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Physics
WAEC / WASSCE 2006
Mechanics and Properties of Matter
An electron of mass m and charge \(\Box\) enters a uniform electric field between two metal plates P and q separated by a distance d. P is maintained at a potential V while Q is earthed. Determine an expression for the magnitude of the acceleration of electron through the field
A
\(\frac{eV}{md}\)
B
\(\frac{d}{meV}\)
C
\(\frac{md}{eV}\)
D
\(\frac{e}{Vmd}\)
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 #34692)
Physics
WAEC / WASSCE 2006
Mechanics and Properties of Matter
The force experienced by a current-carrying conductor move in a magnetic field is employed in the working of the i. moving coil ammeter ii. electric bell iii. electric motor. Which of the statements above are correct?
A
i and ii only
B
ii and iii only
C
i and iii only
D
i, ii and iii
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 #34699)
Physics
WAEC / WASSCE 2006
Mechanics and Properties of Matter
Solution
A 500kV is applied across an X-ray tube. Calculate the maximum velocity of the electrons produced [Me = 9.1 x 10\(^{-31}\)kg, e = 1.6 x 10\(^{-19}\)C]
A
4.2 x 108ms-1
B
1.9 x 108ms-1
C
4.2 x 105ms-1
D
1.8 x 105ms-1
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
500Kv = 5 x 10\(^5\)V
E = Qv = \(\frac{1}{2}\)mv\(^2\)
E = 1.6 x 10\(^{-19}\) x 9.1 x 10\(^{-31}\) = 8 x 10\(^{-14}\)
8 x 10\(^{-14}\) = \(\frac{1}{2}\)mv\(^2\)
8 x 10\(^{-14}\) = \(\frac{1}{2}\) x 9.1 x 10\(^{-31}\) x v\(^2\)
8 x 10\(^{-14}\) = 4.55 x 10\(^{-31}\) x v\(^2\)
v\(^2\) = \(\frac{8 \times 10^{-14}}{4.55 \times 10^{-31}}\) = 1.758 x 10\(^{17}\)
v = \(\sqrt{1.758 \times 10^{17}}\) = 4.2 x 10\(^8\)m/s
Source: Myschool Past Questions Archive (Record #34701)
Physics
WAEC / WASSCE 2006
Mechanics and Properties of Matter
Fusion is not usually used for generating electric power because
A
no energy is released when fusion takes place
B
very high temperatures are required to initiate fusion reaction
C
the raw materials needed is not easily available
D
heavy nuclei are involved
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #34707)
Physics
WAEC / WASSCE 2006
Mechanics and Properties of Matter
Solution
(a) Explain the statement the acceleration of free fall cohesion.
(b) State two factors that can affect the value of the narrow glass tube has a concave meniscus while acceleration of free fall at a place.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) This means that at the equator, the force of gravity on a body causes its velocity to increase at the rate of 9.78m/s per second.
(b)(i) The rotation of the earth on its axis.
(ii) The earth is not a perfect sphere.
(iii) The altitude of a place
Source: Myschool Past Questions Archive (Record #53714)
Physics
WAEC / WASSCE 2006
Mechanics and Properties of Matter
Solution
The diagram below represents the graph of the force applied in stretching a spiral spring against the corresponding extension produced within its elastic limit.
Using the notations on the graph, determine the:
(a) force constant of the spring;
(b) work done in stretching the spring from 10 x 10\(^{-2}\)m to 20 x 10\(^{-2}\)m.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
a) Force constant of the spring = slope of the graph = \(\frac{\Delta force}{\Delta Extension}\)
= \(\frac{60 — 0}{20 - 0 \times 10^{-2}} = \frac{60}{20 \times 10^{-2}}\)
= 300Nm\(^{-2}\)
(b) Work done = Area under the trapezium ABCD
= 1/2 (30 + 60) x (20 — 10) x 10\(^{-2}\) = 4.5J
Source: Myschool Past Questions Archive (Record #53723)
Physics
WAEC / WASSCE 2006
Mechanics and Properties of Matter
Solution
(a) Distinguish between the forces of adhesion and cohension
(b) Give one example each of the forces of adhesion and cohesion.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Forces of adhesion are forces between molecules of different substances while forcea of cohesion are forces between molecules of the same substance.
(b) Examples: Adhesion — force between water molecules and glass molecules.
Cohesion — force between mercury molecules.
Source: Myschool Past Questions Archive (Record #53724)
Physics
WAEC / WASSCE 2006
Mechanics and Properties of Matter
Solution
Explain why water in a narrow glass tube has a concave meniscus while mecury in the same tube, has a convex meniscus
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
For water in a glass tube, the forces of adhesion between the water and glass molecules are greater than the force of cohesion between the water molecules hence water clings to glass forming a concave meniscus. On the other hand, the forces of cohesion between mercury molecules are greater than those between mercury molecules and glass molecules, so mercury molecules cling to each other forming convex meniscus.
Source: Myschool Past Questions Archive (Record #53725)
Physics
WAEC / WASSCE 2006
Mechanics and Properties of Matter
Solution
The uncertinty 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 [Take \(\frac{h}{2\pi} = h = 1.054 \times 10^{-34}\) Js]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
\(\Delta E \geq \frac{h}{2\pi} \times \frac{1}{2t} = 1.054 x 10^{-37} \times \frac{1}{2.0 \times 10^{-9}}\)
\(\Delta E \geq 5.27 \times 10^{-25}\)J
Source: Myschool Past Questions Archive (Record #53727)
Physics
WAEC / WASSCE 2006
Mechanics and Properties of Matter
Solution
(a) State two;
(i) laws of solid friction;
(ii) advantages of friction;
(iii) methods of reducing friction.
(b) Draw and label a diagram of a pulley system with velocity ratio of 5.
(c)(i) Show that the efficiency L, force ratio M.A. and the velocity ratio V.R. of a machine are related by the equation \(E = \frac{M.A.}{V.R}\) x 100%.
(ii) The efficiency of a machine is 80%. Calcuate the work done by a person using the machine to raise a load of 300 kg through a height of 4 m.[ g = 10 ms\(^{-2}\) ]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) Laws of solid friction states that:
— friction opposes motion
— friction depends on the nature of e surfaces
— friction is independent of the area of the surfaces in contact
— frictional force F is proportional to the normal reaction R
— friction is dependent of relative velocity between the surfaces.
ii) Advantages of friction:
— it prevents man from ipping when walking
— it provides gripping effect of beIts in machines
— for grinding when using grinding ones
— used for the production of electric charges used in the brake systems of vehicles etc.
iii) Methods of reducing friction:
— by lubrication by streamlining
— by smoothing
— by using ball bearings or rollers.
b)
(c)(i) Efficiency = \(\frac{\text{Work output}}{\text{work output}}\) x 100%
= \(\frac{\text{load x distance by load}}{\text{effort x distance moved by effort}}\)
= M.A x \(\frac{1}{VR} \times \frac{100 %}{1}\)
(ii) \(e = \frac{\text{Work output}}{\text{Work input}}\) x 100
Work input = \(\frac{\text{Work output}}{E}\) x 100
= \(\frac{300 \times 10 \times 4 \times 100}{80}\)
= 15,000J or 15KJ
Source: Myschool Past Questions Archive (Record #53736)
Physics
WAEC / WASSCE 2006
Mechanics and Properties of Matter
Solution
(a) Define
(i) proton number;
(ii) nucleon number;
(iii) isotopes.
(b) A nuclide \(^A_ZX\) emits \(\beta\)-particle to form a daughter nuclide Y. Write a nuclear equation to illustrate the charge conservation.
(c) The radioactive nuclei \(^{210}_{84}P_o\) emits an \(\alpha\) - particle to produce \(^{206}_{82}P_b\). Calculate the energy, in MeV, released in each disintegration.
Take the masses of \(^{210}_{84}P_o\) = 209.936730 u;
\(^{206}_{82}P_b\) = 205.929421 u;
\(^{4}_{2}He\) = 4.001504 u;
and that 1u = 931 MeV
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) Proton number
— the number of protons in the nucleus of an atom
(ii) Nucleon number — the total number of protons and neutrons in the nucleus of are atom.
(iii) Isotopes — are atoms or nuclides of the samc element having the same number of protons or atomic number but different neutrons or mass numbers
(b) \(^A_Z \to ^A_{Z+}Y + ^0_{-1}\)
\(^{210}_{84}Po \to ^{206}_{82}Pb + ^4_2He + Energy\)
Mass detect = 209.936730u - (205.929421 + 4.001504)u
209.9367330u - 209.930925u
= 5.805 x 10\(^{-3}\)U
Energy released = 5.805 x 10\(^{-3}\) x 931
= 5.40MeV
Source: Myschool Past Questions Archive (Record #53740)
Physics
WAEC / WASSCE 2006
Mechanics and Properties of Matter
Solution
(a) State the conditions for the equilibrium of a rigid body acted upon by parallel forces.
(b)(i) Describe an experiment to determine the mass of a metre rule using the principle of moments.
(ii) State two precautions necessary to obtain accurate results in the experiment described in (b)(i) above.
(c) A bullet of mass 120 g is fired horizontally into a fixed wooden block with a speed of 20 ms\(^{-1}\). If the bullet is brought to rest in the block in 0.1s by a constant resistance, calculate the (i) magnitude of the resistance; (ii) distance moved by the bullet in the wood.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) — The sum of clockwise moments about any point is equal to the sum of anticlockwise moments about the same point
— The sum of the forces in one recticn is equal to the sum of the forces in the opposite direction.
b)
Method: Suspend the metre rule on a thread loop from clamp. Adjust its position until the metre rule balances horizontally and obtain its centre of gravity G. Suspend known mass M at a point on one side of the rule and just the position of the loop until the rule balances horizontally. Read and record the distances y and x in the diagram above. sing the principle of moments Mgx = Mgy
m = \(\frac{My}{x}\)
Therefore the mass M of the metre rule is calculated.
ii) Precautions:
— Firmly fixed the loop
— Avoid draught
— Avoid parallax when reading metre rule.
.
c)(i) Declaration a = \(\frac{v - u}{t} = \frac{0 — 20}{0.1} = — 200m/s^2\)
Resistance F = Ma = 0.12 x 200 = 24N
ii) Distance S = \(\frac{v^2 - u^2}{2a} = \frac{0^2 - 20^0}{2 \times 200}\)
= \(\frac{400}{400} = 1.0m\)
Source: Myschool Past Questions Archive (Record #53741)
Physics
WAEC / WASSCE 2006
Mechanics and Properties of Matter
Solution
You are provided with a constantan wire, a 2\(\Omega\) standard resistor, an accumulator E, an ammeter A, a key K, and other necessary apparatus.
Measure and record the e.m.f.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 I = 90cm, close the key. Read and record the ammeter reading /. Evaluate l\(^{-1}\).
Repeat the procedure for l= 80, 70, 60 and 50cm.
In each case, read and record the ammeter reading and evaluate l\(^{-1}\). Tabulate your readings.
Plot a graph of l on the vertical axis against l\(^{-1}\) 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 / = 55cm.
State two precautions taken to ensure 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 resistance wire depends.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Table of values (Observation)
\(l_o\) (amps)
l(cm)
l\(^{-1}\)
l(amps)
0.65
90
0.011
0.70
0.70
80
0.013
0.75
0.75
70
0.014
0.80
0.80
60
0.017
0.85
0.85
50
0.020
0.90
7.
(8). Slope (s) = \(\frac{\bigtriangleup {I}}{\bigtriangleup {/}^{-1}} = \frac{0.65}{0.024}\) = 27.1
(9). Intercept (c) on the vertical axis = 0.43 amps
(10). Evaluate, K = \(\frac{c}{s} = \frac{0.43}{27.1}\) = 0.016
(11) The value of the current l when l = 55cm, l\(^{-1}\) = 0.018cm,
Therefore; l on the graph = 0.875 amps
Precautions:
The key was removed when the reading were not being teken.
I ensured that terminals were clean.
I ensured that the connetions were tight.
Zero error was noted and corrected in the ammeter/ metre rule
Parallax error was avoided in anmmeter/metre rüle.
(b)i. The potential differenoe between two points in an electric ciroult the work done in Joules when one coulomb of electric charge moves from one point to the other
ii. Factors on which the resistance of a resistance wire depends are:
- resistivity of the wire/ nature of the wire
- length of the wire.
- cross sectional area of the wire /thickness of the wire.
- temperature
Source: Myschool Past Questions Archive (Record #60073)
Physics
WAEC / WASSCE 2006
Mechanics and Properties of Matter
Solution
You have been provided with a metre rule, a clamp, and a set of masses.
Clamp the metre rule to the edge of the bench such that 90cm of the rule projects from the edge as shown in the diagram above. Ensure that the rule is capable of performing oscillatory motion.
Fix a mass M = 50g at the free end of the rule.
Deflect the rule slightly such that it performs vertical oscillation.
Determine the time t for 10 complete oscillations.
Calculate the period T of the oscillations and evaluate T\(^{2}\)
Repeat the procedure for four other values of M = 100, 150, 200, and 250g. In each case determine and record the corresponding values of t, T, and T\(^{2}\). Tabulate your readings.
Plot a graph of T\(^{2}\) on the vertical axis against M on the horizontal axis, starting both axes from the origin (0,0).
Determine the slopes, of the graph and its intercept C on the vertical axis.
Evaluate k = 4\(\pi\)/s. [Take \(\pi\) = \(\frac{22}{7}\)].
From your graph, determine the period T, when M= 180g.
State two precautions taken to ensure accurate results.
(b)i. Explain simple harmonic motion.
ii. Define period and frequency, with respect to a simple harmonic motion.
Precautions:
l ensured that the metre rule was firmly clamped
Readings were repeated
Parallax was avoided when readings on the stopwatch/clock were taken.
zero error was noted and corrected on the stopwatch/clock.
(b)i. Simple harmonic motion is a motion in which the acceleration is proportional to the displacement from a fixed point and is directed towards the point.
ii. Period is the time taken by an oscillatory body to make one complete oscillation.
Frequency: is the number of complete oscillations performed in one second.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Table of values (observation
Mass (M)g
Time (t) sec.
T
T\(^{2}\)
1
3.5
0.35
0.123
2
4.2
0.42
0.176
3
4.9
0.49
0.240
4
5.5
0.55
0.303
5
6.0
0.60
0.360
Slope (s) = \(\frac{y}{x} = \frac{\bigtriangleup {T}^{2}}{\bigtriangleup{M}} = \frac{0.26}{220}\) = 0.001
Intercept (c) on the vertical axis = 0.65
Evaluate, K = \(\frac{4 \pi}{S} = \frac{4 \times {\frac{22}{7}}}{0.001}\)
\(\frac{12.57}{0.0001}\) = 12571
From the graph the value of period T, where M = 180g is;
= \(\sqrt 0.275\) = 0.524, i.e T = 0.524
Source: Myschool Past Questions Archive (Record #60062)
Physics
UTME / JAMB 2005
Mechanics and Properties of Matter
Solution
An object is projected from a height of 80 m above the ground with a velocity of 40ms-1 at an angle of 30^\circ to the horizontal. The time of flight [ g ≈ 10m/s\(^2\)
A
8 s
B
4 s
C
16 s
D
10 s
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
for a body projected at an angle \theta to the horizontal, the time of the flight is given by T = (2u sin \theta)/g =(2x 40)x sin (30)/10 =(2 x 40) x (0.5)/10 = 4.0secs.
Source: Myschool Past Questions Archive (Record #13681)