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Physics
WAEC / WASSCE 2013
Heat Energy & Temperature
Solution
The accelerating potential in a cathode ray oscilloscope is 2.5 kV. Calculate the maximum speed of the accelerated electrons. [ e = 1.6 x 10\(^{-19}\) C; Me = 9.1 x 10\(^{-31}\) kg]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
\(\frac{1}{2}\) MeV\(^2\) = eV
\(V^2 = \frac{2eV}{Me} = \frac{2 \times 1.6 \times 10^{-19} \times 2500}{9.1 \times 10^{-31}}\)
= 8.79 x 10\(^{-14}\)
V = \(\sqrt{8.79 \times 10^{14}}\)
= 2.965 x 10\(^7\) m/s
Source: Myschool Past Questions Archive (Record #53130)
Physics
WAEC / WASSCE 2013
Waves and Optics
Solution
Write down the name of:
(a) two particles used in explaining the wave nature of matter;
(b) one device whose invention is based on the wave nature of matter.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Particles used in explaining the wave nature of matter are: hydrogen atom or proton, helium atom, electrons, neutron and helium nuclei or apha particles.
(b) Inventions based on the wave nature of water devices are:
- Electron microscope and scanning tunneling microscope (STM).
Source: Myschool Past Questions Archive (Record #53131)
Physics
WAEC / WASSCE 2013
Heat Energy & Temperature
Solution
(a) what is Brownian motion?
(b) State the two inferences that can be drawn from Brownian motion experiment.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Brownian motiorbigtcontinuous random movement of particles sus erided in a fluid.
(b) Inferences that can be drawn from Brownian experiment include;
- existence of molecules.
- kinetic motion of molecules.
Source: Myschool Past Questions Archive (Record #53132)
Physics
WAEC / WASSCE 2013
Mechanics and Properties of Matter
Solution
(a) State the triangle law of vector addition.
(b) Name the four physical quantities that are associated with the equationq of linear motion.
(c) Using the same set of axes, sketch and label two graphs to illustrate the variation of potential energy and kinetic energy with time for a body in simple harmonic motion.
(d)
A light spiral spring of force constant K lies on a horizontal frictionless surface and has one end fixed to a vertical wall. A block P of mass 2.0 kg placed against the free end of the spring is pushed a distance 5 cm towards the wall with 10J of energy as illustrated in the diagram above. The block is released and after 0.25s, it collides inelastically with a stationary block Q of mass 4.0 kg. Calculate the:
(i) value of k;
(ii) force used to compress the spring;
(iii) acceleration of the block p after release;
(iv) common speed after collision of the blocks.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Triangle law of vector addition states that if three forces are in equilibrium, they can be represented in magnitude and direction by the three sides of a triangle taken in order.
(b) Physical quantities associated with equations of motion are:
- time
- speed/ velocity
- distance or displacement and acceleration.
(c)
(d)
W = \(\frac{1}{2}Ke^2\)
K = \(\frac{2W}{e^2} = \frac{2 \times 10}{\frac{5}{100}} = 8000 NM^{-1}\)
(ii) F = Ke = 800 x \(\frac{5}{100}\)
= 400N
(iii) F = Ma
a = \(\frac{F}{M} = \frac{400}{2}\) = 200 ms\(^{-2}\)
(iv) Initial velocity U\(_p\) of the block
P = 200 x 0.25
= 50 ms\(^{-1}\)
but \(M_pU_p + M_qU_q = (M_p + M_Q)\)
2 x 50 + 4 x 0 = V(2 + 4)
V = \(\frac{100}{6} = 16.7 ms^{-1}\)
Source: Myschool Past Questions Archive (Record #53138)
Physics
WAEC / WASSCE 2013
Mechanics and Properties of Matter
Solution
(a) On which day would sound wave travel faster: on a hot or cold day? Explain.
(b) Why are megaphones shaped like funnels?
(c) A ray of light is incident on a surface of a ectangular glass prism of refractive index 1.5 illustrated in the diagram below.
(i) Copy the diagram a label the angles of: (\(\alpha\)) Incidence (x); (\(\beta\)) Reflection (y); (\(\gamma\)) refraction (z); with t glass letters indicated.
(ii) Calculate the angle refraction to the nearest whole number.
(d) A sonomesr wire vibrates in simple harmoi motion with a maximum amplitude of 1.0 cm. Calculate the frequency of vibration of the wire, giv that the magnitade of the maximum acceleration of the wire is 980ms\(^{-2}\). [\(\pi \frac{22}{7}\)]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Sound wave travels faster on a hot day because the speed of sound is proportional to the square root of absolute temperature (V \(\alpha \sqrt{T}\)), hence the higher the greater the speed of sound wave.
(b) Megap:iones are shaped like a funnels because at the smaller end of the cone, the sound energy comes into contact will a small mass of air producing greater intensity. As it comes out of the larger open end, the energy sets a larger air mass into vibration, hence the final sound is louder.
(c) (i)
(ii) n = \(\frac{sini}{sinr}\)
sinr = \(\frac{sini}{n} = \frac{sin60}{1.5}\)
= \(\frac{0.8660}{1.5} = 0.5774\)
r = sin\(^{-1}\)0.5774 = 35\(^{o}\)
(d) a = \(\Omega^2r\) but \(\Omega = 2 \pi f\)
a = (2\(\pi f)^2\)r
f = \(\sqrt{\frac{9}{4 \pi^2 r}} = \sqrt{\frac{988}{4(\frac{22}{7})^2} \times \frac{1}{100}}\)
= 50Hz
Source: Myschool Past Questions Archive (Record #53140)
Physics
WAEC / WASSCE 2013
Mechanics and Properties of Matter
Solution
(a) Explain briefly the purpose of earthing electrical appliance.
(b) Why does the light frorr bulb connected to a simple cell dim and eventually goes off after a while?
(c) A coil of incidence 0.007 H, a resistor of resistance 8 \(\Omega\) and a capacitor capacitance 0.001 F are connected in series an a.c. source of frequency \(\frac{500}{\pi}\)Hz. If the r.m.s voltages across the coil, the resistor and capacitor are 30v, 20v and 70v respectively;
(i) draw a vector diagram to illustrate the voltage across the components in the circuit.
(ii) Calculate the: (\(\alpha\)) r.m.s voltage of the source
(\(\beta\)) r.m.s current in the circuit;
(\(\gamma\)) power dissipated in the circuit.
iii) write down the sinusoidal equation for the r.m.s voltage, V, in terms of the time, t.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) The earth wire is used on electrical appliances as a safety device to prevent electric shock when short circuit occurs in such appliances. The dangerous current is then passed into thq ground through the earth wire because the eartIthas a lower resistance than the body.
(b) Light from a bulb connected to a single cell dim and eventually goes off because of the defects of polarization and local action.
(c)
(i) \(V^2\)\(_{\text{rms}}\) + \(V_c - L_L)^2\)
\(V^2\)\( _{\text{rms}}\) = \(\sqrt{V^2 + (V_c - V_L)}^2\)
= \(\sqrt{30^2 + (70 - 30)^2}\)
= \(\sqrt{900 + 1600}\)
= 50v
(ii) (\(\beta\)) = X\(_L = 2\pi fL\)
= 2 \(\pi \times \frac{500}{\pi} \times 0.007\)
= 7 \(\Omega\)
But X\(_c = \frac{1}{2 \pi fC} = \frac{1}{2 \pi \times \frac{500}{\pi} \times 0.001}\)
= L\(\Omega\)
z = \(\sqrt{R^2 + (X_L - X_c)^2}\)
= \(\sqrt{8^2 + (7 - 1)^2} = \sqrt{64 + 36}\)
= \(\sqrt{100}\)
= 10\(\Omega\)
I = \(\frac{v}{z} = \frac{50}{10}\)
= 5.0 A
(iii) (\(\gamma\)) P = I\(^2R = 5^2 \times 8\) = 200W
\(V_o = V_\text{rms} \sqrt{2} = 50\sqrt{2} = 70.7V\)
But V\(_osin2 \pi ft\)
= 70.7\(sin 2 \times 2 \times \pi \times \frac{500}{\pi} \times t\)
= 70.7sin 1000t
Source: Myschool Past Questions Archive (Record #53154)
Physics
WAEC / WASSCE 2013
Mechanics and Properties of Matter
Solution
(a) Define ionization potential.
(b)(i) State the three types of emission spectra.
(ii) Name one source each which produces each of the spectra stated in (b)(i).
(c) In an x-ray tube, electrons are accelerated the target by a potential difference of 80 A Calculate the:
(i) speed of the electron;
ii) threshold wavelength of the electron. [h=6.6 x 10\(^{-34}\) Js; e = 1.6 x 10\(^{-19}\) C; Me = 9.1 x 10\(^{-31}\)
d) An x-ray photon of frequency 4.5 x 10\(^{-18}\) strikes an. electron, assumed to be at rest. If t electron absorbs all the photon energy, calculate the speed acquired by the electron. [ h = 6.6 x 10\(^{-34}\) Js; Me = 9.1 x 10\(^{-31}\) kg ]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
a) Ionization potential V is the potential energy required to remove an electron from an atom completely to infinity from ground state.
(b)(i) The three types of emission are
- band spectrum
- continuous spectrum and
- line spectrum.
(ii) Sources of each emmision spectrum are:
- Band spectrum
- from molecules or carbon(IV) oxides in a discharge tube.
- Continuous spectrum
- from sun, solids and liquids.
- Line spectrum
- from atoms in gases such as hydrogen or neon in a discharge tube.
(c)(i) eV =\(\frac{1}{2} mV^2\)
v = \(\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 80 \times 10^3}{9.1 \times 10^{-31}}}\)
= \(\sqrt{2.81 \times 10^{16}}\)
= 1.68 x 10\(^8 ms^{-1}\)
(ii) \(\pi = \frac{h}{mv} = \frac{6.6 \times 10^{-31}}{9.1 \times 10^{-31} \times 1.68 \times 10^8}\)
= 4.32 x 10\(^{-12}\)m
(d) hf = \(\frac{1}{2}mv^2\)
6.6 x 10\(^{-34} \times 4.5 \times 10^{18}\)
= \(\frac{1}{2} \times 9.1 \times 10^{-31} \times v^2\)
\(v^2 = \frac{2.97 \times 10^{-15}}{4.55 \times 10^{-3}}\)
\(v^2 = \sqrt{6.53 \times 10^{15}}\)
= 8.08 x 10\(^7 ms^{-1}\)
Source: Myschool Past Questions Archive (Record #53156)
Physics
WAEC / WASSCE 2013
Mechanics and Properties of Matter
Solution
(a) Define boiling point of a liquid.
(b) Describe how water in a round bottom flask could be made to boil without heating it. [diagram not necessary]
(c) State three applications of expansion of metals.
(d) A room with floor measurements 7m x 10 m contains air of mass 250 kg at a temperature of 34^\circC. The air is cooled until the temperature falls to 24^\circC. Calculate the: (i) height of the room;
(ii) quantity of energy extracted to cool the room;
(iii) which is higher: the calculated value or the actual energy needed to cool the room? Give a reason for your answer. [ Specific heat capacity of air = 1010 Jkg\(^{-1}\)K\(^{-1}\); density of air = 1.25 kg m\(^{-3}]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Boiling point of a liquid is the temperature at which saturated vapour pressure of a liquid equals the external atmospheric pressure.
(b) Fill a round bottom flask partially with water, then pump out the air gradually with a vacuum pump. At a state when the saturation vapour pressure equal atmospheric pressure, the water is observed to be boiling.
(c) Application of i expansion of metals are in:
- Riveting two or more metal plates.
- Fixing of metal rims on metal wheels.
- Fusing of platinum wire through walls of glass vessels.
- Automatic fire alarms and
- Thermostat.
(d)(i) Volume = \(\frac{Mass}{Density} = \frac{250}{1.25}\)
= 200\(m^2\)
But volume = Ibh
h = \(\frac{v}{Ib} = \frac{200}{7 \times 10}\)
= 2.86m
(ii) Q = mc\(\theta\)
V = 250 x 1010(34 - 24)
= 252500 or 2.525 x 10\(^6\)
(iii) The actual energy value will be higher in value than the calculated value because --Part of heat is extracted. from the walls in the room and other materials. he room also is not airtight, there are heat leakages in and out of the room.
Source: Myschool Past Questions Archive (Record #53162)
Physics
WAEC / WASSCE 2013
Heat Energy & Temperature
Solution
Name the three basic components P, Q and R that make up a cathode ray tube, as illustrated in the diagram above
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
The basic component of the cathode ray tube are:
P = Electron gun. Q = Deflector places or oscillators.
R = Screen.
Source: Myschool Past Questions Archive (Record #53165)
Physics
WAEC / WASSCE 2013
Mechanics and Properties of Matter
Solution
You are provided with a uniform metre rule of mass, M indicated on its reverse side, a knife-edge, a graduated measuring cylinder of known mass, \(M_1\) marked on it and other necessary apparatus.
Read and record with values of M and \(m_1\).
Balance the metre rule horizontally on the knife edge. Read and record the balance point as G.
Tie a loop of thread around the neck of the measuring cylinder.
Fill the cylinder with the sand provided to the 2cm\(^{3}\) mark. Record the volume, V, of the sand.
Hang the cylinder at the 2 cm mark of the metre rule and adjust the position of the knife edge until the rule balances horizontally.
Read and record the new balance position K.
Determine the value of e and f.
Determine the mass, \(m_2\), of the sand in the measuring cylinder. Hint: \(m_2\) = (\(\frac{\text {M x f}}{e}\)) - \(m_1\).
Repeat the procedure by filling the measuring cylinder to the mark V = 4,6,8 and 10 cm\(^{3}\). In each case, ensure that the measuring cylinder is kept constant at the 2 cm mark on the metre rule.
Tabulate your readings.
Plot a graph with \(m_2\) on the vertical axis and V on the horizontal axis.
Determine the slope, s, of the graph.
State two precautions taken to ensure accurate results.
(b)i. Determine the mass of 7.5 cm\(^{3}\) of the sand using your graph.
ii. A gold coin of mass 102.0 g has a uniform cross-sectional area of 10.0 cm\(^{2}\). Calculate its thickness. [Density of gold=19.3 g cm\(^{-3}\)]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
volume of sand v/cm\(^{3}\)
k/cm
\(\theta\)=(k-p)/cm
f=(G-K)/cm
\(m_2\)=\(\frac{mxf}{e}\)-\(m_1\)
2
43.5
41.5
6.9
1.614
4
42.7
40.7
7.7
3.443
6
41.9
39.9
8.5
6.372
8
40.1
39.1
9.3
9.416
10
40.4
38.4
10.0
12.178
Slope = \(\frac{\bigtriangleup {m}_{2}}{\bigtriangleup {v}}\)
= \(\frac{12-8-4}{10.4-4.4} = \frac{(8.8)9}{(6)cm^{3}}\)
= 1.4679/cm\(^{3}\)
Observation;
Mass of uniform metre rule m--> 130g
Mass of graduated measuring cylinder \(m_1\) --20g
Balance point G = 50.4cm
PRECAUTIONS; i
Avoided draught
Avoided parallax error in reading metre rule/ measuring cylinder
Knife-edge firmly clamped
Avoided measuring cylinder from resting on/touching the table
Repeated readings are shown on the table
(b)i. 7.5cm\(^{3}\) correctly shown on graph \(m_2\) = 8.4g
ii. V = \(\frac {\text{Mass}} {\text{Density}} = \frac{102.0}{19.3}\) = 5.28
Volume = thickness x area
thickness = \(\frac {\text{Volume}}{\text{Area}} = \frac{5.28}{10}\)
OR
t = \(\frac{m}{A} = \frac{102}{19.3 \times 10}\) = 0.528cm
Source: Myschool Past Questions Archive (Record #59982)
Physics
WAEC / WASSCE 2013
Waves and Optics
Solution
Using the diagram above as a guide, carry out the following instructions:
Fix a plain sheet of paper on the Celotex board.
Place the rectangular glass prism on the paper and trace its outline, ABCD. Remove the prism.
Draw a normal NMP to meet AB and DC at M and P respectively such that |AM|=DP=2.0cm.
Trace the ray PQ with two pins, \(P_1\) and \(P_2\) at P and Q respectively such that angle MPQ = i =50º
Replace the prism on its outline. Trace the emergent ray with two other pins \(P_3\) and \(P_4\) such that they lie in a straight line with \(P_2\) and the image of \(P_1\) viewed through the glass prism.
Measure and record \(\theta\), the angle between the emergent ray and face AB of the glass prism.
Evaluate cos \(\theta\) and sin i.
Repeat the procedure for four other values of i = 10^\circ, 15^\circ,20^\circ and25^\circ.
Tabulate your readings.
Plot a graph with cos \(\theta\) on the vertical axis and sin i on the horizontal axis.
Determine the slope, s, of the graph
State two precautions taken to ensure accurate results. Attach your traces to your answer booklet
(b)i. State the laws of refraction of light
ii. Explain what is meant by the statement the refractive index of a material is 1.65
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
i/o
\(\theta\)/0
cos\(\theta\)
sini
5.0
83.0
0.137
0.087
10.0
74.0
0.276
0.174
15.0
65.0
0.423
0.259
20.0
56.0
0.559
0.342
25.0
48.0
0.669
0.423
Slope = \(\frac{\bigtriangleup \text{cos}\theta}{\bigtriangleup \text{sin i}}\) = \(\frac{0.720-0.250}{0.450-0.150}\) = \(\frac{0.47}{0.30}\) = 1.567
PRECAUTIONS; i
Ensured pins were erect/upright/metre rule
Avoided parallax error on the protractor/metre rule
Neat traces or sharp pencils shown on traces
Pins reasonably spaced i.e about 4cm
Noted and corrected/avoided zero error on metre rule.
(b)i. The incident ray, refracted ray was normal at the point of incidence. i.e in the same plane. 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 nisini = \(n_2\) sinr where the symbols are defined.
OR \(\frac{\text{sini}}{\text{sinr}}\) = N
for a given pair of media, where symbols are defined.
ii. The ratio of the speed of light in vacuum (air) to the speed of light in the material is 1.65
OR
The ratio of the angle of incidence in air/vacuum to the sine of the angle of refraction in the material is 1.65.
Source: Myschool Past Questions Archive (Record #59986)
Physics
WAEC / WASSCE 2013
Mechanics and Properties of Matter
Solution
You are provided with a constantan wire, a2 \(\Omega\)standard resistor, an accumulator E, an ammeter A, a key K, and other necessary apparatus.
Measure and record the emf of the accumulator provided
Connect a circuit as shown in the diagram above.
Close the key, read and record the ammeter reading \(l_o\) when the crocodile clip is not in contact with the Constantan wire.
Open the key with the clip making contact with the wire, when d=90cm, close the key. Read and record the ammeter reading l.
Evaluate d\(^{_1}\)
Repeat the procedure for four other values of d=80, 70, 60, and 50 cm
In each case, read and record the ammeter reading and evaluatre d\(^{_1}\).
Tabulate your readings.
Plot a graph with I on the vertical axis and d\(^{_1}\) on the horizontal aXIS.
Determine the slope, S of the graph and its intercept, c, on the vertical axis.
Evaluate k = \(\frac{c}{s}\)
Using your graph, determine the current I when d=55cm.
State two precautions taken to obtain accurate results.
(b)i. Explain what is meant by the potential difference between two points in an electric circuit.
ii. State two factors on which the resistance of a wire depends.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
d/cm
I/A
d\(^{-1}\)/cm\(^{-1}\)
80.0
0.23
0.0125
70.0
0.25
0.0143
60.0
0.31
0.0167
50.0
0.34
0.0200
40.0
40.0
0.0250
Slope, SI = \(\frac{\bigtriangleup {I}}{\bigtriangleup {d}^{-1}}\) = \(\frac{0.40-0.13}{0.024-0.006}\)
= \(\frac{(0.27)A}{(0.018)cm^{-1}}\)
= 15A/cm\(^{-1}\)
Intercept C on the vertical axis = 0.04A
k = c/s = \(\frac{0.04}{15}\) = 0.00267 = 2.67 x 10\(^{-3}\)
From the graph, when d = 55cm, I = 0.31A
Precautions; I
- ensured that the keys was open after each reading so that the battery did not run down quickly.
- ensured there was zero error when taking the ammeter reading.
- Key was removed when readings were not being taken
- ensured terminals are clean/tight
- Noted and corrected/avoided zero error on ammeter metre rule
(b)i. The potential difference between two points in an electric circuit is the work done (in joules) when one coulomb of electric charge moves from one point to the other
ii. - Nature of wire/resistivity
- Length of wire
- Cross-sectional area of wire/thickness of wire
- Temperature
Source: Myschool Past Questions Archive (Record #59996)
Physics
WAEC / WASSCE 2012
Current Electricity
Solution
A 12\(\Omega\) resistor dissipates 8kJ of heat in 20s. Calculate the current through the resistor
A
5.8A
B
6.9A
C
30.0A
D
33.3A
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Q = I2Rt
I = \(\sqrt{\frac{Q}{Rt}}\)
= \(\sqrt{\frac{8000}{12 \times 20}}\)
= 5.8
Source: WAEC / JAMB / NECO Classroom Archive (Record #35622)
Physics
WAEC / WASSCE 2012
Reflection of Light
Solution
An object is placed 10cm from a converging lens of foal length 15cm. Calculate the magnification of the image formed
A
3.0
B
1.5
C
0.6
D
0.3
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Concave/converging mirror:
* produces a virtual image when object distance(u) exceeds its focal length(f).
\(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\)
\(\frac{1}{v} = \frac{1}{f} - \frac{1}{u}\)
\(\frac{1}{15} - \frac{1}{10} = \frac{2 - 3}{30}\)
\(\frac{1}{v} = \frac{-1}{30}\)
v = -30cm( virtual image)
but m = \(\frac{v}{u} = \frac{30}{10} = 3\)
Note that the negative sign only shows the placement of the image. It does not affect the magnification.
Source: WAEC / JAMB / NECO Classroom Archive (Record #35598)
Physics
UTME / JAMB 2012
Heat Energy & Temperature
In order to remove the error of parallax when taking measurements with a metre rule, the eye should be focused
A
slantingly towards the right on the markings
B
slantingly towards the left on the markings
C
vertically downwards on the markings
D
vertically upwards on the markings
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 #27374)