WAEC SSCE - Physics - 2007 (Essay)
Question 1 Report
Define
(a) electrolysis;
(b) electrolyte;
(c) electrode.
(a) Electrolysis: the chemical decomposition of an electrolyte (in molten or aqueous solution) brought about by passing an electric current through it.
(b) Electrolyte: a compound which, when molten or dissolved in water, conducts electricity and is chemically decomposed by it (because it contains free-moving ions).
(c) Electrode: a conductor (rod or plate) through which the electric current enters or leaves the electrolyte. The one connected to the positive terminal is the anode; the one connected to the negative terminal is the cathode.
Answer Details
(a) Electrolysis: the chemical decomposition of an electrolyte (in molten or aqueous solution) brought about by passing an electric current through it.
(b) Electrolyte: a compound which, when molten or dissolved in water, conducts electricity and is chemically decomposed by it (because it contains free-moving ions).
(c) Electrode: a conductor (rod or plate) through which the electric current enters or leaves the electrolyte. The one connected to the positive terminal is the anode; the one connected to the negative terminal is the cathode.
Question 2 Report
(a) Define the boiling point of a liquid.
(b) With the aid of a sketch diagram, describe an experiment to determine the boiling point of a small quantity of a Iiquid
(c) A piece of copper of mass 300 g at a temperature of 950°C is quickly transferred into a vessel of negligible thermal capacity containing 250 g of water at 25°C. If the final steady temperature of the mixture is 100°C, calculate the mass of water that will boil away. Especific heat capacity of copper = 4.0 x 10\(^2\) J kg\(^{-1}\)K\(^{-1}\) specific heat capacity of water = 4.2 x 10\(^{3}\) J kg\(^{-1}\)K\(^{-1}\) specific latent heat of vaporization of steam = 2.26 x 10\(^6\)J kg\(^{-1}\)
(a) Boiling point of a liquid
The boiling point of a liquid is the constant temperature at which the liquid changes into vapour throughout its bulk, that is, the temperature at which the saturated vapour pressure of the liquid becomes equal to the external (atmospheric) pressure.
(b) Experiment to determine the boiling point of a small quantity of a liquid
Place the small quantity of the liquid in a clean ignition (test) tube and lower it into a beaker of water so that the liquid is well below the water surface. Support a thermometer with its bulb immersed in the liquid, and place a stirrer in the surrounding water. Stand the beaker on a wire gauze and tripod above a Bunsen burner.
Heat the water bath gently and stir the water continuously so that the heat reaches the small tube evenly. Watch the thermometer: its reading rises steadily and then becomes constant while the liquid is boiling vigorously. Record this steady temperature. The constant temperature at which the thermometer reading remains fixed while the liquid boils is the boiling point of the liquid.
(c) Mass of water that boils away
Data: mass of copper \(m_{Cu}=300\,\text{g}=0.300\,\text{kg}\), specific heat capacity of copper \(c_{Cu}=4.0\times10^{2}\,\text{J kg}^{-1}\text{K}^{-1}\), mass of water \(m_w=250\,\text{g}=0.250\,\text{kg}\), specific heat capacity of water \(c_w=4.2\times10^{3}\,\text{J kg}^{-1}\text{K}^{-1}\), specific latent heat of vaporization \(L=2.26\times10^{6}\,\text{J kg}^{-1}\).
Heat lost by the copper as it cools from \(950^{\circ}\text{C}\) to the final temperature \(100^{\circ}\text{C}\):
\[ Q_{Cu}=m_{Cu}\,c_{Cu}\,\Delta\theta = 0.300\times(4.0\times10^{2})\times(950-100)=0.300\times400\times850=102000\,\text{J} \]Heat needed to raise the \(250\,\text{g}\) of water from \(25^{\circ}\text{C}\) to \(100^{\circ}\text{C}\):
\[ Q_{1}=m_{w}\,c_{w}\,\Delta\theta = 0.250\times(4.2\times10^{3})\times(100-25)=0.250\times4200\times75=78750\,\text{J} \]By conservation of energy, the heat remaining after warming the water is used to boil some of it into steam:
\[ Q_{2}=Q_{Cu}-Q_{1}=102000-78750=23250\,\text{J} \]If \(m_v\) is the mass of water boiled away, then \(Q_{2}=m_v L\):
\[ m_v=\frac{Q_{2}}{L}=\frac{23250}{2.26\times10^{6}}=1.03\times10^{-2}\,\text{kg}\approx 10.3\,\text{g} \]The mass of water that boils away is about \(10.3\,\text{g}\) (\(1.03\times10^{-2}\,\text{kg}\)).
Answer Details
(a) Boiling point of a liquid
The boiling point of a liquid is the constant temperature at which the liquid changes into vapour throughout its bulk, that is, the temperature at which the saturated vapour pressure of the liquid becomes equal to the external (atmospheric) pressure.
(b) Experiment to determine the boiling point of a small quantity of a liquid
Place the small quantity of the liquid in a clean ignition (test) tube and lower it into a beaker of water so that the liquid is well below the water surface. Support a thermometer with its bulb immersed in the liquid, and place a stirrer in the surrounding water. Stand the beaker on a wire gauze and tripod above a Bunsen burner.
Heat the water bath gently and stir the water continuously so that the heat reaches the small tube evenly. Watch the thermometer: its reading rises steadily and then becomes constant while the liquid is boiling vigorously. Record this steady temperature. The constant temperature at which the thermometer reading remains fixed while the liquid boils is the boiling point of the liquid.
(c) Mass of water that boils away
Data: mass of copper \(m_{Cu}=300\,\text{g}=0.300\,\text{kg}\), specific heat capacity of copper \(c_{Cu}=4.0\times10^{2}\,\text{J kg}^{-1}\text{K}^{-1}\), mass of water \(m_w=250\,\text{g}=0.250\,\text{kg}\), specific heat capacity of water \(c_w=4.2\times10^{3}\,\text{J kg}^{-1}\text{K}^{-1}\), specific latent heat of vaporization \(L=2.26\times10^{6}\,\text{J kg}^{-1}\).
Heat lost by the copper as it cools from \(950^{\circ}\text{C}\) to the final temperature \(100^{\circ}\text{C}\):
\[ Q_{Cu}=m_{Cu}\,c_{Cu}\,\Delta\theta = 0.300\times(4.0\times10^{2})\times(950-100)=0.300\times400\times850=102000\,\text{J} \]Heat needed to raise the \(250\,\text{g}\) of water from \(25^{\circ}\text{C}\) to \(100^{\circ}\text{C}\):
\[ Q_{1}=m_{w}\,c_{w}\,\Delta\theta = 0.250\times(4.2\times10^{3})\times(100-25)=0.250\times4200\times75=78750\,\text{J} \]By conservation of energy, the heat remaining after warming the water is used to boil some of it into steam:
\[ Q_{2}=Q_{Cu}-Q_{1}=102000-78750=23250\,\text{J} \]If \(m_v\) is the mass of water boiled away, then \(Q_{2}=m_v L\):
\[ m_v=\frac{Q_{2}}{L}=\frac{23250}{2.26\times10^{6}}=1.03\times10^{-2}\,\text{kg}\approx 10.3\,\text{g} \]The mass of water that boils away is about \(10.3\,\text{g}\) (\(1.03\times10^{-2}\,\text{kg}\)).
Question 3 Report
(a)(i) What is a wave motion?
(ii) State two differences between a radio wave and a sound wave.
(b)(i) Given that you are provided with a tuning fork, a burette and other necessary apparatus, describe with the aid of a diagram, an experiment to determine the frequency of a note emitted by a source of sound. [assume the velocity of sound in air is known]
(ii)State two precautions necessary to obtain accurate result in the experiment described in (b)(i) above
(c) A pipe closed at one end is 100 cm long. If the air in the pipe is set into vibration and a fundamental note is produced, calculate the frequency of the note. [ velocity of sound in air = 340 ms\(^{-1}\)]
(a)(i) Wave motion. A wave motion is a disturbance that travels through a medium (or through space) transferring energy from one point to another without any net transfer of the particles of the medium.
(a)(ii) Two differences between a radio wave and a sound wave.
| Radio wave | Sound wave |
|---|---|
| It is an electromagnetic (transverse) wave. | It is a mechanical (longitudinal) wave. |
| Can travel through a vacuum (needs no material medium). | Requires a material medium for propagation. |
| Travels at the speed of light, \(3\times10^{8}\,\text{ms}^{-1}\). | Travels much more slowly, about \(340\,\text{ms}^{-1}\) in air. |
(b)(i) Experiment to determine the frequency of a note (resonance method using a burette).
The burette is clamped vertically and filled with water, so that the water and the tube walls enclose a short column of air at the open top. The vibrating tuning fork is held horizontally, with its prongs just above (not touching) the open mouth of the burette, as shown below.
The tap of the burette is opened so that water runs out slowly, gradually lengthening the air column. As the length increases, a position is reached where the sound suddenly becomes loudest; this is the first resonance. The length of the air column from the water surface to the open mouth is measured and recorded as \(L_1\).
Water is run out further, keeping the fork sounding over the mouth, until the sound is again loudest; this is the second resonance. This longer air column is measured and recorded as \(L_2\). The whole procedure is repeated a few times and the mean values of \(L_1\) and \(L_2\) taken.
Between the first and second resonances the air column has increased by exactly half a wavelength, so end-correction is eliminated:
\[ L_2-L_1=\frac{\lambda}{2}\;\Rightarrow\;\lambda=2\left(L_2-L_1\right) \]The frequency of the note is then obtained from \(f=\dfrac{v}{\lambda}\), giving
\[ f=\frac{v}{2\left(L_2-L_1\right)} \]where \(v\) is the known velocity of sound in air.
(b)(ii) Two precautions.
- The tuning fork should be struck gently on a soft rubber pad (not on a hard surface) and held with its prongs just above the mouth of the burette without touching it.
- The resonance lengths should be read at the position of loudest sound with the eye level with the water meniscus to avoid parallax error; the readings should be repeated and averaged.
(c) Fundamental frequency of a pipe closed at one end.
For a pipe closed at one end, the fundamental note has an air column equal to a quarter of a wavelength: \(L=\dfrac{\lambda}{4}\), so \(\lambda=4L\). With \(L=100\,\text{cm}=1.0\,\text{m}\),
\[ \lambda=4L=4\times1.0=4.0\,\text{m} \] \[ f=\frac{v}{\lambda}=\frac{340}{4.0}=85\,\text{Hz} \]The frequency of the fundamental note is 85 Hz.
Answer Details
(a)(i) Wave motion. A wave motion is a disturbance that travels through a medium (or through space) transferring energy from one point to another without any net transfer of the particles of the medium.
(a)(ii) Two differences between a radio wave and a sound wave.
| Radio wave | Sound wave |
|---|---|
| It is an electromagnetic (transverse) wave. | It is a mechanical (longitudinal) wave. |
| Can travel through a vacuum (needs no material medium). | Requires a material medium for propagation. |
| Travels at the speed of light, \(3\times10^{8}\,\text{ms}^{-1}\). | Travels much more slowly, about \(340\,\text{ms}^{-1}\) in air. |
(b)(i) Experiment to determine the frequency of a note (resonance method using a burette).
The burette is clamped vertically and filled with water, so that the water and the tube walls enclose a short column of air at the open top. The vibrating tuning fork is held horizontally, with its prongs just above (not touching) the open mouth of the burette, as shown below.
The tap of the burette is opened so that water runs out slowly, gradually lengthening the air column. As the length increases, a position is reached where the sound suddenly becomes loudest; this is the first resonance. The length of the air column from the water surface to the open mouth is measured and recorded as \(L_1\).
Water is run out further, keeping the fork sounding over the mouth, until the sound is again loudest; this is the second resonance. This longer air column is measured and recorded as \(L_2\). The whole procedure is repeated a few times and the mean values of \(L_1\) and \(L_2\) taken.
Between the first and second resonances the air column has increased by exactly half a wavelength, so end-correction is eliminated:
\[ L_2-L_1=\frac{\lambda}{2}\;\Rightarrow\;\lambda=2\left(L_2-L_1\right) \]The frequency of the note is then obtained from \(f=\dfrac{v}{\lambda}\), giving
\[ f=\frac{v}{2\left(L_2-L_1\right)} \]where \(v\) is the known velocity of sound in air.
(b)(ii) Two precautions.
- The tuning fork should be struck gently on a soft rubber pad (not on a hard surface) and held with its prongs just above the mouth of the burette without touching it.
- The resonance lengths should be read at the position of loudest sound with the eye level with the water meniscus to avoid parallax error; the readings should be repeated and averaged.
(c) Fundamental frequency of a pipe closed at one end.
For a pipe closed at one end, the fundamental note has an air column equal to a quarter of a wavelength: \(L=\dfrac{\lambda}{4}\), so \(\lambda=4L\). With \(L=100\,\text{cm}=1.0\,\text{m}\),
\[ \lambda=4L=4\times1.0=4.0\,\text{m} \] \[ f=\frac{v}{\lambda}=\frac{340}{4.0}=85\,\text{Hz} \]The frequency of the fundamental note is 85 Hz.
Question 4 Report
Copper of thickness d is plated on the cathode of a copper voltameter. If the total surface area of the cathode is 60 cm\(^2\) and a steady current of 5.0 A is maintained in the voltameter for 1 hour. calculate the value of d. [density of copper = 8.9 x 10\(^3\)kg m\(^{-3}\) electro chemical equivalent of copper = 3.3 x 10\(^{-7}\) kg C\(^{-1}\)
Mass of copper deposited (Faraday's first law): \(m=zIt\).
\[ m=(3.3\times10^{-7})(5.0)(3600)=5.94\times10^{-3}\,\text{kg} \]Volume of this copper: \(V=\dfrac{m}{\rho}\).
\[ V=\frac{5.94\times10^{-3}}{8.9\times10^{3}}=6.67\times10^{-7}\,\text{m}^3 \]The copper is spread over the cathode area \(A=60\,\text{cm}^2=60\times10^{-4}\,\text{m}^2=6.0\times10^{-3}\,\text{m}^2\), and \(V=A\times d\), so
\[ d=\frac{V}{A}=\frac{6.67\times10^{-7}}{6.0\times10^{-3}}=1.11\times10^{-4}\,\text{m} \]So the plating thickness \(d\approx1.1\times10^{-4}\,\text{m}\) (about \(0.11\,\text{mm}\)).
Answer Details
Mass of copper deposited (Faraday's first law): \(m=zIt\).
\[ m=(3.3\times10^{-7})(5.0)(3600)=5.94\times10^{-3}\,\text{kg} \]Volume of this copper: \(V=\dfrac{m}{\rho}\).
\[ V=\frac{5.94\times10^{-3}}{8.9\times10^{3}}=6.67\times10^{-7}\,\text{m}^3 \]The copper is spread over the cathode area \(A=60\,\text{cm}^2=60\times10^{-4}\,\text{m}^2=6.0\times10^{-3}\,\text{m}^2\), and \(V=A\times d\), so
\[ d=\frac{V}{A}=\frac{6.67\times10^{-7}}{6.0\times10^{-3}}=1.11\times10^{-4}\,\text{m} \]So the plating thickness \(d\approx1.1\times10^{-4}\,\text{m}\) (about \(0.11\,\text{mm}\)).
Question 5 Report
(a)(i) With the aid of a labelled diagram describe the mode of operation of a modern X-ray tube.
(ii)State the energy transformations that take place during the operation of the X-ray tube.
(b) Define, as applied to X-rays, the following terms:
(i) hardness;
(ii) intensity.
(c) State (i) four uses of X-rays;
(ii) one hazard of over-exposure to X-rays in a radiological laboratory.
(a)(i) Mode of operation of a modern (Coolidge) X-ray tube
The tube is a highly evacuated glass envelope carrying a tungsten filament (cathode) at one end and a target-anode at the other, as shown in the labelled diagram below.
A low-voltage supply heats the tungsten filament, which by thermionic emission releases electrons. A concave focusing cup gathers these electrons into a narrow beam and directs them toward the target. A very high potential difference (the E.H.T., tens of kilovolts) is applied between the cathode and the anode, with the anode positive; this accelerates the electrons to a very high speed across the evacuated tube.
The fast electron beam strikes a small tungsten target embedded in a massive copper anode. On being suddenly stopped, the electrons give up their kinetic energy: only about 1 % is emitted as X-rays, which leave the tube through the window, while the greater part appears as heat. The copper anode and its cooling fins (aided in practice by circulating oil or water) conduct this heat away and prevent the target from melting.
The hardness (penetrating power) of the X-rays is controlled by the accelerating voltage across the tube, while the intensity (quantity of X-rays) is controlled by the filament heating current, which fixes the number of electrons produced.
(a)(ii) Energy transformations
\[\text{Electrical energy}\;\rightarrow\;\text{kinetic energy of the accelerated electrons}\;\rightarrow\;\text{X-ray (electromagnetic) energy}\;+\;\text{heat (thermal) energy at the target.}\]
(b) Definitions (as applied to X-rays)
(i) Hardness: the penetrating power of the X-rays. Hard X-rays have short wavelength and high frequency and penetrate deeply; the hardness increases with the accelerating potential difference across the tube.
(ii) Intensity: the energy radiated per unit area per unit time, i.e. the number of X-ray photons arriving per second on unit area. It depends on the number of electrons striking the target, and hence on the filament heating current.
(c)(i) Four uses of X-rays
- Medical diagnosis (radiography) to reveal fractured bones and the condition of internal organs.
- Radiotherapy, to destroy cancerous cells and tumours.
- Detection of hidden flaws, cracks and air bubbles in metals and welded joints (industrial radiography).
- Security screening of luggage at airports, and the study of crystal structure by X-ray crystallography.
(c)(ii) One hazard of over-exposure
Over-exposure to X-rays destroys living body cells, causing skin burns and cancer (for example leukaemia), as well as tissue damage, cataracts, sterility and genetic mutation.
Answer Details
(a)(i) Mode of operation of a modern (Coolidge) X-ray tube
The tube is a highly evacuated glass envelope carrying a tungsten filament (cathode) at one end and a target-anode at the other, as shown in the labelled diagram below.
A low-voltage supply heats the tungsten filament, which by thermionic emission releases electrons. A concave focusing cup gathers these electrons into a narrow beam and directs them toward the target. A very high potential difference (the E.H.T., tens of kilovolts) is applied between the cathode and the anode, with the anode positive; this accelerates the electrons to a very high speed across the evacuated tube.
The fast electron beam strikes a small tungsten target embedded in a massive copper anode. On being suddenly stopped, the electrons give up their kinetic energy: only about 1 % is emitted as X-rays, which leave the tube through the window, while the greater part appears as heat. The copper anode and its cooling fins (aided in practice by circulating oil or water) conduct this heat away and prevent the target from melting.
The hardness (penetrating power) of the X-rays is controlled by the accelerating voltage across the tube, while the intensity (quantity of X-rays) is controlled by the filament heating current, which fixes the number of electrons produced.
(a)(ii) Energy transformations
\[\text{Electrical energy}\;\rightarrow\;\text{kinetic energy of the accelerated electrons}\;\rightarrow\;\text{X-ray (electromagnetic) energy}\;+\;\text{heat (thermal) energy at the target.}\]
(b) Definitions (as applied to X-rays)
(i) Hardness: the penetrating power of the X-rays. Hard X-rays have short wavelength and high frequency and penetrate deeply; the hardness increases with the accelerating potential difference across the tube.
(ii) Intensity: the energy radiated per unit area per unit time, i.e. the number of X-ray photons arriving per second on unit area. It depends on the number of electrons striking the target, and hence on the filament heating current.
(c)(i) Four uses of X-rays
- Medical diagnosis (radiography) to reveal fractured bones and the condition of internal organs.
- Radiotherapy, to destroy cancerous cells and tumours.
- Detection of hidden flaws, cracks and air bubbles in metals and welded joints (industrial radiography).
- Security screening of luggage at airports, and the study of crystal structure by X-ray crystallography.
(c)(ii) One hazard of over-exposure
Over-exposure to X-rays destroys living body cells, causing skin burns and cancer (for example leukaemia), as well as tissue damage, cataracts, sterility and genetic mutation.
Question 6 Report
(a) State the conditions of equilibrium for a number of coplanar parallel forces.
(b) A metre rule is found to balance horizontally at the 48 cm mark. When a body of mass 60 g is suspended at the 6 cm mark, the balance point is found to be at the 30 cm mark. Calculate the;
(i) mass of the metre rule;
(ii) distance of the balance point from the zero end, if the body were moved to the 13 cm mark.
(c) a man pulls up a box of mass 70 kg using an inclined plane of effective length 5 m unto a platform 2.5 m high at a uniform speed. If the frictional force between the box and the plane is 1000 N;
(i) draw a diagram to illustrate all the forces acting on the box while in motion;
(ii) calculate the I. minimum effort applied in pulling up the box; II. velocity ratio of the plane, if it is inclined at 30° to the horizontal; Ill. force ratio of the plane.
(a) Conditions of equilibrium for coplanar parallel forces
A number of coplanar parallel forces are in equilibrium when:
- The algebraic sum of the forces is zero, i.e. the sum of the forces acting in one direction equals the sum of the forces acting in the opposite direction (the resultant force is zero).
- The algebraic sum of the moments of the forces about any point is zero, i.e. the sum of the clockwise moments about the point equals the sum of the anticlockwise moments about that same point (the principle of moments).
(b) Balancing a metre rule
The rule balances by itself at the \(48\,\text{cm}\) mark, so the whole weight of the rule acts at the \(48\,\text{cm}\) mark.
(i) Mass of the metre rule
With the \(60\,\text{g}\) body hung at the \(6\,\text{cm}\) mark, the new balance point (fulcrum) is at the \(30\,\text{cm}\) mark. The body sits on one side of the fulcrum and the weight of the rule acts on the other side. Taking moments about the \(30\,\text{cm}\) fulcrum:
\[ 60\times(30-6)=m\times(48-30) \]\[ 60\times24=m\times18 \]\[ m=\frac{60\times24}{18}=\frac{1440}{18}=80\,\text{g} \]The mass of the metre rule is 80 g.
(ii) New balance point with the body at the 13 cm mark
Let the new balance point be at the \(x\,\text{cm}\) mark. The body (\(60\,\text{g}\)) now acts at \(13\,\text{cm}\) and the rule's weight (\(80\,\text{g}\)) still acts at \(48\,\text{cm}\). Taking moments about the fulcrum at \(x\):
\[ 60\times(x-13)=80\times(48-x) \]\[ 60x-780=3840-80x \]\[ 140x=4620 \]\[ x=\frac{4620}{140}=33\,\text{cm} \]The new balance point is 33 cm from the zero end.
(c) Pulling a box up an inclined plane
Box mass \(=70\,\text{kg}\), so its weight \(W=mg=70\times10=700\,\text{N}\); length of plane \(L=5\,\text{m}\); height of platform \(h=2.5\,\text{m}\); frictional force \(F=1000\,\text{N}\); angle of incline \(\theta=30^{\circ}\).
(i) Diagram of the forces acting on the box while in motion
The four forces acting on the box are: its weight \(W\) acting vertically downwards; the normal (reaction) force \(N\) acting perpendicular to the surface of the plane; the effort \(E\) applied up along the plane; and the frictional force \(F\) acting down along the plane, opposing the upward motion.
(ii) Calculations
I. Minimum effort applied in pulling up the box
Since the box moves up at uniform (constant) speed, the effort must balance the component of the weight along the plane together with the friction acting down the plane:
\[ E=W\sin\theta+F=700\sin30^{\circ}+1000 \]\[ E=700\times0.5+1000=350+1000=1350\,\text{N} \]The minimum effort is 1350 N.
II. Velocity ratio of the plane
\[ V.R.=\frac{1}{\sin\theta}=\frac{1}{\sin30^{\circ}}=\frac{1}{0.5}=2 \](This agrees with \(\dfrac{\text{length}}{\text{height}}=\dfrac{5}{2.5}=2\).) The velocity ratio is 2.
III. Force ratio (mechanical advantage) of the plane
\[ \text{Force ratio}=M.A.=\frac{\text{load}}{\text{effort}}=\frac{700}{1350}=0.52 \]The force ratio of the plane is 0.52.
Answer Details
(a) Conditions of equilibrium for coplanar parallel forces
A number of coplanar parallel forces are in equilibrium when:
- The algebraic sum of the forces is zero, i.e. the sum of the forces acting in one direction equals the sum of the forces acting in the opposite direction (the resultant force is zero).
- The algebraic sum of the moments of the forces about any point is zero, i.e. the sum of the clockwise moments about the point equals the sum of the anticlockwise moments about that same point (the principle of moments).
(b) Balancing a metre rule
The rule balances by itself at the \(48\,\text{cm}\) mark, so the whole weight of the rule acts at the \(48\,\text{cm}\) mark.
(i) Mass of the metre rule
With the \(60\,\text{g}\) body hung at the \(6\,\text{cm}\) mark, the new balance point (fulcrum) is at the \(30\,\text{cm}\) mark. The body sits on one side of the fulcrum and the weight of the rule acts on the other side. Taking moments about the \(30\,\text{cm}\) fulcrum:
\[ 60\times(30-6)=m\times(48-30) \]\[ 60\times24=m\times18 \]\[ m=\frac{60\times24}{18}=\frac{1440}{18}=80\,\text{g} \]The mass of the metre rule is 80 g.
(ii) New balance point with the body at the 13 cm mark
Let the new balance point be at the \(x\,\text{cm}\) mark. The body (\(60\,\text{g}\)) now acts at \(13\,\text{cm}\) and the rule's weight (\(80\,\text{g}\)) still acts at \(48\,\text{cm}\). Taking moments about the fulcrum at \(x\):
\[ 60\times(x-13)=80\times(48-x) \]\[ 60x-780=3840-80x \]\[ 140x=4620 \]\[ x=\frac{4620}{140}=33\,\text{cm} \]The new balance point is 33 cm from the zero end.
(c) Pulling a box up an inclined plane
Box mass \(=70\,\text{kg}\), so its weight \(W=mg=70\times10=700\,\text{N}\); length of plane \(L=5\,\text{m}\); height of platform \(h=2.5\,\text{m}\); frictional force \(F=1000\,\text{N}\); angle of incline \(\theta=30^{\circ}\).
(i) Diagram of the forces acting on the box while in motion
The four forces acting on the box are: its weight \(W\) acting vertically downwards; the normal (reaction) force \(N\) acting perpendicular to the surface of the plane; the effort \(E\) applied up along the plane; and the frictional force \(F\) acting down along the plane, opposing the upward motion.
(ii) Calculations
I. Minimum effort applied in pulling up the box
Since the box moves up at uniform (constant) speed, the effort must balance the component of the weight along the plane together with the friction acting down the plane:
\[ E=W\sin\theta+F=700\sin30^{\circ}+1000 \]\[ E=700\times0.5+1000=350+1000=1350\,\text{N} \]The minimum effort is 1350 N.
II. Velocity ratio of the plane
\[ V.R.=\frac{1}{\sin\theta}=\frac{1}{\sin30^{\circ}}=\frac{1}{0.5}=2 \](This agrees with \(\dfrac{\text{length}}{\text{height}}=\dfrac{5}{2.5}=2\).) The velocity ratio is 2.
III. Force ratio (mechanical advantage) of the plane
\[ \text{Force ratio}=M.A.=\frac{\text{load}}{\text{effort}}=\frac{700}{1350}=0.52 \]The force ratio of the plane is 0.52.
Question 7 Report
Define (a) tensile Stress; (b) tensile strain; (c) yield point.
(a) Tensile stress: the stretching force acting per unit cross-sectional area of a material, \(\text{stress}=\dfrac{F}{A}\). Unit: \(\text{Nm}^{-2}\) (pascal).
(b) Tensile strain: the extension produced per unit original length of the material, \(\text{strain}=\dfrac{e}{L}\). It has no unit.
(c) Yield point: the point on the stress-strain (or load-extension) graph beyond which the material no longer returns to its original length when the load is removed; it begins to undergo permanent (plastic) deformation.
Answer Details
(a) Tensile stress: the stretching force acting per unit cross-sectional area of a material, \(\text{stress}=\dfrac{F}{A}\). Unit: \(\text{Nm}^{-2}\) (pascal).
(b) Tensile strain: the extension produced per unit original length of the material, \(\text{strain}=\dfrac{e}{L}\). It has no unit.
(c) Yield point: the point on the stress-strain (or load-extension) graph beyond which the material no longer returns to its original length when the load is removed; it begins to undergo permanent (plastic) deformation.
Question 8 Report
(a) Explain thermiopic emission
(b) State two applications of electrical conduction through gases.
(a) Thermionic emission: the emission (giving off) of electrons from the surface of a metal when it is heated to a high temperature. The heat gives the free surface electrons enough kinetic energy to overcome the work function (the surface attractive forces) and escape from the metal.
(b) Two applications of electrical conduction through gases:
- Fluorescent tubes and neon/discharge advertising signs.
- Sodium and mercury-vapour street lamps (the cathode-ray/television tube and lightning are also acceptable examples).
Answer Details
(a) Thermionic emission: the emission (giving off) of electrons from the surface of a metal when it is heated to a high temperature. The heat gives the free surface electrons enough kinetic energy to overcome the work function (the surface attractive forces) and escape from the metal.
(b) Two applications of electrical conduction through gases:
- Fluorescent tubes and neon/discharge advertising signs.
- Sodium and mercury-vapour street lamps (the cathode-ray/television tube and lightning are also acceptable examples).
Question 9 Report
State two effects to show (a) the existence of matter waves;
(b) that radiation behaves like particles.
Answer Details
None
Question 10 Report
A wire of length 5.0 m and diameter 2.0 mm extends by 0.25 mm when a force of 50 N was used to stretch it from its end. Calculate the;
(a) stress on the wire;
(b) strain in, the wire. [\(\pi = 3.142\)]
Length \(L=5.0\,\text{m}\), diameter \(=2.0\,\text{mm}\Rightarrow r=1.0\,\text{mm}=1.0\times10^{-3}\,\text{m}\), extension \(e=0.25\,\text{mm}=0.25\times10^{-3}\,\text{m}\), force \(F=50\,\text{N}\).
(a) Stress \(=\dfrac{F}{A}\), where \(A=\pi r^2\).
\[ A=3.142\times(1.0\times10^{-3})^2=3.142\times10^{-6}\,\text{m}^2 \]\[ \text{Stress}=\frac{50}{3.142\times10^{-6}}=1.59\times10^{7}\,\text{Nm}^{-2} \](b) Strain \(=\dfrac{\text{extension}}{\text{original length}}\).
\[ \text{Strain}=\frac{0.25\times10^{-3}}{5.0}=5.0\times10^{-5} \](Strain has no unit as it is a ratio of two lengths.)
Answer Details
Length \(L=5.0\,\text{m}\), diameter \(=2.0\,\text{mm}\Rightarrow r=1.0\,\text{mm}=1.0\times10^{-3}\,\text{m}\), extension \(e=0.25\,\text{mm}=0.25\times10^{-3}\,\text{m}\), force \(F=50\,\text{N}\).
(a) Stress \(=\dfrac{F}{A}\), where \(A=\pi r^2\).
\[ A=3.142\times(1.0\times10^{-3})^2=3.142\times10^{-6}\,\text{m}^2 \]\[ \text{Stress}=\frac{50}{3.142\times10^{-6}}=1.59\times10^{7}\,\text{Nm}^{-2} \](b) Strain \(=\dfrac{\text{extension}}{\text{original length}}\).
\[ \text{Strain}=\frac{0.25\times10^{-3}}{5.0}=5.0\times10^{-5} \](Strain has no unit as it is a ratio of two lengths.)
Question 11 Report
A particle is projected at an angle of 30° to the horizontal with a speed of 250 ms\(^{-1}\). Calculate the;
(a) total time of flight of the particle;
(b) speed of the particle at its maximum height. [ g = 10 ms\(^{-2}\)]
Given \(u=250\,\text{ms}^{-1}\), \(\theta=30^{\circ}\), \(g=10\,\text{ms}^{-2}\).
(a) Total time of flight.
\[ T=\frac{2u\sin\theta}{g}=\frac{2\times250\times\sin30^{\circ}}{10}=\frac{2\times250\times0.5}{10}=\frac{250}{10}=25\,\text{s} \](b) Speed at maximum height. At the highest point the vertical component of velocity is zero, so the speed there is just the (unchanged) horizontal component:
\[ v=u\cos\theta=250\times\cos30^{\circ}=250\times0.866=216.5\,\text{ms}^{-1} \]Answer Details
Given \(u=250\,\text{ms}^{-1}\), \(\theta=30^{\circ}\), \(g=10\,\text{ms}^{-2}\).
(a) Total time of flight.
\[ T=\frac{2u\sin\theta}{g}=\frac{2\times250\times\sin30^{\circ}}{10}=\frac{2\times250\times0.5}{10}=\frac{250}{10}=25\,\text{s} \](b) Speed at maximum height. At the highest point the vertical component of velocity is zero, so the speed there is just the (unchanged) horizontal component:
\[ v=u\cos\theta=250\times\cos30^{\circ}=250\times0.866=216.5\,\text{ms}^{-1} \]Question 12 Report
(a) With the aid of a simple diagram, explain how a step down transformer works.
(b)(i) State three ways by which energy is lost in a transformer
(ii) Mention how each of the losses in (b)(i) above can be minimized
(c) A 95% efficient transformer is used to operate a lamp rated 60W, 220 V from a 4400 V a.c supply. Calculate the;
(i) ratio of the number of turns in the primary coil to the number of turns in the secondary coil of the transformer
(ii) current taken from the main circuit.
(a) How a step-down transformer works
A transformer consists of two coils, a primary (input) coil and a secondary (output) coil, wound on a common laminated soft-iron core. In a step-down transformer the primary coil has more turns than the secondary coil (\(N_p > N_s\)).
When an alternating voltage is applied to the primary coil, the alternating current flowing in it sets up a continuously changing magnetic flux in the soft-iron core. The core carries this changing flux round to the secondary coil, so the secondary is linked by the same changing flux. By electromagnetic induction, the changing flux induces an alternating e.m.f. in the secondary coil. Because the secondary has fewer turns than the primary, the e.m.f. induced in it is smaller than the applied primary voltage, i.e. the voltage is stepped down (while the output current is correspondingly stepped up).
The simple diagram below shows the arrangement:
(b)(i) Three ways energy is lost in a transformer
- Eddy-current loss in the iron core.
- Hysteresis loss in the iron core.
- Copper (resistance/heating) loss in the windings.
(b)(ii) How each loss is minimized
| Energy loss | How it is minimized |
|---|---|
| Eddy-current loss in the core | Use a laminated core (thin iron sheets insulated from one another) to break up the induced eddy-current paths. |
| Hysteresis loss in the core | Make the core of soft iron, which is easily magnetized and demagnetized, so little energy is used up in each magnetization cycle. |
| Copper (heating) loss in the windings | Use thick, low-resistance copper wire for the coils so that \(I^2R\) heating is reduced. |
(c) Calculations
Efficiency \(=95\%=0.95\); lamp (secondary side) rated \(P_{out}=60\,\text{W},\; V_s=220\,\text{V}\); supply \(V_p=4400\,\text{V}\).
(i) Ratio of primary turns to secondary turns.
For a transformer the turns ratio equals the voltage ratio:
\[ \frac{N_p}{N_s}=\frac{V_p}{V_s}=\frac{4400}{220}=20 \]Therefore \(\;N_p:N_s = \mathbf{20:1}.\)
(ii) Current taken from the main circuit.
Using efficiency \(=\dfrac{\text{output power}}{\text{input power}}=\dfrac{V_s I_s}{V_p I_p}\):
\[ 0.95=\frac{P_{out}}{V_p\,I_p}=\frac{60}{4400\times I_p} \]\[ I_p=\frac{60}{0.95\times 4400}=\frac{60}{4180}=0.0144\,\text{A} \]The current taken from the mains is \(I_p \approx \mathbf{0.014\,A}\;(1.44\times10^{-2}\,\text{A},\ \text{about }14.4\,\text{mA}).\)
Answer Details
(a) How a step-down transformer works
A transformer consists of two coils, a primary (input) coil and a secondary (output) coil, wound on a common laminated soft-iron core. In a step-down transformer the primary coil has more turns than the secondary coil (\(N_p > N_s\)).
When an alternating voltage is applied to the primary coil, the alternating current flowing in it sets up a continuously changing magnetic flux in the soft-iron core. The core carries this changing flux round to the secondary coil, so the secondary is linked by the same changing flux. By electromagnetic induction, the changing flux induces an alternating e.m.f. in the secondary coil. Because the secondary has fewer turns than the primary, the e.m.f. induced in it is smaller than the applied primary voltage, i.e. the voltage is stepped down (while the output current is correspondingly stepped up).
The simple diagram below shows the arrangement:
(b)(i) Three ways energy is lost in a transformer
- Eddy-current loss in the iron core.
- Hysteresis loss in the iron core.
- Copper (resistance/heating) loss in the windings.
(b)(ii) How each loss is minimized
| Energy loss | How it is minimized |
|---|---|
| Eddy-current loss in the core | Use a laminated core (thin iron sheets insulated from one another) to break up the induced eddy-current paths. |
| Hysteresis loss in the core | Make the core of soft iron, which is easily magnetized and demagnetized, so little energy is used up in each magnetization cycle. |
| Copper (heating) loss in the windings | Use thick, low-resistance copper wire for the coils so that \(I^2R\) heating is reduced. |
(c) Calculations
Efficiency \(=95\%=0.95\); lamp (secondary side) rated \(P_{out}=60\,\text{W},\; V_s=220\,\text{V}\); supply \(V_p=4400\,\text{V}\).
(i) Ratio of primary turns to secondary turns.
For a transformer the turns ratio equals the voltage ratio:
\[ \frac{N_p}{N_s}=\frac{V_p}{V_s}=\frac{4400}{220}=20 \]Therefore \(\;N_p:N_s = \mathbf{20:1}.\)
(ii) Current taken from the main circuit.
Using efficiency \(=\dfrac{\text{output power}}{\text{input power}}=\dfrac{V_s I_s}{V_p I_p}\):
\[ 0.95=\frac{P_{out}}{V_p\,I_p}=\frac{60}{4400\times I_p} \]\[ I_p=\frac{60}{0.95\times 4400}=\frac{60}{4180}=0.0144\,\text{A} \]The current taken from the mains is \(I_p \approx \mathbf{0.014\,A}\;(1.44\times10^{-2}\,\text{A},\ \text{about }14.4\,\text{mA}).\)
Question 13 Report
State;
(a) the difference between plane polarized light and ordinary light;
(b) two uses of polaroids.
(a) Difference between plane polarized light and ordinary light: in plane (linearly) polarized light the vibrations of the electric field take place in only one plane perpendicular to the direction of travel, whereas in ordinary (unpolarized) light the vibrations occur in all planes perpendicular to the direction of travel.
(b) Two uses of polaroids:
- In sunglasses and camera filters to cut down glare from reflected light.
- In liquid-crystal displays (LCDs); also in polarimeters to measure sugar concentration, in stress analysis of materials, and in 3-D glasses.
Answer Details
(a) Difference between plane polarized light and ordinary light: in plane (linearly) polarized light the vibrations of the electric field take place in only one plane perpendicular to the direction of travel, whereas in ordinary (unpolarized) light the vibrations occur in all planes perpendicular to the direction of travel.
(b) Two uses of polaroids:
- In sunglasses and camera filters to cut down glare from reflected light.
- In liquid-crystal displays (LCDs); also in polarimeters to measure sugar concentration, in stress analysis of materials, and in 3-D glasses.
Question 14 Report
(a) List two factors that can affect the rate of diffusion
(b) State two examples to illustrate the effects of surface tension.
(a) Two factors that affect the rate of diffusion:
- Temperature (higher temperature gives faster diffusion).
- The mass/size of the diffusing molecules (lighter molecules diffuse faster). The concentration gradient is also acceptable.
(b) Two examples illustrating the effects of surface tension:
- Small liquid drops (e.g. water or mercury) take up a spherical shape.
- A dry steel needle or a small insect can rest on the surface of water without sinking. (Capillary rise of liquid in a narrow tube is also acceptable.)
Answer Details
(a) Two factors that affect the rate of diffusion:
- Temperature (higher temperature gives faster diffusion).
- The mass/size of the diffusing molecules (lighter molecules diffuse faster). The concentration gradient is also acceptable.
(b) Two examples illustrating the effects of surface tension:
- Small liquid drops (e.g. water or mercury) take up a spherical shape.
- A dry steel needle or a small insect can rest on the surface of water without sinking. (Capillary rise of liquid in a narrow tube is also acceptable.)
Question 15 Report
A stone projected horizontally from the top of a tower with a speed of 4 ms\(^{-1}\) lands on the level ground at a horizontal distance 25 m from the foot of the tower. Calculate the height of the tower. g = 10 ms\(^{-2}\)]
Horizontal projection: initial horizontal speed \(u=4\,\text{ms}^{-1}\), horizontal range \(x=25\,\text{m}\), \(g=10\,\text{ms}^{-2}\).
Horizontal motion (constant velocity) gives the time of flight:
\[ x=ut\Rightarrow t=\frac{x}{u}=\frac{25}{4}=6.25\,\text{s} \]Vertical motion (falling from rest, initial vertical velocity zero) gives the height of the tower:
\[ h=\tfrac{1}{2}gt^2=\tfrac{1}{2}\times10\times(6.25)^2=5\times39.06=195.3\,\text{m} \]So the tower is about \(195\,\text{m}\) high.
Answer Details
Horizontal projection: initial horizontal speed \(u=4\,\text{ms}^{-1}\), horizontal range \(x=25\,\text{m}\), \(g=10\,\text{ms}^{-2}\).
Horizontal motion (constant velocity) gives the time of flight:
\[ x=ut\Rightarrow t=\frac{x}{u}=\frac{25}{4}=6.25\,\text{s} \]Vertical motion (falling from rest, initial vertical velocity zero) gives the height of the tower:
\[ h=\tfrac{1}{2}gt^2=\tfrac{1}{2}\times10\times(6.25)^2=5\times39.06=195.3\,\text{m} \]So the tower is about \(195\,\text{m}\) high.
