WAEC SSCE - Physics - 2004 (Essay)
Question 1 Report
State three methods of polarising an unpolarised light
Three methods of producing (plane) polarised light from unpolarised light:
- By reflection: reflecting light from a plane surface (e.g. glass or water) at the polarising angle (Brewster's angle) gives a completely plane-polarised reflected beam.
- By selective absorption: passing light through a Polaroid (dichroic material) which transmits vibrations in one plane and absorbs those in the perpendicular plane.
- By double refraction: passing light through a doubly refracting crystal such as calcite (Nicol prism), which separates the light into two plane-polarised beams.
(Polarisation by scattering, e.g. of sunlight by air molecules, is also accepted.)
Answer Details
Three methods of producing (plane) polarised light from unpolarised light:
- By reflection: reflecting light from a plane surface (e.g. glass or water) at the polarising angle (Brewster's angle) gives a completely plane-polarised reflected beam.
- By selective absorption: passing light through a Polaroid (dichroic material) which transmits vibrations in one plane and absorbs those in the perpendicular plane.
- By double refraction: passing light through a doubly refracting crystal such as calcite (Nicol prism), which separates the light into two plane-polarised beams.
(Polarisation by scattering, e.g. of sunlight by air molecules, is also accepted.)
Question 2 Report
The diagram above illustrates the path ABC, in a vertical x z plane, of a bullet shot into the air at an angle above the horizontal. Copy the diagram, and, using arrows, indicate the relative magnitudes and directions of the vertical and horizontal components of the velocities of the bullet at the point A, B and C.
Answer Details
None
Question 3 Report
(a) Define diffusion
(b) State Graham's law of diffusion
(a) Diffusion: the spontaneous movement (spreading) of the molecules of a substance from a region of higher concentration to a region of lower concentration until they are evenly distributed, arising from the random motion of the molecules.
(b) Graham's law of diffusion: at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or of its molar mass):
\[ R \propto \frac{1}{\sqrt{\rho}} \qquad \text{i.e.} \qquad \frac{R_1}{R_2} = \sqrt{\frac{\rho_2}{\rho_1}} \]where \(R\) is the rate of diffusion and \(\rho\) is the density of the gas.
Answer Details
(a) Diffusion: the spontaneous movement (spreading) of the molecules of a substance from a region of higher concentration to a region of lower concentration until they are evenly distributed, arising from the random motion of the molecules.
(b) Graham's law of diffusion: at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or of its molar mass):
\[ R \propto \frac{1}{\sqrt{\rho}} \qquad \text{i.e.} \qquad \frac{R_1}{R_2} = \sqrt{\frac{\rho_2}{\rho_1}} \]where \(R\) is the rate of diffusion and \(\rho\) is the density of the gas.
Question 4 Report
Define upper fixed point and lower fixed point as used in thermometry
(b) The electrical resistances of the element in a platinum restistance thermometer at 100\(^o\)C, 0\(^o\) and room temperature are 75.000, 63.000 and 64.992 \(\Omega\) respectively. Use these data to determine the room temperature.
(c) (i) State Boyle's law
(ii) A uniform capillary tube, closed at one end contained dry air trapped by a thread of mecury 8.5 x 10\(^{-2}\)m long. When the tube was held horinzontally, the length of the air column was 5.0 x 10\(^{-2}\)m, when it was held vertically with the closed end downwards, the length was 4.5 x 10\(^{-2}\)m, Determine the value of the atmospheric pressure. [g = 10ms\(^{-2}\), density of mecury = 1.36 x 10\(^4\) kg m\(^{-3}\)]
(a) Upper fixed point: the temperature of pure boiling water (steam) at standard atmospheric pressure, taken as \(100^{\circ}\text{C}\). Lower fixed point: the temperature of pure melting ice at standard atmospheric pressure, taken as \(0^{\circ}\text{C}\).
(b) For a platinum resistance thermometer, temperature on the resistance scale is:
\[ \theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100 \]With \(R_0 = 63.000\ \Omega\), \(R_{100} = 75.000\ \Omega\), \(R_\theta = 64.992\ \Omega\):
\[ \theta = \frac{64.992 - 63.000}{75.000 - 63.000} \times 100 = \frac{1.992}{12.000} \times 100 \] \[ \theta = 16.6^{\circ}\text{C} \]The room temperature is \(16.6^{\circ}\text{C}\).
(c)(i) Boyle's law: the volume of a fixed mass of gas is inversely proportional to its pressure, provided the temperature is kept constant; i.e. \(PV = \text{constant}\).
(c)(ii) Let the atmospheric pressure be \(H\) (expressed as a length of mercury). The mercury thread is \(8.5\ \text{cm}\) long.
Horizontal: the mercury adds nothing, so the air pressure \(= H\) and the length is \(5.0\ \text{cm}\).
Vertical, closed end downwards: the mercury column presses down on the trapped air, so its pressure \(= H + 8.5\), and the length is \(4.5\ \text{cm}\).
Applying Boyle's law (constant cross-section, so length replaces volume):
\[ H \times 5.0 = (H + 8.5) \times 4.5 \] \[ 5.0H = 4.5H + 38.25 \Rightarrow 0.5H = 38.25 \Rightarrow H = 76.5\ \text{cm of mercury} \]Converting to pascals with \(\rho = 1.36 \times 10^{4}\ \text{kg m}^{-3}\), \(g = 10\ \text{ms}^{-2}\), \(H = 0.765\ \text{m}\):
\[ P = \rho g H = 1.36 \times 10^{4} \times 10 \times 0.765 = 1.04 \times 10^{5}\ \text{Pa} \]The atmospheric pressure is \(76.5\ \text{cmHg} \approx 1.04 \times 10^{5}\ \text{Pa}\).
Answer Details
(a) Upper fixed point: the temperature of pure boiling water (steam) at standard atmospheric pressure, taken as \(100^{\circ}\text{C}\). Lower fixed point: the temperature of pure melting ice at standard atmospheric pressure, taken as \(0^{\circ}\text{C}\).
(b) For a platinum resistance thermometer, temperature on the resistance scale is:
\[ \theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100 \]With \(R_0 = 63.000\ \Omega\), \(R_{100} = 75.000\ \Omega\), \(R_\theta = 64.992\ \Omega\):
\[ \theta = \frac{64.992 - 63.000}{75.000 - 63.000} \times 100 = \frac{1.992}{12.000} \times 100 \] \[ \theta = 16.6^{\circ}\text{C} \]The room temperature is \(16.6^{\circ}\text{C}\).
(c)(i) Boyle's law: the volume of a fixed mass of gas is inversely proportional to its pressure, provided the temperature is kept constant; i.e. \(PV = \text{constant}\).
(c)(ii) Let the atmospheric pressure be \(H\) (expressed as a length of mercury). The mercury thread is \(8.5\ \text{cm}\) long.
Horizontal: the mercury adds nothing, so the air pressure \(= H\) and the length is \(5.0\ \text{cm}\).
Vertical, closed end downwards: the mercury column presses down on the trapped air, so its pressure \(= H + 8.5\), and the length is \(4.5\ \text{cm}\).
Applying Boyle's law (constant cross-section, so length replaces volume):
\[ H \times 5.0 = (H + 8.5) \times 4.5 \] \[ 5.0H = 4.5H + 38.25 \Rightarrow 0.5H = 38.25 \Rightarrow H = 76.5\ \text{cm of mercury} \]Converting to pascals with \(\rho = 1.36 \times 10^{4}\ \text{kg m}^{-3}\), \(g = 10\ \text{ms}^{-2}\), \(H = 0.765\ \text{m}\):
\[ P = \rho g H = 1.36 \times 10^{4} \times 10 \times 0.765 = 1.04 \times 10^{5}\ \text{Pa} \]The atmospheric pressure is \(76.5\ \text{cmHg} \approx 1.04 \times 10^{5}\ \text{Pa}\).
Question 5 Report
(a) Define Young's modulus
(b) When a force of 50 N is applied to the free end of an elastic cord, an extension of 4 cm is produced in the cord. Calculate the work done on the cord.
(a) Young's modulus: the ratio of the tensile stress to the tensile strain of a material, within the limit of proportionality (elastic limit). That is:
\[ E = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/l} = \frac{Fl}{Ae} \]where \(F\) is the applied force, \(A\) the cross-sectional area, \(e\) the extension and \(l\) the original length.
(b) Force \(F = 50\ \text{N}\), extension \(e = 4\ \text{cm} = 0.04\ \text{m}\). For an elastic (Hooke's-law) cord the force builds up uniformly from 0 to \(F\), so the work done (energy stored) is:
\[ W = \frac{1}{2} F e = \frac{1}{2} \times 50 \times 0.04 \] \[ W = 1.0\ \text{J} \]The work done on the cord is \(1.0\ \text{J}\).
Answer Details
(a) Young's modulus: the ratio of the tensile stress to the tensile strain of a material, within the limit of proportionality (elastic limit). That is:
\[ E = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/l} = \frac{Fl}{Ae} \]where \(F\) is the applied force, \(A\) the cross-sectional area, \(e\) the extension and \(l\) the original length.
(b) Force \(F = 50\ \text{N}\), extension \(e = 4\ \text{cm} = 0.04\ \text{m}\). For an elastic (Hooke's-law) cord the force builds up uniformly from 0 to \(F\), so the work done (energy stored) is:
\[ W = \frac{1}{2} F e = \frac{1}{2} \times 50 \times 0.04 \] \[ W = 1.0\ \text{J} \]The work done on the cord is \(1.0\ \text{J}\).
Question 6 Report
(a) For a water voltameter, identify the;
(i) electrolyte;
(ii) electrodes;
(iii) substances deposited on the electrodes
(b)State the ratio of the volume of the substances deposited in (a) (iii) above
(a)(i) Electrolyte: acidulated water (water containing a few drops of tetraoxosulphate(VI) acid, i.e. dilute sulphuric acid).
(ii) Electrodes: platinum electrodes (an anode and a cathode).
(iii) Substances deposited (liberated): hydrogen gas is liberated at the cathode and oxygen gas is liberated at the anode.
(b) Ratio of volumes: volume of hydrogen : volume of oxygen \(= 2 : 1\) (twice as much hydrogen as oxygen).
Answer Details
(a)(i) Electrolyte: acidulated water (water containing a few drops of tetraoxosulphate(VI) acid, i.e. dilute sulphuric acid).
(ii) Electrodes: platinum electrodes (an anode and a cathode).
(iii) Substances deposited (liberated): hydrogen gas is liberated at the cathode and oxygen gas is liberated at the anode.
(b) Ratio of volumes: volume of hydrogen : volume of oxygen \(= 2 : 1\) (twice as much hydrogen as oxygen).
Question 7 Report
Give three observations in support of de Broglie's assumption that moving particles behave like waves.
Question 8 Report
In his first attempt, a long jumper took off from the spring board with a speed of 8 ms\(^{-1}\) at 30° to the horizontal. He makes a second attempt with the same speed at 45° to the horizontal. Given that the expression for the horizontal range of a projectile is \(\frac{u^2 sin 2\theta}{g}\) where all the symbols have their usual meanings, show that he gains a distance of 0.8576 m in his second attempt. [g = 10ms\(^{-2}\)]
Using \(R = \dfrac{u^2 \sin 2\theta}{g}\) with \(u = 8\ \text{ms}^{-1}\) and \(g = 10\ \text{ms}^{-2}\), so \(u^2 = 64\).
First attempt (\(\theta = 30^{\circ}\), \(2\theta = 60^{\circ}\), \(\sin 60^{\circ} = 0.8660\)):
\[ R_1 = \frac{64 \times 0.8660}{10} = \frac{55.425}{10} = 5.5425\ \text{m} \]Second attempt (\(\theta = 45^{\circ}\), \(2\theta = 90^{\circ}\), \(\sin 90^{\circ} = 1\)):
\[ R_2 = \frac{64 \times 1}{10} = 6.4\ \text{m} \]Gain in distance:
\[ R_2 - R_1 = 6.4 - 5.5425 = 0.8575 \approx 0.8576\ \text{m} \]Hence he gains about \(0.8576\ \text{m}\) in the second attempt, as required.
Answer Details
Using \(R = \dfrac{u^2 \sin 2\theta}{g}\) with \(u = 8\ \text{ms}^{-1}\) and \(g = 10\ \text{ms}^{-2}\), so \(u^2 = 64\).
First attempt (\(\theta = 30^{\circ}\), \(2\theta = 60^{\circ}\), \(\sin 60^{\circ} = 0.8660\)):
\[ R_1 = \frac{64 \times 0.8660}{10} = \frac{55.425}{10} = 5.5425\ \text{m} \]Second attempt (\(\theta = 45^{\circ}\), \(2\theta = 90^{\circ}\), \(\sin 90^{\circ} = 1\)):
\[ R_2 = \frac{64 \times 1}{10} = 6.4\ \text{m} \]Gain in distance:
\[ R_2 - R_1 = 6.4 - 5.5425 = 0.8575 \approx 0.8576\ \text{m} \]Hence he gains about \(0.8576\ \text{m}\) in the second attempt, as required.
Question 9 Report
(a) List two properties of cathode rays
(b) Explain how the intensity and energy of cathode rays may be increased
(a) Two properties of cathode rays:
- They travel in straight lines and cast sharp shadows of objects placed in their path.
- They are negatively charged, so they are deflected by both electric and magnetic fields.
- They possess kinetic energy (produce heat) and cause certain substances to fluoresce; they travel from cathode to anode.
(b) Increasing the intensity and energy of cathode rays:
- Intensity (the number of electrons emitted per second) is increased by raising the temperature of the cathode, i.e. increasing the heating (filament) current, so that more electrons are emitted by thermionic emission.
- Energy (the speed/kinetic energy of the electrons) is increased by raising the accelerating potential difference between the cathode and the anode, since each electron gains kinetic energy \( \tfrac{1}{2}mv^2 = eV\).
Answer Details
(a) Two properties of cathode rays:
- They travel in straight lines and cast sharp shadows of objects placed in their path.
- They are negatively charged, so they are deflected by both electric and magnetic fields.
- They possess kinetic energy (produce heat) and cause certain substances to fluoresce; they travel from cathode to anode.
(b) Increasing the intensity and energy of cathode rays:
- Intensity (the number of electrons emitted per second) is increased by raising the temperature of the cathode, i.e. increasing the heating (filament) current, so that more electrons are emitted by thermionic emission.
- Energy (the speed/kinetic energy of the electrons) is increased by raising the accelerating potential difference between the cathode and the anode, since each electron gains kinetic energy \( \tfrac{1}{2}mv^2 = eV\).
Question 10 Report
A Spiral spring of natural length 20.00cm has a scale hanging freely in its lower end. When an object ass 40 g is placed in the pan. its length becomes cm. When the object is replaced with another of 60g, the length becomes 22.05cm. Calculate the mass scale pan. [g = 10 ms\(^{-2}\)]
Using Hooke’s law: extension of the spring is proportional to the total mass attached.
For the 40 g mass, the spring length is \(21.80\text{ cm}\), hence extension:
\[ e_1 = 21.80 - 20.00 = 1.80\text{ cm} \] \[ (m + 40)g = 1.80k \qquad \text{... (i)} \]For the 60 g mass, the extension is:
\[ e_2 = 22.05 - 20.00 = 2.05\text{ cm} \] \[ (m + 60)g = 2.05k \qquad \text{... (ii)} \]Dividing equation (i) by equation (ii):
\[ \frac{m+40}{m+60}=\frac{1.80}{2.05} \] \[ 1.80(m+60)=2.05(m+40) \] \[ 1.80m+108=2.05m+82 \] \[ 26=0.25m \] \[ m=104\text{ g} \]Mass of the scale pan \(= 104\text{ g}\).
Answer Details
Using Hooke’s law: extension of the spring is proportional to the total mass attached.
For the 40 g mass, the spring length is \(21.80\text{ cm}\), hence extension:
\[ e_1 = 21.80 - 20.00 = 1.80\text{ cm} \] \[ (m + 40)g = 1.80k \qquad \text{... (i)} \]For the 60 g mass, the extension is:
\[ e_2 = 22.05 - 20.00 = 2.05\text{ cm} \] \[ (m + 60)g = 2.05k \qquad \text{... (ii)} \]Dividing equation (i) by equation (ii):
\[ \frac{m+40}{m+60}=\frac{1.80}{2.05} \] \[ 1.80(m+60)=2.05(m+40) \] \[ 1.80m+108=2.05m+82 \] \[ 26=0.25m \] \[ m=104\text{ g} \]Mass of the scale pan \(= 104\text{ g}\).
Question 11 Report
(a) State two;
(i) properties of x-rays
(ii) reasons to show that x-rays are waves
(iii) uses of x-rays other than those in medicine;
(iv) hazards of x-rays.
(b) The potential difference between the cathode and target of an x-ray tube is 5.00 x 10\(^4\)V and the current in the tube is 2.00 x 10\(^{-2}\)A. Given that only one percent of the total energy supplied is emitted as x-radiation, determine the ospheric pressure.
(i) maximum frequency of the emitted radiation
(ii) rate at which heat is removed from the target in order to keep it at steady temperature. [Planck's constant, h = 6.63 x 10\(^{-34}\) Js, electronic charge e = 1.60 x 10\(^{-19}\) C]
Answer Details
None
Question 12 Report
(a) Explain;
(i) wave motion
(ii) stationary wave
(b)(i) List four physical properties of a wave
(ii) Define amplitude and use it to distinguish between the node and antinode of a stationary wave
(iii) List the factors on which the frequency of vibration in a stretched string depends
(c) The equation, y = 5 sin (3x - 4t), where y is in millimeters, x is in meres and t is in seconds represents a wave motion . Determine the;
(i) frequency
(ii) period
(iii) speed of the wave
(a)(i) Wave motion: a disturbance that travels through a medium (or space) transferring energy from one point to another without any bulk transfer of the particles of the medium; the particles merely vibrate about their fixed mean positions.
(ii) Stationary (standing) wave: a wave formed when two progressive waves of the same frequency and amplitude travelling in opposite directions superpose; it has fixed points of zero amplitude (nodes) and points of maximum amplitude (antinodes), and the wave profile does not advance.
(b)(i) Four physical properties of a wave: amplitude, wavelength, frequency (or period), and speed (velocity).
(ii) Amplitude is the maximum displacement of a particle from its equilibrium (mean) position. In a stationary wave, a node is a point of zero amplitude (the particles do not move), while an antinode is a point of maximum amplitude.
(iii) The frequency of vibration of a stretched string depends on its length, the tension in the string, and its mass per unit length (linear density).
(c) Comparing \(y = 5\sin(3x - 4t)\) with \(y = A\sin(kx - \omega t)\): \(A = 5\ \text{mm}\), \(k = 3\ \text{m}^{-1}\), \(\omega = 4\ \text{rad s}^{-1}\).
(i) Frequency:
\[ f = \frac{\omega}{2\pi} = \frac{4}{2\pi} = 0.637\ \text{Hz} \](ii) Period:
\[ T = \frac{1}{f} = \frac{2\pi}{\omega} = \frac{2\pi}{4} = 1.57\ \text{s} \](iii) Speed:
\[ v = \frac{\omega}{k} = \frac{4}{3} = 1.33\ \text{ms}^{-1} \](Equivalently \(v = f\lambda\), with \(\lambda = \tfrac{2\pi}{k} = 2.09\ \text{m}\).)
Answer Details
(a)(i) Wave motion: a disturbance that travels through a medium (or space) transferring energy from one point to another without any bulk transfer of the particles of the medium; the particles merely vibrate about their fixed mean positions.
(ii) Stationary (standing) wave: a wave formed when two progressive waves of the same frequency and amplitude travelling in opposite directions superpose; it has fixed points of zero amplitude (nodes) and points of maximum amplitude (antinodes), and the wave profile does not advance.
(b)(i) Four physical properties of a wave: amplitude, wavelength, frequency (or period), and speed (velocity).
(ii) Amplitude is the maximum displacement of a particle from its equilibrium (mean) position. In a stationary wave, a node is a point of zero amplitude (the particles do not move), while an antinode is a point of maximum amplitude.
(iii) The frequency of vibration of a stretched string depends on its length, the tension in the string, and its mass per unit length (linear density).
(c) Comparing \(y = 5\sin(3x - 4t)\) with \(y = A\sin(kx - \omega t)\): \(A = 5\ \text{mm}\), \(k = 3\ \text{m}^{-1}\), \(\omega = 4\ \text{rad s}^{-1}\).
(i) Frequency:
\[ f = \frac{\omega}{2\pi} = \frac{4}{2\pi} = 0.637\ \text{Hz} \](ii) Period:
\[ T = \frac{1}{f} = \frac{2\pi}{\omega} = \frac{2\pi}{4} = 1.57\ \text{s} \](iii) Speed:
\[ v = \frac{\omega}{k} = \frac{4}{3} = 1.33\ \text{ms}^{-1} \](Equivalently \(v = f\lambda\), with \(\lambda = \tfrac{2\pi}{k} = 2.09\ \text{m}\).)
Question 13 Report
(a) (i) Explain electromotive force
(ii) list two sources of electromotive force other than a chemical cell
(b) A chemical cell of electromotive force, E, and internal resistance, r, is connected in series with an ammeter, a plug key a plug key and an external load of resistance R. A volumeter is connected across the cell. Draw a circuit diagram to illustrate the arrangement,
(c) for the arrangement in (b) above, with the key opened and closed, the voltmeter readings are V\(_o\) and V respectively.
(i) Explain the physical meanings of V\(_o\) and V
(ii) Find an expression for the (I) current passing through the circuit (II) maximum power dissipated in the cell and external load respectively; (III) efficiency of the cell
(a)(i) Electromotive force (e.m.f.): the total energy supplied by a source in driving one coulomb of charge completely round the whole circuit (both the external load and the internal resistance). Equivalently, it is the terminal potential difference of the source when no current is being drawn from it (open circuit). It is measured in volts (V).
(a)(ii) Two sources of e.m.f. other than a chemical cell:
- a dynamo / electrical generator (electromagnetic induction);
- a thermocouple (thermoelectric effect).
(A solar cell / photocell is also acceptable.)
(b) Circuit diagram. The cell (e.m.f. \(E\), internal resistance \(r\)) is joined in series with a plug key, an ammeter and the external load of resistance \(R\), while the voltmeter is connected across the terminals of the cell:
(c)(i) Physical meaning of \(V_o\) and \(V\):
- With the key open, no current flows, so there is no voltage drop across the internal resistance \(r\). The voltmeter reading \(V_o\) is therefore the full e.m.f. of the cell, \(V_o = E\).
- With the key closed, a current \(I\) flows and a voltage \(Ir\) is lost inside the cell. The voltmeter reading \(V\) is then the terminal potential difference of the cell, \(V = E - Ir\), which is less than \(V_o\).
(c)(ii)(I) Current in the circuit. Applying the circuit equation \(E = I(R+r)\):
\[ I = \frac{E}{R + r} \](c)(ii)(II) Maximum power dissipated. The power delivered to the external load is
\[ P_R = I^2 R = \frac{E^2 R}{(R + r)^2}. \]Differentiating \(P_R\) with respect to \(R\) and setting \(\dfrac{dP_R}{dR}=0\) shows that this is greatest when the load is matched to the cell, i.e. when \(R = r\) (the maximum-power-transfer condition). Substituting \(R = r\):
\[ P_{R(\max)} = \frac{E^2 r}{(r + r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}. \]At this same condition the power dissipated inside the cell is \(P_r = I^2 r = \dfrac{E^2 r}{(2r)^2} = \dfrac{E^2}{4r}\) as well, so the total maximum power drawn from the cell is
\[ P_{\text{total}} = P_{R(\max)} + P_r = \frac{E^2}{4r} + \frac{E^2}{4r} = \frac{E^2}{2r}. \](c)(ii)(III) Efficiency of the cell. The efficiency is the ratio of the useful power delivered to the external load to the total power produced by the cell:
\[ \eta = \frac{\text{power in } R}{\text{total power}} = \frac{I^2 R}{I^2 (R + r)} = \frac{R}{R + r} \times 100\% = \frac{V}{V_o} \times 100\%. \]In particular, at the maximum-power-transfer condition \(R = r\), the efficiency is only \(\dfrac{r}{r+r}\times100\% = 50\%\): half of the energy is wasted inside the cell when the greatest power is transferred to the load.
Answer Details
(a)(i) Electromotive force (e.m.f.): the total energy supplied by a source in driving one coulomb of charge completely round the whole circuit (both the external load and the internal resistance). Equivalently, it is the terminal potential difference of the source when no current is being drawn from it (open circuit). It is measured in volts (V).
(a)(ii) Two sources of e.m.f. other than a chemical cell:
- a dynamo / electrical generator (electromagnetic induction);
- a thermocouple (thermoelectric effect).
(A solar cell / photocell is also acceptable.)
(b) Circuit diagram. The cell (e.m.f. \(E\), internal resistance \(r\)) is joined in series with a plug key, an ammeter and the external load of resistance \(R\), while the voltmeter is connected across the terminals of the cell:
(c)(i) Physical meaning of \(V_o\) and \(V\):
- With the key open, no current flows, so there is no voltage drop across the internal resistance \(r\). The voltmeter reading \(V_o\) is therefore the full e.m.f. of the cell, \(V_o = E\).
- With the key closed, a current \(I\) flows and a voltage \(Ir\) is lost inside the cell. The voltmeter reading \(V\) is then the terminal potential difference of the cell, \(V = E - Ir\), which is less than \(V_o\).
(c)(ii)(I) Current in the circuit. Applying the circuit equation \(E = I(R+r)\):
\[ I = \frac{E}{R + r} \](c)(ii)(II) Maximum power dissipated. The power delivered to the external load is
\[ P_R = I^2 R = \frac{E^2 R}{(R + r)^2}. \]Differentiating \(P_R\) with respect to \(R\) and setting \(\dfrac{dP_R}{dR}=0\) shows that this is greatest when the load is matched to the cell, i.e. when \(R = r\) (the maximum-power-transfer condition). Substituting \(R = r\):
\[ P_{R(\max)} = \frac{E^2 r}{(r + r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}. \]At this same condition the power dissipated inside the cell is \(P_r = I^2 r = \dfrac{E^2 r}{(2r)^2} = \dfrac{E^2}{4r}\) as well, so the total maximum power drawn from the cell is
\[ P_{\text{total}} = P_{R(\max)} + P_r = \frac{E^2}{4r} + \frac{E^2}{4r} = \frac{E^2}{2r}. \](c)(ii)(III) Efficiency of the cell. The efficiency is the ratio of the useful power delivered to the external load to the total power produced by the cell:
\[ \eta = \frac{\text{power in } R}{\text{total power}} = \frac{I^2 R}{I^2 (R + r)} = \frac{R}{R + r} \times 100\% = \frac{V}{V_o} \times 100\%. \]In particular, at the maximum-power-transfer condition \(R = r\), the efficiency is only \(\dfrac{r}{r+r}\times100\% = 50\%\): half of the energy is wasted inside the cell when the greatest power is transferred to the load.
Question 14 Report
(a) Define gravitational field intensity
(b) In an experiment to determine the acceleration of free-fall due to gravity, g, using a simple pendulum of length I, six different values of I were used to obtain six corresponding values of period T. If a graph of I along the vertical axis is plotted against T\(^2\) on the horizontal axis;
(i) make a sketch to show the nature of the graph,
(ii) write down the equation that relates T, I and g hence obtain an expression for the slope of the graph
(iii) given that the slope of the graph is 0.25, determine the value for g [Take \(\pi\) = 3.142]
(c) A stone, thrown horizontally from the top of a vertical wall with a velocity of 15 ms\(^{-1}\), hits the horizontal ground at a point 45m from the base of the wall. Calculate the
(i) times of light of the stone
(ii) height of the wall [g = 10ms\(^{-2}\)]
(a) Gravitational field intensity at a point is the gravitational force experienced per unit mass placed at that point; \(g = \dfrac{F}{m}\). It is a vector quantity directed towards the centre of the attracting body and is measured in \(\text{N kg}^{-1}\).
(b)(i) Since \(l = \dfrac{g}{4\pi^2}\,T^2\), the graph of \(l\) (vertical axis) against \(T^2\) (horizontal axis) is a straight line passing through the origin with a positive, constant slope:
(b)(ii) For a simple pendulum of length \(l\) oscillating through a small angle, the period is
\[ T = 2\pi \sqrt{\frac{l}{g}} \]Squaring both sides,
\[ T^2 = \frac{4\pi^2}{g}\, l \quad\Rightarrow\quad l = \frac{g}{4\pi^2}\, T^2 \]Comparing with \(l = (\text{slope})\,T^2\), the slope of the \(l\)-against-\(T^2\) graph is
\[ \text{slope} = \frac{g}{4\pi^2} \](b)(iii) Given slope \(= 0.25\):
\[ g = 4\pi^2 \times \text{slope} = 4 \times (3.142)^2 \times 0.25 \] \[ g = 4 \times 9.872 \times 0.25 = 9.87\ \text{ms}^{-2} \](c) Horizontal projection: initial horizontal velocity \(u = 15\ \text{ms}^{-1}\), horizontal range \(R = 45\ \text{m}\), \(g = 10\ \text{ms}^{-2}\).
(c)(i) The horizontal motion is at constant velocity, so the time of flight is
\[ R = u\,t \quad\Rightarrow\quad t = \frac{R}{u} = \frac{45}{15} = 3\ \text{s} \](c)(ii) Vertically the stone starts from rest and falls freely, so the height of the wall is
\[ h = \frac{1}{2}\,g\,t^2 = \frac{1}{2} \times 10 \times 3^2 = \frac{1}{2} \times 10 \times 9 = 45\ \text{m} \]Therefore the time of flight is \(3\ \text{s}\) and the wall is \(45\ \text{m}\) high.
Answer Details
(a) Gravitational field intensity at a point is the gravitational force experienced per unit mass placed at that point; \(g = \dfrac{F}{m}\). It is a vector quantity directed towards the centre of the attracting body and is measured in \(\text{N kg}^{-1}\).
(b)(i) Since \(l = \dfrac{g}{4\pi^2}\,T^2\), the graph of \(l\) (vertical axis) against \(T^2\) (horizontal axis) is a straight line passing through the origin with a positive, constant slope:
(b)(ii) For a simple pendulum of length \(l\) oscillating through a small angle, the period is
\[ T = 2\pi \sqrt{\frac{l}{g}} \]Squaring both sides,
\[ T^2 = \frac{4\pi^2}{g}\, l \quad\Rightarrow\quad l = \frac{g}{4\pi^2}\, T^2 \]Comparing with \(l = (\text{slope})\,T^2\), the slope of the \(l\)-against-\(T^2\) graph is
\[ \text{slope} = \frac{g}{4\pi^2} \](b)(iii) Given slope \(= 0.25\):
\[ g = 4\pi^2 \times \text{slope} = 4 \times (3.142)^2 \times 0.25 \] \[ g = 4 \times 9.872 \times 0.25 = 9.87\ \text{ms}^{-2} \](c) Horizontal projection: initial horizontal velocity \(u = 15\ \text{ms}^{-1}\), horizontal range \(R = 45\ \text{m}\), \(g = 10\ \text{ms}^{-2}\).
(c)(i) The horizontal motion is at constant velocity, so the time of flight is
\[ R = u\,t \quad\Rightarrow\quad t = \frac{R}{u} = \frac{45}{15} = 3\ \text{s} \](c)(ii) Vertically the stone starts from rest and falls freely, so the height of the wall is
\[ h = \frac{1}{2}\,g\,t^2 = \frac{1}{2} \times 10 \times 3^2 = \frac{1}{2} \times 10 \times 9 = 45\ \text{m} \]Therefore the time of flight is \(3\ \text{s}\) and the wall is \(45\ \text{m}\) high.
Question 15 Report
A parallel beam of unpolarised light is incident on a plane glass surface at an angle of 58\(^{o}\) to the normal. If the reflected beam is completely polarised, calculate the refractive index of the glass.
When the reflected beam is completely polarised, the angle of incidence is the polarising (Brewster) angle \(\theta_p = 58^{\circ}\), and by Brewster's law:
\[ n = \tan \theta_p = \tan 58^{\circ} \] \[ n = 1.60 \]The refractive index of the glass is about \(1.60\).
Answer Details
When the reflected beam is completely polarised, the angle of incidence is the polarising (Brewster) angle \(\theta_p = 58^{\circ}\), and by Brewster's law:
\[ n = \tan \theta_p = \tan 58^{\circ} \] \[ n = 1.60 \]The refractive index of the glass is about \(1.60\).
