WAEC SSCE - Physics - 2015 (Essay)
Ajụjụ 1 Ripọtì
The circuit above consists of an a. c. voltage input, a diode, a resistor and a voltmeter.
(a) Identify the circuit.
(b) Draw the waveform for the output voltage.
(a) Identification of the circuit
An a.c. supply feeding a single diode in series with a load resistor, with a voltmeter connected across the resistor, is a half-wave rectifier. The diode conducts only when it is forward-biased. During the positive half-cycle of the a.c. input the diode conducts and the current passes through the resistor, so the input voltage appears across it. During the negative half-cycle the diode is reverse-biased and blocks the current, so the output falls to zero. Only one half of every complete a.c. cycle therefore reaches the load, which is why the arrangement is called a half-wave rectifier.
(b) Waveform for the output voltage
The output consists of the positive half-cycles only. During each positive half-cycle the voltage follows the sine curve of the input, rising to a peak value \(V_{p}\) and returning to zero; during each negative half-cycle the diode does not conduct, so the output stays at zero, leaving a flat gap before the next positive hump appears. The graph below shows the sinusoidal a.c. input (dashed) together with the resulting half-wave rectified output (solid) over two complete cycles.
Key features of the output waveform:
- It is a series of separated positive half-sine humps (the negative half-cycles are removed).
- The peak of each hump equals the peak input voltage \(V_{p}\).
- Between successive humps the output is zero for a full half-cycle.
- The output is a pulsating d.c. (unidirectional) voltage with the same period as the a.c. input.
Akọwa Nkọwa
(a) Identification of the circuit
An a.c. supply feeding a single diode in series with a load resistor, with a voltmeter connected across the resistor, is a half-wave rectifier. The diode conducts only when it is forward-biased. During the positive half-cycle of the a.c. input the diode conducts and the current passes through the resistor, so the input voltage appears across it. During the negative half-cycle the diode is reverse-biased and blocks the current, so the output falls to zero. Only one half of every complete a.c. cycle therefore reaches the load, which is why the arrangement is called a half-wave rectifier.
(b) Waveform for the output voltage
The output consists of the positive half-cycles only. During each positive half-cycle the voltage follows the sine curve of the input, rising to a peak value \(V_{p}\) and returning to zero; during each negative half-cycle the diode does not conduct, so the output stays at zero, leaving a flat gap before the next positive hump appears. The graph below shows the sinusoidal a.c. input (dashed) together with the resulting half-wave rectified output (solid) over two complete cycles.
Key features of the output waveform:
- It is a series of separated positive half-sine humps (the negative half-cycles are removed).
- The peak of each hump equals the peak input voltage \(V_{p}\).
- Between successive humps the output is zero for a full half-cycle.
- The output is a pulsating d.c. (unidirectional) voltage with the same period as the a.c. input.
Ajụjụ 2 Ripọtì
(a) Define specific latent heat of vaporization.
(b) (i) What are renewable energy sources?
(ii) List four renewable energy sources.
(c) Explain why tomatoes keep longer when kept in a moist jute bag in a clay pot.
(d) A box has a volume of 0.28 m\(^3\) and is 70% filled with iron fillings at 25°C. Calculate the:
(i) total mass of the iron fillings;
(ii) energy required to melt 10% of the iron fillings. [Density of iron = 8.00 x 10\(^3\) kgm\(^{-3}\); specific latent heat of fusion of iron = 1.38 x 10\(^5\) Jkg\(^{-1}\), specific heat capacity of iron = 460 Jkg\(^{-1}\) k\(^{-1}\), melting point of iron = 1500°C] 10.
(a) Specific latent heat of vaporisation is the quantity of heat required to change unit mass of a liquid into vapour at constant temperature (its boiling point).
(b)(i) Renewable energy sources are sources of energy that are naturally replenished and are not used up (do not run out) when used.
(b)(ii) Four renewable sources: solar energy, wind energy, hydro (water/tidal) energy, and biomass (biogas). Geothermal energy is also acceptable.
(c) The moist jute bag round a clay (earthenware) pot stays wet; water continually evaporates from its surface. Evaporation takes latent heat from the pot and its contents, so the tomatoes inside are cooled and kept below the surrounding temperature, slowing ripening and decay, so they keep longer.
(d) Volume of iron fillings \(= 70\% \times 0.28 = 0.196\ \text{m}^3\).
(i) Total mass \(= \rho V = 8.00\times10^{3} \times 0.196 = 1568\ \text{kg} \approx 1.57\times10^{3}\ \text{kg}.\)
(ii) Mass to be melted \(= 10\% \times 1568 = 156.8\ \text{kg}.\) It must first be raised from 25°C to the melting point 1500°C, then melted:
\[ Q = mc\,\Delta\theta + mL_f. \] \[ mc\,\Delta\theta = 156.8 \times 460 \times (1500-25) = 156.8 \times 460 \times 1475 = 1.06\times10^{8}\ \text{J}. \] \[ mL_f = 156.8 \times 1.38\times10^{5} = 2.16\times10^{7}\ \text{J}. \] \[ Q = 1.06\times10^{8} + 2.16\times10^{7} = 1.28\times10^{8}\ \text{J}. \]Energy required \(\approx 1.28\times10^{8}\ \text{J}\).
Akọwa Nkọwa
(a) Specific latent heat of vaporisation is the quantity of heat required to change unit mass of a liquid into vapour at constant temperature (its boiling point).
(b)(i) Renewable energy sources are sources of energy that are naturally replenished and are not used up (do not run out) when used.
(b)(ii) Four renewable sources: solar energy, wind energy, hydro (water/tidal) energy, and biomass (biogas). Geothermal energy is also acceptable.
(c) The moist jute bag round a clay (earthenware) pot stays wet; water continually evaporates from its surface. Evaporation takes latent heat from the pot and its contents, so the tomatoes inside are cooled and kept below the surrounding temperature, slowing ripening and decay, so they keep longer.
(d) Volume of iron fillings \(= 70\% \times 0.28 = 0.196\ \text{m}^3\).
(i) Total mass \(= \rho V = 8.00\times10^{3} \times 0.196 = 1568\ \text{kg} \approx 1.57\times10^{3}\ \text{kg}.\)
(ii) Mass to be melted \(= 10\% \times 1568 = 156.8\ \text{kg}.\) It must first be raised from 25°C to the melting point 1500°C, then melted:
\[ Q = mc\,\Delta\theta + mL_f. \] \[ mc\,\Delta\theta = 156.8 \times 460 \times (1500-25) = 156.8 \times 460 \times 1475 = 1.06\times10^{8}\ \text{J}. \] \[ mL_f = 156.8 \times 1.38\times10^{5} = 2.16\times10^{7}\ \text{J}. \] \[ Q = 1.06\times10^{8} + 2.16\times10^{7} = 1.28\times10^{8}\ \text{J}. \]Energy required \(\approx 1.28\times10^{8}\ \text{J}\).
Ajụjụ 3 Ripọtì
(a) What is doping in a semi conductor?
(b) Draw the symbol for OR gate.
(a) Doping in a semiconductor
Doping is the deliberate addition of a very small, controlled amount of impurity to a pure semiconductor, such as silicon, to increase the number of charge carriers and therefore increase its electrical conductivity.
- A pentavalent impurity provides extra electrons, producing an n-type semiconductor.
- A trivalent impurity produces holes, producing a p-type semiconductor.
(b) Symbol for an OR gate
An OR gate gives an output of 1 when input A, input B, or both inputs are 1: \(Q=A\text{ OR }B\).
Examination reminder: Do not confuse an OR gate with an AND gate. The OR-gate symbol has a curved, inward left-hand side and a pointed curved right-hand side. Its output is 1 if at least one input is 1.
Akọwa Nkọwa
(a) Doping in a semiconductor
Doping is the deliberate addition of a very small, controlled amount of impurity to a pure semiconductor, such as silicon, to increase the number of charge carriers and therefore increase its electrical conductivity.
- A pentavalent impurity provides extra electrons, producing an n-type semiconductor.
- A trivalent impurity produces holes, producing a p-type semiconductor.
(b) Symbol for an OR gate
An OR gate gives an output of 1 when input A, input B, or both inputs are 1: \(Q=A\text{ OR }B\).
Examination reminder: Do not confuse an OR gate with an AND gate. The OR-gate symbol has a curved, inward left-hand side and a pointed curved right-hand side. Its output is 1 if at least one input is 1.
Ajụjụ 4 Ripọtì
State three ways of increasing the rate of cooling of a cup of hot tea.
Three ways of increasing the rate of cooling of a cup of hot tea:
- Increase the exposed surface area, for example by pouring the tea into a wider, shallower cup or into a saucer, so that more evaporation and heat loss can take place.
- Blow air across the surface (or place it in a draught/fan), which speeds up evaporation and convection by removing the warm layer of air/vapour above the tea.
- Lower the temperature of the surroundings, for example by placing the cup in a cool place or standing it in cold water, so that the temperature difference driving heat loss is larger.
(Stirring the tea and using an uncovered, poorly lagged cup also increase the cooling rate.)
Akọwa Nkọwa
Three ways of increasing the rate of cooling of a cup of hot tea:
- Increase the exposed surface area, for example by pouring the tea into a wider, shallower cup or into a saucer, so that more evaporation and heat loss can take place.
- Blow air across the surface (or place it in a draught/fan), which speeds up evaporation and convection by removing the warm layer of air/vapour above the tea.
- Lower the temperature of the surroundings, for example by placing the cup in a cool place or standing it in cold water, so that the temperature difference driving heat loss is larger.
(Stirring the tea and using an uncovered, poorly lagged cup also increase the cooling rate.)
Ajụjụ 5 Ripọtì
(a) Define nucleon number
(b) A radioactive isotope of Americium (Am —241) decays into a nucleus of Neptunium (Np — 237) and an alpha (\(\alpha\)) particle as shown in the nuclear equation below. \(^{241}_{95}Am\) \(\to\) \(^{237}_{c} + ^b_a \alpha\)
(i) State the number of neutrons in the nucleus of Americium — 241.
(ii) Determine the values of a,b and c.
(c)(i) Why are y-rays not deflected by electromagnetic field?
(ii) State two properties of gamma rays that make them suitable for sterilizing medical equipment.
(d) A sample of radioactive substance was found to be left with h of its initial count rate after 110 years. Calculate its decay constant.
(a) Nucleon number (mass number) is the total number of protons and neutrons contained in the nucleus of an atom.
(b) The decay is \({}^{241}_{95}\text{Am} \to {}^{237}_{c}\text{Np} + {}^{b}_{a}\alpha\).
(i) Number of neutrons in \(\text{Am}\text{-}241 = \text{mass number} - \text{proton number} = 241 - 95 = 146\).
(ii) An alpha particle is a helium nucleus, \({}^{4}_{2}\text{He}\), so \(a = 2\) and \(b = 4\).
Conserving proton (atomic) number: \(95 = c + 2 \Rightarrow c = 93\).
Check with mass number: \(241 = 237 + 4\), which balances. Hence \(a = 2,\ b = 4,\ c = 93\).
(c)(i) Gamma rays are not deflected by an electric or magnetic field because they are electromagnetic waves and carry no electric charge; only charged particles experience a deflecting force in such fields.
(c)(ii) Two properties that make gamma rays suitable for sterilising medical equipment:
- They have very high penetrating power, so they can reach and destroy micro-organisms even inside sealed or packaged equipment.
- They are strongly lethal to bacteria and other micro-organisms, and they do so without leaving any residue or making the equipment radioactive.
(d) The sample is left with \(\dfrac{1}{32}\) of its initial count rate after \(110\) years. Writing the fraction as a power of one half:
\[\frac{N}{N_0} = \frac{1}{32} = \left(\frac{1}{2}\right)^{5},\]so \(n = 5\) half-lives have elapsed. The half-life is
\[t_{1/2} = \frac{\text{total time}}{n} = \frac{110}{5} = 22\ \text{years}.\]The decay constant is
\[\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{22} = 3.15\times10^{-2}\ \text{year}^{-1}\ (\approx 3.2\times10^{-2}\ \text{year}^{-1}).\]Examination note: the neutron count is always mass number minus proton number, so use \(241 - 95 = 146\), not \(241 - 90\). For the half-life step, express the remaining fraction as \(\left(\tfrac{1}{2}\right)^{n}\) to read off the number of half-lives before using \(\lambda = \dfrac{\ln 2}{t_{1/2}}\).
Akọwa Nkọwa
(a) Nucleon number (mass number) is the total number of protons and neutrons contained in the nucleus of an atom.
(b) The decay is \({}^{241}_{95}\text{Am} \to {}^{237}_{c}\text{Np} + {}^{b}_{a}\alpha\).
(i) Number of neutrons in \(\text{Am}\text{-}241 = \text{mass number} - \text{proton number} = 241 - 95 = 146\).
(ii) An alpha particle is a helium nucleus, \({}^{4}_{2}\text{He}\), so \(a = 2\) and \(b = 4\).
Conserving proton (atomic) number: \(95 = c + 2 \Rightarrow c = 93\).
Check with mass number: \(241 = 237 + 4\), which balances. Hence \(a = 2,\ b = 4,\ c = 93\).
(c)(i) Gamma rays are not deflected by an electric or magnetic field because they are electromagnetic waves and carry no electric charge; only charged particles experience a deflecting force in such fields.
(c)(ii) Two properties that make gamma rays suitable for sterilising medical equipment:
- They have very high penetrating power, so they can reach and destroy micro-organisms even inside sealed or packaged equipment.
- They are strongly lethal to bacteria and other micro-organisms, and they do so without leaving any residue or making the equipment radioactive.
(d) The sample is left with \(\dfrac{1}{32}\) of its initial count rate after \(110\) years. Writing the fraction as a power of one half:
\[\frac{N}{N_0} = \frac{1}{32} = \left(\frac{1}{2}\right)^{5},\]so \(n = 5\) half-lives have elapsed. The half-life is
\[t_{1/2} = \frac{\text{total time}}{n} = \frac{110}{5} = 22\ \text{years}.\]The decay constant is
\[\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{22} = 3.15\times10^{-2}\ \text{year}^{-1}\ (\approx 3.2\times10^{-2}\ \text{year}^{-1}).\]Examination note: the neutron count is always mass number minus proton number, so use \(241 - 95 = 146\), not \(241 - 90\). For the half-life step, express the remaining fraction as \(\left(\tfrac{1}{2}\right)^{n}\) to read off the number of half-lives before using \(\lambda = \dfrac{\ln 2}{t_{1/2}}\).
Ajụjụ 6 Ripọtì
(a) Write the de Broglie equation.
(b) Explain the significance of the equation.
(a) de Broglie equation:
\[ \lambda = \frac{h}{p} = \frac{h}{mv}, \]where \(\lambda\) is the wavelength associated with a particle, \(h\) is Planck's constant, \(p = mv\) is the momentum, \(m\) the mass and \(v\) the velocity of the particle.
(b) Significance: the equation expresses the wave-particle duality of matter. It shows that every moving particle (such as an electron) has a wavelength and can therefore behave like a wave, exhibiting effects such as diffraction and interference. The wavelength is inversely proportional to momentum, so only very small (light, slow-enough) particles have wavelengths large enough for wave behaviour to be observable; ordinary large bodies have wavelengths far too small to detect. This idea underlies electron diffraction and the electron microscope.
Akọwa Nkọwa
(a) de Broglie equation:
\[ \lambda = \frac{h}{p} = \frac{h}{mv}, \]where \(\lambda\) is the wavelength associated with a particle, \(h\) is Planck's constant, \(p = mv\) is the momentum, \(m\) the mass and \(v\) the velocity of the particle.
(b) Significance: the equation expresses the wave-particle duality of matter. It shows that every moving particle (such as an electron) has a wavelength and can therefore behave like a wave, exhibiting effects such as diffraction and interference. The wavelength is inversely proportional to momentum, so only very small (light, slow-enough) particles have wavelengths large enough for wave behaviour to be observable; ordinary large bodies have wavelengths far too small to detect. This idea underlies electron diffraction and the electron microscope.
Ajụjụ 7 Ripọtì
(a) Define:(i) the coulomb; (ii) resistance.
(b) State the three effects of an electric current.
(c) State the standard international colour convention for the insulating material covering the following electrical wires in a three-pin plug: (i) live; (ii) neutral; (iii) earth.
(d)
A water melon of mass 5.0 kg is suspended on a uniform rod of mass 4.0 kg and 4.0 in long as illustrated in the diagram above. If the rod is in equilibrium by the action of the force between the charges +q and -q, calculate the:
(i) anti-clockwise moment;
(ii) value of q. [ g = 10 ms\(^{-2}\)]
(a)(i) The coulomb is the quantity of electric charge that passes a point in a circuit when a steady current of one ampere flows for one second: \( 1\ \text{C} = 1\ \text{A}\times1\ \text{s} \).
(a)(ii) Resistance is the opposition offered by a conductor to the flow of electric current through it. It is the ratio of the potential difference across the conductor to the current through it, \( R = \dfrac{V}{I} \), measured in ohms \( (\Omega) \).
(b) Three effects of an electric current:
- Heating (thermal) effect.
- Magnetic effect.
- Chemical effect (electrolysis).
(c) Standard international colour convention:
- (i) Live: brown.
- (ii) Neutral: blue.
- (iii) Earth: green-and-yellow (striped).
(d) From the diagram, the fulcrum divides the uniform rod so that the \( 5.0\ \text{kg} \) watermelon hangs at the left end, a distance \( L_1 = 1.0\ \text{m} \) to the left of the pivot. The rod is \( 4.0\ \text{m} \) long and uniform, so its weight (\( 4.0\ \text{kg} \)) acts at its centre, which is \( L_2 = 1.0\ \text{m} \) to the right of the pivot. The right end carries \( +q \), a distance \( 3.0\ \text{m} \) to the right of the pivot, with \( -q \) fixed \( 2.5\ \text{m} \) directly below it. Take \( g = 10\ \text{m s}^{-2} \).
(i) Anti-clockwise moment (produced by the watermelon at the left end):
\[ W_{\text{melon}} = 5.0\times10 = 50\ \text{N} \]\[ \tau_{\text{acw}} = 50\times1.0 = 50\ \text{N m} \](ii) Value of \( q \): the clockwise moments are produced by the rod's weight and by the attractive electrostatic force \( F \) which pulls the right end downward.
\[ W_{\text{rod}} = 4.0\times10 = 40\ \text{N},\qquad \tau_{\text{rod}} = 40\times1.0 = 40\ \text{N m} \]Taking moments about the pivot (anti-clockwise = clockwise):
\[ 50 = 40 + F\times3.0 \]\[ F = \frac{50-40}{3.0} = \frac{10}{3.0} = 3.33\ \text{N} \]This is the Coulomb force between the two charges \( 2.5\ \text{m} \) apart:
\[ F = \frac{1}{4\pi\varepsilon_0}\frac{q^{2}}{r^{2}} = \frac{9\times10^{9}\,q^{2}}{(2.5)^{2}} \]\[ 3.33 = \frac{9\times10^{9}\,q^{2}}{6.25} \]\[ q^{2} = \frac{3.33\times6.25}{9\times10^{9}} = \frac{20.8}{9\times10^{9}} = 2.31\times10^{-9} \]\[ q = \sqrt{2.31\times10^{-9}} = 4.8\times10^{-5}\ \text{C} \]Therefore \( q \approx 4.8\times10^{-5}\ \text{C}\ (48\ \mu\text{C}) \).
Akọwa Nkọwa
(a)(i) The coulomb is the quantity of electric charge that passes a point in a circuit when a steady current of one ampere flows for one second: \( 1\ \text{C} = 1\ \text{A}\times1\ \text{s} \).
(a)(ii) Resistance is the opposition offered by a conductor to the flow of electric current through it. It is the ratio of the potential difference across the conductor to the current through it, \( R = \dfrac{V}{I} \), measured in ohms \( (\Omega) \).
(b) Three effects of an electric current:
- Heating (thermal) effect.
- Magnetic effect.
- Chemical effect (electrolysis).
(c) Standard international colour convention:
- (i) Live: brown.
- (ii) Neutral: blue.
- (iii) Earth: green-and-yellow (striped).
(d) From the diagram, the fulcrum divides the uniform rod so that the \( 5.0\ \text{kg} \) watermelon hangs at the left end, a distance \( L_1 = 1.0\ \text{m} \) to the left of the pivot. The rod is \( 4.0\ \text{m} \) long and uniform, so its weight (\( 4.0\ \text{kg} \)) acts at its centre, which is \( L_2 = 1.0\ \text{m} \) to the right of the pivot. The right end carries \( +q \), a distance \( 3.0\ \text{m} \) to the right of the pivot, with \( -q \) fixed \( 2.5\ \text{m} \) directly below it. Take \( g = 10\ \text{m s}^{-2} \).
(i) Anti-clockwise moment (produced by the watermelon at the left end):
\[ W_{\text{melon}} = 5.0\times10 = 50\ \text{N} \]\[ \tau_{\text{acw}} = 50\times1.0 = 50\ \text{N m} \](ii) Value of \( q \): the clockwise moments are produced by the rod's weight and by the attractive electrostatic force \( F \) which pulls the right end downward.
\[ W_{\text{rod}} = 4.0\times10 = 40\ \text{N},\qquad \tau_{\text{rod}} = 40\times1.0 = 40\ \text{N m} \]Taking moments about the pivot (anti-clockwise = clockwise):
\[ 50 = 40 + F\times3.0 \]\[ F = \frac{50-40}{3.0} = \frac{10}{3.0} = 3.33\ \text{N} \]This is the Coulomb force between the two charges \( 2.5\ \text{m} \) apart:
\[ F = \frac{1}{4\pi\varepsilon_0}\frac{q^{2}}{r^{2}} = \frac{9\times10^{9}\,q^{2}}{(2.5)^{2}} \]\[ 3.33 = \frac{9\times10^{9}\,q^{2}}{6.25} \]\[ q^{2} = \frac{3.33\times6.25}{9\times10^{9}} = \frac{20.8}{9\times10^{9}} = 2.31\times10^{-9} \]\[ q = \sqrt{2.31\times10^{-9}} = 4.8\times10^{-5}\ \text{C} \]Therefore \( q \approx 4.8\times10^{-5}\ \text{C}\ (48\ \mu\text{C}) \).
Ajụjụ 8 Ripọtì
(a)(i) what is resonance?
(ii) State two eXwilples of resonance.
(iii) Differentiate between loudness and intensity of sound.
(b) When a ray is refracted through a rectangular glass prism, which of the following properties of the ray will change? Wavelength, frequency and speed.
(c)
(i) Copy the diagram above in your answer booklet.
(iii) On the copied diagram, sketch the pattern of the waves immediately after passing through the opening.
(d) A diverging lens of focal length 18.0 in is used to view a shark that is 90.0 in away from the lens. If the image formed is 1.0 m long, calculate the:
(i) image distance;
(ii) length of the shark.
(a)(i) Resonance. Resonance occurs when a body is set into vibration by a periodic force (or by impulses from a nearby vibrating body) whose frequency equals the natural frequency of the body, so that the body vibrates with maximum (very large) amplitude.
(a)(ii) Two examples of resonance.
- The loud sound produced when the vibrating prongs of a tuning fork are held over an air column (resonance tube) whose length is adjusted until the sound is loudest.
- A bridge vibrating strongly when soldiers march across it in step at a frequency equal to the natural frequency of the bridge (sympathetic vibration).
(a)(iii) Loudness versus intensity of sound. Intensity is a measurable physical quantity: the sound energy flowing per second through unit area held perpendicular to the direction of travel (unit \(\text{W m}^{-2}\)); it depends only on physical quantities (energy, time and area). Loudness is the physiological sensation produced in the ear by the sound; it depends on the intensity but also on the sensitivity and response of the individual ear, so it is subjective and not directly measurable.
(b) Refraction through a rectangular glass prism. When light passes from air into glass, the speed changes (it decreases) and the wavelength changes (it decreases in the same ratio). The frequency does not change. So: speed changes, wavelength changes, frequency unchanged.
(c) Diffraction through the wide opening. The given diagram shows straight (plane) wavefronts, drawn as parallel vertical lines, travelling towards a barrier that has a wide opening. Because the width of the opening is much larger than the wavelength, the wavefronts emerging on the far side remain almost straight and parallel across the middle of the gap, and only curve (bend) slightly round the two edges of the opening. The required sketch of the pattern immediately after passing through the opening is shown below.
(If the opening were instead made very narrow, comparable with the wavelength, the emerging waves would spread out strongly as almost circular arcs.)
(d) Diverging lens forming the image of a shark. Data: focal length \(f = -18.0\ \text{cm}\) (negative for a diverging lens), object distance \(u = 90.0\ \text{cm}\), image length \(= 1.0\ \text{m}\).
(d)(i) Image distance. Using the lens formula with \(f\) negative for a diverging lens:
\[ \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \]\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{90} = \frac{-5 - 1}{90} = \frac{-6}{90} = -\frac{1}{15} \]\[ v = -15\ \text{cm} \]The image distance is \(15\ \text{cm}\); the negative sign shows the image is virtual, upright and on the same side as the object, as expected for a diverging lens.
(d)(ii) Length of the shark. The magnification links the image and object sizes to their distances:
\[ m = \left|\frac{v}{u}\right| = \frac{\text{length of image}}{\text{length of shark}} \]\[ \text{length of shark} = \frac{\text{length of image} \times u}{|v|} = \frac{1.0 \times 90}{15} = 6.0\ \text{m} \]The shark is \(6.0\ \text{m}\) long.
Akọwa Nkọwa
(a)(i) Resonance. Resonance occurs when a body is set into vibration by a periodic force (or by impulses from a nearby vibrating body) whose frequency equals the natural frequency of the body, so that the body vibrates with maximum (very large) amplitude.
(a)(ii) Two examples of resonance.
- The loud sound produced when the vibrating prongs of a tuning fork are held over an air column (resonance tube) whose length is adjusted until the sound is loudest.
- A bridge vibrating strongly when soldiers march across it in step at a frequency equal to the natural frequency of the bridge (sympathetic vibration).
(a)(iii) Loudness versus intensity of sound. Intensity is a measurable physical quantity: the sound energy flowing per second through unit area held perpendicular to the direction of travel (unit \(\text{W m}^{-2}\)); it depends only on physical quantities (energy, time and area). Loudness is the physiological sensation produced in the ear by the sound; it depends on the intensity but also on the sensitivity and response of the individual ear, so it is subjective and not directly measurable.
(b) Refraction through a rectangular glass prism. When light passes from air into glass, the speed changes (it decreases) and the wavelength changes (it decreases in the same ratio). The frequency does not change. So: speed changes, wavelength changes, frequency unchanged.
(c) Diffraction through the wide opening. The given diagram shows straight (plane) wavefronts, drawn as parallel vertical lines, travelling towards a barrier that has a wide opening. Because the width of the opening is much larger than the wavelength, the wavefronts emerging on the far side remain almost straight and parallel across the middle of the gap, and only curve (bend) slightly round the two edges of the opening. The required sketch of the pattern immediately after passing through the opening is shown below.
(If the opening were instead made very narrow, comparable with the wavelength, the emerging waves would spread out strongly as almost circular arcs.)
(d) Diverging lens forming the image of a shark. Data: focal length \(f = -18.0\ \text{cm}\) (negative for a diverging lens), object distance \(u = 90.0\ \text{cm}\), image length \(= 1.0\ \text{m}\).
(d)(i) Image distance. Using the lens formula with \(f\) negative for a diverging lens:
\[ \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \]\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{90} = \frac{-5 - 1}{90} = \frac{-6}{90} = -\frac{1}{15} \]\[ v = -15\ \text{cm} \]The image distance is \(15\ \text{cm}\); the negative sign shows the image is virtual, upright and on the same side as the object, as expected for a diverging lens.
(d)(ii) Length of the shark. The magnification links the image and object sizes to their distances:
\[ m = \left|\frac{v}{u}\right| = \frac{\text{length of image}}{\text{length of shark}} \]\[ \text{length of shark} = \frac{\text{length of image} \times u}{|v|} = \frac{1.0 \times 90}{15} = 6.0\ \text{m} \]The shark is \(6.0\ \text{m}\) long.
Ajụjụ 9 Ripọtì
A body is projected at an angle of 30° to the horizontal with a velocity of 150 ms\(^{-1.}\). Calculate the time it takes to reach the greatest height [Take g = 10 ms\(^2\) and neglect air resistance]
At the greatest height the vertical component of velocity is zero. Time to reach the greatest height:
\[ t = \frac{u\sin\theta}{g}. \]With \(u = 150\ \text{m s}^{-1}\), \(\theta = 30^\circ\), \(g = 10\ \text{m s}^{-2}\):
\[ t = \frac{150 \times \sin 30^\circ}{10} = \frac{150 \times 0.5}{10} = \frac{75}{10} = 7.5\ \text{s}. \]Time to reach greatest height \(= 7.5\ \text{s}\).
Akọwa Nkọwa
At the greatest height the vertical component of velocity is zero. Time to reach the greatest height:
\[ t = \frac{u\sin\theta}{g}. \]With \(u = 150\ \text{m s}^{-1}\), \(\theta = 30^\circ\), \(g = 10\ \text{m s}^{-2}\):
\[ t = \frac{150 \times \sin 30^\circ}{10} = \frac{150 \times 0.5}{10} = \frac{75}{10} = 7.5\ \text{s}. \]Time to reach greatest height \(= 7.5\ \text{s}\).
Ajụjụ 10 Ripọtì
(a) (i) Define uniform acceleration.
(ii) Write an equation that relates linear velocity, angular velocity and radius
of path in circular motion.
(b) Two forces 30 N and 40 N act at right angles to each other. Determine by scale drawing, the magnitude and direction of the resultant force,.using a scale of 1 cm to 5 N.
(c) Explain why ships are usually refilled with sand and water after they have been emptied of their cargo.
(d) A crate of drinks of mass 20 kg is placed on a plane inclined at 30° to the horizontal. If the crate slides down with a constant speed, calculate the:
(i) co-efficient of kinetic friction;
(ii) magnitude of the frictional force acting on the crate. [ g = 10 ms\(^{-2}\)]
(a)(i) Uniform acceleration is a constant rate of change of velocity with time; equal changes of velocity occur in equal intervals of time.
(a)(ii) In circular motion the linear (tangential) velocity, angular velocity and radius are related by \( v = \omega r \).
(b) The two forces are perpendicular, so the resultant is the diagonal of the rectangle (right-angled triangle):
\[ R = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \text{N}. \]Direction (angle \(\theta\) the resultant makes with the 40 N force):
\[ \tan\theta = \frac{30}{40} = 0.75 \;\Rightarrow\; \theta = 36.9^\circ. \]By scale drawing (1 cm to 5 N), the 40 N force is 8 cm and the 30 N force is 6 cm at right angles; the diagonal measures 10 cm, i.e. 50 N at about 37° to the 40 N force.
(c) An empty ship floats high, raising its centre of gravity and making it unstable and easily rolled over by waves. Refilling with sand and water (ballast) lowers the centre of gravity and increases stability, and gives the ship enough weight and draught to sit properly in the water so it is not tossed about.
(d) Sliding down at constant speed means the friction force balances the component of weight along the plane.
(i) Coefficient of kinetic friction: for motion at constant speed on an incline, \(\mu = \tan\theta = \tan 30^\circ = 0.577\).
(ii) Frictional force \(= mg\sin\theta = 20 \times 10 \times \sin 30^\circ = 20 \times 10 \times 0.5 = 100\ \text{N}\).
Akọwa Nkọwa
(a)(i) Uniform acceleration is a constant rate of change of velocity with time; equal changes of velocity occur in equal intervals of time.
(a)(ii) In circular motion the linear (tangential) velocity, angular velocity and radius are related by \( v = \omega r \).
(b) The two forces are perpendicular, so the resultant is the diagonal of the rectangle (right-angled triangle):
\[ R = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \text{N}. \]Direction (angle \(\theta\) the resultant makes with the 40 N force):
\[ \tan\theta = \frac{30}{40} = 0.75 \;\Rightarrow\; \theta = 36.9^\circ. \]By scale drawing (1 cm to 5 N), the 40 N force is 8 cm and the 30 N force is 6 cm at right angles; the diagonal measures 10 cm, i.e. 50 N at about 37° to the 40 N force.
(c) An empty ship floats high, raising its centre of gravity and making it unstable and easily rolled over by waves. Refilling with sand and water (ballast) lowers the centre of gravity and increases stability, and gives the ship enough weight and draught to sit properly in the water so it is not tossed about.
(d) Sliding down at constant speed means the friction force balances the component of weight along the plane.
(i) Coefficient of kinetic friction: for motion at constant speed on an incline, \(\mu = \tan\theta = \tan 30^\circ = 0.577\).
(ii) Frictional force \(= mg\sin\theta = 20 \times 10 \times \sin 30^\circ = 20 \times 10 \times 0.5 = 100\ \text{N}\).
Ajụjụ 11 Ripọtì
(a) What is a projectile?
(b) Give the reason why the horizontal component of the velocity of a projectile remains the same at every point of its flight.
(a) A projectile is a body that is given an initial velocity and then moves freely under the action of gravity alone (its own weight), following a curved (parabolic) path. Examples are a thrown ball or a fired bullet.
(b) The horizontal component of the velocity remains constant throughout the flight because, neglecting air resistance, there is no horizontal force acting on the projectile. Gravity acts vertically downward only, so it changes the vertical component of velocity but has no component along the horizontal direction; hence the horizontal velocity is unchanged (zero horizontal acceleration).
Akọwa Nkọwa
(a) A projectile is a body that is given an initial velocity and then moves freely under the action of gravity alone (its own weight), following a curved (parabolic) path. Examples are a thrown ball or a fired bullet.
(b) The horizontal component of the velocity remains constant throughout the flight because, neglecting air resistance, there is no horizontal force acting on the projectile. Gravity acts vertically downward only, so it changes the vertical component of velocity but has no component along the horizontal direction; hence the horizontal velocity is unchanged (zero horizontal acceleration).
Ajụjụ 12 Ripọtì
A wire of length 2.00 m and radius 1.0 mm is stretched by 25.0 mm on application of a force of 103 N. Calculate the Young's modulus for the wire.
Young's modulus \(Y = \dfrac{\text{stress}}{\text{strain}} = \dfrac{F/A}{e/L} = \dfrac{FL}{Ae}\).
Data: \(L = 2.00\ \text{m}\), radius \(r = 1.0\ \text{mm} = 1.0\times10^{-3}\ \text{m}\), extension \(e = 25.0\ \text{mm} = 2.5\times10^{-2}\ \text{m}\), force \(F = 103\ \text{N}\).
Cross-sectional area:
\[ A = \pi r^2 = \pi (1.0\times10^{-3})^2 = 3.14\times10^{-6}\ \text{m}^2. \]Therefore
\[ Y = \frac{FL}{Ae} = \frac{103 \times 2.00}{3.14\times10^{-6} \times 2.5\times10^{-2}} = \frac{206}{7.85\times10^{-8}}. \] \[ Y = 2.62\times10^{9}\ \text{N m}^{-2}. \]Young's modulus \(\approx 2.6\times10^{9}\ \text{N m}^{-2}\).
Akọwa Nkọwa
Young's modulus \(Y = \dfrac{\text{stress}}{\text{strain}} = \dfrac{F/A}{e/L} = \dfrac{FL}{Ae}\).
Data: \(L = 2.00\ \text{m}\), radius \(r = 1.0\ \text{mm} = 1.0\times10^{-3}\ \text{m}\), extension \(e = 25.0\ \text{mm} = 2.5\times10^{-2}\ \text{m}\), force \(F = 103\ \text{N}\).
Cross-sectional area:
\[ A = \pi r^2 = \pi (1.0\times10^{-3})^2 = 3.14\times10^{-6}\ \text{m}^2. \]Therefore
\[ Y = \frac{FL}{Ae} = \frac{103 \times 2.00}{3.14\times10^{-6} \times 2.5\times10^{-2}} = \frac{206}{7.85\times10^{-8}}. \] \[ Y = 2.62\times10^{9}\ \text{N m}^{-2}. \]Young's modulus \(\approx 2.6\times10^{9}\ \text{N m}^{-2}\).
