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 466–480 of 1709 verified questions
Sort by:
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
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
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
UTME / JAMB 2012
Mechanics and Properties of Matter
Solution
A load is pulled at a uniform speed along a horizontal floor by a rope at 45o to floor. If the force in the rope is 1500N, what is the frictional force on the load ?
A
1524N
B
1350N
C
1260N
D
1061N
medium Difficulty
Verified Correct Answer: Option D
Physical Principles & Numerical Solution
F = mg cos 45o = 1500 x 0.708
Source: Myschool Past Questions Archive (Record #27375)
Physics
UTME / JAMB 2012
Mechanics and Properties of Matter
Solution
A car starts from rest and moves with a uniform acceleration of 30ms-2 for 20s. Calculate the distance covered at the end of the motion
A
6km
B
12km
C
18km
D
24km
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
S = \(ut + \frac{1}{2}at^2\)
S = \(0 \times 2 + \frac{1}{2} \times 30 \times 20^2\)
Source: Myschool Past Questions Archive (Record #27377)
Physics
UTME / JAMB 2012
Mechanics and Properties of Matter
I. Small size.
II. Low power requirement
III. Not easily damaged by high temperature
IV. Highly durable.
Which of the above are the advantages of semiconductors?
A
I,II and lll only
B
ll,lll and lV only
C
l, II and lV only
D
l, ll, lll and lV
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: Official Exam / Past Question Archive
Physics
UTME / JAMB 2012
Mechanics and Properties of Matter
A rocket is fired from the earth's surface to a distant planet. By Newton's law of universal gravitation, the force F will
A
increase as r reduces
B
increase as G varies
C
remains constant
D
increases as r increases
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 #27378)
Physics
UTME / JAMB 2012
Mechanics and Properties of Matter
Solution
An object moves in a circular path of radius 0.5m with a speed of 1ms-1 What is its angular velocity?
A
8 rads-1
B
4 rads-1
C
2 rads-1
D
1 rads-1
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
W = \(\frac{v}{r} = \frac{1}{0.5}\)
Source: Myschool Past Questions Archive (Record #27383)
Physics
UTME / JAMB 2012
Mechanics and Properties of Matter
Solution
A block and tackle is used to raise a load of 250N through a vertical distance of 30m. What is the efficiency of the system if the work done against friction is 1500J?
A
62.5%
B
73.3%
C
83.3%
D
94.3%
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
Y = \(\frac{work output}{work input} \times \frac{100}{1}\)
Y = \(\frac{250 \times 30}{250 \times 30 + 1500} \times \frac{100}{1}\)
Source: Myschool Past Questions Archive (Record #27385)
Physics
UTME / JAMB 2012
Mechanics and Properties of Matter
Solution
If a load of 1 kg stretches a cord by 1.2cm, what is the force constant of the cord? {g=10ms-2}
A
866Nm-1
B
833Nm-1
C
769Nm-1
D
667Nm-1
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
K = \(\frac{F}{e}\)
K = \(\frac{1 \times 10}{\frac{0.2}{100}}\)
K = \(\frac{10}{12} \times \frac{1000}{1}\)
Source: Myschool Past Questions Archive (Record #27386)
Physics
UTME / JAMB 2012
Mechanics and Properties of Matter
Solution
An object of volume 1 m 3 and mass 2kg is totally immersed in a liquid of density 1 kgm-3 . Calculate its apparent weight.
A
20N
B
10N
C
2N
D
1 N
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
M = DV = 1 x 1 = 1kg
App Wt = (2 - 1)g = 1 x 10
= 10
Source: Myschool Past Questions Archive (Record #27387)
Physics
UTME / JAMB 2012
Mechanics and Properties of Matter
The pressure at any point in a liquid at rest depends only on the
A
depth and the density
B
mass and the volume
C
quantity and the surface area
D
surface area and the viscosity
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 #27388)
Physics
UTME / JAMB 2012
Mechanics and Properties of Matter
Solution
A balloon whose volume is 300m3 is filled with hydrogen. If the density of air is 1.3 kgm-3, find the upthrust on the balloon. [g = 10ms-2]
A
3000N
B
3800N
C
3900N
D
4200N
medium Difficulty
Verified Correct Answer: Option C
Physical Principles & Numerical Solution
F = mg = D x V x g = 1.3 x 300 x 10 = 3900
Source: Myschool Past Questions Archive (Record #27389)