WAEC SSCE - Physics - 1992 (Essay)
Question 1 Rapport
(a)(i) What is meant by resonance?
(ii) Outline the necessary steps taken in a simple experiment to illustrate top resonance
(iii) Explain why a vibrating tuning fork sounds louder when its stem is pressed against a table top than when held in air.
(b) Explain with the aid of diagrams, how a concave mirror could be used to: (i) Ignite a piece of carbon paper; (ii) produce an exact copy of a picture on a screen.
(a)(i) Resonance occurs when a body is made to vibrate at its own natural frequency by a periodic force having the same frequency; the amplitude of vibration then becomes a maximum.
(ii) Experiment to illustrate resonance (resonance tube):
- Hold a vibrating tuning fork of known frequency just above the open end of a tube whose air-column length can be varied (e.g. a tube dipping in water, raised or lowered).
- Slowly change the length of the air column.
- At a particular length the sound suddenly becomes very loud; here the natural frequency of the air column equals the frequency of the fork, i.e. resonance occurs.
- Measure that resonating length.
(iii) When the stem of the vibrating fork is pressed on a table top, it forces the large table surface to vibrate at the same frequency (forced vibration / resonance). The large surface sets a much greater volume of air in motion than the thin prongs alone, so more sound energy reaches the ear and the note sounds louder. In air, only the small prongs move the air, so the sound is faint.
(b) Concave-mirror uses.
(i) To ignite carbon paper: the mirror faces the sun, whose rays are parallel to the axis; the mirror converges them to its principal focus \(F\). The paper placed at \(F\) receives concentrated heat and burns.
(ii) To produce an exact copy (same size) of a picture on a screen: place the object at the centre of curvature \(C\) (i.e. at \(2f\)). The mirror then forms a real, inverted image, of exactly the same size, also at \(C\), which can be received on a screen.
Détails de la réponse
(a)(i) Resonance occurs when a body is made to vibrate at its own natural frequency by a periodic force having the same frequency; the amplitude of vibration then becomes a maximum.
(ii) Experiment to illustrate resonance (resonance tube):
- Hold a vibrating tuning fork of known frequency just above the open end of a tube whose air-column length can be varied (e.g. a tube dipping in water, raised or lowered).
- Slowly change the length of the air column.
- At a particular length the sound suddenly becomes very loud; here the natural frequency of the air column equals the frequency of the fork, i.e. resonance occurs.
- Measure that resonating length.
(iii) When the stem of the vibrating fork is pressed on a table top, it forces the large table surface to vibrate at the same frequency (forced vibration / resonance). The large surface sets a much greater volume of air in motion than the thin prongs alone, so more sound energy reaches the ear and the note sounds louder. In air, only the small prongs move the air, so the sound is faint.
(b) Concave-mirror uses.
(i) To ignite carbon paper: the mirror faces the sun, whose rays are parallel to the axis; the mirror converges them to its principal focus \(F\). The paper placed at \(F\) receives concentrated heat and burns.
(ii) To produce an exact copy (same size) of a picture on a screen: place the object at the centre of curvature \(C\) (i.e. at \(2f\)). The mirror then forms a real, inverted image, of exactly the same size, also at \(C\), which can be received on a screen.
Question 2 Rapport
(a) Draw a simple labelled diagram illustrating the principle of a step-down transformer and explain how it works
(b) State three ways by which energy is lost in a transformer and how they can be minimized.
(c) If a transformer is used to light a lamp rated at 60W, 220V from a 4400V a.c. supply, calculate the; (i) ratio of the number of turns of the primary coil to the secondary coil in the transformer (ii) current taken from the main circuit if the efficiency of the transforme is 95%.
(a) Step-down transformer
A step-down transformer has more turns in the primary coil than in the secondary coil, that is, \(N_p > N_s\). When an alternating e.m.f. is applied to the primary coil, an alternating current flows in it and produces a changing magnetic flux in the laminated soft-iron core. This changing flux links the secondary coil and induces an alternating e.m.f. in it by electromagnetic induction.
Since the induced e.m.f. is proportional to the number of turns,
\[\frac{E_s}{E_p}=\frac{N_s}{N_p}\]
As \(N_s < N_p\), \(E_s < E_p\). Hence, the output voltage is lower than the input voltage.
(b) Energy losses and their reduction
- Copper loss: Heat is produced in the windings because of their resistance. This is minimized by using thick, low-resistance copper wire.
- Eddy-current loss: Eddy currents are induced in the iron core and produce heating. This is minimized by laminating the core into thin insulated sheets.
- Hysteresis loss: Energy is dissipated during repeated magnetisation and demagnetisation of the core. This is minimized by using soft iron or a soft magnetic material with low hysteresis loss.
(c)
Given: \(V_p=4400\text{ V}\), \(V_s=220\text{ V}\), lamp power \(P_{out}=60\text{ W}\), and efficiency \(\eta=95\%=0.95\).
(i) Turns ratio
\[\frac{N_p}{N_s}=\frac{V_p}{V_s}=\frac{4400}{220}=20\]
Therefore, the ratio of primary turns to secondary turns is:
\[\boxed{N_p:N_s=20:1}\]
(ii) Current taken from the mains
\[\eta=\frac{P_{out}}{P_{in}}\]
\[0.95=\frac{60}{P_{in}}\]
\[P_{in}=\frac{60}{0.95}=63.16\text{ W}\]
\[I_p=\frac{P_{in}}{V_p}=\frac{63.16}{4400}=0.0144\text{ A}\]
\[\boxed{I_p=0.0144\text{ A}=14.4\text{ mA}}\]
Détails de la réponse
(a) Step-down transformer
A step-down transformer has more turns in the primary coil than in the secondary coil, that is, \(N_p > N_s\). When an alternating e.m.f. is applied to the primary coil, an alternating current flows in it and produces a changing magnetic flux in the laminated soft-iron core. This changing flux links the secondary coil and induces an alternating e.m.f. in it by electromagnetic induction.
Since the induced e.m.f. is proportional to the number of turns,
\[\frac{E_s}{E_p}=\frac{N_s}{N_p}\]
As \(N_s < N_p\), \(E_s < E_p\). Hence, the output voltage is lower than the input voltage.
(b) Energy losses and their reduction
- Copper loss: Heat is produced in the windings because of their resistance. This is minimized by using thick, low-resistance copper wire.
- Eddy-current loss: Eddy currents are induced in the iron core and produce heating. This is minimized by laminating the core into thin insulated sheets.
- Hysteresis loss: Energy is dissipated during repeated magnetisation and demagnetisation of the core. This is minimized by using soft iron or a soft magnetic material with low hysteresis loss.
(c)
Given: \(V_p=4400\text{ V}\), \(V_s=220\text{ V}\), lamp power \(P_{out}=60\text{ W}\), and efficiency \(\eta=95\%=0.95\).
(i) Turns ratio
\[\frac{N_p}{N_s}=\frac{V_p}{V_s}=\frac{4400}{220}=20\]
Therefore, the ratio of primary turns to secondary turns is:
\[\boxed{N_p:N_s=20:1}\]
(ii) Current taken from the mains
\[\eta=\frac{P_{out}}{P_{in}}\]
\[0.95=\frac{60}{P_{in}}\]
\[P_{in}=\frac{60}{0.95}=63.16\text{ W}\]
\[I_p=\frac{P_{in}}{V_p}=\frac{63.16}{4400}=0.0144\text{ A}\]
\[\boxed{I_p=0.0144\text{ A}=14.4\text{ mA}}\]
Question 3 Rapport
(a) Explain what is meant by the following statement. The specific latent heat of fusion of ice is \(3.4 \times 10^{5}Jkg^{-1}\).
(b) Describe an experiment to determine the specific latent heat of fusion of ice. State two precautions necessary to obtain an accurate result.
(c) Using the kinetic theory of matter, explain why ice can change to water at 0°C without any change in temperature.
(a) The statement means that \(3.4\times10^{5}\,\text{J}\) of heat is required to change \(1\,\text{kg}\) of ice at \(0^\circ\text{C}\) completely into water at \(0^\circ\text{C}\), without any change in temperature.
(b) Experiment (method of mixtures):
- Half-fill a lagged calorimeter (of known mass \(m_c\) and specific heat \(c_c\)) with warm water; measure the mass of water \(m_w\) and the initial temperature \(\theta_1\).
- Dry some small pieces of pure melting ice at \(0^\circ\text{C}\) with blotting paper and drop them into the water; stir until all the ice just melts.
- Record the lowest (final) temperature \(\theta_2\), and find the mass of ice added \(m_i\) by reweighing.
- Heat lost by water and calorimeter = heat used to melt the ice + heat to warm the melted ice from \(0^\circ\text{C}\) to \(\theta_2\):
from which the specific latent heat of fusion \(L\) is calculated.
Two precautions: dry the ice thoroughly and use ice already at \(0^\circ\text{C}\); lag the calorimeter and stir gently to reduce heat exchange with the surroundings.
(c) Kinetic-theory explanation. When ice melts at \(0^\circ\text{C}\), the heat supplied is used to break down the rigid bonds of the ice lattice, increasing the potential energy of the molecules and letting them move past one another as a liquid. It does not increase their average kinetic energy. Since temperature depends on the average kinetic energy of the molecules, the temperature stays constant at \(0^\circ\text{C}\) throughout melting.
Détails de la réponse
(a) The statement means that \(3.4\times10^{5}\,\text{J}\) of heat is required to change \(1\,\text{kg}\) of ice at \(0^\circ\text{C}\) completely into water at \(0^\circ\text{C}\), without any change in temperature.
(b) Experiment (method of mixtures):
- Half-fill a lagged calorimeter (of known mass \(m_c\) and specific heat \(c_c\)) with warm water; measure the mass of water \(m_w\) and the initial temperature \(\theta_1\).
- Dry some small pieces of pure melting ice at \(0^\circ\text{C}\) with blotting paper and drop them into the water; stir until all the ice just melts.
- Record the lowest (final) temperature \(\theta_2\), and find the mass of ice added \(m_i\) by reweighing.
- Heat lost by water and calorimeter = heat used to melt the ice + heat to warm the melted ice from \(0^\circ\text{C}\) to \(\theta_2\):
from which the specific latent heat of fusion \(L\) is calculated.
Two precautions: dry the ice thoroughly and use ice already at \(0^\circ\text{C}\); lag the calorimeter and stir gently to reduce heat exchange with the surroundings.
(c) Kinetic-theory explanation. When ice melts at \(0^\circ\text{C}\), the heat supplied is used to break down the rigid bonds of the ice lattice, increasing the potential energy of the molecules and letting them move past one another as a liquid. It does not increase their average kinetic energy. Since temperature depends on the average kinetic energy of the molecules, the temperature stays constant at \(0^\circ\text{C}\) throughout melting.
Question 4 Rapport
(a) Explain what is meant by photoelectric emission
(b) Draw a labelled diagram showing the structure of a simple type of a photocell and explain its mode of operation.
(c) State four applications of photoelectric emission.
(d) In a photocell, no electrons are emitted until the threshold frequency of light is reached. Explain what happens to the energy of the light before emission of electrons begins. State one factor that may affect the number of emitted electrons.
(a) Photoelectric emission
Photoelectric emission is the emission of electrons from the surface of a metal when electromagnetic radiation of sufficiently high frequency falls on it.
(b) Simple photocell
The photocell consists of an evacuated glass or quartz bulb containing a curved photosensitive cathode, usually coated with caesium, and a collecting anode. When light of frequency equal to or greater than the threshold frequency falls on the caesium-coated cathode, photoelectrons are emitted. The positively charged anode attracts and collects these electrons. Electrons then flow through the external circuit and a current is registered by the microammeter. Increasing the intensity of light increases the number of electrons emitted per second and hence increases the photocurrent.
(c) Applications of photoelectric emission
- Automatic switching on of street lights at dusk.
- Burglar and security alarm systems.
- Television camera tubes.
- Industrial counting and control devices, for example counting objects moving on a conveyor belt.
(d)
Before photoelectric emission begins, the light energy is absorbed by electrons in the metal. For light below the threshold frequency, the energy supplied to each electron is less than the work function, so the electron cannot escape from the metal surface; the absorbed energy is dissipated mainly as heat. At the threshold frequency, each photon supplies just enough energy to overcome the binding energy of an electron, and emission begins.
One factor affecting the number of emitted electrons is the intensity of the incident light, provided that its frequency is at least the threshold frequency. Greater intensity produces more emitted electrons per second.
Détails de la réponse
(a) Photoelectric emission
Photoelectric emission is the emission of electrons from the surface of a metal when electromagnetic radiation of sufficiently high frequency falls on it.
(b) Simple photocell
The photocell consists of an evacuated glass or quartz bulb containing a curved photosensitive cathode, usually coated with caesium, and a collecting anode. When light of frequency equal to or greater than the threshold frequency falls on the caesium-coated cathode, photoelectrons are emitted. The positively charged anode attracts and collects these electrons. Electrons then flow through the external circuit and a current is registered by the microammeter. Increasing the intensity of light increases the number of electrons emitted per second and hence increases the photocurrent.
(c) Applications of photoelectric emission
- Automatic switching on of street lights at dusk.
- Burglar and security alarm systems.
- Television camera tubes.
- Industrial counting and control devices, for example counting objects moving on a conveyor belt.
(d)
Before photoelectric emission begins, the light energy is absorbed by electrons in the metal. For light below the threshold frequency, the energy supplied to each electron is less than the work function, so the electron cannot escape from the metal surface; the absorbed energy is dissipated mainly as heat. At the threshold frequency, each photon supplies just enough energy to overcome the binding energy of an electron, and emission begins.
One factor affecting the number of emitted electrons is the intensity of the incident light, provided that its frequency is at least the threshold frequency. Greater intensity produces more emitted electrons per second.
