WAEC SSCE - Physics - 1988 (Essay)
Tambaya 1 Rahoto
(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.
Bayanin Amsa
(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.
Tambaya 2 Rahoto
(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 is its turning effect about a pivot or point. It is given by:
\[ \text{Moment} = F \times d \]
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, \(\text{N m}\).
The distance used is not simply any distance from the pivot to the force: it must be measured at right angles to the force's line of action.
(b) Conditions for equilibrium of coplanar parallel forces
- The total force in one direction equals the total force in the opposite direction, so the resultant force is zero.
- The sum of clockwise moments about any point equals the sum of anticlockwise moments about that point, so the resultant moment is zero.
Metre rule calculation
Since the metre rule balances by itself at the \(48\,\text{cm}\) mark, its weight acts at the \(48\,\text{cm}\) mark.
When the \(60\,\text{g}\) body is at \(6\,\text{cm}\), the pivot is at \(30\,\text{cm}\).
- Distance of the \(60\,\text{g}\) mass from the pivot: \(30-6=24\,\text{cm}\).
- Distance of the rule's weight from the pivot: \(48-30=18\,\text{cm}\).
Taking moments about the balance point:
\[ 60 \times 24 = M \times 18 \]
\[ M=\frac{60\times24}{18}=80\,\text{g} \]
(i) Mass of the metre rule \(=80\,\text{g}\).
For the body at the \(13\,\text{cm}\) mark, let the new balance point be \(x\,\text{cm}\) from the zero end. The body is to the left of the pivot and the rule's weight acts to the right:
\[ 60(x-13)=80(48-x) \]
\[ 60x-780=3840-80x \]
\[ 140x=4620 \]
\[ x=33\,\text{cm} \]
(ii) The balance point is \(33\,\text{cm}\) from the zero end.
(c) Relationship between efficiency, mechanical advantage and velocity ratio
Efficiency is:
\[ E=\frac{\text{work output}}{\text{work input}}\times100\% \]
If \(L\) is the load, \(P\) is the effort, \(d_L\) is the distance moved by the load, and \(d_P\) is the distance moved by the effort:
\[ E=\frac{L\times d_L}{P\times d_P}\times100\% \]
Rearranging the factors:
\[ E=\frac{L}{P}\times\frac{d_L}{d_P}\times100\% \]
But:
\[ M.A=\frac{L}{P} \qquad \text{and} \qquad V.R=\frac{d_P}{d_L} \]
Therefore:
\[ \frac{d_L}{d_P}=\frac{1}{V.R} \]
Hence:
\[ \boxed{E=\frac{M.A}{V.R}\times100\%} \]
For the machine, the useful work done on the load is:
\[ W_{\text{output}}=mgh=200\times10\times3.0=6000\,\text{J} \]
Since the efficiency is \(80\%=0.80\):
\[ 0.80=\frac{W_{\text{output}}}{W_{\text{input}}} \]
\[ W_{\text{input}}=\frac{6000}{0.80}=7500\,\text{J} \]
The work done by the person is \(7500\,\text{J}\) (or \(7.5\,\text{kJ}\)).
Bayanin Amsa
(a) Moment of a force about a point
The moment of a force is its turning effect about a pivot or point. It is given by:
\[ \text{Moment} = F \times d \]
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, \(\text{N m}\).
The distance used is not simply any distance from the pivot to the force: it must be measured at right angles to the force's line of action.
(b) Conditions for equilibrium of coplanar parallel forces
- The total force in one direction equals the total force in the opposite direction, so the resultant force is zero.
- The sum of clockwise moments about any point equals the sum of anticlockwise moments about that point, so the resultant moment is zero.
Metre rule calculation
Since the metre rule balances by itself at the \(48\,\text{cm}\) mark, its weight acts at the \(48\,\text{cm}\) mark.
When the \(60\,\text{g}\) body is at \(6\,\text{cm}\), the pivot is at \(30\,\text{cm}\).
- Distance of the \(60\,\text{g}\) mass from the pivot: \(30-6=24\,\text{cm}\).
- Distance of the rule's weight from the pivot: \(48-30=18\,\text{cm}\).
Taking moments about the balance point:
\[ 60 \times 24 = M \times 18 \]
\[ M=\frac{60\times24}{18}=80\,\text{g} \]
(i) Mass of the metre rule \(=80\,\text{g}\).
For the body at the \(13\,\text{cm}\) mark, let the new balance point be \(x\,\text{cm}\) from the zero end. The body is to the left of the pivot and the rule's weight acts to the right:
\[ 60(x-13)=80(48-x) \]
\[ 60x-780=3840-80x \]
\[ 140x=4620 \]
\[ x=33\,\text{cm} \]
(ii) The balance point is \(33\,\text{cm}\) from the zero end.
(c) Relationship between efficiency, mechanical advantage and velocity ratio
Efficiency is:
\[ E=\frac{\text{work output}}{\text{work input}}\times100\% \]
If \(L\) is the load, \(P\) is the effort, \(d_L\) is the distance moved by the load, and \(d_P\) is the distance moved by the effort:
\[ E=\frac{L\times d_L}{P\times d_P}\times100\% \]
Rearranging the factors:
\[ E=\frac{L}{P}\times\frac{d_L}{d_P}\times100\% \]
But:
\[ M.A=\frac{L}{P} \qquad \text{and} \qquad V.R=\frac{d_P}{d_L} \]
Therefore:
\[ \frac{d_L}{d_P}=\frac{1}{V.R} \]
Hence:
\[ \boxed{E=\frac{M.A}{V.R}\times100\%} \]
For the machine, the useful work done on the load is:
\[ W_{\text{output}}=mgh=200\times10\times3.0=6000\,\text{J} \]
Since the efficiency is \(80\%=0.80\):
\[ 0.80=\frac{W_{\text{output}}}{W_{\text{input}}} \]
\[ W_{\text{input}}=\frac{6000}{0.80}=7500\,\text{J} \]
The work done by the person is \(7500\,\text{J}\) (or \(7.5\,\text{kJ}\)).
Tambaya 3 Rahoto
(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
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 direction is the direction of the force on a positive test charge. Its SI unit is \(\text{N C}^{-1}\), equivalently \(\text{V m}^{-1}\).
(a)(ii) Electric lines of force
Electric lines of force are imaginary lines used to show an electric field pattern. The tangent to a line at any point gives the direction of the electric field at that point. Lines leave positive charges and enter negative charges. They never cross, because the field cannot have two different directions at one point. Where the lines are closer together, the electric field is stronger.
(b) Resultant electric field intensity at \(P\)
The distances shown are \(0.03\ \text{m}\) from \(+q\) and \(0.05\ \text{m}\) from \(-q\). Since these add to \(0.08\ \text{m}\), point \(P\) lies between the charges.
The field due to a positive charge points away from it. At \(P\), this is towards the negative charge. The field due to a negative charge points towards it, which is also towards the negative charge. Therefore, the two fields act in the same direction and must be added.
For the positive charge:
\[ E_{+}=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} =\frac{(9.0\times10^9)(5.0\times10^{-8})}{(0.03)^2} =5.0\times10^5\ \text{N C}^{-1} \]
For the negative charge:
\[ E_{-}=\frac{(9.0\times10^9)(5.0\times10^{-8})}{(0.05)^2} =1.8\times10^5\ \text{N C}^{-1} \]
\[ E=E_{+}+E_{-} =5.0\times10^5+1.8\times10^5 =6.8\times10^5\ \text{N C}^{-1} \]
Result: \(\boxed{6.8\times10^5\ \text{N C}^{-1}}\), directed from \(+q\) towards \(-q\).
The supplied reference answer subtracts the two field magnitudes. That would only be appropriate if the fields were in opposite directions. Here \(P\) is between opposite charges, so both fields point towards \(-q\).
(c) Series \(R\)-\(C\) circuit
Using the circuit values shown in the supplied reference material: \(R=1000\ \Omega\), \(C=10\times10^{-6}\ \text{F}\), \(f=25\ \text{Hz}\), and \(V=90\ \text{V}_{\rm rms}\).
-
Capacitive reactance
\[ X_C=\frac{1}{2\pi fC} =\frac{1}{2\pi(25)(10\times10^{-6})} =\frac{2000}{\pi} =636.6\ \Omega \]
-
Impedance
In a series \(R\)-\(C\) circuit, resistance and capacitive reactance are at right angles in the phasor diagram:
\[ Z=\sqrt{R^2+X_C^2} =\sqrt{(1000)^2+(636.6)^2} =1.19\times10^3\ \Omega \]
-
Current
\[ I=\frac{V}{Z} =\frac{90}{1185.5} =0.0759\ \text{A}_{\rm rms} \]
-
Voltage across the capacitor
\[ V_C=IX_C =(0.0759)(636.6) =48.3\ \text{V}_{\rm rms} \]
-
Average power used
Only the resistor dissipates average power. An ideal capacitor stores and returns energy, so its average power is zero.
\[ P=I^2R =(0.0759)^2(1000) =5.76\ \text{W} \]
The value \(I^2Z=6.83\) is not the average power in watts; it is associated with apparent power, \(VI\), measured in volt-amperes. For average power in an \(R\)-\(C\) circuit, use \(I^2R\) or \(VI\cos\phi\).
Examination reminder: decide whether electric-field vectors point in the same or opposite directions before adding their magnitudes. In an a.c. circuit containing a capacitor, use \(R\), not \(Z\), when calculating real average power.
Bayanin Amsa
(a)(i) Electric field intensity
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 direction is the direction of the force on a positive test charge. Its SI unit is \(\text{N C}^{-1}\), equivalently \(\text{V m}^{-1}\).
(a)(ii) Electric lines of force
Electric lines of force are imaginary lines used to show an electric field pattern. The tangent to a line at any point gives the direction of the electric field at that point. Lines leave positive charges and enter negative charges. They never cross, because the field cannot have two different directions at one point. Where the lines are closer together, the electric field is stronger.
(b) Resultant electric field intensity at \(P\)
The distances shown are \(0.03\ \text{m}\) from \(+q\) and \(0.05\ \text{m}\) from \(-q\). Since these add to \(0.08\ \text{m}\), point \(P\) lies between the charges.
The field due to a positive charge points away from it. At \(P\), this is towards the negative charge. The field due to a negative charge points towards it, which is also towards the negative charge. Therefore, the two fields act in the same direction and must be added.
For the positive charge:
\[ E_{+}=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} =\frac{(9.0\times10^9)(5.0\times10^{-8})}{(0.03)^2} =5.0\times10^5\ \text{N C}^{-1} \]
For the negative charge:
\[ E_{-}=\frac{(9.0\times10^9)(5.0\times10^{-8})}{(0.05)^2} =1.8\times10^5\ \text{N C}^{-1} \]
\[ E=E_{+}+E_{-} =5.0\times10^5+1.8\times10^5 =6.8\times10^5\ \text{N C}^{-1} \]
Result: \(\boxed{6.8\times10^5\ \text{N C}^{-1}}\), directed from \(+q\) towards \(-q\).
The supplied reference answer subtracts the two field magnitudes. That would only be appropriate if the fields were in opposite directions. Here \(P\) is between opposite charges, so both fields point towards \(-q\).
(c) Series \(R\)-\(C\) circuit
Using the circuit values shown in the supplied reference material: \(R=1000\ \Omega\), \(C=10\times10^{-6}\ \text{F}\), \(f=25\ \text{Hz}\), and \(V=90\ \text{V}_{\rm rms}\).
-
Capacitive reactance
\[ X_C=\frac{1}{2\pi fC} =\frac{1}{2\pi(25)(10\times10^{-6})} =\frac{2000}{\pi} =636.6\ \Omega \]
-
Impedance
In a series \(R\)-\(C\) circuit, resistance and capacitive reactance are at right angles in the phasor diagram:
\[ Z=\sqrt{R^2+X_C^2} =\sqrt{(1000)^2+(636.6)^2} =1.19\times10^3\ \Omega \]
-
Current
\[ I=\frac{V}{Z} =\frac{90}{1185.5} =0.0759\ \text{A}_{\rm rms} \]
-
Voltage across the capacitor
\[ V_C=IX_C =(0.0759)(636.6) =48.3\ \text{V}_{\rm rms} \]
-
Average power used
Only the resistor dissipates average power. An ideal capacitor stores and returns energy, so its average power is zero.
\[ P=I^2R =(0.0759)^2(1000) =5.76\ \text{W} \]
The value \(I^2Z=6.83\) is not the average power in watts; it is associated with apparent power, \(VI\), measured in volt-amperes. For average power in an \(R\)-\(C\) circuit, use \(I^2R\) or \(VI\cos\phi\).
Examination reminder: decide whether electric-field vectors point in the same or opposite directions before adding their magnitudes. In an a.c. circuit containing a capacitor, use \(R\), not \(Z\), when calculating real average power.
Tambaya 4 Rahoto
(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.
Bayanin Amsa
(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.
