WAEC SSCE - Physics - 1993 (Essay)
Question 1 Report
(a)(i) Explain what is meant by a machine (ii) Define the terms: mechanical advantage, velocity ratio and efficiency as applied to a machine. Derive the equation connecting the three terms.
(b) Explain why the efficiency of a machine is usually less than 100%
(c) A screw jack whose pitch is 4.4mm is used to raise a body of mass 8000 kg through a height of 20cm. The length of the tommy bar of the jack is 70cm. If the efficiency of the jack is 80%, calcuate the: (i) velocity ratio of the jack; (ii) mechanical advantage of the jack (iii) effort required in raising the body, (iv) work done by the effort in raising the body \((g = 10ms^{-2}; \pi = \frac{22}{7}\))
(a)(i) A machine is a device by means of which a force (the effort) applied at one point is used to overcome another force (the load) at some other point, usually giving a mechanical advantage or a change in the direction of the force.
(a)(ii)
- Mechanical advantage (M.A.) is the ratio of the load to the effort: \(M.A.=\dfrac{L}{E}\).
- Velocity ratio (V.R.) is the ratio of the distance moved by the effort to the distance moved by the load in the same time: \(V.R.=\dfrac{\text{effort distance}}{\text{load distance}}\).
- Efficiency is the ratio of the useful work output to the work input, expressed as a percentage.
Derivation. \(\text{Efficiency}=\dfrac{\text{work output}}{\text{work input}}=\dfrac{L\times d_L}{E\times d_E}=\dfrac{L/E}{d_E/d_L}=\dfrac{M.A.}{V.R.}\). Hence \(\text{Efficiency}=\dfrac{M.A.}{V.R.}\times100\%\).
(b) Efficiency is less than 100% because part of the work input is used to overcome friction between moving parts and to lift the useless (movable) parts of the machine; this energy is lost mainly as heat and is not delivered to the load.
(c) Pitch = 4.4 mm = \(4.4\times10^{-3}\,\text{m}\); tommy bar L = 70 cm = 0.70 m; height h = 20 cm = 0.20 m; mass = 8000 kg.
(i) Velocity ratio: \(V.R.=\dfrac{2\pi L}{\text{pitch}}=\dfrac{2\times\frac{22}{7}\times0.70}{4.4\times10^{-3}}=\dfrac{4.4}{4.4\times10^{-3}}=1000\).
(ii) Mechanical advantage: \(M.A.=\text{efficiency}\times V.R.=0.80\times1000=800\).
(iii) Effort: Load \(=mg=8000\times10=80000\,\text{N}\). \(E=\dfrac{L}{M.A.}=\dfrac{80000}{800}=100\,\text{N}\).
(iv) Work done by the effort: \(=\dfrac{\text{work output}}{\text{efficiency}}=\dfrac{mgh}{0.80}=\dfrac{80000\times0.20}{0.80}=\dfrac{16000}{0.80}=20000\,\text{J}=20\,\text{kJ}\).
Answer Details
(a)(i) A machine is a device by means of which a force (the effort) applied at one point is used to overcome another force (the load) at some other point, usually giving a mechanical advantage or a change in the direction of the force.
(a)(ii)
- Mechanical advantage (M.A.) is the ratio of the load to the effort: \(M.A.=\dfrac{L}{E}\).
- Velocity ratio (V.R.) is the ratio of the distance moved by the effort to the distance moved by the load in the same time: \(V.R.=\dfrac{\text{effort distance}}{\text{load distance}}\).
- Efficiency is the ratio of the useful work output to the work input, expressed as a percentage.
Derivation. \(\text{Efficiency}=\dfrac{\text{work output}}{\text{work input}}=\dfrac{L\times d_L}{E\times d_E}=\dfrac{L/E}{d_E/d_L}=\dfrac{M.A.}{V.R.}\). Hence \(\text{Efficiency}=\dfrac{M.A.}{V.R.}\times100\%\).
(b) Efficiency is less than 100% because part of the work input is used to overcome friction between moving parts and to lift the useless (movable) parts of the machine; this energy is lost mainly as heat and is not delivered to the load.
(c) Pitch = 4.4 mm = \(4.4\times10^{-3}\,\text{m}\); tommy bar L = 70 cm = 0.70 m; height h = 20 cm = 0.20 m; mass = 8000 kg.
(i) Velocity ratio: \(V.R.=\dfrac{2\pi L}{\text{pitch}}=\dfrac{2\times\frac{22}{7}\times0.70}{4.4\times10^{-3}}=\dfrac{4.4}{4.4\times10^{-3}}=1000\).
(ii) Mechanical advantage: \(M.A.=\text{efficiency}\times V.R.=0.80\times1000=800\).
(iii) Effort: Load \(=mg=8000\times10=80000\,\text{N}\). \(E=\dfrac{L}{M.A.}=\dfrac{80000}{800}=100\,\text{N}\).
(iv) Work done by the effort: \(=\dfrac{\text{work output}}{\text{efficiency}}=\dfrac{mgh}{0.80}=\dfrac{80000\times0.20}{0.80}=\dfrac{16000}{0.80}=20000\,\text{J}=20\,\text{kJ}\).
Question 2 Report
(a) With the aid of a labelled diagram, describe an experiment to illustrate the relationship between the volume and the temperature of a given mass of air at constant pressure.
(b) A uniform capillary tube of negligible expansivity sealed at one end, contains air trapped by a pellet of mercury. The trapped air column is 13.7cm long at 0°C and 18.7cm long at 100°C. Calculate the cubical expansivity of the air at constant pressure.
(c) Using the kinetic theory of gases, explain why the volume of a fixed mass of gas at constant pressure increases with increase in temperature.
(a) Experiment
A uniform capillary tube is sealed at one end. A short pellet of mercury traps a fixed mass of dry air between the sealed end and the mercury pellet. The tube is held horizontally in a stirred water bath beside a half-metre rule, with a thermometer in the bath. The mercury pellet is free to move, and the pressure on its outer face remains constant.
At a steady temperature \(\theta\), read the temperature from the thermometer and the length \(L\) of the trapped air column from the rule. Heat the water bath slowly, stirring continuously, and repeat the readings at several steady temperatures. Since the capillary tube has a uniform bore, \(V\propto L\).
| Temperature, \(\theta\) (°C) | 0 | 20 | 40 | 60 | 80 | 100 |
|---|---|---|---|---|---|---|
| Length, \(L\) (cm) | 13.7 | 14.7 | 15.7 | 16.7 | 17.7 | 18.7 |
A plot of \(L\) against \(\theta\) is a straight line. When produced backwards, it cuts the temperature axis at about \(-274\ ^\circ\mathrm{C}\). Thus, for a fixed mass of air at constant pressure, its volume increases uniformly with temperature.
Precautions: Use dry air; stir the water to maintain a uniform temperature; take readings only when the temperature is steady; and avoid parallax when reading the thermometer and rule.
(b) Since the tube is uniform, \(V\propto L\). Therefore, the cubical expansivity is
\[\gamma=\frac{V_{100}-V_0}{V_0(100-0)}=\frac{L_{100}-L_0}{L_0\times100}\]
\[\gamma=\frac{18.7-13.7}{13.7\times100}=\frac{5.0}{1370}=3.65\times10^{-3}\ \mathrm{K^{-1}}.\]
(c) Increasing the temperature increases the average kinetic energy and speed of the gas molecules. The molecules would then strike the walls more forcefully and tend to increase the pressure. At constant pressure, the gas expands so that the molecules travel greater distances between collisions with the walls. Hence the volume of the fixed mass of gas increases.
Answer Details
(a) Experiment
A uniform capillary tube is sealed at one end. A short pellet of mercury traps a fixed mass of dry air between the sealed end and the mercury pellet. The tube is held horizontally in a stirred water bath beside a half-metre rule, with a thermometer in the bath. The mercury pellet is free to move, and the pressure on its outer face remains constant.
At a steady temperature \(\theta\), read the temperature from the thermometer and the length \(L\) of the trapped air column from the rule. Heat the water bath slowly, stirring continuously, and repeat the readings at several steady temperatures. Since the capillary tube has a uniform bore, \(V\propto L\).
| Temperature, \(\theta\) (°C) | 0 | 20 | 40 | 60 | 80 | 100 |
|---|---|---|---|---|---|---|
| Length, \(L\) (cm) | 13.7 | 14.7 | 15.7 | 16.7 | 17.7 | 18.7 |
A plot of \(L\) against \(\theta\) is a straight line. When produced backwards, it cuts the temperature axis at about \(-274\ ^\circ\mathrm{C}\). Thus, for a fixed mass of air at constant pressure, its volume increases uniformly with temperature.
Precautions: Use dry air; stir the water to maintain a uniform temperature; take readings only when the temperature is steady; and avoid parallax when reading the thermometer and rule.
(b) Since the tube is uniform, \(V\propto L\). Therefore, the cubical expansivity is
\[\gamma=\frac{V_{100}-V_0}{V_0(100-0)}=\frac{L_{100}-L_0}{L_0\times100}\]
\[\gamma=\frac{18.7-13.7}{13.7\times100}=\frac{5.0}{1370}=3.65\times10^{-3}\ \mathrm{K^{-1}}.\]
(c) Increasing the temperature increases the average kinetic energy and speed of the gas molecules. The molecules would then strike the walls more forcefully and tend to increase the pressure. At constant pressure, the gas expands so that the molecules travel greater distances between collisions with the walls. Hence the volume of the fixed mass of gas increases.
Question 3 Report
(a) State the laws of electromagnetic induction.
(b) Draw a labelled diagram of a simple d.c. generator and explain how it works.
(c) State three methods by which higher e.m.f. could be obtained from the generator.
(a) Laws of electromagnetic induction
- Whenever the magnetic flux linking a conductor or coil changes, an e.m.f. is induced in it. The magnitude of the induced e.m.f. is proportional to the rate of change of flux linkage:
\[E=-N\frac{d\Phi}{dt}.\] - The induced current flows in such a direction as to oppose the change in magnetic flux that produces it. This is Lenz's law.
(b) Labelled diagram of a simple d.c. generator
How it works: A rectangular coil is rotated mechanically between the poles of a magnet. As the coil rotates, the magnetic flux linking it changes and an e.m.f. is induced in the coil. The induced e.m.f. in the coil reverses after every half revolution. However, at the same instant, the split-ring commutator reverses contact with the carbon brushes. Thus, the connections to the external circuit are reversed whenever the coil current reverses, so current in the external circuit always flows in one direction. The output is therefore a pulsating direct current.
(c) Three methods of obtaining a higher e.m.f.
- Increase the number of turns of wire in the coil.
- Use a stronger magnetic field, for example by using stronger magnets or a soft-iron core.
- Rotate the coil faster.
Answer Details
(a) Laws of electromagnetic induction
- Whenever the magnetic flux linking a conductor or coil changes, an e.m.f. is induced in it. The magnitude of the induced e.m.f. is proportional to the rate of change of flux linkage:
\[E=-N\frac{d\Phi}{dt}.\] - The induced current flows in such a direction as to oppose the change in magnetic flux that produces it. This is Lenz's law.
(b) Labelled diagram of a simple d.c. generator
How it works: A rectangular coil is rotated mechanically between the poles of a magnet. As the coil rotates, the magnetic flux linking it changes and an e.m.f. is induced in the coil. The induced e.m.f. in the coil reverses after every half revolution. However, at the same instant, the split-ring commutator reverses contact with the carbon brushes. Thus, the connections to the external circuit are reversed whenever the coil current reverses, so current in the external circuit always flows in one direction. The output is therefore a pulsating direct current.
(c) Three methods of obtaining a higher e.m.f.
- Increase the number of turns of wire in the coil.
- Use a stronger magnetic field, for example by using stronger magnets or a soft-iron core.
- Rotate the coil faster.
Question 4 Report
(a) Explain the following, illustrating your answer with one example in each case: (i) nuclear fusion: (ii) nuclear fission: (iii) radiation hazards.
(b) State two advantages of fusion over fission and explain briefly why, in spite of these advantages, fusion is not normally used for the generation of power.
(c) The current, I in an a.c. circuit is given by the equation: \(I = 30 sin 100\pi t\), where t is the time in seconds. Deduce the following from this equation: (i) frequency of the current (ii) peak value of the current, (iii) r.m.s value of the current.
(a)(i) Nuclear fusion: the joining together of two light atomic nuclei to form a single heavier nucleus, with the release of a large amount of energy. Example: the fusion of hydrogen (deuterium) nuclei to form helium in the Sun.
(a)(ii) Nuclear fission: the splitting of a heavy atomic nucleus into two lighter nuclei of comparable mass, accompanied by the release of neutrons and a large amount of energy. Example: the splitting of a uranium-235 nucleus when it captures a slow neutron.
(a)(iii) Radiation hazards: the harmful effects that nuclear radiations (alpha, beta, gamma) have on living tissue. Example: exposure to gamma rays can damage or kill body cells and cause cancer or genetic mutations.
(b) Two advantages of fusion over fission:
- Fusion produces more energy per unit mass of fuel and its fuel (hydrogen) is cheap and abundant.
- Fusion produces little or no long-lived radioactive waste, so it is cleaner.
In spite of these, fusion is not normally used for power generation because it requires extremely high temperatures (millions of degrees) and pressures to bring the nuclei close enough to fuse, and no ordinary container can withstand or confine such conditions economically.
(c) Given \(I=30\sin100\pi t\), compare with \(I=I_o\sin(2\pi f t)\).
(i) Frequency: \(2\pi f=100\pi\Rightarrow f=50\,\text{Hz}\).
(ii) Peak value: \(I_o=30\,\text{A}\).
(iii) r.m.s. value: \(I_{rms}=\dfrac{I_o}{\sqrt{2}}=\dfrac{30}{\sqrt{2}}=21.2\,\text{A}\).
Answer Details
(a)(i) Nuclear fusion: the joining together of two light atomic nuclei to form a single heavier nucleus, with the release of a large amount of energy. Example: the fusion of hydrogen (deuterium) nuclei to form helium in the Sun.
(a)(ii) Nuclear fission: the splitting of a heavy atomic nucleus into two lighter nuclei of comparable mass, accompanied by the release of neutrons and a large amount of energy. Example: the splitting of a uranium-235 nucleus when it captures a slow neutron.
(a)(iii) Radiation hazards: the harmful effects that nuclear radiations (alpha, beta, gamma) have on living tissue. Example: exposure to gamma rays can damage or kill body cells and cause cancer or genetic mutations.
(b) Two advantages of fusion over fission:
- Fusion produces more energy per unit mass of fuel and its fuel (hydrogen) is cheap and abundant.
- Fusion produces little or no long-lived radioactive waste, so it is cleaner.
In spite of these, fusion is not normally used for power generation because it requires extremely high temperatures (millions of degrees) and pressures to bring the nuclei close enough to fuse, and no ordinary container can withstand or confine such conditions economically.
(c) Given \(I=30\sin100\pi t\), compare with \(I=I_o\sin(2\pi f t)\).
(i) Frequency: \(2\pi f=100\pi\Rightarrow f=50\,\text{Hz}\).
(ii) Peak value: \(I_o=30\,\text{A}\).
(iii) r.m.s. value: \(I_{rms}=\dfrac{I_o}{\sqrt{2}}=\dfrac{30}{\sqrt{2}}=21.2\,\text{A}\).
