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Physics
WAEC / WASSCE 2008
Mechanics and Properties of Matter
Solution
The horizontal component of the initial speed of a particle projected at 30^\circ to the horizontal is 50 m\(^{-1}\). If the acceleration of free fall due to gravity is 10 ms\(^{-2}\), determine its;
(a) initial speed;
(b) speed at the maximum height reached.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) Horizontal component
\(U_x = U cos \theta\)
U = \(\frac{U_x}{cos0} = \frac{50}{cos 30^o}\) = 57.7 m/s
(b) At maximum height
\(v_y = O\) and \(V_x = U_x\) = 50m/s
V = \(\sqrt{V^2_x + V^2_v} = \sqrt{50^2 + 0^2}\)
= 50m/s
Source: Myschool Past Questions Archive (Record #53537)
Physics
WAEC / WASSCE 2008
Mechanics and Properties of Matter
Solution
A ball bearing falls through a viscous liquid
(a) Using a labelled diagram of a tall vessel, show all the forces acting on it
(b) When will it attain terminal velocity?
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)
(b) Terminal velocity is attained when W = U + V
Source: Myschool Past Questions Archive (Record #53539)
Physics
WAEC / WASSCE 2008
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
Energy of electron = eV = 1.60 x 10\(^{-19}\) x 1.0 x 10\(^{5}\)
= 1.60 x 1 0\(^{-14}\)J
Source: Myschool Past Questions Archive (Record #53549)
Physics
WAEC / WASSCE 2008
Mechanics and Properties of Matter
Solution
a) On what principle does lighting in a fluorescent tube operate?
(b) State two factors which determine the colour of light produced in a fluorescent tube.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) The princle on which lighting in a fluorescent operation is electrical discharge through a gas at low pressure and at high voltage.
(b) The colour of light produced in a fluorescent tube depends on factors;
(i) coating inside the tube; and (ii) the nature ype of gas in the tube.
Source: Myschool Past Questions Archive (Record #53551)
Physics
WAEC / WASSCE 2008
Mechanics and Properties of Matter
Solution
(a) State Heisenberg's Uncertainty Principle.
(b) State one phenomenon that can only be explained in terms of the wave nature of light.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Heiscnberg's uncertainty principle states that it ipossible to determine with accuracy both the ion and the momentum of a particle simultaneously i.e; \(\Delta p.\Delta x \geq \frac{h}{2} \pi\) .
(b) The phenomenon that can only explained in terms of the wave nature of light are ference and diffraction.
Source: Myschool Past Questions Archive (Record #53553)
Physics
WAEC / WASSCE 2008
Mechanics and Properties of Matter
Solution
(a) A particle moves on a straight path with an initial speed u and final speed v in time t. Show that the total distance X covered by the particle is given by
\(x = ut + \frac{1}{2}at^2\)
where a is the magnitude of acceleration
(b) State;
(i) Newton's second law of motion
(ii) the principle of conservation of energy
(iii) the law of floatation.
(c) Consider a balloon of mass 0.030 kg being inflated with a gas of density 0.54 kg m\(^{-3}\). What will be the volume of the balloon when it just begins to rise in air of density 1.29 kg m\(^{-3}\)? [ g = 10 ms\(^{-2}\)]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
a) Average velocity = \(\frac{u + u}{2}\)
From equation V = u + at
Aeverage velocity = \(\frac{u + u + at}{2}\) = \(u + \frac{1}{2} at\)
Distance covered = Ave. vel. x time
x = (u + \(\frac{1}{2}\)at)t
x = ut +s
(b)i) Newton's second law of motion states that the of change of momentum is proportional to the lied force and takes place in tho direction of the force.
(ii) The principle of conservation of energy es that energy is neither created nor destroyed can be changed from one form to another. The law of floatation states that an object wil t in a fluid when the upthrust exerted upon it by fluid in which it floats equals the weight of the
(c) Mass of air displaced = mass of ballon on and contents V x density of A air
= V x density of gas + content
V x 1.29 = v x 0.54 + 0.030
V = \(\frac{0.03}{0.75}\) = 0.04m\(^3}\)
Source: Myschool Past Questions Archive (Record #53554)
Physics
WAEC / WASSCE 2008
Mechanics and Properties of Matter
Solution
(a) State two differences between boiling and evaporation.
(b) A closed inexpansibie vessel contains air saturated with water vapour at 77^\circC. The total pressure in the vessel is 1007 mmHg. Calculate the new pressure in the vessel if the temperature is reduced 1.2 27^\circC. [ The s.v.p. of water at 77^\circC and 27^\circC respectively are 314 mmHg and 27 mmHg. Treat the air in the vessel as an ideal gas]
(c) The lengtn of a zinc rod at 23^\circC is 200 m. Calculate the increase in length of the rod when its temperature rises to 33^\circC. [ expansivity of zinc = 2.6 x 10\(^{-5}\) K\(^{-1}\)]
(d) Explain with there is no temperature change when a solid being heated changes into liquid at its melting point.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
a)
Boiling
Evaporation
Takes place at a particular temperature
Takes place through-out the whole volume of the liquid.
Takes place at a steady temperature
Takes place at all temperatures
Takes place only on the surface of the liquid.
Can take place at reduced temperature.
(b) \(P_1 = 1007 - 314 = 693 mmH_g\)
\(\frac{P_1}{T_1} - \frac{P_2}{T_2} = \frac{693}{350} = \frac{P_2}{300}\)
\(p_2 = \frac{963}{350} \times \frac{300}{1}\) = 594 mmH\(_g\)
Pressure in vessel at 27^\circC = pressure of air + S.V.P of water at 27^\circC
= 594 + 27 = 621mm Hg
(c) \(\alpha = \frac{L_2 - L_1}{L_1(\theta_1 - \theta_2)}\)
= 2.6 x 10-5 x 200 (33-23) = 0.052m
(d) At the solid state the molecules are tightly bounded, therefore at the melting point the heat supplied is absorbed and used to break the bonds holding the molecules together.
Source: Myschool Past Questions Archive (Record #53555)
Physics
WAEC / WASSCE 2008
Mechanics and Properties of Matter
Solution
(a) What is a magnetic field?
(b) With the aid of a labelled diagram describe an experiment to show that a magnetic field exists around a straight wire carrying current.
(c) A 40 \(\mu\)F capacitor in series with a 40 Q resistor is connected to a 100 V, 50 Hz a.c. supply.
(i) Draw a circuit diagram of the arrangement
(ii) Calculate the: I. impedance in the circuit; II. current in the circuit III. potential difference across the capacitor.
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a) A magnetic field is a region or space where a magnetic force is experienced.
(b)
Connect a straight wire to the circuit as shown above, sprinkle iron fillings on the cardboard paper round the straight wire; and tap gently. Then close the key and gently tap again, the iron fillings form a concentric circles round the wire. This shows that there is magnetic field round a current carrying conductor.
(c)(i)
\(x_c = \frac{1}{2 \pi fc} = \frac{1}{2 \pi \times 50(40 \times 10^{-6})}\)
= 79.567\(\Omega\)
z = \(\sqrt{R^2 + X^2_c} = \sqrt{40^2 + 79,580^2}\)
(ii) I = \(\frac{v}{2} = \frac{100}{89.1} = 1.12A\)
(iii) V = IX\(_c\) = 1.12 x 79.567
= 89.1V
Source: Myschool Past Questions Archive (Record #53561)
Physics
WAEC / WASSCE 2008
Mechanics and Properties of Matter
Solution
(a)(i) State Einstein's equation of photoelectric effect
(ii) What conservation principle does the equation represent?
(b) List three applications of photocells
(c) A photo emissive surface has a threshold frequency of 4.02 x 10\(^{14}\)Hz. If the surface is illuminated by light of frequency 5.0 x 10\(^{15}\)Hz, calculate the:
(i) threshold wavelength;
(ii) work function;
(iii) kinetic energy of the emitted photoelectrons. [ c = 3.0 x 10\(^{8}\) ms\(^{-1}\), h = 6.63 x 10\(^{-34}\) Js]
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
(a)(i) E = W + K.E, where
E = energy of incident radiation
W = work function of metal
K.E. = Kinetic energy of ejected electrons
(ii) The equation represents the principle of conservation of energy.
(b) Applications of photocells:
1. for burglar alarm
2. for television cameras
3. automatic doors
4. automatic street lighting
5. automatic digital counters.
(c)(i) V = \(f_o\lambda_o\)
\(\lambda_o = \frac{v}{f_o} = \frac{3 \times10^8}{4.02 \times 10^{14}}\)
= 7.46 x 10\(^{-7}\)m
(ii) W = \(hf_o = 6.63 \times 10^{-34} \times 4.02 \times 10^{14}\)
= 2.67 x 10\(^{-19}\)J
(iii) K.E. = \(h (f - f_o) = 6.63 \times10^{-34} (5.0 \times 10^{15} - 4.02 \times 10^{14})\)
= 3.05 x 10\(^{-15}\)J
Source: Myschool Past Questions Archive (Record #53603)
Physics
WAEC / WASSCE 2008
Mechanics and Properties of Matter
Solution
You are provided with a uniform meter rule, a knife edge, masses and other necessary apparatus.
Suspend the metre rule horizontally on the knife edge. Read and record the point of balance G of the métre rule. Keep the knife edge at this point throughout the experiment.
Using the thread provided, suspend the object labelled W at the 15cm mark of the metre rule.
Suspend a mass M= 20g on the other side of G. Adjust the position of the mass until the metre rule balances horizontally again.
Read and record the position Y of the mass M on the metre rule.
Determine and record the distance L between the mass and G. Also determine and record the distance D between W and G.
Repeat the procedure for four other values of M = 30, 40, 50 and 60 g. In each casse ensure that W is kept Constant at the 15 cm mark and the knife edge at G.
Evaluate L\(^{-1}\) in each case. Tabulate your readings.
Plot a graph of M on the vertical axis against L\(^{-1}\) on the horizontal axis.
Determine the slope S, of the graph.
Evaluate \(\frac{4}{D}\)
State two precautions taken to obtain accurate results.
(b)i. State the principle of moments.
ii. Define centre of gravity
medium Difficulty
Verified Correct Answer: Option
Physical Principles & Numerical Solution
Table of values (observation)
G = 48cm
D = 48 - 15 = 33cm
S/N
MCg
YCcm
LCcm
DCcm
L\(^{-1}\)Ccm
1
20
-
-
33.00
-
2
30
81.00
33.00
33.00
0.03
3
40
72.75
24.75
33.00
0.04
4
50
67.80
19.80
33.00
0.05
5
60
64.58
16.50
33.00
0.06
Slope = dm = \(m_2\) - \(m_1\) = 60 - 10 = 50g
d(1\(^{-1}\))\(_{2}\) - (1\(^{-1}\))\(_{1}\) = 0.06 - 0.01 = 0.05cm\(^{-1}\)
= 1000gcm
\(\frac{S}{D} = \frac{1000}{33}\) 30.3g
Precautions: i
Draught was avoided.
Parallax error on reading metre rule/was avoided.
Repeated readings (must be shown on table).
Ensured suspended masses did not touch the table.
(b) The principle of moments states that for a body in equilibrium, the sum of the clockwise moments about a point is equal to the sum of the anticlockwise moments about the same point.
ii. The centre of gravity of a body is the point through which its resultant weight acts.
Source: Myschool Past Questions Archive (Record #60034)
Physics
UTME / JAMB 2007
Mechanics and Properties of Matter
Solution
A gramophone record takes 5s to reach its constant angular velocity of 4\(\pi\) rads-1 from rest. Find its constant angular acceleration.
A
0.4\(\pi\) rads-2
B
0.8\(\pi\) rads-2
C
1.3\(\pi\) rads-2
D
20.0\(\pi\) rads-2
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
From the radiation:
V = +at, and if it starts from N & T, U = 0
And for angular motion, W = O + at
4R = aχ5
∴ a = (4a)/5 = 0. 8\(\pi\) rads-2
Source: Myschool Past Questions Archive (Record #13903)
Physics
UTME / JAMB 2007
Mechanics and Properties of Matter
Solution
A car of mass 1500 kg goes round a circular curve of radius 50m at a speed of 40 ms-1. The magnitude of the centripetal force on the car is ?
A
4.8 x 104 N
B
4.8 x 103 N
C
1.2 x 103 N
D
1.2 x 102 N
medium Difficulty
Verified Correct Answer: Option A
Physical Principles & Numerical Solution
Centipetal force of the car = ma = \(\frac{(MV^2)}{\text{r}}\)
= \(\frac{(1500 \times 40 \times 40}{50}\)
= 48000 = 4.8 x 10\(^4\) N
Source: Myschool Past Questions Archive (Record #13904)
Physics
UTME / JAMB 2007
Mechanics and Properties of Matter
Solution
Two spheres of masses 5.0 kg and 10.0 kg are 0.3 m apart. Calculate the force of attraction between them?
[G = 6.67 X 10-11 Nm2 Kg\(^{-2}\)]
A
3.50 x 10-10N
B
3.71 x 10-8 N
C
3.57 x 10-2N
D
4.00 x 102N
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
The force of attraction between the two spheres is given by F (GM1 M2)/r2 = (6.67X 10-11 X5X10)/(0.3)2
= 3.71 X 10-8N.
Source: Myschool Past Questions Archive (Record #13909)
Physics
UTME / JAMB 2007
Mechanics and Properties of Matter
Solution
A machine has a velocity ratio of 4. if it requires 800 N to overcome a load of 1600 N,what is the efficiency of the machine?
A
60%
B
50%
C
40%
D
2%
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
Efficiency = (M.A)/V.R X (100)/1
But M.A = (Load)/Effort = (1600)/800 = 2
Efficiency = (2)/4 x (100)/1 = 50%
Source: Myschool Past Questions Archive (Record #13916)
Physics
UTME / JAMB 2007
Mechanics and Properties of Matter
Solution
A reservoir 500 m deep is filled with a fluid of density 850 kgm-3. If the atmospheric pressure is 1.05 x 105 Nm-2, the pressure at the bottom of the reservoir is.
[g = 10ms-2]
A
4.72 x 106 Nm-2
B
4.36 x 106 Nm-2
C
4.28 x 106 Nm-2
D
4.25 x 106 Nm-2
medium Difficulty
Verified Correct Answer: Option B
Physical Principles & Numerical Solution
In general, pressure varies with depth in the relation p = \(\rho\)gh
= 850 x 10 x 500
= 4250000
= 4.25 x 10\(^6\) Nm\(^{-2}\)
Therefore, Total pressure = 1.05 x 10\(^5\) + 4.25 x 10\(^6\) = 4.36 x 10\(^6\) Nm\(^{-2}\)
Source: Myschool Past Questions Archive (Record #13917)