WAEC SSCE - Physics - 2003 (Essay)
Frage 1 Bericht
(a) What is the principle upon which the lighting in fluorescent tubes operate?
(b) State two factors which determine the colour of projection, g the acceleration of free fall due to gravity light from a fluorescent tube
(a) Principle of the fluorescent tube
It works on the principle of fluorescence. An electric discharge is passed through mercury vapour in the tube; the excited mercury atoms emit mostly ultraviolet radiation. This ultraviolet light strikes the phosphor (fluorescent) powder coating the inside of the tube, which absorbs it and re-emits the energy as visible light.
(b) Two factors that determine the colour of the emitted light
- The nature (chemical composition) of the phosphor coating on the inside of the tube.
- The type of gas or vapour enclosed in the tube. (The energy/wavelength of the exciting radiation is also acceptable.)
Antwortdetails
(a) Principle of the fluorescent tube
It works on the principle of fluorescence. An electric discharge is passed through mercury vapour in the tube; the excited mercury atoms emit mostly ultraviolet radiation. This ultraviolet light strikes the phosphor (fluorescent) powder coating the inside of the tube, which absorbs it and re-emits the energy as visible light.
(b) Two factors that determine the colour of the emitted light
- The nature (chemical composition) of the phosphor coating on the inside of the tube.
- The type of gas or vapour enclosed in the tube. (The energy/wavelength of the exciting radiation is also acceptable.)
Frage 2 Bericht
Explain the rise of water in a glass capillary tube using the kinetic theory.
Rise of water in a glass capillary tube (kinetic theory)
The molecules of water and the molecules of glass are in constant motion and exert intermolecular forces. Two forces are involved: the adhesive force between water molecules and glass molecules, and the cohesive force between water molecules themselves.
For water and glass the adhesive force is stronger than the cohesive force. The water molecules are therefore attracted more strongly to the glass wall than to one another, so they climb up the sides of the tube and the water wets the glass, giving a concave meniscus.
This upward pull produces a surface-tension force acting all round the line of contact between the water and the glass. This force lifts the column of water, and the liquid rises up the narrow tube until the weight of the raised column exactly balances the upward surface-tension force. The narrower the tube, the smaller the weight to be supported, so the higher the water rises.
(For a liquid such as mercury, where cohesion exceeds adhesion, the reverse happens and the liquid is depressed below the outside level.)
Antwortdetails
Rise of water in a glass capillary tube (kinetic theory)
The molecules of water and the molecules of glass are in constant motion and exert intermolecular forces. Two forces are involved: the adhesive force between water molecules and glass molecules, and the cohesive force between water molecules themselves.
For water and glass the adhesive force is stronger than the cohesive force. The water molecules are therefore attracted more strongly to the glass wall than to one another, so they climb up the sides of the tube and the water wets the glass, giving a concave meniscus.
This upward pull produces a surface-tension force acting all round the line of contact between the water and the glass. This force lifts the column of water, and the liquid rises up the narrow tube until the weight of the raised column exactly balances the upward surface-tension force. The narrower the tube, the smaller the weight to be supported, so the higher the water rises.
(For a liquid such as mercury, where cohesion exceeds adhesion, the reverse happens and the liquid is depressed below the outside level.)
Frage 3 Bericht
(a) Using the kinetic theory of matter, explain why;
(i) Evaporation causes cooling
(ii) Boiling water changes to steam without any change in temperature, although heat is being supplied to the water.
(b) (i) State Boyle's law.
(ii) With the aid of a labelled diagram, describe an experiment to illustrate the relationship between the volume and pressure of a given mass of gas at constant temperature. (iii) State two precautions necessary to obtain accurate results.
(a) Kinetic theory of matter
(i) Evaporation causes cooling: Molecules in a liquid are in continual random motion and possess different kinetic energies. The molecules with the greatest kinetic energy near the surface escape from the liquid during evaporation. Their escape reduces the average kinetic energy of the molecules left in the liquid. Since temperature is proportional to the average kinetic energy of the molecules, the temperature of the liquid falls; hence evaporation causes cooling.
(ii) Water boils without a rise in temperature: At its boiling point, the heat supplied to water is used as latent heat of vaporization. It is used to overcome the attractive intermolecular forces and to separate the water molecules to form steam. It does not increase the average kinetic energy of the molecules. Therefore, the temperature remains constant until all the water has changed to steam.
(b)(i) Boyle's law
For a fixed mass of gas at constant temperature, the volume of the gas is inversely proportional to its pressure.
Thus,
\[V\propto\frac{1}{P}\qquad\text{or}\qquad PV=\text{constant}.\]
(b)(ii) Experiment
The apparatus is arranged as shown below. A quantity of dry air is trapped in the graduated burette B by closing the tap. The burette is kept vertical. The mercury reservoir R is raised slowly to compress the trapped air. For each setting, read the volume, \(V\), of trapped air in the burette and the difference, \(h\), between the mercury levels. The barometer reading, \(H\), is also noted.
When the mercury level in the reservoir is higher than that in the burette, the pressure of the trapped air is:
\[P=H+h.\]
For example, if \(H=760\ \text{mmHg}\), the following readings may be obtained:
| Mercury head, \(h\) (mm) | Pressure, \(P=760+h\) (mmHg) | Volume, \(V\) (cm3) | \(1/V\) (cm−3) | \(PV\) (mmHg cm3) |
|---|---|---|---|---|
| 0 | 760 | 30.00 | 0.03333 | 22 800 |
| 95 | 855 | 26.67 | 0.03750 | 22 800 |
| 190 | 950 | 24.00 | 0.04167 | 22 800 |
| 380 | 1140 | 20.00 | 0.05000 | 22 800 |
| 760 | 1520 | 15.00 | 0.06667 | 22 800 |
A plot of \(P\) against \(1/V\) is a straight line through the origin:
The gradient of the graph is \(22\,800\ \text{mmHg cm}^3\). Hence \(P=(22\,800)(1/V)\), so \(PV=22\,800\ \text{mmHg cm}^3\), a constant. This verifies Boyle's law.
(b)(iii) Precautions
- Use dry air and ensure that the apparatus is airtight, so that the mass of trapped gas remains constant.
- Raise or lower the reservoir slowly and wait for the gas to return to room temperature before taking each reading.
- Take all scale readings at eye level to avoid parallax error.
Antwortdetails
(a) Kinetic theory of matter
(i) Evaporation causes cooling: Molecules in a liquid are in continual random motion and possess different kinetic energies. The molecules with the greatest kinetic energy near the surface escape from the liquid during evaporation. Their escape reduces the average kinetic energy of the molecules left in the liquid. Since temperature is proportional to the average kinetic energy of the molecules, the temperature of the liquid falls; hence evaporation causes cooling.
(ii) Water boils without a rise in temperature: At its boiling point, the heat supplied to water is used as latent heat of vaporization. It is used to overcome the attractive intermolecular forces and to separate the water molecules to form steam. It does not increase the average kinetic energy of the molecules. Therefore, the temperature remains constant until all the water has changed to steam.
(b)(i) Boyle's law
For a fixed mass of gas at constant temperature, the volume of the gas is inversely proportional to its pressure.
Thus,
\[V\propto\frac{1}{P}\qquad\text{or}\qquad PV=\text{constant}.\]
(b)(ii) Experiment
The apparatus is arranged as shown below. A quantity of dry air is trapped in the graduated burette B by closing the tap. The burette is kept vertical. The mercury reservoir R is raised slowly to compress the trapped air. For each setting, read the volume, \(V\), of trapped air in the burette and the difference, \(h\), between the mercury levels. The barometer reading, \(H\), is also noted.
When the mercury level in the reservoir is higher than that in the burette, the pressure of the trapped air is:
\[P=H+h.\]
For example, if \(H=760\ \text{mmHg}\), the following readings may be obtained:
| Mercury head, \(h\) (mm) | Pressure, \(P=760+h\) (mmHg) | Volume, \(V\) (cm3) | \(1/V\) (cm−3) | \(PV\) (mmHg cm3) |
|---|---|---|---|---|
| 0 | 760 | 30.00 | 0.03333 | 22 800 |
| 95 | 855 | 26.67 | 0.03750 | 22 800 |
| 190 | 950 | 24.00 | 0.04167 | 22 800 |
| 380 | 1140 | 20.00 | 0.05000 | 22 800 |
| 760 | 1520 | 15.00 | 0.06667 | 22 800 |
A plot of \(P\) against \(1/V\) is a straight line through the origin:
The gradient of the graph is \(22\,800\ \text{mmHg cm}^3\). Hence \(P=(22\,800)(1/V)\), so \(PV=22\,800\ \text{mmHg cm}^3\), a constant. This verifies Boyle's law.
(b)(iii) Precautions
- Use dry air and ensure that the apparatus is airtight, so that the mass of trapped gas remains constant.
- Raise or lower the reservoir slowly and wait for the gas to return to room temperature before taking each reading.
- Take all scale readings at eye level to avoid parallax error.
Frage 4 Bericht
(a) What is surface tension?
(b) State two methods by which the surface tension of a liquid can be reduced.
(a) Surface tension
Surface tension is the property of the free surface of a liquid by which it behaves like a stretched elastic membrane, tending to contract to the smallest possible area. It is measured as the force acting at right angles per unit length along a line drawn in the surface: \[ \gamma = \frac{F}{L} \] with S.I. unit newton per metre (N m-1). It arises from the net inward cohesive pull on molecules at the surface.
(b) Two methods of reducing surface tension
- Adding a detergent, soap or other surface-active impurity to the liquid.
- Raising the temperature of the liquid.
Antwortdetails
(a) Surface tension
Surface tension is the property of the free surface of a liquid by which it behaves like a stretched elastic membrane, tending to contract to the smallest possible area. It is measured as the force acting at right angles per unit length along a line drawn in the surface: \[ \gamma = \frac{F}{L} \] with S.I. unit newton per metre (N m-1). It arises from the net inward cohesive pull on molecules at the surface.
(b) Two methods of reducing surface tension
- Adding a detergent, soap or other surface-active impurity to the liquid.
- Raising the temperature of the liquid.
Frage 5 Bericht
(a) Define (i) Linear momentum; (ii) Impulse
(b) State the principle of conservation of linear momentum.
(c) A tractor of mass 5.0 x 10\(^{3}\)kg is used a tow a car of mass 2.5 x 103 kg. The tractor moved with a speed of 3.0 ms\(^{-1}\) just before the towing rope becomes taut. Calculate the:
(i) Speed of the tractor immediately the rope becomes taut
(ii) Loss in kinetic energy of the system just after the car has started Moving;
(iii) Impulse in the rope when it jerks the car into motion.
(a) Definitions
(i) Linear momentum is the product of the mass of a body and its velocity: \(p = mv\). Its S.I. unit is kg m s-1 (N s).
(ii) Impulse is the product of a force and the time for which it acts, and it equals the change in momentum it produces: \(J = Ft = \Delta(mv)\). Its S.I. unit is N s.
(b) Principle of conservation of linear momentum
In a system of colliding bodies on which no external resultant force acts, the total linear momentum before impact is equal to the total linear momentum after impact.
(c) Calculations
Tractor: \(m_1 = 5.0\times10^{3}\) kg at \(u_1 = 3.0\) m s-1; car: \(m_2 = 2.5\times10^{3}\) kg at rest. When the rope becomes taut they move together with common velocity v.
(i) Common speed after the rope is taut
\[ m_1 u_1 = (m_1 + m_2)v \] \[ v = \frac{5.0\times10^{3}\times3.0}{(5.0\times10^{3}+2.5\times10^{3})} = \frac{15\,000}{7\,500} = 2.0\ \text{m s}^{-1} \]
(ii) Loss in kinetic energy
\[ KE_i = \tfrac{1}{2}m_1 u_1^2 = \tfrac{1}{2}\times5.0\times10^{3}\times3.0^2 = 22\,500\ \text{J} \] \[ KE_f = \tfrac{1}{2}(m_1+m_2)v^2 = \tfrac{1}{2}\times7.5\times10^{3}\times2.0^2 = 15\,000\ \text{J} \] \[ \text{Loss} = 22\,500 - 15\,000 = 7\,500\ \text{J} \]
(iii) Impulse in the rope
Impulse equals the change in momentum of the car: \[ J = m_2 v - 0 = 2.5\times10^{3}\times2.0 = 5.0\times10^{3}\ \text{N s} \]
Antwortdetails
(a) Definitions
(i) Linear momentum is the product of the mass of a body and its velocity: \(p = mv\). Its S.I. unit is kg m s-1 (N s).
(ii) Impulse is the product of a force and the time for which it acts, and it equals the change in momentum it produces: \(J = Ft = \Delta(mv)\). Its S.I. unit is N s.
(b) Principle of conservation of linear momentum
In a system of colliding bodies on which no external resultant force acts, the total linear momentum before impact is equal to the total linear momentum after impact.
(c) Calculations
Tractor: \(m_1 = 5.0\times10^{3}\) kg at \(u_1 = 3.0\) m s-1; car: \(m_2 = 2.5\times10^{3}\) kg at rest. When the rope becomes taut they move together with common velocity v.
(i) Common speed after the rope is taut
\[ m_1 u_1 = (m_1 + m_2)v \] \[ v = \frac{5.0\times10^{3}\times3.0}{(5.0\times10^{3}+2.5\times10^{3})} = \frac{15\,000}{7\,500} = 2.0\ \text{m s}^{-1} \]
(ii) Loss in kinetic energy
\[ KE_i = \tfrac{1}{2}m_1 u_1^2 = \tfrac{1}{2}\times5.0\times10^{3}\times3.0^2 = 22\,500\ \text{J} \] \[ KE_f = \tfrac{1}{2}(m_1+m_2)v^2 = \tfrac{1}{2}\times7.5\times10^{3}\times2.0^2 = 15\,000\ \text{J} \] \[ \text{Loss} = 22\,500 - 15\,000 = 7\,500\ \text{J} \]
(iii) Impulse in the rope
Impulse equals the change in momentum of the car: \[ J = m_2 v - 0 = 2.5\times10^{3}\times2.0 = 5.0\times10^{3}\ \text{N s} \]
Frage 6 Bericht
a) (i) Illustrate, using a ray diagram, how an image can be formed by a convex mirror.
(ii) State one advantage rid one disadvantage of using a convex mirror as a iving mirror.
(iii) Explain the action of a compound microscope.
(b) Illustrate using labelled diagrams only, sonometer wire of length I, vibrating at its fundamental (ii) first overtone (iii) second overtone
(c) A tuning fork vibrating at a frequency of 512 Hz is held over the top of a jar filled with water and fitted with a tap at the buttom. If the jar is 60 cm tall and the speed of sound is 350 ms\(^{-1}\), determine the possible resonance position(s).
(a)(i) Formation of an image by a convex mirror
The reflected rays diverge, but their backward extensions meet behind the mirror. Hence the image is virtual, erect and diminished, and is formed between the pole, P, and the principal focus, F.
(a)(ii) Convex mirror as a driving mirror
- Advantage: It provides a wide field of view and gives an erect image of vehicles behind.
- Disadvantage: The image is diminished, so objects appear farther away than they actually are and distance may be misjudged.
(a)(iii) Action of a compound microscope
A compound microscope consists of two converging lenses of short focal lengths: the objective lens and the eyepiece lens. The object is placed just beyond the focal point of the objective lens. The objective forms a real, inverted and magnified intermediate image within the focal length of the eyepiece. The eyepiece acts as a simple magnifying glass and produces a final image which is virtual, highly magnified and inverted relative to the original object.
(b) Modes of vibration of a sonometer wire
(c) Resonance positions
The air column above the water is closed at the water surface and open at the top. Thus, resonance occurs at air-column lengths:
\[L=\frac{\lambda}{4},\quad \frac{3\lambda}{4},\quad \frac{5\lambda}{4},\ldots\]
\[\lambda=\frac{v}{f}=\frac{350}{512}=0.6836\ \text{m}\approx0.684\ \text{m}\]
First resonance position:
\[L_1=\frac{\lambda}{4}=\frac{0.6836}{4}=0.1709\ \text{m}\approx17.1\ \text{cm}\]
Second resonance position:
\[L_2=\frac{3\lambda}{4}=\frac{3(0.6836)}{4}=0.5127\ \text{m}\approx51.3\ \text{cm}\]
Third resonance position:
\[L_3=\frac{5\lambda}{4}=\frac{5(0.6836)}{4}=0.8545\ \text{m}\approx85.5\ \text{cm}\]
Since the jar is only \(60\ \text{cm}\) high, the third resonance is not possible. Therefore, the possible resonance air-column lengths are 17.1 cm and 51.3 cm, measured downwards from the open top of the jar.
Antwortdetails
(a)(i) Formation of an image by a convex mirror
The reflected rays diverge, but their backward extensions meet behind the mirror. Hence the image is virtual, erect and diminished, and is formed between the pole, P, and the principal focus, F.
(a)(ii) Convex mirror as a driving mirror
- Advantage: It provides a wide field of view and gives an erect image of vehicles behind.
- Disadvantage: The image is diminished, so objects appear farther away than they actually are and distance may be misjudged.
(a)(iii) Action of a compound microscope
A compound microscope consists of two converging lenses of short focal lengths: the objective lens and the eyepiece lens. The object is placed just beyond the focal point of the objective lens. The objective forms a real, inverted and magnified intermediate image within the focal length of the eyepiece. The eyepiece acts as a simple magnifying glass and produces a final image which is virtual, highly magnified and inverted relative to the original object.
(b) Modes of vibration of a sonometer wire
(c) Resonance positions
The air column above the water is closed at the water surface and open at the top. Thus, resonance occurs at air-column lengths:
\[L=\frac{\lambda}{4},\quad \frac{3\lambda}{4},\quad \frac{5\lambda}{4},\ldots\]
\[\lambda=\frac{v}{f}=\frac{350}{512}=0.6836\ \text{m}\approx0.684\ \text{m}\]
First resonance position:
\[L_1=\frac{\lambda}{4}=\frac{0.6836}{4}=0.1709\ \text{m}\approx17.1\ \text{cm}\]
Second resonance position:
\[L_2=\frac{3\lambda}{4}=\frac{3(0.6836)}{4}=0.5127\ \text{m}\approx51.3\ \text{cm}\]
Third resonance position:
\[L_3=\frac{5\lambda}{4}=\frac{5(0.6836)}{4}=0.8545\ \text{m}\approx85.5\ \text{cm}\]
Since the jar is only \(60\ \text{cm}\) high, the third resonance is not possible. Therefore, the possible resonance air-column lengths are 17.1 cm and 51.3 cm, measured downwards from the open top of the jar.
Frage 7 Bericht
(a)(i) Explain photoelectric emission.
(ii) State four applications of photoelectric emission.
(b) Draw and label a diagram showing the structure of a simple type of a photocell and explain its mode of operation.
(c) In a photocell, no electrons are emitted until the threshold frequency of light is reached.
(i) Explain what happens to the energy of the light before emission of electrons begin.
(ii) State one factor that may affect the number of emitted electrons.
(a)(i) 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. The minimum frequency capable of producing emission is called the threshold frequency.
(a)(ii) Applications of photoelectric emission
- Burglar alarm systems.
- Photographic exposure meters.
- Automatic doors and automatic street-light switches.
- Reproduction of sound from the sound track of films.
(b) Structure and operation of a simple photoelectric cell
The photocell consists of an evacuated glass bulb containing a curved photosensitive cathode and a central anode. The cathode is coated with a photosensitive material such as caesium. The anode is connected to the positive terminal of a battery through a microammeter.
When light of frequency at least equal to the threshold frequency falls on the cathode, photoelectrons are emitted. The positively charged anode attracts and collects these electrons. Their movement through the external circuit constitutes a photoelectric current, which is indicated by the microammeter. For radiation above the threshold frequency, increasing the intensity of the light increases the photoelectric current.
(c)(i) Before photoelectric emission begins, the light energy is absorbed by electrons in the photosensitive surface. If the frequency is below the threshold frequency, the energy supplied to each electron is insufficient to overcome the work function or potential barrier of the metal. The absorbed energy is therefore dissipated, mainly as heat, and no electrons are emitted.
(c)(ii) One factor affecting the number of electrons emitted is the intensity of the incident light. Greater intensity supplies more photons per second and hence emits more electrons, provided that the frequency is above the threshold frequency.
Antwortdetails
(a)(i) 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. The minimum frequency capable of producing emission is called the threshold frequency.
(a)(ii) Applications of photoelectric emission
- Burglar alarm systems.
- Photographic exposure meters.
- Automatic doors and automatic street-light switches.
- Reproduction of sound from the sound track of films.
(b) Structure and operation of a simple photoelectric cell
The photocell consists of an evacuated glass bulb containing a curved photosensitive cathode and a central anode. The cathode is coated with a photosensitive material such as caesium. The anode is connected to the positive terminal of a battery through a microammeter.
When light of frequency at least equal to the threshold frequency falls on the cathode, photoelectrons are emitted. The positively charged anode attracts and collects these electrons. Their movement through the external circuit constitutes a photoelectric current, which is indicated by the microammeter. For radiation above the threshold frequency, increasing the intensity of the light increases the photoelectric current.
(c)(i) Before photoelectric emission begins, the light energy is absorbed by electrons in the photosensitive surface. If the frequency is below the threshold frequency, the energy supplied to each electron is insufficient to overcome the work function or potential barrier of the metal. The absorbed energy is therefore dissipated, mainly as heat, and no electrons are emitted.
(c)(ii) One factor affecting the number of electrons emitted is the intensity of the incident light. Greater intensity supplies more photons per second and hence emits more electrons, provided that the frequency is above the threshold frequency.
Frage 8 Bericht
A force of 40 N applied at the end of a wire of length 4m and diameter 2.00 mm produces an extension of 0.24 mm. Calculate the;
(a) stress on the wire;
(b) strain in the wire (\(\pi = 3.142\))
Given: F = 40 N, L = 4 m, diameter d = 2.00 mm = \(2.00\times10^{-3}\) m, extension e = 0.24 mm = \(0.24\times10^{-3}\) m, \(\pi = 3.142\).
(a) Stress on the wire
Cross-sectional area of the wire: \[ A = \frac{\pi d^2}{4} = \frac{3.142 \times (2.00\times10^{-3})^2}{4} = \frac{3.142 \times 4.00\times10^{-6}}{4} = 3.142\times10^{-6}\ \text{m}^2 \]
Stress: \[ \sigma = \frac{F}{A} = \frac{40}{3.142\times10^{-6}} = 1.27\times10^{7}\ \text{N m}^{-2} \]
(b) Strain in the wire
\[ \text{strain} = \frac{e}{L} = \frac{0.24\times10^{-3}}{4} = 6.0\times10^{-5} \]
Strain has no unit (it is a ratio).
Antwortdetails
Given: F = 40 N, L = 4 m, diameter d = 2.00 mm = \(2.00\times10^{-3}\) m, extension e = 0.24 mm = \(0.24\times10^{-3}\) m, \(\pi = 3.142\).
(a) Stress on the wire
Cross-sectional area of the wire: \[ A = \frac{\pi d^2}{4} = \frac{3.142 \times (2.00\times10^{-3})^2}{4} = \frac{3.142 \times 4.00\times10^{-6}}{4} = 3.142\times10^{-6}\ \text{m}^2 \]
Stress: \[ \sigma = \frac{F}{A} = \frac{40}{3.142\times10^{-6}} = 1.27\times10^{7}\ \text{N m}^{-2} \]
(b) Strain in the wire
\[ \text{strain} = \frac{e}{L} = \frac{0.24\times10^{-3}}{4} = 6.0\times10^{-5} \]
Strain has no unit (it is a ratio).
Frage 9 Bericht
Define;
(i) Elasticity
(ii) Young's modulus
(iii) Force constant
(i) Elasticity
Elasticity is the property of a material by which it returns to its original size and shape after the deforming force applied to it has been removed.
(ii) Young's modulus
Young's modulus (the modulus of elasticity) is the ratio of the tensile stress to the tensile strain of a material, provided the elastic limit is not exceeded: \[ E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{e/L} = \frac{FL}{Ae} \] Its S.I. unit is the pascal (N m-2).
(iii) Force constant
The force constant (stiffness) of an elastic material is the force required to produce unit extension in it: \[ k = \frac{F}{e} \] Its S.I. unit is newton per metre (N m-1).
Antwortdetails
(i) Elasticity
Elasticity is the property of a material by which it returns to its original size and shape after the deforming force applied to it has been removed.
(ii) Young's modulus
Young's modulus (the modulus of elasticity) is the ratio of the tensile stress to the tensile strain of a material, provided the elastic limit is not exceeded: \[ E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{e/L} = \frac{FL}{Ae} \] Its S.I. unit is the pascal (N m-2).
(iii) Force constant
The force constant (stiffness) of an elastic material is the force required to produce unit extension in it: \[ k = \frac{F}{e} \] Its S.I. unit is newton per metre (N m-1).
Frage 10 Bericht
(a) Explain the term electrolyte and give two examples.
(b) State the relationship between the mass of a substance liberated during electrolysis and the charge passed.
(a) Electrolyte
An electrolyte is a substance (usually a molten ionic compound or an aqueous solution of an acid, base or salt) that conducts electric current by the movement of its free ions and is chemically decomposed in the process.
Two examples:
- Dilute tetraoxosulphate(VI) acid (dilute H2SO4).
- Copper(II) tetraoxosulphate(VI) solution (CuSO4 solution); acidified water or sodium chloride solution are also acceptable.
(b) Mass liberated and charge passed
By Faraday's first law of electrolysis, the mass m of a substance liberated at an electrode is directly proportional to the quantity of electric charge Q that passes through the electrolyte: \[ m \propto Q \Rightarrow m = z Q = z I t \]
where z is the electrochemical equivalent of the substance, I the current and t the time.
Antwortdetails
(a) Electrolyte
An electrolyte is a substance (usually a molten ionic compound or an aqueous solution of an acid, base or salt) that conducts electric current by the movement of its free ions and is chemically decomposed in the process.
Two examples:
- Dilute tetraoxosulphate(VI) acid (dilute H2SO4).
- Copper(II) tetraoxosulphate(VI) solution (CuSO4 solution); acidified water or sodium chloride solution are also acceptable.
(b) Mass liberated and charge passed
By Faraday's first law of electrolysis, the mass m of a substance liberated at an electrode is directly proportional to the quantity of electric charge Q that passes through the electrolyte: \[ m \propto Q \Rightarrow m = z Q = z I t \]
where z is the electrochemical equivalent of the substance, I the current and t the time.
Frage 11 Bericht
(a) List two types of waves, apart from light, that can be plane polarized.
(b) State two applications of plane polarized light.
(a) Waves (other than light) that can be plane polarized
Only transverse waves can be polarized. Examples apart from light are:
- Radio waves (and microwaves) - transverse electromagnetic waves.
- Transverse mechanical waves on a stretched rope or string (X-rays are also acceptable).
(b) Two applications of plane polarized light
- In polaroid sunglasses and camera filters to cut down glare from reflected light.
- In liquid-crystal displays (LCDs); also used in stress analysis of transparent models (photoelasticity) and in the polarimeter to measure sugar concentration.
Antwortdetails
(a) Waves (other than light) that can be plane polarized
Only transverse waves can be polarized. Examples apart from light are:
- Radio waves (and microwaves) - transverse electromagnetic waves.
- Transverse mechanical waves on a stretched rope or string (X-rays are also acceptable).
(b) Two applications of plane polarized light
- In polaroid sunglasses and camera filters to cut down glare from reflected light.
- In liquid-crystal displays (LCDs); also used in stress analysis of transparent models (photoelasticity) and in the polarimeter to measure sugar concentration.
Frage 12 Bericht
(a) Explain diffusion
(b) State one factor that can affect the rate of diffusion.
(a) Diffusion
Diffusion is the net movement of the molecules of a substance from a region of higher concentration to a region of lower concentration until they are evenly spread out. It occurs because molecules are in continuous random motion; the greater number in the crowded region means more of them, on average, drift towards the less crowded region, so the substance gradually spreads throughout the space available.
(b) One factor affecting the rate of diffusion
- Temperature (higher temperature gives faster molecular motion and quicker diffusion). Other acceptable factors: the concentration gradient, or the mass/density of the diffusing molecules.
Antwortdetails
(a) Diffusion
Diffusion is the net movement of the molecules of a substance from a region of higher concentration to a region of lower concentration until they are evenly spread out. It occurs because molecules are in continuous random motion; the greater number in the crowded region means more of them, on average, drift towards the less crowded region, so the substance gradually spreads throughout the space available.
(b) One factor affecting the rate of diffusion
- Temperature (higher temperature gives faster molecular motion and quicker diffusion). Other acceptable factors: the concentration gradient, or the mass/density of the diffusing molecules.
Frage 13 Bericht
(a)(i) What is meant by neutral point in a magnetic field?
(ii) Draw and label a diagram to show the pattern and direction of the magnetic field produced around a straight current-carrying wire.
(b) When is an ammeter said to be (i) Sensitive (ii) accurate?
(c)(i) Explain, using a labelled diagram, how a delicate magnetic material could be protected, from the Earth's magnetic field.
(ii) A charge of 1.6 x 10\(^{-19}\) C enters a magnetic field of flux density 2.0 T with a velocity of 2.5 x 10\(^7\) ms\(^{-1}\) at an angle of 30° with the field. Calculate the magnitude of the force exerted on the charge by the field.
(d) State the laws of electro-magnetic induction.
(a)(i) Neutral point
A neutral point is a point at which the resultant magnetic flux density is zero. At this point, the magnetic fields acting there are equal in magnitude and opposite in direction.
(a)(ii) Magnetic field around a straight current-carrying wire
The field consists of concentric circles centred on the wire. For a current out of the page, the direction of the field is anticlockwise, as shown.
(b) Ammeter
(i) Sensitive: An ammeter is sensitive when it produces an appreciable or large deflection for a very small current.
(ii) Accurate: An ammeter is accurate when the current it measures is very close to the true value of the current flowing in the circuit.
(c)(i) Protection from the Earth's magnetic field
The delicate magnetic material is enclosed in a thick soft-iron or other high-permeability magnetic shield. The shield provides an easier path for the Earth's magnetic field lines, so that very little magnetic field exists in the cavity containing the material.
(c)(ii) Force on the charge
The magnetic force is given by:
\[F=qvB\sin\theta\]
\[F=(1.6\times10^{-19})(2.5\times10^{7})(2.0)\sin30^\circ\]
\[F=(1.6\times10^{-19})(2.5\times10^{7})(2.0)(0.5)\]
\[\boxed{F=4.0\times10^{-12}\ \text{N}}\]
(d) Laws of electromagnetic induction
- Whenever the magnetic flux linkage through a circuit changes, an e.m.f. is induced in the circuit. The magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linkage.
- The induced current flows in such a direction as to oppose the change in magnetic flux that produces it.
Antwortdetails
(a)(i) Neutral point
A neutral point is a point at which the resultant magnetic flux density is zero. At this point, the magnetic fields acting there are equal in magnitude and opposite in direction.
(a)(ii) Magnetic field around a straight current-carrying wire
The field consists of concentric circles centred on the wire. For a current out of the page, the direction of the field is anticlockwise, as shown.
(b) Ammeter
(i) Sensitive: An ammeter is sensitive when it produces an appreciable or large deflection for a very small current.
(ii) Accurate: An ammeter is accurate when the current it measures is very close to the true value of the current flowing in the circuit.
(c)(i) Protection from the Earth's magnetic field
The delicate magnetic material is enclosed in a thick soft-iron or other high-permeability magnetic shield. The shield provides an easier path for the Earth's magnetic field lines, so that very little magnetic field exists in the cavity containing the material.
(c)(ii) Force on the charge
The magnetic force is given by:
\[F=qvB\sin\theta\]
\[F=(1.6\times10^{-19})(2.5\times10^{7})(2.0)\sin30^\circ\]
\[F=(1.6\times10^{-19})(2.5\times10^{7})(2.0)(0.5)\]
\[\boxed{F=4.0\times10^{-12}\ \text{N}}\]
(d) Laws of electromagnetic induction
- Whenever the magnetic flux linkage through a circuit changes, an e.m.f. is induced in the circuit. The magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linkage.
- The induced current flows in such a direction as to oppose the change in magnetic flux that produces it.
Frage 14 Bericht
(a) What is meant by the wave-particles duality of matter?
(b) mention one physical phenomenon, in each case, that can be explained in terms of the wave nature and particle nature of light.
(a) Wave-particle duality of matter
Wave-particle duality is the concept that matter (and radiation) can exhibit both wave-like and particle-like properties, depending on the experiment used to observe it. A moving particle of momentum p has an associated (de Broglie) wavelength \[ \lambda = \frac{h}{p} = \frac{h}{mv} \] so that the same entity behaves as a stream of particles in some situations and as a wave in others.
(b) One phenomenon for each nature of light
- Wave nature: interference (or diffraction, or polarization) of light.
- Particle nature: the photoelectric effect (the Compton effect is also acceptable).
Antwortdetails
(a) Wave-particle duality of matter
Wave-particle duality is the concept that matter (and radiation) can exhibit both wave-like and particle-like properties, depending on the experiment used to observe it. A moving particle of momentum p has an associated (de Broglie) wavelength \[ \lambda = \frac{h}{p} = \frac{h}{mv} \] so that the same entity behaves as a stream of particles in some situations and as a wave in others.
(b) One phenomenon for each nature of light
- Wave nature: interference (or diffraction, or polarization) of light.
- Particle nature: the photoelectric effect (the Compton effect is also acceptable).
Frage 15 Bericht
(a) State two objects each used in sports and warfare which may be considered as projectiles.
(b) the horizontal range R, of a projectile is given by the expression;
R = \(\frac{u^2 sin 2\theta}{g}\) where \(\theta\) is the angle of projection, g the acceleration of free fall due to gravity and u the initial velocity. At what value of \(\theta\) will R be maximium?
(a) Objects that behave as projectiles
- In sports: a thrown javelin, a kicked football (also a shot-put, discus, or golf ball).
- In warfare: a bullet fired from a gun, a launched bomb or missile (also a cannon ball or arrow).
(b) Angle for maximum range
The range is \[ R = \frac{u^2 \sin 2\theta}{g} \] For a fixed initial speed u and fixed g, R is greatest when \(\sin 2\theta\) is greatest. The maximum value of a sine function is 1, so \[ \sin 2\theta = 1 \Rightarrow 2\theta = 90^{\circ} \Rightarrow \theta = 45^{\circ} \]
The range is maximum when the angle of projection is 45°.
Antwortdetails
(a) Objects that behave as projectiles
- In sports: a thrown javelin, a kicked football (also a shot-put, discus, or golf ball).
- In warfare: a bullet fired from a gun, a launched bomb or missile (also a cannon ball or arrow).
(b) Angle for maximum range
The range is \[ R = \frac{u^2 \sin 2\theta}{g} \] For a fixed initial speed u and fixed g, R is greatest when \(\sin 2\theta\) is greatest. The maximum value of a sine function is 1, so \[ \sin 2\theta = 1 \Rightarrow 2\theta = 90^{\circ} \Rightarrow \theta = 45^{\circ} \]
The range is maximum when the angle of projection is 45°.
