WAEC SSCE - Physics - 1988 (Essay)
Question 1 Report
(a) Explain with the aid of a diagram what is meant by the moment of a force about a point.
(b) State the conditions of equilibrium for a number of coplanar parallel forces.
A metre rule is found to balance at the 48cm mark. When a body of mass 60g is suspended at the 6cm mark, the balance point is found to be at the 30cm mark. Calculate:
(i) the mass of the metre rule; (ii) the distance of the balance point from the zero end, if the body were moved to the 13cm mark.
(c) Show that the efficiency E, the force ratio M.A and the velocity ratio V.R of a machine are related by the equation \(E = \frac{M.A}{V.R} \times 100%\)
The efficiency of a machine is 80%. Determine the work done by a person using this machine to raise a load of 200kg through a vertical distance of 3.0m.
[Take g = 10ms\(^{-2}\)]
(a) Moment of a force about a point
The moment of a force about a point is the turning effect of the force about that point. It is equal to the product of the force and the perpendicular distance of its line of action from the point.
\[\text{Moment}=F\times d\]
where \(F\) is the force and \(d\) is the perpendicular distance from the pivot to the line of action of the force. The SI unit is newton metre, \(\text{N m}\).
(b) Conditions for equilibrium of coplanar parallel forces
- The algebraic sum of the forces must be zero, that is, total upward force equals total downward force.
- The algebraic sum of moments about any point must be zero, that is, total clockwise moment equals total anticlockwise moment.
The metre rule balances at the 48 cm mark, so its weight acts at the 48 cm mark.
(i) Mass of the metre rule
Taking moments about the balance point at 30 cm:
\[60(30-6)=M(48-30)\]
\[60\times24=M\times18\]
\[M=\frac{60\times24}{18}=80\text{ g}\]
Therefore, the mass of the metre rule is 80 g.
(ii) New balance point
Let the new balance point be \(x\) cm from the zero end. The 60 g body is at 13 cm and the rule's weight acts at 48 cm.
Taking moments about the new balance point:
\[60(x-13)=80(48-x)\]
\[60x-780=3840-80x\]
\[140x=4620\]
\[x=33\text{ cm}\]
Therefore, the balance point is 33 cm from the zero end.
(c) Relation between efficiency, mechanical advantage and velocity ratio
\[E=\frac{\text{work output}}{\text{work input}}\times100\%\]
\[E=\frac{\text{Load}\times\text{distance moved by load}}{\text{Effort}\times\text{distance moved by effort}}\times100\%\]
\[E=\left(\frac{\text{Load}}{\text{Effort}}\right)\left(\frac{\text{distance moved by load}}{\text{distance moved by effort}}\right)\times100\%\]
Since \(\text{M.A.}=\frac{\text{Load}}{\text{Effort}}\) and \(\text{V.R.}=\frac{\text{distance moved by effort}}{\text{distance moved by load}}\),
\[\boxed{E=\frac{\text{M.A.}}{\text{V.R.}}\times100\%}\]
Work done by the person
Work output in raising the load is:
\[W_{\text{out}}=mgh=200\times10\times3.0=6000\text{ J}\]
\[80\%=\frac{W_{\text{out}}}{W_{\text{in}}}\times100\%\]
\[0.80=\frac{6000}{W_{\text{in}}}\]
\[W_{\text{in}}=\frac{6000}{0.80}=7500\text{ J}\]
Therefore, the work done by the person is \(7.5\times10^3\text{ J}\).
Answer Details
(a) Moment of a force about a point
The moment of a force about a point is the turning effect of the force about that point. It is equal to the product of the force and the perpendicular distance of its line of action from the point.
\[\text{Moment}=F\times d\]
where \(F\) is the force and \(d\) is the perpendicular distance from the pivot to the line of action of the force. The SI unit is newton metre, \(\text{N m}\).
(b) Conditions for equilibrium of coplanar parallel forces
- The algebraic sum of the forces must be zero, that is, total upward force equals total downward force.
- The algebraic sum of moments about any point must be zero, that is, total clockwise moment equals total anticlockwise moment.
The metre rule balances at the 48 cm mark, so its weight acts at the 48 cm mark.
(i) Mass of the metre rule
Taking moments about the balance point at 30 cm:
\[60(30-6)=M(48-30)\]
\[60\times24=M\times18\]
\[M=\frac{60\times24}{18}=80\text{ g}\]
Therefore, the mass of the metre rule is 80 g.
(ii) New balance point
Let the new balance point be \(x\) cm from the zero end. The 60 g body is at 13 cm and the rule's weight acts at 48 cm.
Taking moments about the new balance point:
\[60(x-13)=80(48-x)\]
\[60x-780=3840-80x\]
\[140x=4620\]
\[x=33\text{ cm}\]
Therefore, the balance point is 33 cm from the zero end.
(c) Relation between efficiency, mechanical advantage and velocity ratio
\[E=\frac{\text{work output}}{\text{work input}}\times100\%\]
\[E=\frac{\text{Load}\times\text{distance moved by load}}{\text{Effort}\times\text{distance moved by effort}}\times100\%\]
\[E=\left(\frac{\text{Load}}{\text{Effort}}\right)\left(\frac{\text{distance moved by load}}{\text{distance moved by effort}}\right)\times100\%\]
Since \(\text{M.A.}=\frac{\text{Load}}{\text{Effort}}\) and \(\text{V.R.}=\frac{\text{distance moved by effort}}{\text{distance moved by load}}\),
\[\boxed{E=\frac{\text{M.A.}}{\text{V.R.}}\times100\%}\]
Work done by the person
Work output in raising the load is:
\[W_{\text{out}}=mgh=200\times10\times3.0=6000\text{ J}\]
\[80\%=\frac{W_{\text{out}}}{W_{\text{in}}}\times100\%\]
\[0.80=\frac{6000}{W_{\text{in}}}\]
\[W_{\text{in}}=\frac{6000}{0.80}=7500\text{ J}\]
Therefore, the work done by the person is \(7.5\times10^3\text{ J}\).
Question 2 Report
(a) Draw a labelled diagram of a vacuum flask. Explain how its construction minimizes heat exchange with the surroundings.
(b) State Boyle's law. A thread of mercury of length 15cm is used to trap some air in a capillary tube with uniform cross- sectional area and closed at one end with the tube vertical and the open end uppermost, the length of the trapped air column is 20cm. Calculate the length of the air column when the tube is held:
(i) Horizontally ; (ii) vertically with the open end underneath. [Atmospheric pressure = 76cm Hg]
(c) Explain why it is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure.
(a) Labelled diagram of a vacuum flask
A vacuum flask minimizes heat exchange as follows:
- The vacuum between the double glass walls prevents heat transfer by conduction and convection.
- The silvered surfaces reflect thermal radiation, thereby reducing heat transfer by radiation.
- The cork or plastic stopper is a poor conductor of heat, while the narrow neck reduces convection and evaporation through the top.
(b) Boyle's law
Boyle's law states that the volume of a fixed mass of gas is inversely proportional to its pressure, provided that its temperature remains constant.
Thus,
\[PV=\text{constant}\qquad\text{or}\qquad P_1V_1=P_2V_2.\]
Since the capillary tube has a uniform cross-sectional area, the volume of the trapped air is proportional to its length. Therefore,
\[P_1L_1=P_2L_2.\]
Initially, with the open end uppermost, the pressure of the trapped air is:
\[P_1=76+15=91\ \text{cm Hg},\qquad L_1=20\ \text{cm}.\]
(i) Tube held horizontally
There is no vertical mercury head acting on the trapped air, so:
\[P_2=76\ \text{cm Hg}.\]
\[91\times20=76\times L_2\]
\[L_2=\frac{1820}{76}=23.95\ \text{cm}\approx24\ \text{cm}.\]
Length of air column = \(24\ \text{cm}\).
(ii) Tube held vertically with the open end underneath
The mercury thread now reduces the pressure on the trapped air:
\[P_2=76-15=61\ \text{cm Hg}.\]
\[91\times20=61\times L_2\]
\[L_2=\frac{1820}{61}=29.84\ \text{cm}\approx30\ \text{cm}.\]
Length of air column = \(30\ \text{cm}\).
(c) A clinical thermometer has a limited range, usually about \(35^\circ\text{C}\) to \(42^\circ\text{C}\). In boiling water at \(100^\circ\text{C}\), the mercury expands excessively and may force its way into the upper end of the capillary tube, cracking or damaging the thermometer. It should therefore not be sterilized in boiling water.
Answer Details
(a) Labelled diagram of a vacuum flask
A vacuum flask minimizes heat exchange as follows:
- The vacuum between the double glass walls prevents heat transfer by conduction and convection.
- The silvered surfaces reflect thermal radiation, thereby reducing heat transfer by radiation.
- The cork or plastic stopper is a poor conductor of heat, while the narrow neck reduces convection and evaporation through the top.
(b) Boyle's law
Boyle's law states that the volume of a fixed mass of gas is inversely proportional to its pressure, provided that its temperature remains constant.
Thus,
\[PV=\text{constant}\qquad\text{or}\qquad P_1V_1=P_2V_2.\]
Since the capillary tube has a uniform cross-sectional area, the volume of the trapped air is proportional to its length. Therefore,
\[P_1L_1=P_2L_2.\]
Initially, with the open end uppermost, the pressure of the trapped air is:
\[P_1=76+15=91\ \text{cm Hg},\qquad L_1=20\ \text{cm}.\]
(i) Tube held horizontally
There is no vertical mercury head acting on the trapped air, so:
\[P_2=76\ \text{cm Hg}.\]
\[91\times20=76\times L_2\]
\[L_2=\frac{1820}{76}=23.95\ \text{cm}\approx24\ \text{cm}.\]
Length of air column = \(24\ \text{cm}\).
(ii) Tube held vertically with the open end underneath
The mercury thread now reduces the pressure on the trapped air:
\[P_2=76-15=61\ \text{cm Hg}.\]
\[91\times20=61\times L_2\]
\[L_2=\frac{1820}{61}=29.84\ \text{cm}\approx30\ \text{cm}.\]
Length of air column = \(30\ \text{cm}\).
(c) A clinical thermometer has a limited range, usually about \(35^\circ\text{C}\) to \(42^\circ\text{C}\). In boiling water at \(100^\circ\text{C}\), the mercury expands excessively and may force its way into the upper end of the capillary tube, cracking or damaging the thermometer. It should therefore not be sterilized in boiling water.
Question 3 Report
(a) Explain with the aid of a diagram how a converging lens could be used to : (i) ignite a piece of carbon paper ; (ii) produce an enlarged picture on a screen ; (iii) correct an eye defect.
(b) What is a mechanical wave? Describe with the aid of a diagram, an experiment to show that sound needs a material medium for transmission.
State 3 characteristics of sound and mention the factor on which each depends.
(a) Uses of a converging lens
(i) Igniting carbon paper: Hold the carbon paper at the principal focus, F, of the converging lens and direct the lens towards the Sun. The parallel rays from the Sun are brought to a very small bright spot at F. The heat energy is concentrated there and ignites the carbon paper.
(ii) Producing an enlarged picture on a screen: Place the object or slide between F and 2F of the converging lens. A real, inverted and magnified image is formed beyond 2F; a screen is placed at this image position.
(iii) Correcting an eye defect: A converging lens is used to correct long-sightedness (hypermetropia). In a long-sighted eye, rays from a near object would be focused behind the retina. The spectacle lens converges the rays before they enter the eye, so that the eye lens forms the image on the retina.
(b) Mechanical wave: A mechanical wave is a disturbance that requires a material medium, such as a solid, liquid or gas, for its propagation. It cannot travel through a vacuum. Examples are sound waves and water waves.
Experiment showing that sound needs a material medium: An electric bell is placed inside an airtight bell jar connected to a vacuum pump. When the switch is closed, the bell rings and its hammer is seen striking the gong. While air is present in the jar, the sound is heard clearly. As the pump removes the air, the sound becomes fainter. When the jar is nearly evacuated, the hammer is still seen striking the gong but little or no sound is heard. Therefore, sound requires a material medium, in this case air, for transmission.
Characteristics of sound
- Loudness (intensity): depends on the amplitude of vibration.
- Pitch: depends on the frequency of vibration.
- Quality (timbre): depends on the number and relative strengths of the overtones or harmonics present.
Answer Details
(a) Uses of a converging lens
(i) Igniting carbon paper: Hold the carbon paper at the principal focus, F, of the converging lens and direct the lens towards the Sun. The parallel rays from the Sun are brought to a very small bright spot at F. The heat energy is concentrated there and ignites the carbon paper.
(ii) Producing an enlarged picture on a screen: Place the object or slide between F and 2F of the converging lens. A real, inverted and magnified image is formed beyond 2F; a screen is placed at this image position.
(iii) Correcting an eye defect: A converging lens is used to correct long-sightedness (hypermetropia). In a long-sighted eye, rays from a near object would be focused behind the retina. The spectacle lens converges the rays before they enter the eye, so that the eye lens forms the image on the retina.
(b) Mechanical wave: A mechanical wave is a disturbance that requires a material medium, such as a solid, liquid or gas, for its propagation. It cannot travel through a vacuum. Examples are sound waves and water waves.
Experiment showing that sound needs a material medium: An electric bell is placed inside an airtight bell jar connected to a vacuum pump. When the switch is closed, the bell rings and its hammer is seen striking the gong. While air is present in the jar, the sound is heard clearly. As the pump removes the air, the sound becomes fainter. When the jar is nearly evacuated, the hammer is still seen striking the gong but little or no sound is heard. Therefore, sound requires a material medium, in this case air, for transmission.
Characteristics of sound
- Loudness (intensity): depends on the amplitude of vibration.
- Pitch: depends on the frequency of vibration.
- Quality (timbre): depends on the number and relative strengths of the overtones or harmonics present.
Question 4 Report
(a) Explain what is meant by: (i) electric field intensity ; (ii) electric lines of force.
(b) Two similar but opposite point charges -q and +q each of magnitude \(5 \times 10^{-8} C\) are seperated by a distance of 8.0cm in vacuum as shown in the diagram below.
Calculate the magnitude and direction of the resultant electric field intensity E at the point P. Draw the lines of force due to this system of charges. [Take \(\frac{1}{4 \pi \varepsilon _{0}}\)]
(c) 
Calculate the following in the series circuit shown above: (i) reactance of the capacitor ; (ii) impedance of the circuit ; (iii) current through the circuit ; (iv) voltage across the capacitor ; (v) average power used in the circuit.
(a)(i) Electric field intensity, E, at a point is the force experienced per unit positive test charge placed at that point.
\[E=\frac{F}{q}\]
It is a vector quantity. Its SI unit is \(\mathrm{N\,C^{-1}}\) (or \(\mathrm{V\,m^{-1}}\)), and its direction is the direction of the force on a positive test charge.
(ii) Electric lines of force are imaginary lines used to show an electric field. The tangent to a field line at any point gives the direction of the field. They emerge from positive charges and terminate on negative charges; closer spacing indicates a stronger field, and field lines never intersect.
(b) Point \(P\) lies between the charges, \(0.050\,\mathrm{m}\) from \(-q\) and \(0.030\,\mathrm{m}\) from \(+q\). Taking \(k=\dfrac{1}{4\pi\varepsilon_0}=9.0\times10^9\,\mathrm{N\,m^2\,C^{-2}}\):
Field at \(P\) due to \(-q\):
\[E_{-}=\frac{kq}{(0.050)^2}=\frac{(9.0\times10^9)(5.0\times10^{-8})}{(0.050)^2}=1.80\times10^5\,\mathrm{N\,C^{-1}}\]
Field at \(P\) due to \(+q\):
\[E_{+}=\frac{kq}{(0.030)^2}=\frac{(9.0\times10^9)(5.0\times10^{-8})}{(0.030)^2}=5.00\times10^5\,\mathrm{N\,C^{-1}}\]
At \(P\), both fields are directed from \(+q\) towards \(-q\). Therefore,
\[E=E_{+}+E_{-}=5.00\times10^5+1.80\times10^5=6.80\times10^5\,\mathrm{N\,C^{-1}}\]
Resultant field intensity at \(P\) = \(6.8\times10^5\,\mathrm{N\,C^{-1}}\), directed towards the negative charge (from \(+q\) to \(-q\)).
The electric lines of force for the two unlike point charges are shown below.
(c)
Given: \(f=25\,\mathrm{Hz}\), \(C=10\,\mu\mathrm{F}=10\times10^{-6}\,\mathrm{F}\), \(R=1000\,\Omega\), and \(V=90\,\mathrm{V_{rms}}\).
(i) Capacitive reactance
\[X_C=\frac{1}{2\pi fC}=\frac{1}{2\pi(25)(10\times10^{-6})}=636.6\,\Omega\]
(ii) Impedance
\[Z=\sqrt{R^2+X_C^2}=\sqrt{1000^2+636.6^2}=1185.4\,\Omega\]
(iii) Current through the circuit
\[I=\frac{V}{Z}=\frac{90}{1185.4}=7.59\times10^{-2}\,\mathrm{A}=0.0759\,\mathrm{A_{rms}}\]
(iv) Voltage across the capacitor
\[V_C=IX_C=(0.0759)(636.6)=48.3\,\mathrm{V_{rms}}\]
(v) Average power used in the circuit
Only the resistor dissipates average power:
\[P=I^2R=(0.0759)^2(1000)=5.76\,\mathrm{W}\]
Answer Details
(a)(i) Electric field intensity, E, at a point is the force experienced per unit positive test charge placed at that point.
\[E=\frac{F}{q}\]
It is a vector quantity. Its SI unit is \(\mathrm{N\,C^{-1}}\) (or \(\mathrm{V\,m^{-1}}\)), and its direction is the direction of the force on a positive test charge.
(ii) Electric lines of force are imaginary lines used to show an electric field. The tangent to a field line at any point gives the direction of the field. They emerge from positive charges and terminate on negative charges; closer spacing indicates a stronger field, and field lines never intersect.
(b) Point \(P\) lies between the charges, \(0.050\,\mathrm{m}\) from \(-q\) and \(0.030\,\mathrm{m}\) from \(+q\). Taking \(k=\dfrac{1}{4\pi\varepsilon_0}=9.0\times10^9\,\mathrm{N\,m^2\,C^{-2}}\):
Field at \(P\) due to \(-q\):
\[E_{-}=\frac{kq}{(0.050)^2}=\frac{(9.0\times10^9)(5.0\times10^{-8})}{(0.050)^2}=1.80\times10^5\,\mathrm{N\,C^{-1}}\]
Field at \(P\) due to \(+q\):
\[E_{+}=\frac{kq}{(0.030)^2}=\frac{(9.0\times10^9)(5.0\times10^{-8})}{(0.030)^2}=5.00\times10^5\,\mathrm{N\,C^{-1}}\]
At \(P\), both fields are directed from \(+q\) towards \(-q\). Therefore,
\[E=E_{+}+E_{-}=5.00\times10^5+1.80\times10^5=6.80\times10^5\,\mathrm{N\,C^{-1}}\]
Resultant field intensity at \(P\) = \(6.8\times10^5\,\mathrm{N\,C^{-1}}\), directed towards the negative charge (from \(+q\) to \(-q\)).
The electric lines of force for the two unlike point charges are shown below.
(c)
Given: \(f=25\,\mathrm{Hz}\), \(C=10\,\mu\mathrm{F}=10\times10^{-6}\,\mathrm{F}\), \(R=1000\,\Omega\), and \(V=90\,\mathrm{V_{rms}}\).
(i) Capacitive reactance
\[X_C=\frac{1}{2\pi fC}=\frac{1}{2\pi(25)(10\times10^{-6})}=636.6\,\Omega\]
(ii) Impedance
\[Z=\sqrt{R^2+X_C^2}=\sqrt{1000^2+636.6^2}=1185.4\,\Omega\]
(iii) Current through the circuit
\[I=\frac{V}{Z}=\frac{90}{1185.4}=7.59\times10^{-2}\,\mathrm{A}=0.0759\,\mathrm{A_{rms}}\]
(iv) Voltage across the capacitor
\[V_C=IX_C=(0.0759)(636.6)=48.3\,\mathrm{V_{rms}}\]
(v) Average power used in the circuit
Only the resistor dissipates average power:
\[P=I^2R=(0.0759)^2(1000)=5.76\,\mathrm{W}\]
