WAEC SSCE - Physics - 2010 (Essay)
Question 1 Report
(a) State three conclusions that can be drawn from Rutherford's experiment on the scattering of alpha particles by a thin metal foil in relation to the structure of the atom
The diagram above illustrates th3 energy levels of an electron in an atom. If an excited electron moves from \(n_2\) to \(n_\theta\), calculate the:
(i) frequency;
(ii) wavelength of the emitted radiation. [ \(h = 6.6 \times 10^{-34}\) Js; le V = .6 \(\times 10^{-19}\) J; C = 3.0 \(\times 10^8\) ms\(^{-1}\)]
(c) The following nuclear equations represent two types of radioactivity.
\({}^{226}_{88}R_a \to {}^{222}_{86}R_n + {}^a_2a\) (Equation A)
\({}^{14}_7N + {}^4_2a \to {}^{17}_8O + {}^1_1p\) (Equation B)
Identify each type and explain briefly the difference between them
(a) Three conclusions from Rutherford's alpha-particle scattering experiment
- Most of the atom is empty space, because the vast majority of alpha particles passed straight through the thin metal foil undeflected.
- Each atom has a very small, dense core (the nucleus) which carries the whole positive charge, because a few alpha particles were deflected through large angles and a very small number were bounced almost straight back.
- Nearly all the mass of the atom is concentrated in this tiny nucleus, while the electrons occupy the large space surrounding it.
(b) Transition from \(n_2\) to \(n_0\)
From the energy-level diagram: \(n_2 = -2.0\ \text{eV}\) and \(n_0 = -12.0\ \text{eV}\). The energy of the emitted photon is the difference between these levels:
\[ E = E_{n_2} - E_{n_0} = (-2.0) - (-12.0) = 10.0\ \text{eV} \]
Convert to joules using \(1\ \text{eV} = 1.6\times10^{-19}\ \text{J}\):
\[ E = 10.0 \times 1.6\times10^{-19} = 1.6\times10^{-18}\ \text{J} \]
(i) Frequency from \(E = hf\):
\[ f = \frac{E}{h} = \frac{1.6\times10^{-18}}{6.6\times10^{-34}} = 2.42\times10^{15}\ \text{Hz} \]
(ii) Wavelength from \(c = f\lambda\):
\[ \lambda = \frac{c}{f} = \frac{3.0\times10^{8}}{2.42\times10^{15}} = 1.24\times10^{-7}\ \text{m} \]
(about 124 nm, in the ultraviolet region.)
(c) Types of radioactivity
Equation A: \(^{226}_{88}\text{Ra} \to\ ^{222}_{86}\text{Rn} + ^{4}_{2}\alpha\). This is natural (spontaneous) radioactivity, specifically alpha decay: an unstable nucleus disintegrates on its own, emitting an alpha particle and forming a new element.
Equation B: \(^{14}_{7}\text{N} + ^{4}_{2}\alpha \to\ ^{17}_{8}\text{O} + ^{1}_{1}p\). This is artificial (induced) transmutation: a stable nucleus is deliberately bombarded by an incoming particle (an alpha particle), changing it into a different nucleus.
Difference: In A the disintegration is spontaneous, happening by itself with no external cause; in B the nuclear change is artificially induced by bombarding the target nucleus with a fast-moving particle.
Answer Details
(a) Three conclusions from Rutherford's alpha-particle scattering experiment
- Most of the atom is empty space, because the vast majority of alpha particles passed straight through the thin metal foil undeflected.
- Each atom has a very small, dense core (the nucleus) which carries the whole positive charge, because a few alpha particles were deflected through large angles and a very small number were bounced almost straight back.
- Nearly all the mass of the atom is concentrated in this tiny nucleus, while the electrons occupy the large space surrounding it.
(b) Transition from \(n_2\) to \(n_0\)
From the energy-level diagram: \(n_2 = -2.0\ \text{eV}\) and \(n_0 = -12.0\ \text{eV}\). The energy of the emitted photon is the difference between these levels:
\[ E = E_{n_2} - E_{n_0} = (-2.0) - (-12.0) = 10.0\ \text{eV} \]
Convert to joules using \(1\ \text{eV} = 1.6\times10^{-19}\ \text{J}\):
\[ E = 10.0 \times 1.6\times10^{-19} = 1.6\times10^{-18}\ \text{J} \]
(i) Frequency from \(E = hf\):
\[ f = \frac{E}{h} = \frac{1.6\times10^{-18}}{6.6\times10^{-34}} = 2.42\times10^{15}\ \text{Hz} \]
(ii) Wavelength from \(c = f\lambda\):
\[ \lambda = \frac{c}{f} = \frac{3.0\times10^{8}}{2.42\times10^{15}} = 1.24\times10^{-7}\ \text{m} \]
(about 124 nm, in the ultraviolet region.)
(c) Types of radioactivity
Equation A: \(^{226}_{88}\text{Ra} \to\ ^{222}_{86}\text{Rn} + ^{4}_{2}\alpha\). This is natural (spontaneous) radioactivity, specifically alpha decay: an unstable nucleus disintegrates on its own, emitting an alpha particle and forming a new element.
Equation B: \(^{14}_{7}\text{N} + ^{4}_{2}\alpha \to\ ^{17}_{8}\text{O} + ^{1}_{1}p\). This is artificial (induced) transmutation: a stable nucleus is deliberately bombarded by an incoming particle (an alpha particle), changing it into a different nucleus.
Difference: In A the disintegration is spontaneous, happening by itself with no external cause; in B the nuclear change is artificially induced by bombarding the target nucleus with a fast-moving particle.
Question 2 Report
(a) Differentiate between plane polarization and interference as applied to waves.
(b) List two uses of polaroids.
(a) Plane polarisation compared with interference
- Plane polarisation is the restriction of the vibrations of a transverse wave to a single plane (one direction) perpendicular to the direction of travel. It shows that light is a transverse wave and occurs only with transverse waves.
- Interference is the superposition of two waves of the same frequency travelling in the same region, producing points of reinforcement (constructive) and cancellation (destructive). It occurs with both transverse and longitudinal waves and demonstrates the wave nature of light.
In short, polarisation concerns the plane of vibration of a single wave, while interference concerns the combined effect of two coherent waves.
(b) Two uses of polaroids
- In sunglasses (polaroid spectacles) to reduce glare from reflected sunlight.
- In camera filters and in the control of light intensity, e.g. in liquid-crystal displays and photographic work.
Answer Details
(a) Plane polarisation compared with interference
- Plane polarisation is the restriction of the vibrations of a transverse wave to a single plane (one direction) perpendicular to the direction of travel. It shows that light is a transverse wave and occurs only with transverse waves.
- Interference is the superposition of two waves of the same frequency travelling in the same region, producing points of reinforcement (constructive) and cancellation (destructive). It occurs with both transverse and longitudinal waves and demonstrates the wave nature of light.
In short, polarisation concerns the plane of vibration of a single wave, while interference concerns the combined effect of two coherent waves.
(b) Two uses of polaroids
- In sunglasses (polaroid spectacles) to reduce glare from reflected sunlight.
- In camera filters and in the control of light intensity, e.g. in liquid-crystal displays and photographic work.
Question 3 Report
(a) Distinguish between perfectly elastic collision and perfectly inelastic collision.
(b) Sketch a distance — time graph for a particle moving in a straight line with:
(i) uniform speed;
(ii) variable speed.
(c) A body starts from rest and travels distances of 120, 300 and 180m in successive equal time intervals of 12 s. During each interval the body is uniformly accelerated. (i) Calculate the velocity of the body at the end of each successive time interval.
(ii) Sketch a velocity-time graph for the motion.
(a) Perfectly elastic vs perfectly inelastic collision
| Perfectly elastic collision | Perfectly inelastic collision |
|---|---|
| Both momentum and kinetic energy are conserved. | Momentum is conserved but kinetic energy is not; some is converted to heat, sound and deformation. |
| The colliding bodies separate (do not stick) after impact. | The colliding bodies stick together and move as one body after impact. |
| The relative velocity of separation equals the relative velocity of approach. | The bodies move with a single common velocity after impact (relative velocity of separation = 0). |
(b) Distance–time graphs
For uniform speed the distance increases by equal amounts in equal times, so the graph is a straight line of constant gradient through the origin. For variable speed the speed keeps changing, so the gradient keeps changing and the graph is a curve (here the gradient increases, showing increasing speed).
(c) Body starting from rest
The body covers 120 m, 300 m and 180 m in three successive equal intervals of \(t = 12\ \text{s}\), being uniformly accelerated within each interval. For uniform acceleration the average velocity over an interval equals the mean of the velocities at its two ends, so the end-of-interval velocities follow directly.
(i) Velocity at the end of each interval
Interval 1 (\(u = 0\)): \(s = ut + \tfrac{1}{2}at^2\)
\[120 = 0\times 12 + \tfrac{1}{2}a(12)^2 = 72a \;\Rightarrow\; a = \tfrac{5}{3}\ \text{m s}^{-2}\] \[v_1 = u + at = 0 + \tfrac{5}{3}\times 12 = 20\ \text{m s}^{-1}\]Interval 2 (\(u = v_1 = 20\ \text{m s}^{-1}\)):
\[300 = 20\times 12 + \tfrac{1}{2}a(12)^2 = 240 + 72a \;\Rightarrow\; a = \tfrac{5}{6}\ \text{m s}^{-2}\] \[v_2 = 20 + \tfrac{5}{6}\times 12 = 30\ \text{m s}^{-1}\]Interval 3 (\(u = v_2 = 30\ \text{m s}^{-1}\)):
\[180 = 30\times 12 + \tfrac{1}{2}a(12)^2 = 360 + 72a \;\Rightarrow\; a = -2.5\ \text{m s}^{-2}\] \[v_3 = 30 + (-2.5)\times 12 = 0\ \text{m s}^{-1}\]So the velocities at the end of the successive intervals are 20 m s\(^{-1}\), 30 m s\(^{-1}\) and 0 m s\(^{-1}\) (at \(t = 12\ \text{s}, 24\ \text{s}\) and \(36\ \text{s}\) respectively). Each result checks against the average speed: \(120/12 = 10 = \tfrac{0+20}{2}\), \(300/12 = 25 = \tfrac{20+30}{2}\), \(180/12 = 15 = \tfrac{30+0}{2}\).
(ii) Velocity–time graph
The graph is made of three straight segments (constant acceleration in each interval): rising from 0 to 20 m s\(^{-1}\) in the first 12 s, rising more gently to 30 m s\(^{-1}\) at 24 s, then falling to rest at 36 s.
Answer Details
(a) Perfectly elastic vs perfectly inelastic collision
| Perfectly elastic collision | Perfectly inelastic collision |
|---|---|
| Both momentum and kinetic energy are conserved. | Momentum is conserved but kinetic energy is not; some is converted to heat, sound and deformation. |
| The colliding bodies separate (do not stick) after impact. | The colliding bodies stick together and move as one body after impact. |
| The relative velocity of separation equals the relative velocity of approach. | The bodies move with a single common velocity after impact (relative velocity of separation = 0). |
(b) Distance–time graphs
For uniform speed the distance increases by equal amounts in equal times, so the graph is a straight line of constant gradient through the origin. For variable speed the speed keeps changing, so the gradient keeps changing and the graph is a curve (here the gradient increases, showing increasing speed).
(c) Body starting from rest
The body covers 120 m, 300 m and 180 m in three successive equal intervals of \(t = 12\ \text{s}\), being uniformly accelerated within each interval. For uniform acceleration the average velocity over an interval equals the mean of the velocities at its two ends, so the end-of-interval velocities follow directly.
(i) Velocity at the end of each interval
Interval 1 (\(u = 0\)): \(s = ut + \tfrac{1}{2}at^2\)
\[120 = 0\times 12 + \tfrac{1}{2}a(12)^2 = 72a \;\Rightarrow\; a = \tfrac{5}{3}\ \text{m s}^{-2}\] \[v_1 = u + at = 0 + \tfrac{5}{3}\times 12 = 20\ \text{m s}^{-1}\]Interval 2 (\(u = v_1 = 20\ \text{m s}^{-1}\)):
\[300 = 20\times 12 + \tfrac{1}{2}a(12)^2 = 240 + 72a \;\Rightarrow\; a = \tfrac{5}{6}\ \text{m s}^{-2}\] \[v_2 = 20 + \tfrac{5}{6}\times 12 = 30\ \text{m s}^{-1}\]Interval 3 (\(u = v_2 = 30\ \text{m s}^{-1}\)):
\[180 = 30\times 12 + \tfrac{1}{2}a(12)^2 = 360 + 72a \;\Rightarrow\; a = -2.5\ \text{m s}^{-2}\] \[v_3 = 30 + (-2.5)\times 12 = 0\ \text{m s}^{-1}\]So the velocities at the end of the successive intervals are 20 m s\(^{-1}\), 30 m s\(^{-1}\) and 0 m s\(^{-1}\) (at \(t = 12\ \text{s}, 24\ \text{s}\) and \(36\ \text{s}\) respectively). Each result checks against the average speed: \(120/12 = 10 = \tfrac{0+20}{2}\), \(300/12 = 25 = \tfrac{20+30}{2}\), \(180/12 = 15 = \tfrac{30+0}{2}\).
(ii) Velocity–time graph
The graph is made of three straight segments (constant acceleration in each interval): rising from 0 to 20 m s\(^{-1}\) in the first 12 s, rising more gently to 30 m s\(^{-1}\) at 24 s, then falling to rest at 36 s.
Question 4 Report
(a) State two factors which affect the mass of elements deposited during electrolysis.
(b) List two non-electrolysis.
(a) Two factors that affect the mass of an element deposited during electrolysis (Faraday's laws):
- The quantity of electricity passed, i.e. the product of current and time (\(Q = It\)); the mass is directly proportional to it.
- The electrochemical equivalent of the element (which depends on its relative atomic mass and valency).
Other acceptable factor: the duration of current flow at a fixed current.
(b) Two non-electrolytes (substances that do not conduct electricity by ionic dissociation):
- Distilled (pure) water.
- Sugar solution (also acceptable: kerosene, ethanol, pure alcohol).
Answer Details
(a) Two factors that affect the mass of an element deposited during electrolysis (Faraday's laws):
- The quantity of electricity passed, i.e. the product of current and time (\(Q = It\)); the mass is directly proportional to it.
- The electrochemical equivalent of the element (which depends on its relative atomic mass and valency).
Other acceptable factor: the duration of current flow at a fixed current.
(b) Two non-electrolytes (substances that do not conduct electricity by ionic dissociation):
- Distilled (pure) water.
- Sugar solution (also acceptable: kerosene, ethanol, pure alcohol).
Question 5 Report
(a) List two examples each of substances with:
(i) low viscosity;
(ii) high viscosity.
(b) When is a liquid said to be viscostatic?
(a) Examples
- (i) Low viscosity: water and kerosene (also petrol, ethanol).
- (ii) High viscosity: glycerine and honey (also engine oil, palm oil, tar).
(b) Viscostatic liquid
A liquid is said to be viscostatic when its viscosity remains almost constant (does not change appreciably) with change in temperature. Such liquids, e.g. certain multigrade lubricating oils, keep a steady flow behaviour whether hot or cold.
Answer Details
(a) Examples
- (i) Low viscosity: water and kerosene (also petrol, ethanol).
- (ii) High viscosity: glycerine and honey (also engine oil, palm oil, tar).
(b) Viscostatic liquid
A liquid is said to be viscostatic when its viscosity remains almost constant (does not change appreciably) with change in temperature. Such liquids, e.g. certain multigrade lubricating oils, keep a steady flow behaviour whether hot or cold.
Question 6 Report
(a) State two factors which affect the angle of deviation of a ray of light through a triangular glass prism.
(b) Seven virtual images of an object are formed when two plane mirrors are inclined at an angle 0 to each other. Calculate the value of 0.
(c) By means of a ripple tank, a student was able to generate series of transverse waves by varying the frequency of the dipper and all the waves so generated covered a distance of 0.80 m in 0.2s.
(i) Determine the speed, v, of the waves.
Copy and complete the table given in your answer booklet.
(iii) Plot a graph with f on the vertical axis and \(\lambda ^{-1}\) on the horizontal axis.
(iv) What does the slope of the graph represent?
(a) Two factors which affect the angle of deviation of a ray through a triangular glass prism:
- The angle of incidence of the ray on the first refracting face.
- The refracting (apex) angle of the prism together with the refractive index of the glass (which itself depends on the colour/wavelength of the light).
(b) Angle between the two plane mirrors.
The number of images formed by two plane mirrors inclined at an angle \(\theta\) is
\[N=\frac{360^\circ}{\theta}-1.\]With \(N=7\):
\[7=\frac{360^\circ}{\theta}-1 \;\Rightarrow\; \frac{360^\circ}{\theta}=8 \;\Rightarrow\; \theta=\frac{360^\circ}{8}=45^\circ.\](c)(i) Speed of the waves.
\[v=\frac{\text{distance}}{\text{time}}=\frac{0.80\ \text{m}}{0.2\ \text{s}}=4.0\ \text{m s}^{-1}.\](ii) Completed table. Using \(\lambda=\dfrac{v}{f}=\dfrac{4.0}{f}\) and \(\lambda^{-1}=\dfrac{1}{\lambda}\):
| \(f\) /Hz | \(\lambda\) /m | \(\lambda^{-1}\) /m\(^{-1}\) |
|---|---|---|
| 2.0 | 2.00 | 0.50 |
| 4.0 | 1.00 | 1.00 |
| 6.0 | 0.67 | 1.50 |
| 8.0 | 0.50 | 2.00 |
| 10.0 | 0.40 | 2.50 |
(iii) Graph of \(f\) (vertical axis) against \(\lambda^{-1}\) (horizontal axis).
The points \((0.50,2.0),(1.00,4.0),(1.50,6.0),(2.00,8.0),(2.50,10.0)\) lie on a straight line passing through the origin, since \(f=v\,(\lambda^{-1})\).
(iv) Meaning of the slope.
Reading two points on the line of best fit, e.g. \((2.50,10.0)\) and \((0.50,2.0)\):
\[\text{slope}=\frac{f}{\lambda^{-1}}=\frac{10.0-2.0}{2.50-0.50}=\frac{8.0}{2.0}=4.0.\]Since \(f=v\times\lambda^{-1}\), the slope \(=f\lambda=v=4.0\ \text{m s}^{-1}\). The slope of the graph represents the speed (velocity) of the waves.
Answer Details
(a) Two factors which affect the angle of deviation of a ray through a triangular glass prism:
- The angle of incidence of the ray on the first refracting face.
- The refracting (apex) angle of the prism together with the refractive index of the glass (which itself depends on the colour/wavelength of the light).
(b) Angle between the two plane mirrors.
The number of images formed by two plane mirrors inclined at an angle \(\theta\) is
\[N=\frac{360^\circ}{\theta}-1.\]With \(N=7\):
\[7=\frac{360^\circ}{\theta}-1 \;\Rightarrow\; \frac{360^\circ}{\theta}=8 \;\Rightarrow\; \theta=\frac{360^\circ}{8}=45^\circ.\](c)(i) Speed of the waves.
\[v=\frac{\text{distance}}{\text{time}}=\frac{0.80\ \text{m}}{0.2\ \text{s}}=4.0\ \text{m s}^{-1}.\](ii) Completed table. Using \(\lambda=\dfrac{v}{f}=\dfrac{4.0}{f}\) and \(\lambda^{-1}=\dfrac{1}{\lambda}\):
| \(f\) /Hz | \(\lambda\) /m | \(\lambda^{-1}\) /m\(^{-1}\) |
|---|---|---|
| 2.0 | 2.00 | 0.50 |
| 4.0 | 1.00 | 1.00 |
| 6.0 | 0.67 | 1.50 |
| 8.0 | 0.50 | 2.00 |
| 10.0 | 0.40 | 2.50 |
(iii) Graph of \(f\) (vertical axis) against \(\lambda^{-1}\) (horizontal axis).
The points \((0.50,2.0),(1.00,4.0),(1.50,6.0),(2.00,8.0),(2.50,10.0)\) lie on a straight line passing through the origin, since \(f=v\,(\lambda^{-1})\).
(iv) Meaning of the slope.
Reading two points on the line of best fit, e.g. \((2.50,10.0)\) and \((0.50,2.0)\):
\[\text{slope}=\frac{f}{\lambda^{-1}}=\frac{10.0-2.0}{2.50-0.50}=\frac{8.0}{2.0}=4.0.\]Since \(f=v\times\lambda^{-1}\), the slope \(=f\lambda=v=4.0\ \text{m s}^{-1}\). The slope of the graph represents the speed (velocity) of the waves.
Question 7 Report
(a) Explain the terms: (i) inertia; (ii) inertial mass.
(b) List three factors which affect the rate of evaporation of water in a pond.
(c) Two ice cubes pressed together for some time were found to stick together when the pressure was removed. Explain this observation.
(d) Two vertical capillary tubes of the same diameter are lowered into beakers situated at the same level, containing liquids A and B of densities 9.2 x 10\(^2\) kgm\(^{-3}\) and 1.30 x 10\(^3\) kgm\(^{-3}\) respectively. A suction pump is used to withdraw air from the top of the liquid columns in the tubes by means of a T-piece arrangement until the liquid in A rises to a height of 26.0 cm. Calculate the height of the liquid in tube B.
(a)(i) Inertia is the property of a body by which it resists any change in its state of rest or of uniform motion in a straight line.
(a)(ii) Inertial mass is the measure of the inertia of a body; it is the ratio of the resultant force acting on the body to the acceleration it produces, \(m = \dfrac{F}{a}\).
(b) Three factors affecting the rate of evaporation of water in a pond:
- Temperature of the water (and of the surroundings).
- Surface area exposed to the air.
- Movement of air (wind) and the humidity of the surrounding air.
(c) Ice cubes sticking together (regelation): the pressure between the two cubes lowers the melting point of the ice at the contact surfaces, so a thin film of water forms there. When the pressure is removed the melting point rises again and this water re-freezes, joining the two cubes into one.
(d) Capillary tubes calculation. The two tubes have the same diameter and are joined to the same reduced pressure through the T-piece, so the pressure supporting each column is the same. Hence \(\rho_A g h_A = \rho_B g h_B\), giving
\[ h_B = \frac{\rho_A h_A}{\rho_B} = \frac{(9.2\times10^{2})(26.0)}{1.30\times10^{3}} \] \[ h_B = \frac{23920}{1300} = 18.4\ \text{cm} \]The liquid in tube B rises to 18.4 cm.
Answer Details
(a)(i) Inertia is the property of a body by which it resists any change in its state of rest or of uniform motion in a straight line.
(a)(ii) Inertial mass is the measure of the inertia of a body; it is the ratio of the resultant force acting on the body to the acceleration it produces, \(m = \dfrac{F}{a}\).
(b) Three factors affecting the rate of evaporation of water in a pond:
- Temperature of the water (and of the surroundings).
- Surface area exposed to the air.
- Movement of air (wind) and the humidity of the surrounding air.
(c) Ice cubes sticking together (regelation): the pressure between the two cubes lowers the melting point of the ice at the contact surfaces, so a thin film of water forms there. When the pressure is removed the melting point rises again and this water re-freezes, joining the two cubes into one.
(d) Capillary tubes calculation. The two tubes have the same diameter and are joined to the same reduced pressure through the T-piece, so the pressure supporting each column is the same. Hence \(\rho_A g h_A = \rho_B g h_B\), giving
\[ h_B = \frac{\rho_A h_A}{\rho_B} = \frac{(9.2\times10^{2})(26.0)}{1.30\times10^{3}} \] \[ h_B = \frac{23920}{1300} = 18.4\ \text{cm} \]The liquid in tube B rises to 18.4 cm.
Question 8 Report
(a)
When a positively charged conductor is placed near a candle flame, the flame spreads out as shown in the diagram above. Explain this observation.
(b) A proton moving with a speed of 5.0 x 10\(^{5}\) ms\(^{-1}\) enters a magnetic field of flux density 0.2 T at an angle of 30° to the field. Calculate the magnitude of the magnetic fcrce exerted on the proton. [Proton charge = 1.6 x 10\(^{-19}\) C]
(c)
The diagram above illustrates a 9.0 V battery of internal resistance 0.5 \(\Omega\) connected to two resistors of values 2.0 \(\Omega\) and R \(\Omega\). A\(_1\) A\(_2\) and A\(_3\) are ammeters of negligible internal resistances. If Al reads 4.0 A, calculate the:
(i) equivalent resistance of the combined resistors 2.0 \(\Omega\) and R \(\Omega\);
(ii) currents through A\(_1\) and A\(_3\) ; (iii) value of R.
Answer Details
None
Question 9 Report
State one reason each why cathode rays:
(a) are not electromagnetic waves;
(b) cast sharp shadows of objects in their path;
(c) can rotate a light paddle wheel inside a discharge tube.
(a) Cathode rays are not electromagnetic waves because they are deflected by both electric and magnetic fields, showing that they are streams of negatively charged particles (electrons); electromagnetic waves carry no charge and are undeflected by such fields.
(b) They cast sharp shadows of objects in their path because they travel in straight lines from the cathode.
(c) They can rotate a light paddle wheel inside a discharge tube because they are material particles possessing mass and momentum (kinetic energy), which they transfer to the vanes on striking them.
Answer Details
(a) Cathode rays are not electromagnetic waves because they are deflected by both electric and magnetic fields, showing that they are streams of negatively charged particles (electrons); electromagnetic waves carry no charge and are undeflected by such fields.
(b) They cast sharp shadows of objects in their path because they travel in straight lines from the cathode.
(c) They can rotate a light paddle wheel inside a discharge tube because they are material particles possessing mass and momentum (kinetic energy), which they transfer to the vanes on striking them.
Question 10 Report
When a lead-acid accumulator is fully charged, evolution of gases occurs at the electrodes. Name these gases and the respective electrodes at which thcy are given off.
Gases evolved are:
(i) Hydrogen at the cathode (negative lead plate).
(ii) Oxygen at the anode (positive lead peroxide plate).
Answer Details
Gases evolved are:
(i) Hydrogen at the cathode (negative lead plate).
(ii) Oxygen at the anode (positive lead peroxide plate).
Question 11 Report
A stone is projected vertically upward with a speed of 30ms\(^{1}\) from the top of a tower of height 50 m. Neglecting air resistance, determine the maximum height it reached from the ground. [g = 10 ms\(^-2\)]
Given: \(u = 30\ \text{m s}^{-1}\) (upward), tower height \(= 50\ \text{m}\), \(g = 10\ \text{m s}^{-2}\).
Step 1 - rise above the top of the tower. At the highest point the velocity is zero, so using \(v^{2} = u^{2} - 2gh\):
\[ 0 = 30^{2} - 2(10)h \] \[ h = \frac{900}{20} = 45\ \text{m} \]Step 2 - height above the ground. Add the height of the tower:
\[ H = 50 + 45 = 95\ \text{m} \]The stone reaches a maximum height of 95 m above the ground.
Answer Details
Given: \(u = 30\ \text{m s}^{-1}\) (upward), tower height \(= 50\ \text{m}\), \(g = 10\ \text{m s}^{-2}\).
Step 1 - rise above the top of the tower. At the highest point the velocity is zero, so using \(v^{2} = u^{2} - 2gh\):
\[ 0 = 30^{2} - 2(10)h \] \[ h = \frac{900}{20} = 45\ \text{m} \]Step 2 - height above the ground. Add the height of the tower:
\[ H = 50 + 45 = 95\ \text{m} \]The stone reaches a maximum height of 95 m above the ground.
Question 12 Report
An X-ray tube operates at a potential of 2500 V. If the power of the tube is 750 W, calculate the speed of the electron striking the target. [e = 1.6 x 10\(^{-19}\) C; mass of electron = 9.1 x 10\(^{-3}\) kg]
Given: accelerating p.d. \(V = 2500\ \text{V}\), \(e = 1.6\times10^{-19}\ \text{C}\), electron mass \(m = 9.1\times10^{-31}\ \text{kg}\). (The mass is taken as \(9.1\times10^{-31}\ \text{kg}\), the standard electron mass; the exponent printed as \(10^{-3}\) is a typographical slip.)
The work done by the accelerating field is converted to kinetic energy of the electron:
\[ eV = \tfrac{1}{2}mv^{2} \] \[ v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2(1.6\times10^{-19})(2500)}{9.1\times10^{-31}}} \] \[ v = \sqrt{\frac{8.0\times10^{-16}}{9.1\times10^{-31}}} = \sqrt{8.79\times10^{14}} \] \[ v \approx 2.97\times10^{7}\ \text{m s}^{-1} \]The electron strikes the target at about \(3.0\times10^{7}\ \text{m s}^{-1}\). (The power rating of 750 W is not needed for the speed; it would fix the tube current.)
Answer Details
Given: accelerating p.d. \(V = 2500\ \text{V}\), \(e = 1.6\times10^{-19}\ \text{C}\), electron mass \(m = 9.1\times10^{-31}\ \text{kg}\). (The mass is taken as \(9.1\times10^{-31}\ \text{kg}\), the standard electron mass; the exponent printed as \(10^{-3}\) is a typographical slip.)
The work done by the accelerating field is converted to kinetic energy of the electron:
\[ eV = \tfrac{1}{2}mv^{2} \] \[ v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2(1.6\times10^{-19})(2500)}{9.1\times10^{-31}}} \] \[ v = \sqrt{\frac{8.0\times10^{-16}}{9.1\times10^{-31}}} = \sqrt{8.79\times10^{14}} \] \[ v \approx 2.97\times10^{7}\ \text{m s}^{-1} \]The electron strikes the target at about \(3.0\times10^{7}\ \text{m s}^{-1}\). (The power rating of 750 W is not needed for the speed; it would fix the tube current.)
Question 13 Report
A force of 40 N is applied at the free end of a wire fixed at one end to produce an extension of 0.24 mm. If the original length and diameter of the wire art., 3 m and 2.0 mm respectively, calculate the: (a) stress on the wire; (b) strain in the wire.
Given: \(F = 40\ \text{N}\), extension \(e = 0.24\ \text{mm} = 0.24\times10^{-3}\ \text{m}\), original length \(L = 3\ \text{m}\), diameter \(d = 2.0\ \text{mm} = 2.0\times10^{-3}\ \text{m}\).
(a) Stress \(= \dfrac{F}{A}\), where the cross-sectional area \(A = \dfrac{\pi d^{2}}{4}\).
\[ A = \frac{\pi (2.0\times10^{-3})^{2}}{4} = 3.14\times10^{-6}\ \text{m}^{2} \] \[ \text{Stress} = \frac{40}{3.14\times10^{-6}} = 1.27\times10^{7}\ \text{N m}^{-2} \](b) Strain \(= \dfrac{\text{extension}}{\text{original length}}\):
\[ \text{Strain} = \frac{0.24\times10^{-3}}{3} = 8.0\times10^{-5} \]Strain has no unit.
Answer Details
Given: \(F = 40\ \text{N}\), extension \(e = 0.24\ \text{mm} = 0.24\times10^{-3}\ \text{m}\), original length \(L = 3\ \text{m}\), diameter \(d = 2.0\ \text{mm} = 2.0\times10^{-3}\ \text{m}\).
(a) Stress \(= \dfrac{F}{A}\), where the cross-sectional area \(A = \dfrac{\pi d^{2}}{4}\).
\[ A = \frac{\pi (2.0\times10^{-3})^{2}}{4} = 3.14\times10^{-6}\ \text{m}^{2} \] \[ \text{Stress} = \frac{40}{3.14\times10^{-6}} = 1.27\times10^{7}\ \text{N m}^{-2} \](b) Strain \(= \dfrac{\text{extension}}{\text{original length}}\):
\[ \text{Strain} = \frac{0.24\times10^{-3}}{3} = 8.0\times10^{-5} \]Strain has no unit.
Question 14 Report
(a) Write down the names of two particles used in explaining the wave nature of matter.
(b) State the wave characteristics which are exhibited by the particles named in (a) above.
(a) Two particles used to explain the wave nature of matter:
- The electron.
- The neutron (the proton is also acceptable).
(b) Wave characteristics exhibited by these particles:
- Diffraction (they bend around obstacles/through gaps, as in electron diffraction).
- Interference (they can superpose to give reinforcement and cancellation).
These effects confirm that moving particles have an associated de Broglie wavelength \(\lambda = \dfrac{h}{mv}\).
Answer Details
(a) Two particles used to explain the wave nature of matter:
- The electron.
- The neutron (the proton is also acceptable).
(b) Wave characteristics exhibited by these particles:
- Diffraction (they bend around obstacles/through gaps, as in electron diffraction).
- Interference (they can superpose to give reinforcement and cancellation).
These effects confirm that moving particles have an associated de Broglie wavelength \(\lambda = \dfrac{h}{mv}\).
Question 15 Report
(s) State the Heisenberg Uncertainty Principle.
(b) The product of the uncertainties in Heisenberg Uncertainty Principle is equal to or greater than a constant. State the mathematical expression for this constant.
