WAEC SSCE - Physics - 2008 (Essay)
Question 1 Report
Explain plane polarized light
Plane polarized light
Ordinary (unpolarized) light consists of transverse waves whose electric-field vibrations occur in all possible directions perpendicular to the direction of travel. Plane polarized light is light in which these vibrations have been restricted so that they take place in only one plane containing the direction of propagation.
In other words, the electric vector of a plane polarized beam oscillates along a single fixed direction (perpendicular to the direction of travel), rather than in every transverse direction. Such light can be produced by passing ordinary light through a polaroid (or by reflection at the polarizing angle), and its state can be detected using a second polaroid (analyser) which cuts off the light when its axis is crossed with the first. The fact that light can be polarized is direct evidence that light is a transverse wave.
Answer Details
Plane polarized light
Ordinary (unpolarized) light consists of transverse waves whose electric-field vibrations occur in all possible directions perpendicular to the direction of travel. Plane polarized light is light in which these vibrations have been restricted so that they take place in only one plane containing the direction of propagation.
In other words, the electric vector of a plane polarized beam oscillates along a single fixed direction (perpendicular to the direction of travel), rather than in every transverse direction. Such light can be produced by passing ordinary light through a polaroid (or by reflection at the polarizing angle), and its state can be detected using a second polaroid (analyser) which cuts off the light when its axis is crossed with the first. The fact that light can be polarized is direct evidence that light is a transverse wave.
Question 2 Report
(a) The equation y = \(\alpha\) sin (wt - kx) represents a plane wave travelling in a medium along the x-direction, being the displacement at the point x at time t.
(i) Given that x is in metres and t is in seconds, state the units of k and w
(ii) What physical quantity does \(\frac{W}{k}\) represent? Justify your answer
(iii) State whether the wave is travelling in the positive or negative x-direction.
(b)(i) What are beats?
(ii) A sonometer wire has a frequency of 259 Hz. It is sounded alongside a tuning fork of frequency 256 Hz. Calculate the beat frequency
(iii) The sonometer wire in (b)(ii) above is under a tension of 1200 N. If a metre of the wire has a mass of 0.03 kg, calculate the length l of the wire when it is vibrating in the fundamental mode.
(c) List two similarities between the human eye and the photographic camera.
(a)(i) Units in \(y = a\sin(\omega t - kx)\): the argument of sine must be a pure number. Since \(t\) is in seconds, \(\omega\) has unit rad s\(^{-1}\) (per second); since \(x\) is in metres, \(k\) has unit rad m\(^{-1}\) (per metre).
(a)(ii) \(\dfrac{\omega}{k}\) represents the speed (velocity) of the wave. Justification: \(\omega = 2\pi f\) and \(k = \dfrac{2\pi}{\lambda}\), so \(\dfrac{\omega}{k} = \dfrac{2\pi f}{2\pi/\lambda} = f\lambda = v\), which is the wave speed.
(a)(iii) The form \((\omega t - kx)\) represents a wave travelling in the positive x-direction.
(b)(i) Beats are the periodic rise and fall in the loudness (amplitude) of sound heard when two notes of slightly different frequencies are sounded together, caused by their alternate reinforcement and cancellation.
(b)(ii) Beat frequency:
\[ f_{\text{beat}} = |259 - 256| = 3\,\text{Hz} \](b)(iii) Length of the sonometer wire (fundamental mode): the wave speed on the wire is
\[ v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{1200}{0.03}} = \sqrt{40000} = 200\,\text{ms}^{-1} \]For the fundamental, \(f = \dfrac{v}{2l}\), so:
\[ l = \frac{v}{2f} = \frac{200}{2\times 259} = 0.386\,\text{m}\ (\approx 38.6\,\text{cm}) \](c) Two similarities between the human eye and the photographic camera:
- Both use a converging (convex) lens to form a real, inverted, diminished image of the object.
- Both form the image on a light-sensitive surface (the retina in the eye, the film/sensor in the camera) and both control the amount of light entering through an adjustable aperture (the iris/pupil in the eye, the diaphragm in the camera).
Answer Details
(a)(i) Units in \(y = a\sin(\omega t - kx)\): the argument of sine must be a pure number. Since \(t\) is in seconds, \(\omega\) has unit rad s\(^{-1}\) (per second); since \(x\) is in metres, \(k\) has unit rad m\(^{-1}\) (per metre).
(a)(ii) \(\dfrac{\omega}{k}\) represents the speed (velocity) of the wave. Justification: \(\omega = 2\pi f\) and \(k = \dfrac{2\pi}{\lambda}\), so \(\dfrac{\omega}{k} = \dfrac{2\pi f}{2\pi/\lambda} = f\lambda = v\), which is the wave speed.
(a)(iii) The form \((\omega t - kx)\) represents a wave travelling in the positive x-direction.
(b)(i) Beats are the periodic rise and fall in the loudness (amplitude) of sound heard when two notes of slightly different frequencies are sounded together, caused by their alternate reinforcement and cancellation.
(b)(ii) Beat frequency:
\[ f_{\text{beat}} = |259 - 256| = 3\,\text{Hz} \](b)(iii) Length of the sonometer wire (fundamental mode): the wave speed on the wire is
\[ v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{1200}{0.03}} = \sqrt{40000} = 200\,\text{ms}^{-1} \]For the fundamental, \(f = \dfrac{v}{2l}\), so:
\[ l = \frac{v}{2f} = \frac{200}{2\times 259} = 0.386\,\text{m}\ (\approx 38.6\,\text{cm}) \](c) Two similarities between the human eye and the photographic camera:
- Both use a converging (convex) lens to form a real, inverted, diminished image of the object.
- Both form the image on a light-sensitive surface (the retina in the eye, the film/sensor in the camera) and both control the amount of light entering through an adjustable aperture (the iris/pupil in the eye, the diaphragm in the camera).
Question 3 Report
(a) Define electrolysis.
(b) Classify each of the following substances as an electrolyte or a non-electrolyte
(i) sugar solution;
(ii) kerosene;
(iii) alkaline solution
(iv) lemon fruit juice.
(a) Electrolysis is the chemical decomposition of a substance (an electrolyte) brought about by passing an electric current through it when in the molten state or in aqueous solution.
(b) Classification
| Substance | Classification |
|---|---|
| (i) Sugar solution | Non-electrolyte |
| (ii) Kerosene | Non-electrolyte |
| (iii) Alkaline solution | Electrolyte |
| (iv) Lemon fruit juice | Electrolyte |
Electrolytes (alkaline solution and lemon juice, which is acidic) conduct electricity because they contain free mobile ions, whereas sugar solution and kerosene contain no free ions and so do not conduct.
Answer Details
(a) Electrolysis is the chemical decomposition of a substance (an electrolyte) brought about by passing an electric current through it when in the molten state or in aqueous solution.
(b) Classification
| Substance | Classification |
|---|---|
| (i) Sugar solution | Non-electrolyte |
| (ii) Kerosene | Non-electrolyte |
| (iii) Alkaline solution | Electrolyte |
| (iv) Lemon fruit juice | Electrolyte |
Electrolytes (alkaline solution and lemon juice, which is acidic) conduct electricity because they contain free mobile ions, whereas sugar solution and kerosene contain no free ions and so do not conduct.
Question 4 Report
(a) State Heisenberg's Uncertainty Principle.
(b) State one phenomenon that can only be explained in terms of the wave nature of light.
(a) Heisenberg's Uncertainty Principle
It is impossible to determine simultaneously, with perfect accuracy, both the position and momentum of a particle. Mathematically,
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
(b) Interference of light is a phenomenon that can only be explained in terms of the wave nature of light. Diffraction is also acceptable.
Answer Details
(a) Heisenberg's Uncertainty Principle
It is impossible to determine simultaneously, with perfect accuracy, both the position and momentum of a particle. Mathematically,
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
(b) Interference of light is a phenomenon that can only be explained in terms of the wave nature of light. Diffraction is also acceptable.
Question 5 Report
In an electrolysis experiment, the ammeter records a steady current of 1 A. The mass of copper deposited in 30 minutes is 0.66 g. Calculate the error in the ammeter reading. [ Electrochemical equivalent of copper = 0.00033 g C\(^{-1}\)]
Error in the ammeter reading (electrolysis of copper)
By Faraday's first law, the true charge that passed is found from the mass deposited and the electrochemical equivalent \(z\):
\[ m = zQ \;\Rightarrow\; Q = \frac{m}{z} = \frac{0.66}{0.00033} = 2000\,\text{C} \]The time of the experiment is \(t = 30\,\text{min} = 1800\,\text{s}\), so the true (actual) current is:
\[ I_{\text{true}} = \frac{Q}{t} = \frac{2000}{1800} = 1.11\,\text{A} \]The ammeter reads \(1.0\,\text{A}\), so the error is:
\[ \text{error} = I_{\text{true}} - I_{\text{read}} = 1.11 - 1.00 = 0.11\,\text{A} \]The ammeter reads about \(0.11\,\text{A}\) too low (a percentage error of roughly \(10\%\)).
Answer Details
Error in the ammeter reading (electrolysis of copper)
By Faraday's first law, the true charge that passed is found from the mass deposited and the electrochemical equivalent \(z\):
\[ m = zQ \;\Rightarrow\; Q = \frac{m}{z} = \frac{0.66}{0.00033} = 2000\,\text{C} \]The time of the experiment is \(t = 30\,\text{min} = 1800\,\text{s}\), so the true (actual) current is:
\[ I_{\text{true}} = \frac{Q}{t} = \frac{2000}{1800} = 1.11\,\text{A} \]The ammeter reads \(1.0\,\text{A}\), so the error is:
\[ \text{error} = I_{\text{true}} - I_{\text{read}} = 1.11 - 1.00 = 0.11\,\text{A} \]The ammeter reads about \(0.11\,\text{A}\) too low (a percentage error of roughly \(10\%\)).
Question 6 Report
The horizontal component of the initial speed of a particle projected at 30° to the horizontal is 50 m\(^{-1}\). If the acceleration of free fall due to gravity is 10 ms\(^{-2}\), determine its;
(a) initial speed;
(b) speed at the maximum height reached.
Projectile launched at 30 degrees with horizontal component 50 ms\(^{-1}\)
(a) Initial speed
The horizontal component is \(v_x = v\cos\theta\), so:
\[ v = \frac{v_x}{\cos\theta} = \frac{50}{\cos 30^\circ} = \frac{50}{0.8660} = 57.7\,\text{ms}^{-1} \](b) Speed at maximum height
At the highest point of the flight the vertical component of velocity is zero; only the horizontal component remains (it is unchanged throughout the motion). Therefore:
\[ v_{\text{top}} = v_x = 50\,\text{ms}^{-1} \]The speed at maximum height is \(50\,\text{ms}^{-1}\).
Answer Details
Projectile launched at 30 degrees with horizontal component 50 ms\(^{-1}\)
(a) Initial speed
The horizontal component is \(v_x = v\cos\theta\), so:
\[ v = \frac{v_x}{\cos\theta} = \frac{50}{\cos 30^\circ} = \frac{50}{0.8660} = 57.7\,\text{ms}^{-1} \](b) Speed at maximum height
At the highest point of the flight the vertical component of velocity is zero; only the horizontal component remains (it is unchanged throughout the motion). Therefore:
\[ v_{\text{top}} = v_x = 50\,\text{ms}^{-1} \]The speed at maximum height is \(50\,\text{ms}^{-1}\).
Question 7 Report
(a)(i) State Einstein's equation of photoelectric effect
(ii) What conservation principle does the equation represent?
(b) List three applications of photocells
(c) A photo emissive surface has a threshold frequency of 4.02 x 10\(^{14}\)Hz. If the surface is illuminated by light of frequency 5.0 x 10\(^{15}\)Hz, calculate the:
(i) threshold wavelength;
(ii) work function;
(iii) kinetic energy of the emitted photoelectrons. [ c = 3.0 x 10\(^{8}\) ms\(^{-1}\), h = 6.63 x 10\(^{-34}\) Js]
Question 8 Report
An electron of charge 1.60 x 10\(^{-19}\)C is accelerated under a potential difference of 1.0 x 10\(^{5}\) V. Calculate the energy of the electron in joules.
Energy gained by an accelerated electron
When a charge \(q\) is accelerated through a potential difference \(V\), the energy it gains is the work done on it:
\[ E = qV \]Substituting \(q = 1.60\times10^{-19}\,\text{C}\) and \(V = 1.0\times10^{5}\,\text{V}\):
\[ E = (1.60\times10^{-19})(1.0\times10^{5}) = 1.6\times10^{-14}\,\text{J} \]The energy of the electron is \(1.6\times10^{-14}\,\text{J}\).
Answer Details
Energy gained by an accelerated electron
When a charge \(q\) is accelerated through a potential difference \(V\), the energy it gains is the work done on it:
\[ E = qV \]Substituting \(q = 1.60\times10^{-19}\,\text{C}\) and \(V = 1.0\times10^{5}\,\text{V}\):
\[ E = (1.60\times10^{-19})(1.0\times10^{5}) = 1.6\times10^{-14}\,\text{J} \]The energy of the electron is \(1.6\times10^{-14}\,\text{J}\).
Question 9 Report
(a) What is a magnetic field?
(b) With the aid of a labelled diagram describe an experiment to show that a magnetic field exists around a straight wire carrying current.
(c) A 40 \(\mu\)F capacitor in series with a 40 Q resistor is connected to a 100 V, 50 Hz a.c. supply.
(i) Draw a circuit diagram of the arrangement
(ii) Calculate the: I. impedance in the circuit; II. current in the circuit III. potential difference across the capacitor.
(a) A magnetic field is the region of space around a magnet or a current-carrying conductor within which a magnetic force is experienced by another magnet, a magnetic material, or a moving charge.
(b) Experiment to show that a magnetic field exists around a straight current-carrying wire (Oersted's experiment)
A stiff, straight copper wire is passed vertically through a hole in the centre of a horizontal cardboard sheet. The wire is connected in series with a battery, a switch (key) and a rheostat so that a large current can flow through it. Iron filings are sprinkled evenly over the card, and a small plotting compass is placed on the card near the wire.
When the switch is closed and a strong current flows, the card is gently tapped. The iron filings arrange themselves into a pattern of concentric circles centred on the wire, and the plotting compass needle sets itself tangential to these circles. This shows that a magnetic field exists in the space around the wire. When the direction of the current is reversed, the compass needle turns to point the opposite way, showing that the direction of the field also reverses.
(c) Series R-C circuit on a.c. supply: \(C = 40\,\mu\text{F} = 40\times10^{-6}\,\text{F}\), \(R = 40\,\Omega\), \(V = 100\,\text{V}\), \(f = 50\,\text{Hz}\).
(i) Circuit diagram — the 40 Ω resistor and the 40 µF capacitor connected in series across the 100 V, 50 Hz a.c. supply:
(ii) I. Impedance
Capacitive reactance:
\[ X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi(50)(40\times10^{-6})} = 79.6\,\Omega \]Impedance:
\[ Z = \sqrt{R^2 + X_C^2} = \sqrt{40^2 + 79.6^2} = \sqrt{1600 + 6337} = \sqrt{7937} = 89.1\,\Omega \]II. Current in the circuit
\[ I = \frac{V}{Z} = \frac{100}{89.1} = 1.12\,\text{A} \]III. Potential difference across the capacitor
\[ V_C = I\,X_C = 1.12 \times 79.6 = 89.2\,\text{V} \]Answer Details
(a) A magnetic field is the region of space around a magnet or a current-carrying conductor within which a magnetic force is experienced by another magnet, a magnetic material, or a moving charge.
(b) Experiment to show that a magnetic field exists around a straight current-carrying wire (Oersted's experiment)
A stiff, straight copper wire is passed vertically through a hole in the centre of a horizontal cardboard sheet. The wire is connected in series with a battery, a switch (key) and a rheostat so that a large current can flow through it. Iron filings are sprinkled evenly over the card, and a small plotting compass is placed on the card near the wire.
When the switch is closed and a strong current flows, the card is gently tapped. The iron filings arrange themselves into a pattern of concentric circles centred on the wire, and the plotting compass needle sets itself tangential to these circles. This shows that a magnetic field exists in the space around the wire. When the direction of the current is reversed, the compass needle turns to point the opposite way, showing that the direction of the field also reverses.
(c) Series R-C circuit on a.c. supply: \(C = 40\,\mu\text{F} = 40\times10^{-6}\,\text{F}\), \(R = 40\,\Omega\), \(V = 100\,\text{V}\), \(f = 50\,\text{Hz}\).
(i) Circuit diagram — the 40 Ω resistor and the 40 µF capacitor connected in series across the 100 V, 50 Hz a.c. supply:
(ii) I. Impedance
Capacitive reactance:
\[ X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi(50)(40\times10^{-6})} = 79.6\,\Omega \]Impedance:
\[ Z = \sqrt{R^2 + X_C^2} = \sqrt{40^2 + 79.6^2} = \sqrt{1600 + 6337} = \sqrt{7937} = 89.1\,\Omega \]II. Current in the circuit
\[ I = \frac{V}{Z} = \frac{100}{89.1} = 1.12\,\text{A} \]III. Potential difference across the capacitor
\[ V_C = I\,X_C = 1.12 \times 79.6 = 89.2\,\text{V} \]Question 10 Report
Use the diagram to answer question;
Camphor is a chemical that sub-limes and interacts with water reducing the surface tension where it is put. Explain why the toy boat illustrated above would move forward with the camphor placed at the back of it but would be stationary before the placement of the camphor.
What the diagram shows. The pointed toy boat floats on water. Camphor is fixed at the flat rear end. The label \(F_1\) is a leftward (backward) arrow acting at the rear where the camphor is, and \(F_2\) is a rightward (forward) arrow acting at the pointed front. \(F_1\) and \(F_2\) represent the surface-tension pulls of the water on the two ends of the boat.
Key idea: forces due to surface tension. The free surface of water behaves like a stretched elastic skin. Wherever this skin touches the edge of the floating boat, it pulls the boat outward (away from the boat) along the surface. So the water pulls the rear of the boat backward with force \(F_1\) and pulls the front of the boat forward with force \(F_2\). The size of each pull is proportional to the surface tension of the water touching that edge.
Before the camphor is placed (boat stationary). The surface tension of the water is the same all around the boat, so the backward pull equals the forward pull:
\[ F_1 = F_2 \]The two pulls cancel, the resultant horizontal force is zero, and the boat stays at rest.
After the camphor is placed at the back. The camphor sublimes and dissolves into the water immediately behind the boat, and it reduces the surface tension of the water there. The water in front of the boat still has its full (higher) surface tension. Therefore the backward pull becomes small while the forward pull stays large:
\[ F_2 > F_1 \]There is now an unbalanced resultant force acting forward:
\[ F_{net} = F_2 - F_1 \quad (\text{directed forward}) \]By Newton's second law this net force accelerates the boat, so it moves forward (in the direction of \(F_2\)). Equivalently, the higher-tension water in front pulls harder than the weakened-tension water behind, and the boat is dragged toward the region of greater surface tension. As soon as the camphor is used up and the surface tension becomes uniform again, \(F_1\) once more equals \(F_2\) and the boat stops.
Answer Details
What the diagram shows. The pointed toy boat floats on water. Camphor is fixed at the flat rear end. The label \(F_1\) is a leftward (backward) arrow acting at the rear where the camphor is, and \(F_2\) is a rightward (forward) arrow acting at the pointed front. \(F_1\) and \(F_2\) represent the surface-tension pulls of the water on the two ends of the boat.
Key idea: forces due to surface tension. The free surface of water behaves like a stretched elastic skin. Wherever this skin touches the edge of the floating boat, it pulls the boat outward (away from the boat) along the surface. So the water pulls the rear of the boat backward with force \(F_1\) and pulls the front of the boat forward with force \(F_2\). The size of each pull is proportional to the surface tension of the water touching that edge.
Before the camphor is placed (boat stationary). The surface tension of the water is the same all around the boat, so the backward pull equals the forward pull:
\[ F_1 = F_2 \]The two pulls cancel, the resultant horizontal force is zero, and the boat stays at rest.
After the camphor is placed at the back. The camphor sublimes and dissolves into the water immediately behind the boat, and it reduces the surface tension of the water there. The water in front of the boat still has its full (higher) surface tension. Therefore the backward pull becomes small while the forward pull stays large:
\[ F_2 > F_1 \]There is now an unbalanced resultant force acting forward:
\[ F_{net} = F_2 - F_1 \quad (\text{directed forward}) \]By Newton's second law this net force accelerates the boat, so it moves forward (in the direction of \(F_2\)). Equivalently, the higher-tension water in front pulls harder than the weakened-tension water behind, and the boat is dragged toward the region of greater surface tension. As soon as the camphor is used up and the surface tension becomes uniform again, \(F_1\) once more equals \(F_2\) and the boat stops.
Question 11 Report
(a) A particle moves on a straight path with an initial speed u and final speed v in time t. Show that the total distance X covered by the particle is given by
\(x = ut + \frac{1}{2}at^2\)
where a is the magnitude of acceleration
(b) State;
(i) Newton's second law of motion
(ii) the principle of conservation of energy
(iii) the law of floatation.
(c) Consider a balloon of mass 0.030 kg being inflated with a gas of density 0.54 kg m\(^{-3}\). What will be the volume of the balloon when it just begins to rise in air of density 1.29 kg m\(^{-3}\)? [ g = 10 ms\(^{-2}\)]
(a) Derivation of \(x = ut + \tfrac{1}{2}at^2\)
For uniform acceleration \(a\), initial speed \(u\), final speed \(v\) after time \(t\), the distance is the area under the velocity-time graph (a trapezium):
\[ x = \left(\frac{u+v}{2}\right)t \]But \(v = u + at\). Substituting:
\[ x = \left(\frac{u + (u+at)}{2}\right)t = \left(\frac{2u + at}{2}\right)t = ut + \tfrac{1}{2}at^2 \quad\text{(shown)} \](b)(i) Newton's second law: The rate of change of momentum of a body is directly proportional to the applied (resultant) force and takes place in the direction of the force; hence \(F = ma\).
(b)(ii) Principle of conservation of energy: Energy can neither be created nor destroyed but only changed from one form to another; the total energy of an isolated system remains constant.
(b)(iii) Law of floatation: A floating body displaces its own weight of the fluid in which it floats.
(c) Volume of the balloon when it just begins to rise
The balloon just begins to rise when the upthrust from the air equals the total weight (of the balloon fabric plus the gas):
\[ \rho_{\text{air}}\,V g = (m + \rho_{\text{gas}}\,V)g \] \[ 1.29\,V = 0.030 + 0.54\,V \] \[ (1.29 - 0.54)V = 0.030 \;\Rightarrow\; 0.75\,V = 0.030 \] \[ V = \frac{0.030}{0.75} = 0.04\,\text{m}^3 \]The volume of the balloon is \(0.04\,\text{m}^3\).
Answer Details
(a) Derivation of \(x = ut + \tfrac{1}{2}at^2\)
For uniform acceleration \(a\), initial speed \(u\), final speed \(v\) after time \(t\), the distance is the area under the velocity-time graph (a trapezium):
\[ x = \left(\frac{u+v}{2}\right)t \]But \(v = u + at\). Substituting:
\[ x = \left(\frac{u + (u+at)}{2}\right)t = \left(\frac{2u + at}{2}\right)t = ut + \tfrac{1}{2}at^2 \quad\text{(shown)} \](b)(i) Newton's second law: The rate of change of momentum of a body is directly proportional to the applied (resultant) force and takes place in the direction of the force; hence \(F = ma\).
(b)(ii) Principle of conservation of energy: Energy can neither be created nor destroyed but only changed from one form to another; the total energy of an isolated system remains constant.
(b)(iii) Law of floatation: A floating body displaces its own weight of the fluid in which it floats.
(c) Volume of the balloon when it just begins to rise
The balloon just begins to rise when the upthrust from the air equals the total weight (of the balloon fabric plus the gas):
\[ \rho_{\text{air}}\,V g = (m + \rho_{\text{gas}}\,V)g \] \[ 1.29\,V = 0.030 + 0.54\,V \] \[ (1.29 - 0.54)V = 0.030 \;\Rightarrow\; 0.75\,V = 0.030 \] \[ V = \frac{0.030}{0.75} = 0.04\,\text{m}^3 \]The volume of the balloon is \(0.04\,\text{m}^3\).
Question 12 Report
(a) State two differences between boiling and evaporation.
(b) A closed inexpansibie vessel contains air saturated with water vapour at 77°C. The total pressure in the vessel is 1007 mmHg. Calculate the new pressure in the vessel if the temperature is reduced 1.2 27°C. [ The s.v.p. of water at 77°C and 27°C respectively are 314 mmHg and 27 mmHg. Treat the air in the vessel as an ideal gas]
(c) The lengtn of a zinc rod at 23°C is 200 m. Calculate the increase in length of the rod when its temperature rises to 33°C. [ expansivity of zinc = 2.6 x 10\(^{-5}\) K\(^{-1}\)]
(d) Explain with there is no temperature change when a solid being heated changes into liquid at its melting point.
(a) Two differences between boiling and evaporation
| Boiling | Evaporation |
|---|---|
| Occurs at a fixed temperature (the boiling point) | Occurs at all temperatures |
| Takes place throughout the whole liquid (bubbles form within it) | Takes place only at the exposed surface of the liquid |
(b) New pressure in the vessel after cooling
At \(77^\circ\text{C}\ (350\,\text{K})\), total pressure = 1007 mmHg and s.v.p. of water = 314 mmHg, so the partial pressure of the dry air is:
\[ P_{\text{air}} = 1007 - 314 = 693\,\text{mmHg} \]The air (ideal, constant volume) is cooled to \(27^\circ\text{C}\ (300\,\text{K})\):
\[ P'_{\text{air}} = 693\times\frac{300}{350} = 594\,\text{mmHg} \]Adding the s.v.p. of water at \(27^\circ\text{C}\) (27 mmHg):
\[ P_{\text{total}} = 594 + 27 = 621\,\text{mmHg} \](c) Increase in length of the zinc rod
\[ \Delta L = L\,\alpha\,\Delta\theta = 200 \times (2.6\times10^{-5}) \times (33-23) \] \[ \Delta L = 200 \times 2.6\times10^{-5} \times 10 = 0.052\,\text{m} \ (5.2\,\text{cm}) \](d) When a solid melts at its melting point, the heat supplied (the latent heat of fusion) is used entirely to break the bonds/forces holding the particles in the fixed solid lattice, not to increase their kinetic energy. Since temperature depends on the average kinetic energy of the particles, and this does not change while the solid is turning to liquid, the temperature stays constant until melting is complete.
Answer Details
(a) Two differences between boiling and evaporation
| Boiling | Evaporation |
|---|---|
| Occurs at a fixed temperature (the boiling point) | Occurs at all temperatures |
| Takes place throughout the whole liquid (bubbles form within it) | Takes place only at the exposed surface of the liquid |
(b) New pressure in the vessel after cooling
At \(77^\circ\text{C}\ (350\,\text{K})\), total pressure = 1007 mmHg and s.v.p. of water = 314 mmHg, so the partial pressure of the dry air is:
\[ P_{\text{air}} = 1007 - 314 = 693\,\text{mmHg} \]The air (ideal, constant volume) is cooled to \(27^\circ\text{C}\ (300\,\text{K})\):
\[ P'_{\text{air}} = 693\times\frac{300}{350} = 594\,\text{mmHg} \]Adding the s.v.p. of water at \(27^\circ\text{C}\) (27 mmHg):
\[ P_{\text{total}} = 594 + 27 = 621\,\text{mmHg} \](c) Increase in length of the zinc rod
\[ \Delta L = L\,\alpha\,\Delta\theta = 200 \times (2.6\times10^{-5}) \times (33-23) \] \[ \Delta L = 200 \times 2.6\times10^{-5} \times 10 = 0.052\,\text{m} \ (5.2\,\text{cm}) \](d) When a solid melts at its melting point, the heat supplied (the latent heat of fusion) is used entirely to break the bonds/forces holding the particles in the fixed solid lattice, not to increase their kinetic energy. Since temperature depends on the average kinetic energy of the particles, and this does not change while the solid is turning to liquid, the temperature stays constant until melting is complete.
Question 13 Report
a) On what principle does lighting in a fluorescent tube operate?
(b) State two factors which determine the colour of light produced in a fluorescent tube.
(a) Principle of the fluorescent tube
A fluorescent tube works on the principle of gas discharge and fluorescence. When a high voltage is applied, electrons flow through the low-pressure mercury vapour in the tube, exciting the mercury atoms; these emit ultraviolet radiation. The ultraviolet light strikes the fluorescent (phosphor) coating on the inside of the tube, which absorbs it and re-emits the energy as visible light (fluorescence).
(b) Two factors determining the colour of the light produced
- The nature (chemical composition) of the phosphor / fluorescent powder coating the inside of the tube.
- The type of gas or vapour (and its pressure) contained in the tube.
Answer Details
(a) Principle of the fluorescent tube
A fluorescent tube works on the principle of gas discharge and fluorescence. When a high voltage is applied, electrons flow through the low-pressure mercury vapour in the tube, exciting the mercury atoms; these emit ultraviolet radiation. The ultraviolet light strikes the fluorescent (phosphor) coating on the inside of the tube, which absorbs it and re-emits the energy as visible light (fluorescence).
(b) Two factors determining the colour of the light produced
- The nature (chemical composition) of the phosphor / fluorescent powder coating the inside of the tube.
- The type of gas or vapour (and its pressure) contained in the tube.
Question 14 Report
A ball bearing falls through a viscous liquid
(a) Using a labelled diagram of a tall vessel, show all the forces acting on it
(b) When will it attain terminal velocity?
Ball bearing falling through a viscous liquid
(a) Labelled diagram showing the forces acting on the ball
Inside a tall vessel of viscous liquid, a spherical ball bearing falling vertically downwards experiences three forces:
- Weight, \(W = mg\), acting vertically downwards through the centre of the ball.
- Upthrust (buoyancy), \(U\), acting vertically upwards, equal to the weight of liquid displaced.
- Viscous drag, \(F\), a fluid-friction force given by Stokes' law \(F = 6\pi\eta r v\), acting vertically upwards, opposing the motion.
(b) When terminal velocity is attained
As the ball accelerates downwards its speed \(v\) increases, and because the viscous drag \(F = 6\pi\eta r v\) grows with speed, the total upward force \((U + F)\) increases too. The ball attains its terminal velocity at the instant the upward forces exactly balance the downward weight, so the resultant force and hence the acceleration become zero:
\[ W = U + F \quad\Rightarrow\quad mg = U + 6\pi\eta r v \]From that moment onwards there is no further acceleration, and the ball continues to fall through the liquid with a constant (maximum, uniform) velocity, which is the terminal velocity.
Answer Details
Ball bearing falling through a viscous liquid
(a) Labelled diagram showing the forces acting on the ball
Inside a tall vessel of viscous liquid, a spherical ball bearing falling vertically downwards experiences three forces:
- Weight, \(W = mg\), acting vertically downwards through the centre of the ball.
- Upthrust (buoyancy), \(U\), acting vertically upwards, equal to the weight of liquid displaced.
- Viscous drag, \(F\), a fluid-friction force given by Stokes' law \(F = 6\pi\eta r v\), acting vertically upwards, opposing the motion.
(b) When terminal velocity is attained
As the ball accelerates downwards its speed \(v\) increases, and because the viscous drag \(F = 6\pi\eta r v\) grows with speed, the total upward force \((U + F)\) increases too. The ball attains its terminal velocity at the instant the upward forces exactly balance the downward weight, so the resultant force and hence the acceleration become zero:
\[ W = U + F \quad\Rightarrow\quad mg = U + 6\pi\eta r v \]From that moment onwards there is no further acceleration, and the ball continues to fall through the liquid with a constant (maximum, uniform) velocity, which is the terminal velocity.
Question 15 Report
A wire gradually stretched by loading it until it snaps.
(a) Sketch a load-extension graph for the wire
(b) Indicate on the graph the
(i) elastic limit (E);
(ii) yield point (Y);
(iii) breaking point (B).
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Load-extension graph for a wire stretched to breaking
Plot load (force) on the vertical axis against extension on the horizontal axis. The wire extends elastically at first. Beyond the elastic limit, permanent extension begins. At the yield point, the wire extends greatly with little increase in load. Eventually, the wire breaks at point B.
E is the elastic limit: up to this point, the wire returns to its original length when the load is removed.
Y is the yield point: the wire begins to undergo noticeable permanent deformation.
B is the breaking point: the wire snaps.
```Answer Details
```html
Load-extension graph for a wire stretched to breaking
Plot load (force) on the vertical axis against extension on the horizontal axis. The wire extends elastically at first. Beyond the elastic limit, permanent extension begins. At the yield point, the wire extends greatly with little increase in load. Eventually, the wire breaks at point B.
E is the elastic limit: up to this point, the wire returns to its original length when the load is removed.
Y is the yield point: the wire begins to undergo noticeable permanent deformation.
B is the breaking point: the wire snaps.
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