\frac{m}{1000}\)(10-a)
X=\(\frac{a}{R}\)
1.0
50.0
5.00
5.00
\(\frac{5.0+5.0}{2}\)=5
25.000
0.110
0.490
1.380
2.0
70.0
4.80
4.80
\(\frac{4.8+4.8}{2}\)=4.8
23.040
0.120
0.690
1.50
3.0
90.0
4.60
4.60
\(\frac{4.6+4.6}{2}\)=4.6
21.160
0.130
0.890
1.630
4.0
110.0
4.40
4.40
\(\frac{4.4+4.4}{2}\)=4.4
19.360
0.140
1.080
1.750
5.0
130.0
4.20
4.20
\(\frac{4.2+4.2}{2}\)=4.2
17.640
0.150
1.280
1.880
h = 1.4m = 140cm
R = 0.08m = 8cm
h = 1.4m
R = 8cm = 0.08cm
xvi. Precautions; i
- ensured that no parallax error when talcen my reading with metre rule.
- made sure, the clamp was properly tightened.
- ensured that the experiment was not affected by air interference.
(b)i. Centripetal force (\(F_T\)) is defined as that inward force required to keep an object moving with a constant speed in a circular path \(F_T\) = \(\frac{mo^{2}}{r}\)
ii. K.e = P.e
½ mv\(^{2}\) = mgh
\(\frac{V^{2}}{2}\) = gh
V\(^{2}\) = 2gh
V\(^{2}\) = 2 x 10 x 2
V = \(\sqrt 40\) = 6.37mls.
Source: Myschool Past Questions Archive (Record #59960)
Physics
WAEC / WASSCE 2015
Waves and Optics
Diagram
Solution
You are provided with a triangular glass prism, four optical pins, and other necessary materials.
Place the triangular glass prism on a drawing paper and draw its outline UMR. Remove the prism.
Measure and record the value of the angle at U.
Draw a normal to the line UM at N. Also, draw another line TN to the normal such that \(\phi\)-60. Fix two pins at \(P_1\) and \(P_2\).
Replace the prism and fix two other pains at \(P_3\), and \(P_4\), such that the pins appear to be in a straight line with the images of the pins at \(P_1\) and \(P_2\), when viewed from the side UR. Remove the prism.
Join points \(P_3\), and \(P_4\), producing the line to meet TN produced at Z. Draw the normal XY.
Measure and record the angle of emergence e and that of deviation d.
Repeat the experiment with \(\phi\) = 55^\circ, 50^\circ, 40^\circ and 35^\circ.
In each case, measure and record the Corresponding values of e and d.
Tabulate your readings.
Plot a graph with d on the vertical axis and e on the horizontal axis starting both axes from the origin (0.0) Join your points with a smooth curve.
From your graph, obtain the minimum deviation \(d_m\) and the corresponding angle of emergence \(e_m\) . Hence, calculate the refractive index n of the prism using the formula:
n = \(\frac{sin (\frac{d_{m}+U}{2})}{sin{(\frac{u}{2})}}\)
State two precautions taken to obtain accurate results. Attach your traces to your answer booklet.
(b)i. State the conditions necessary for total internal reflection of light to occur.
ii. The critical angle for a transparent substance is 39^\circ. Calculate the refractive index of the substance.
Attached Figure / Diagram
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
U = 60^\circ
S/N
\(\phi\)/^\circ
e/^\circ
d/^\circ
1
60.0
30.00
2.000
2
55.0
35.00
10.000
3
50.0
40.00
21.000
4
40.0
50.00
26.000
5
35.0
55.00
30.00
Slope = \(\frac{57-30}{38.2} = \frac{27}{33.5}\)
xi. u = 60^\circ
dm = 25^\circ
n = [\(\frac{25^\circ = 60^\circ}{2}\)]
Sin[\(\frac{60^\circ}{2}\)]
n = \(\frac{sin(\frac{85}{2})}{sin30}\)
n = \(\frac{sin47.5}{sin30^\circ} = \frac{0.7373}{0.5000}\)
n = 1.5
PRECAUTIONS;
- ensured neat traces/sharp pencil (Shown)
- ensured that pins were reasonably Spaced.
- avoided parallax error in reading protractor.
(b)i. - Light must travel from an optically dense medium to optically less dense medium.
- The critical angle must be exceeded.
ii. n = \(\frac{1}{Sinc}\)
n = \(\frac{1}{Sin30} = \frac{1}{0.6293}\) = 1.59
see graph above
Source: Myschool Past Questions Archive (Record #59963)
Physics
WAEC / WASSCE 2015
Mechanics and Properties of Matter
Solution
You have been provided with a resistance box, a voltmeter, a key, a battery, and other necessary materials.
Connect the circuit as shown in the diagram above.
With the key K closed, read and record the voltmeter reading \(V_0\).
Set the resistance R in the resistance box equal to I\(\Omega\).
Close the key, read and record the potential difference V on the voltmeter.
EvaluateR\(^{-1}\) and V\(^{-1}\) .
Repeat the procedure for five other values or R = 2\(\Omega\), 3\(\Omega\), 4\(\Omega\), 5\(\Omega\) and 6\(\Omega\).
Tabulate your readings.
Plot a graph with V\(^{-1}\) on the vertical axis and R on the horizontal axis.
Determine the slope, s, of the graph and the intercept, C, on the vertical axis.
Evaluate c\(^{-1}\)
State two precautions taken to obtain accurate results.
(b)i. Define the potential difference between two points in an electric circuit.
ii. Explain why the emf of a cell is greater than the p.d. across the call when it is supplying Current through an external resistance.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
S/N
R/W
V/vat
R\(^{-1}\)/W\(^{-1}\)
V\(^{-1}\)/vat
1
1.0
1.80
1.00
0.56
2
2.0
2.10
0.50
0.48
3
3.0
2.30
0.33
0.43
4
4.0
2.40
0.25
0.42
5
5.0
2.50
0.20
0.40
6
6.0
2.60
0.17
0.30
Slope = \(\frac{0.65 - 0.40}{1.10 - 0.20} = \frac{0.25}{0.9}\)
= 0.3
X = c = 0.35 = C\(^{-1}\) = 2.86
Precautions; I
ensured that the key was open after each reading so that the battery will not run down quickly.
ensured there was no Zero error when taking the voltmeter reading.
ensured the electrical wire was tightly connected to the terminals in order to avoid current fluctuations.
(b) i. Potential difference is the amount of work done in moving a unit positive charge from a point to another, mostly from a point of lower potential to the point of higher potential. Its density p.d. measured in Volts.
ii. The e.m.f. of a cell is the p.d. across the terminals when it is in an open circuit i.e. hot supplying current to an external circuit. While terminal p.d. 5 is defined as the p.d, between the terminals of a cell when it is delivering current to an external circuit. Therefore, emf is greater because it has not yet been connected to an external circuit.
Source: Myschool Past Questions Archive (Record #59971)
Physics
WAEC / WASSCE 2014
Heat Energy & Temperature
Solution
The mass of an empty density bottle is 30g. When filled with a liquid, the mass of the bottle and the liquid is 40g. On heating the filled bottle through 40ºc, the mass reduces to 38g. Calculate the apparent cubic expansivity of the liquid
A
6.25 x 10K-1
B
6.25 x 10-3K-1
C
2.50 x 10-3K-1
D
2.50 x 10-4K-1
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Apparent cubic expansivity = \(\frac{\text{mass of hot liquid expelled}}{\text{mass of hot liquid left x temp. rise}}\)
= \(\frac{2}{8 \times 40} = 0.00625\)
Source: WAEC / JAMB / NECO Classroom Archive (Record #35758)
Physics
UTME / JAMB 2014
Heat Energy & Temperature
What is the least possible error encountered when taking measurement with a meter rule?
A
0.1mm
B
1.0mm
C
0.5mm
D
0.2mm
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 #26954)
Physics
UTME / JAMB 2014
Heat Energy & Temperature
Solution
A quantity which requires magnitude and direction to be specified is
A
Temperature
B
Distance
C
Displacement
D
Mass
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Displacement is a vector quantity
Source: Myschool Past Questions Archive (Record #26955)
Physics
UTME / JAMB 2014
Mechanics and Properties of Matter
I Electrical potential, II Torque, III Kinetic Energy, IV Momentum. Which of the quantities listed are vectors?
A
II and IV
B
I and II
C
I and III
D
II and III
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 #26956)
Physics
UTME / JAMB 2014
Heat Energy & Temperature
Solution
Which type of motion do the wheels of a moving car undergo?
A
Vibratory and translational motion
B
Random and translational motion
C
Rotational mand oscillatory motion
D
Translational and rotational motion
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
The tyre in motion makes a rotation and it translates in a straight line alongside. if the wheels is turned it continued further in a translational straight line through rotational motion.
Source: Myschool Past Questions Archive (Record #26957)
Physics
UTME / JAMB 2014
Mechanics and Properties of Matter
Solution
A car accelerates uniformly from rest at 3ms-2. its velocity after traveling a distance of 24m is
A
12ms-1
B
144ms-1
C
72ms-1
D
36ms-1
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
V2 = U2 x 2ax
V2 = 0 x 2(3)(24)
V2 = 144
V = \(\sqrt{144}\)
V = 12ms-1
Source: Myschool Past Questions Archive (Record #26958)
Physics
UTME / JAMB 2014
Mechanics and Properties of Matter
Solution
Calculate the escape velocity of a satellite launched from the earth's surface if the radius of the earth is 6.4x106m
A
25.3kms-1
B
4.2kms-1
C
4.0kms-1
D
11.3kms-1
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
Vescape = \(\sqrt{2gR}\)
Vescape = \(\sqrt{2\times10\times6.4\times10^6}\)
Vescape = \(\sqrt{128000000}\)
Vescape = 11313.7085mS-1
Vescape = 11.313kmS-1
Source: Myschool Past Questions Archive (Record #26959)
Physics
UTME / JAMB 2014
Mechanics and Properties of Matter
An object of weight 8kg on earth is taken to a planet where acceleration due to gravity is one-third of its value on earth. The weight of the object on the planet is
A
48N
B
12N
C
27N
D
36N
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 #26960)