WAEC SSCE - Physics - 2013 (Essay)
Question 1 Report
(a) Define ionization potential.
(b)(i) State the three types of emission spectra.
(ii) Name one source each which produces each of the spectra stated in (b)(i).
(c) In an x-ray tube, electrons are accelerated the target by a potential difference of 80 A Calculate the:
(i) speed of the electron;
ii) threshold wavelength of the electron. [h=6.6 x 10\(^{-34}\) Js; e = 1.6 x 10\(^{-19}\) C; Me = 9.1 x 10\(^{-31}\)
d) An x-ray photon of frequency 4.5 x 10\(^{-18}\) strikes an. electron, assumed to be at rest. If t electron absorbs all the photon energy, calculate the speed acquired by the electron. [ h = 6.6 x 10\(^{-34}\) Js; Me = 9.1 x 10\(^{-31}\) kg ]
Question 2 Report
The accelerating potential in a cathode ray oscilloscope is 2.5 kV. Calculate the maximum speed of the accelerated electrons. [ e = 1.6 x 10\(^{-19}\) C; Me = 9.1 x 10\(^{-31}\) kg]
An electron accelerated through a potential difference \(V\) gains kinetic energy equal to the electrical work done on it:
\[ eV = \frac{1}{2}m v^{2} \Rightarrow v = \sqrt{\frac{2eV}{m}} \]Data: \(V = 2.5\,\text{kV} = 2.5\times10^{3}\,\text{V}\); \(e = 1.6\times10^{-19}\,\text{C}\); \(m = 9.1\times10^{-31}\,\text{kg}\).
\[ v = \sqrt{\frac{2(1.6\times10^{-19})(2.5\times10^{3})}{9.1\times10^{-31}}} = \sqrt{\frac{8.0\times10^{-16}}{9.1\times10^{-31}}} \] \[ v = \sqrt{8.79\times10^{14}} = 2.97\times10^{7}\,\text{ms}^{-1} \]The maximum speed of the accelerated electrons is about \(3.0\times10^{7}\,\text{ms}^{-1}\).
Answer Details
An electron accelerated through a potential difference \(V\) gains kinetic energy equal to the electrical work done on it:
\[ eV = \frac{1}{2}m v^{2} \Rightarrow v = \sqrt{\frac{2eV}{m}} \]Data: \(V = 2.5\,\text{kV} = 2.5\times10^{3}\,\text{V}\); \(e = 1.6\times10^{-19}\,\text{C}\); \(m = 9.1\times10^{-31}\,\text{kg}\).
\[ v = \sqrt{\frac{2(1.6\times10^{-19})(2.5\times10^{3})}{9.1\times10^{-31}}} = \sqrt{\frac{8.0\times10^{-16}}{9.1\times10^{-31}}} \] \[ v = \sqrt{8.79\times10^{14}} = 2.97\times10^{7}\,\text{ms}^{-1} \]The maximum speed of the accelerated electrons is about \(3.0\times10^{7}\,\text{ms}^{-1}\).
Question 3 Report
The horizontal component of the initial speed of a particle projected at 30° to the horizontal is 50 ms\(^{-1}\). If the acceleration cf free fall due to gravity is 10ms\(^{-2}\), determine its: (a) initial speed; (b) speed at maximum height reached.
The particle is projected at \(30^\circ\) to the horizontal, and its horizontal component of speed is \(u_x = 50\,\text{ms}^{-1}\).
(a) Initial speed \(u\)
The horizontal component is \(u_x = u\cos\theta\), so:
\[ u = \frac{u_x}{\cos\theta} = \frac{50}{\cos 30^\circ} = \frac{50}{0.866} = 57.7\,\text{ms}^{-1} \](b) Speed at maximum height
At the maximum height the vertical component of velocity is zero, so the speed there equals the (unchanged) horizontal component:
\[ v = u\cos\theta = 50\,\text{ms}^{-1} \]Answer Details
The particle is projected at \(30^\circ\) to the horizontal, and its horizontal component of speed is \(u_x = 50\,\text{ms}^{-1}\).
(a) Initial speed \(u\)
The horizontal component is \(u_x = u\cos\theta\), so:
\[ u = \frac{u_x}{\cos\theta} = \frac{50}{\cos 30^\circ} = \frac{50}{0.866} = 57.7\,\text{ms}^{-1} \](b) Speed at maximum height
At the maximum height the vertical component of velocity is zero, so the speed there equals the (unchanged) horizontal component:
\[ v = u\cos\theta = 50\,\text{ms}^{-1} \]Question 4 Report
(a) Define boiling point of a liquid.
(b) Describe how water in a round bottom flask could be made to boil without heating it. [diagram not necessary]
(c) State three applications of expansion of metals.
(d) A room with floor measurements 7m x 10 m contains air of mass 250 kg at a temperature of 34°C. The air is cooled until the temperature falls to 24°C. Calculate the: (i) height of the room;
(ii) quantity of energy extracted to cool the room;
(iii) which is higher: the calculated value or the actual energy needed to cool the room? Give a reason for your answer. [ Specific heat capacity of air = 1010 Jkg\(^{-1}\)K\(^{-1}\); density of air = 1.25 kg m\(^{-3}]
(a) Boiling point
The boiling point of a liquid is the constant temperature at which the liquid boils, i.e. at which its saturated vapour pressure equals the external atmospheric pressure.
(b) Boiling water without heating it
Put warm water in a round-bottomed flask, cork it, and connect it to a vacuum pump. As the pump lowers the pressure above the water, the boiling point falls; when the reduced pressure equals the water's vapour pressure at that temperature, the water boils without any further heating. (Alternatively, boil the water, cork the flask, invert it and pour cold water over the top: the vapour inside condenses, pressure drops, and the water boils again.)
(c) Three applications of expansion of metals
- The bimetallic strip in thermostats and fire alarms.
- Fitting (shrink-fitting) of metal tyres/rims onto wheels and rivets.
- Expansion gaps/rollers left in railway lines and bridges.
(d) Cooling the room air from 34°C to 24°C
Floor area \(= 7\times10 = 70\,\text{m}^{2}\); air mass \(= 250\,\text{kg}\); \(c = 1010\,\text{Jkg}^{-1}\text{K}^{-1}\); density \(= 1.25\,\text{kg m}^{-3}\).
(i) Height of the room: volume of air \(= \dfrac{m}{\rho} = \dfrac{250}{1.25} = 200\,\text{m}^{3}\).
\[ \text{height} = \frac{\text{volume}}{\text{floor area}} = \frac{200}{70} = 2.86\,\text{m} \](ii) Energy extracted:
\[ Q = mc\,\Delta\theta = 250\times1010\times(34-24) = 250\times1010\times10 = 2.525\times10^{6}\,\text{J} \](iii) Which is higher? The actual energy that must be extracted is higher than the calculated value. The calculation accounts only for cooling the air; in practice heat must also be removed from the walls, furniture and occupants, and heat continually leaks in from the warmer surroundings, so more energy is needed than the value found above.
Answer Details
(a) Boiling point
The boiling point of a liquid is the constant temperature at which the liquid boils, i.e. at which its saturated vapour pressure equals the external atmospheric pressure.
(b) Boiling water without heating it
Put warm water in a round-bottomed flask, cork it, and connect it to a vacuum pump. As the pump lowers the pressure above the water, the boiling point falls; when the reduced pressure equals the water's vapour pressure at that temperature, the water boils without any further heating. (Alternatively, boil the water, cork the flask, invert it and pour cold water over the top: the vapour inside condenses, pressure drops, and the water boils again.)
(c) Three applications of expansion of metals
- The bimetallic strip in thermostats and fire alarms.
- Fitting (shrink-fitting) of metal tyres/rims onto wheels and rivets.
- Expansion gaps/rollers left in railway lines and bridges.
(d) Cooling the room air from 34°C to 24°C
Floor area \(= 7\times10 = 70\,\text{m}^{2}\); air mass \(= 250\,\text{kg}\); \(c = 1010\,\text{Jkg}^{-1}\text{K}^{-1}\); density \(= 1.25\,\text{kg m}^{-3}\).
(i) Height of the room: volume of air \(= \dfrac{m}{\rho} = \dfrac{250}{1.25} = 200\,\text{m}^{3}\).
\[ \text{height} = \frac{\text{volume}}{\text{floor area}} = \frac{200}{70} = 2.86\,\text{m} \](ii) Energy extracted:
\[ Q = mc\,\Delta\theta = 250\times1010\times(34-24) = 250\times1010\times10 = 2.525\times10^{6}\,\text{J} \](iii) Which is higher? The actual energy that must be extracted is higher than the calculated value. The calculation accounts only for cooling the air; in practice heat must also be removed from the walls, furniture and occupants, and heat continually leaks in from the warmer surroundings, so more energy is needed than the value found above.
Question 5 Report
(a) On which day would sound wave travel faster: on a hot or cold day? Explain.
(b) Why are megaphones shaped like funnels?
(c) A ray of light is incident on a surface of a ectangular glass prism of refractive index 1.5 illustrated in the diagram below.
(i) Copy the diagram a label the angles of: (\(\alpha\)) Incidence (x); (\(\beta\)) Reflection (y); (\(\gamma\)) refraction (z); with t glass letters indicated.
(ii) Calculate the angle refraction to the nearest whole number.
(d) A sonomesr wire vibrates in simple harmoi motion with a maximum amplitude of 1.0 cm. Calculate the frequency of vibration of the wire, giv that the magnitade of the maximum acceleration of the wire is 980ms\(^{-2}\). [\(\pi \frac{22}{7}\)]
(a) Faster on a hot or cold day?
Sound travels faster on a hot day. The speed of sound in air increases with temperature because the air molecules move faster and transmit the disturbance more rapidly (the speed is proportional to \(\sqrt{T}\), where \(T\) is the absolute temperature).
(b) Why megaphones are shaped like funnels
The funnel (conical) shape channels the sound energy and directs it forward in a narrow beam instead of allowing it to spread out in all directions. This concentrates the sound waves, so the sound is louder and carries farther in the intended direction.
(c) Refraction at the air-glass surface (refractive index \(n = 1.5\), angle of incidence read from the diagram \(= 30^{\circ}\))
(i) In the copied diagram: the angle of incidence (x) is between the incident ray and the normal in the air; the angle of reflection (y) is between the reflected ray and the normal (in air, equal to x); the angle of refraction (z) is between the refracted ray and the normal inside the glass.
(ii) Applying Snell's law from air into glass:
\[ n = \frac{\sin i}{\sin r} \;\Rightarrow\; \sin r = \frac{\sin i}{n} = \frac{\sin 30^{\circ}}{1.5} = \frac{0.5}{1.5} = 0.3333 \]
\[ r = \sin^{-1}(0.3333) = 19.47^{\circ} \approx 19^{\circ} \]
(d) Frequency of the vibrating sonometer wire (SHM)
Amplitude \(A = 1.0\ \text{cm} = 0.01\ \text{m}\); maximum acceleration \(a_{max} = 980\ \text{m s}^{-2}\). For SHM, \(a_{max} = \omega^{2} A\):
\[ \omega^{2} = \frac{a_{max}}{A} = \frac{980}{0.01} = 98000 \]
\[ \omega = \sqrt{98000} = 313.05\ \text{rad s}^{-1} \]
Since \(\omega = 2\pi f\), with \(\pi = \dfrac{22}{7}\):
\[ f = \frac{\omega}{2\pi} = \frac{313.05}{2 \times \tfrac{22}{7}} = \frac{313.05}{6.286} \approx 49.8\ \text{Hz} \approx 50\ \text{Hz} \]
Answer Details
(a) Faster on a hot or cold day?
Sound travels faster on a hot day. The speed of sound in air increases with temperature because the air molecules move faster and transmit the disturbance more rapidly (the speed is proportional to \(\sqrt{T}\), where \(T\) is the absolute temperature).
(b) Why megaphones are shaped like funnels
The funnel (conical) shape channels the sound energy and directs it forward in a narrow beam instead of allowing it to spread out in all directions. This concentrates the sound waves, so the sound is louder and carries farther in the intended direction.
(c) Refraction at the air-glass surface (refractive index \(n = 1.5\), angle of incidence read from the diagram \(= 30^{\circ}\))
(i) In the copied diagram: the angle of incidence (x) is between the incident ray and the normal in the air; the angle of reflection (y) is between the reflected ray and the normal (in air, equal to x); the angle of refraction (z) is between the refracted ray and the normal inside the glass.
(ii) Applying Snell's law from air into glass:
\[ n = \frac{\sin i}{\sin r} \;\Rightarrow\; \sin r = \frac{\sin i}{n} = \frac{\sin 30^{\circ}}{1.5} = \frac{0.5}{1.5} = 0.3333 \]
\[ r = \sin^{-1}(0.3333) = 19.47^{\circ} \approx 19^{\circ} \]
(d) Frequency of the vibrating sonometer wire (SHM)
Amplitude \(A = 1.0\ \text{cm} = 0.01\ \text{m}\); maximum acceleration \(a_{max} = 980\ \text{m s}^{-2}\). For SHM, \(a_{max} = \omega^{2} A\):
\[ \omega^{2} = \frac{a_{max}}{A} = \frac{980}{0.01} = 98000 \]
\[ \omega = \sqrt{98000} = 313.05\ \text{rad s}^{-1} \]
Since \(\omega = 2\pi f\), with \(\pi = \dfrac{22}{7}\):
\[ f = \frac{\omega}{2\pi} = \frac{313.05}{2 \times \tfrac{22}{7}} = \frac{313.05}{6.286} \approx 49.8\ \text{Hz} \approx 50\ \text{Hz} \]
Question 6 Report
(a) What is a polarizer?
(b) With the aid of a diagram, explain how a polarizer can be used to polarize a beam of unpolarized light.
(a) What is a polarizer?
A polarizer is an optical device (for example a Polaroid sheet or a tourmaline crystal) that has a single characteristic direction called its transmission axis. When light falls on it, it transmits only the component of the light vibrations that is parallel to this transmission axis and absorbs (or blocks) all the other components. It therefore converts an unpolarized beam, whose vibrations occur in all planes perpendicular to the direction of travel, into a plane-polarized beam whose vibrations lie in only one plane.
(b) How a polarizer polarizes a beam of unpolarized light
In an unpolarized beam the electric-field vibrations take place in every direction in the plane perpendicular to the direction of propagation. When this beam strikes the polarizer, only the vibration component that is parallel to the polarizer's transmission axis is allowed through; every component perpendicular to that axis is absorbed. The light emerging from the polarizer therefore vibrates in one plane only, that is, it is plane-polarized. Because only one component is transmitted, the emerging intensity is about half that of the incident unpolarized light \(\left(I = \tfrac{1}{2}I_0\right)\).
The action is shown in the diagram below. The unpolarized light (vibrating in all directions) passes through the first polarizer and emerges polarized in the vertical plane. When a second polarizer (the analyser) is placed with its transmission axis crossed at \(90^\circ\) to the first, it blocks these vertical vibrations and no light passes, which confirms that the beam leaving the first polarizer is indeed plane-polarized.
Answer Details
(a) What is a polarizer?
A polarizer is an optical device (for example a Polaroid sheet or a tourmaline crystal) that has a single characteristic direction called its transmission axis. When light falls on it, it transmits only the component of the light vibrations that is parallel to this transmission axis and absorbs (or blocks) all the other components. It therefore converts an unpolarized beam, whose vibrations occur in all planes perpendicular to the direction of travel, into a plane-polarized beam whose vibrations lie in only one plane.
(b) How a polarizer polarizes a beam of unpolarized light
In an unpolarized beam the electric-field vibrations take place in every direction in the plane perpendicular to the direction of propagation. When this beam strikes the polarizer, only the vibration component that is parallel to the polarizer's transmission axis is allowed through; every component perpendicular to that axis is absorbed. The light emerging from the polarizer therefore vibrates in one plane only, that is, it is plane-polarized. Because only one component is transmitted, the emerging intensity is about half that of the incident unpolarized light \(\left(I = \tfrac{1}{2}I_0\right)\).
The action is shown in the diagram below. The unpolarized light (vibrating in all directions) passes through the first polarizer and emerges polarized in the vertical plane. When a second polarizer (the analyser) is placed with its transmission axis crossed at \(90^\circ\) to the first, it blocks these vertical vibrations and no light passes, which confirms that the beam leaving the first polarizer is indeed plane-polarized.
Question 7 Report
A projectile is released with a speed u at an angle \(\theta\) to the horizontal. With the aid of a diagram, show that the time of flight is equal to \(\frac{2uSin\theta}{g}\), where g is the acceleration of free fall.
Projectile motion and time of flight
A body is projected from ground level with speed \(u\) at an angle \(\theta\) to the horizontal. The diagram shows the parabolic path and the initial velocity resolved into its horizontal and vertical components.
Resolving the initial velocity:
Horizontal component: \(u_x = u\cos\theta\) (this stays constant because there is no horizontal force).
Vertical component: \(u_y = u\sin\theta\) (this is decelerated by gravity \(g\) acting downward).
Time to reach maximum height
Taking upward as positive, the vertical velocity at time \(t\) is
\[ v_y = u\sin\theta - g t \]At the maximum height the vertical velocity is momentarily zero, \(v_y = 0\). Let \(t\) be the time taken to reach this point:
\[ 0 = u\sin\theta - g t \]\[ t = \frac{u\sin\theta}{g} \]Time of flight
By the symmetry of the path, the time taken to fall from the maximum height back to the projection level equals the time taken to rise to it. Hence the total time of flight \(T\) is twice the time to reach maximum height:
\[ T = 2t = \frac{2u\sin\theta}{g} \]This is the required result: the time of flight \(T = \dfrac{2u\sin\theta}{g}\), where \(g\) is the acceleration of free fall.
Answer Details
Projectile motion and time of flight
A body is projected from ground level with speed \(u\) at an angle \(\theta\) to the horizontal. The diagram shows the parabolic path and the initial velocity resolved into its horizontal and vertical components.
Resolving the initial velocity:
Horizontal component: \(u_x = u\cos\theta\) (this stays constant because there is no horizontal force).
Vertical component: \(u_y = u\sin\theta\) (this is decelerated by gravity \(g\) acting downward).
Time to reach maximum height
Taking upward as positive, the vertical velocity at time \(t\) is
\[ v_y = u\sin\theta - g t \]At the maximum height the vertical velocity is momentarily zero, \(v_y = 0\). Let \(t\) be the time taken to reach this point:
\[ 0 = u\sin\theta - g t \]\[ t = \frac{u\sin\theta}{g} \]Time of flight
By the symmetry of the path, the time taken to fall from the maximum height back to the projection level equals the time taken to rise to it. Hence the total time of flight \(T\) is twice the time to reach maximum height:
\[ T = 2t = \frac{2u\sin\theta}{g} \]This is the required result: the time of flight \(T = \dfrac{2u\sin\theta}{g}\), where \(g\) is the acceleration of free fall.
Question 8 Report
In an electrolysis experiment, the ammeter records a steady current of 1 A. The mass of copper deposited in 30 minutes is 0.66 g. Calculate the error in the ammeter reading. [Electrochemical equivalent of copper = 0.00033 g C\(^{-1}\)]
By Faraday's law of electrolysis, the mass deposited is \(m = zIt\), where \(z\) is the electrochemical equivalent, \(I\) the true current and \(t\) the time.
Data: \(m = 0.66\,\text{g}\), \(z = 0.00033\,\text{g C}^{-1}\), \(t = 30\,\text{min} = 1800\,\text{s}\).
True charge that passed:
\[ Q = \frac{m}{z} = \frac{0.66}{0.00033} = 2000\,\text{C} \]True (actual) current:
\[ I_{true} = \frac{Q}{t} = \frac{2000}{1800} = 1.11\,\text{A} \]Error in the ammeter reading:
\[ \text{Error} = I_{true} - I_{reading} = 1.11 - 1.00 = 0.11\,\text{A} \]The ammeter reads about \(0.11\,\text{A}\) too low (a percentage error of about \(\dfrac{0.11}{1.11}\times100 \approx 10\%\)).
Answer Details
By Faraday's law of electrolysis, the mass deposited is \(m = zIt\), where \(z\) is the electrochemical equivalent, \(I\) the true current and \(t\) the time.
Data: \(m = 0.66\,\text{g}\), \(z = 0.00033\,\text{g C}^{-1}\), \(t = 30\,\text{min} = 1800\,\text{s}\).
True charge that passed:
\[ Q = \frac{m}{z} = \frac{0.66}{0.00033} = 2000\,\text{C} \]True (actual) current:
\[ I_{true} = \frac{Q}{t} = \frac{2000}{1800} = 1.11\,\text{A} \]Error in the ammeter reading:
\[ \text{Error} = I_{true} - I_{reading} = 1.11 - 1.00 = 0.11\,\text{A} \]The ammeter reads about \(0.11\,\text{A}\) too low (a percentage error of about \(\dfrac{0.11}{1.11}\times100 \approx 10\%\)).
Question 9 Report
Name the three basic components P, Q and R that make up a cathode ray tube, as illustrated in the diagram above
The diagram is a cathode-ray tube (CRT), and the three labelled sections are its three basic components:
- P - the electron gun: the assembly at the narrow end containing the heated filament/cathode and the anodes (focusing and accelerating electrodes). It produces, focuses and accelerates a narrow beam of electrons.
- Q - the deflecting system (deflecting plates): the pairs of parallel plates (the X- and Y-plates) through which the beam passes. They deflect the electron beam horizontally and vertically by means of the electric fields between the plates.
- R - the fluorescent screen: the flared front face of the tube, coated on the inside with a fluorescent (phosphor) material. It glows to produce a visible spot where the electron beam strikes it.
Answer Details
The diagram is a cathode-ray tube (CRT), and the three labelled sections are its three basic components:
- P - the electron gun: the assembly at the narrow end containing the heated filament/cathode and the anodes (focusing and accelerating electrodes). It produces, focuses and accelerates a narrow beam of electrons.
- Q - the deflecting system (deflecting plates): the pairs of parallel plates (the X- and Y-plates) through which the beam passes. They deflect the electron beam horizontally and vertically by means of the electric fields between the plates.
- R - the fluorescent screen: the flared front face of the tube, coated on the inside with a fluorescent (phosphor) material. It glows to produce a visible spot where the electron beam strikes it.
Question 10 Report
(a) what is Brownian motion?
(b) State the two inferences that can be drawn from Brownian motion experiment.
(a) Brownian motion
Brownian motion is the continuous, random (zig-zag) movement of tiny particles suspended in a fluid (such as smoke particles in air or pollen grains in water), caused by their being bombarded unevenly by the fast-moving molecules of the surrounding fluid.
(b) Two inferences from the Brownian-motion experiment
- Matter (the fluid) is made up of tiny particles (molecules).
- These molecules are in constant, random motion and possess kinetic energy (the motion becomes more vigorous as temperature rises).
Answer Details
(a) Brownian motion
Brownian motion is the continuous, random (zig-zag) movement of tiny particles suspended in a fluid (such as smoke particles in air or pollen grains in water), caused by their being bombarded unevenly by the fast-moving molecules of the surrounding fluid.
(b) Two inferences from the Brownian-motion experiment
- Matter (the fluid) is made up of tiny particles (molecules).
- These molecules are in constant, random motion and possess kinetic energy (the motion becomes more vigorous as temperature rises).
Question 11 Report
(a) Explain briefly the purpose of earthing electrical appliance.
(b) Why does the light frorr bulb connected to a simple cell dim and eventually goes off after a while?
(c) A coil of incidence 0.007 H, a resistor of resistance 8 \(\Omega\) and a capacitor capacitance 0.001 F are connected in series an a.c. source of frequency \(\frac{500}{\pi}\)Hz. If the r.m.s voltages across the coil, the resistor and capacitor are 30v, 20v and 70v respectively;
(i) draw a vector diagram to illustrate the voltage across the components in the circuit.
(ii) Calculate the: (\(\alpha\)) r.m.s voltage of the source
(\(\beta\)) r.m.s current in the circuit;
(\(\gamma\)) power dissipated in the circuit.
iii) write down the sinusoidal equation for the r.m.s voltage, V, in terms of the time, t.
(a) Purpose of earthing an electrical appliance
Earthing means connecting the metal casing of an appliance to the ground through a low-resistance earth wire. Its purpose is safety: if a live wire becomes loose and touches the metal casing (a fault), the casing would otherwise become live and give a fatal shock to anyone who touches it. The earth wire provides a path of very low resistance straight to the ground, so the large fault current flows to earth (rather than through the user's body) and blows the fuse, cutting off the supply. Thus earthing protects the user from electric shock.
(b) Why the bulb of a simple cell dims and goes off
The light produced by a bulb connected to a simple (voltaic) cell gradually dims and finally goes off because of two defects of the simple cell:
- Polarisation: hydrogen gas liberated during the chemical reaction collects as a layer of bubbles on the copper (positive) plate. This layer increases the internal resistance and also sets up a back e.m.f. that opposes the current, so the current falls.
- Local action: impurities in the zinc plate form tiny local cells with the zinc, so the zinc is eaten away (and current is wasted) even when the cell is not in use, weakening the cell.
Together these defects steadily reduce the current, so the bulb dims and eventually goes out.
(c) Series R-L-C circuit on an a.c. source
Given: coil \(L = 0.007\,\text{H}\), resistor \(R = 8\,\Omega\), capacitor \(C = 0.001\,\text{F}\), frequency \(f = \dfrac{500}{\pi}\,\text{Hz}\). The r.m.s voltages are: across the coil \(V_L = 30\,\text{V}\), across the resistor \(V_R = 20\,\text{V}\), across the capacitor \(V_C = 70\,\text{V}\).
(i) Vector (phasor) diagram of the voltages
Taking the current \(I\) as the reference (horizontal) direction: \(V_R\) is in phase with the current, \(V_L\) leads the current by \(90^\circ\) (drawn upward) and \(V_C\) lags the current by \(90^\circ\) (drawn downward). The source voltage \(V\) is the resultant.
(ii) Calculations
(\(\alpha\)) r.m.s voltage of the source
From the phasor diagram, the resultant of the perpendicular components gives
\[ V = \sqrt{V_R^{2} + (V_L - V_C)^{2}} = \sqrt{20^{2} + (30 - 70)^{2}} \]\[ V = \sqrt{400 + 1600} = \sqrt{2000} \approx 44.7\,\text{V} \](\(\beta\)) r.m.s current in the circuit
First find the reactances and the impedance.
\[ X_L = 2\pi f L = 2\pi \times \frac{500}{\pi} \times 0.007 = 1000 \times 0.007 = 7\,\Omega \]\[ X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times \frac{500}{\pi} \times 0.001} = \frac{1}{1000 \times 0.001} = 1\,\Omega \]\[ Z = \sqrt{R^{2} + (X_L - X_C)^{2}} = \sqrt{8^{2} + (7 - 1)^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10\,\Omega \]\[ I = \frac{V}{Z} = \frac{44.7}{10} \approx 4.47\,\text{A} \](\(\gamma\)) power dissipated in the circuit
Power is dissipated only in the resistor:
\[ P = I^{2}R = (4.47)^{2} \times 8 = 20 \times 8 = 160\,\text{W} \](iii) Sinusoidal equation for the source voltage
The peak (maximum) voltage is
\[ V_o = V\sqrt{2} = 44.7 \times \sqrt{2} \approx 63.2\,\text{V} \]The angular frequency is
\[ \omega = 2\pi f = 2\pi \times \frac{500}{\pi} = 1000\,\text{rad s}^{-1} \]Therefore the instantaneous source voltage is
\[ V = V_o \sin(\omega t) = 63.2\,\sin(1000\,t)\ \text{volts} \]Answer Details
(a) Purpose of earthing an electrical appliance
Earthing means connecting the metal casing of an appliance to the ground through a low-resistance earth wire. Its purpose is safety: if a live wire becomes loose and touches the metal casing (a fault), the casing would otherwise become live and give a fatal shock to anyone who touches it. The earth wire provides a path of very low resistance straight to the ground, so the large fault current flows to earth (rather than through the user's body) and blows the fuse, cutting off the supply. Thus earthing protects the user from electric shock.
(b) Why the bulb of a simple cell dims and goes off
The light produced by a bulb connected to a simple (voltaic) cell gradually dims and finally goes off because of two defects of the simple cell:
- Polarisation: hydrogen gas liberated during the chemical reaction collects as a layer of bubbles on the copper (positive) plate. This layer increases the internal resistance and also sets up a back e.m.f. that opposes the current, so the current falls.
- Local action: impurities in the zinc plate form tiny local cells with the zinc, so the zinc is eaten away (and current is wasted) even when the cell is not in use, weakening the cell.
Together these defects steadily reduce the current, so the bulb dims and eventually goes out.
(c) Series R-L-C circuit on an a.c. source
Given: coil \(L = 0.007\,\text{H}\), resistor \(R = 8\,\Omega\), capacitor \(C = 0.001\,\text{F}\), frequency \(f = \dfrac{500}{\pi}\,\text{Hz}\). The r.m.s voltages are: across the coil \(V_L = 30\,\text{V}\), across the resistor \(V_R = 20\,\text{V}\), across the capacitor \(V_C = 70\,\text{V}\).
(i) Vector (phasor) diagram of the voltages
Taking the current \(I\) as the reference (horizontal) direction: \(V_R\) is in phase with the current, \(V_L\) leads the current by \(90^\circ\) (drawn upward) and \(V_C\) lags the current by \(90^\circ\) (drawn downward). The source voltage \(V\) is the resultant.
(ii) Calculations
(\(\alpha\)) r.m.s voltage of the source
From the phasor diagram, the resultant of the perpendicular components gives
\[ V = \sqrt{V_R^{2} + (V_L - V_C)^{2}} = \sqrt{20^{2} + (30 - 70)^{2}} \]\[ V = \sqrt{400 + 1600} = \sqrt{2000} \approx 44.7\,\text{V} \](\(\beta\)) r.m.s current in the circuit
First find the reactances and the impedance.
\[ X_L = 2\pi f L = 2\pi \times \frac{500}{\pi} \times 0.007 = 1000 \times 0.007 = 7\,\Omega \]\[ X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times \frac{500}{\pi} \times 0.001} = \frac{1}{1000 \times 0.001} = 1\,\Omega \]\[ Z = \sqrt{R^{2} + (X_L - X_C)^{2}} = \sqrt{8^{2} + (7 - 1)^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10\,\Omega \]\[ I = \frac{V}{Z} = \frac{44.7}{10} \approx 4.47\,\text{A} \](\(\gamma\)) power dissipated in the circuit
Power is dissipated only in the resistor:
\[ P = I^{2}R = (4.47)^{2} \times 8 = 20 \times 8 = 160\,\text{W} \](iii) Sinusoidal equation for the source voltage
The peak (maximum) voltage is
\[ V_o = V\sqrt{2} = 44.7 \times \sqrt{2} \approx 63.2\,\text{V} \]The angular frequency is
\[ \omega = 2\pi f = 2\pi \times \frac{500}{\pi} = 1000\,\text{rad s}^{-1} \]Therefore the instantaneous source voltage is
\[ V = V_o \sin(\omega t) = 63.2\,\sin(1000\,t)\ \text{volts} \]Question 12 Report
(a) State the triangle law of vector addition.
(b) Name the four physical quantities that are associated with the equationq of linear motion.
(c) Using the same set of axes, sketch and label two graphs to illustrate the variation of potential energy and kinetic energy with time for a body in simple harmonic motion.
(d)
A light spiral spring of force constant K lies on a horizontal frictionless surface and has one end fixed to a vertical wall. A block P of mass 2.0 kg placed against the free end of the spring is pushed a distance 5 cm towards the wall with 10J of energy as illustrated in the diagram above. The block is released and after 0.25s, it collides inelastically with a stationary block Q of mass 4.0 kg. Calculate the:
(i) value of k;
(ii) force used to compress the spring;
(iii) acceleration of the block p after release;
(iv) common speed after collision of the blocks.
(a) Triangle law of vector addition
The triangle law of vector addition states that if two vectors are represented in both magnitude and direction by the two adjacent sides of a triangle taken in order, then their resultant (sum) is represented in magnitude and direction by the third side of the triangle taken in the reverse order.
(b) Physical quantities associated with the equations of linear motion
The four physical quantities are:
- Displacement (or distance), \(s\)
- Velocity/speed (initial \(u\) and final \(v\))
- Acceleration, \(a\)
- Time, \(t\)
(c) Variation of potential energy and kinetic energy with time in S.H.M.
In simple harmonic motion the total mechanical energy \(E\) stays constant, so the potential energy (P.E.) and kinetic energy (K.E.) continuously interchange. When the body is at the equilibrium position the K.E. is maximum and the P.E. is zero; at the extreme positions the P.E. is maximum and the K.E. is zero. Both curves are \(\sin^2\)/\(\cos^2\) shapes that always add up to the constant total energy \(E\).
Key features of the sketch: both curves lie between \(0\) and \(E\); K.E. starts at its maximum \(E\) (body passing through equilibrium) while P.E. starts at zero; wherever one curve peaks the other is zero, and for every instant \(\text{K.E.} + \text{P.E.} = E\).
(d) The spring-and-block problem
Given: compression \(e = 5\,\text{cm} = 0.05\,\text{m}\); energy stored \(W = 10\,\text{J}\); mass of P, \(m_P = 2.0\,\text{kg}\); mass of Q, \(m_Q = 4.0\,\text{kg}\); time after release \(t = 0.25\,\text{s}\).
(i) Value of the force constant \(k\)
The energy stored in a compressed spring is \(W = \tfrac{1}{2}k e^{2}\), so
\[ k = \frac{2W}{e^{2}} = \frac{2 \times 10}{(0.05)^{2}} = \frac{20}{0.0025} = 8000\ \text{N m}^{-1}. \](ii) Force used to compress the spring
\[ F = k e = 8000 \times 0.05 = 400\ \text{N}. \](iii) Acceleration of block P after release
Using Newton's second law with the spring force acting on P:
\[ a = \frac{F}{m_P} = \frac{400}{2.0} = 200\ \text{m s}^{-2}. \](iv) Common speed after the inelastic collision
First find the velocity of P just before impact. Taking the acceleration to act over \(t = 0.25\,\text{s}\):
\[ u_P = a t = 200 \times 0.25 = 50\ \text{m s}^{-1}. \]The collision is inelastic, so the blocks move together. Applying conservation of linear momentum:
\[ m_P u_P + m_Q u_Q = (m_P + m_Q)v \] \[ (2.0 \times 50) + (4.0 \times 0) = (2.0 + 4.0)\,v \] \[ 100 = 6v \quad\Rightarrow\quad v = \frac{100}{6} = 16.7\ \text{m s}^{-1}. \]The common speed of the two blocks after collision is \(16.7\ \text{m s}^{-1}\).
Answer Details
(a) Triangle law of vector addition
The triangle law of vector addition states that if two vectors are represented in both magnitude and direction by the two adjacent sides of a triangle taken in order, then their resultant (sum) is represented in magnitude and direction by the third side of the triangle taken in the reverse order.
(b) Physical quantities associated with the equations of linear motion
The four physical quantities are:
- Displacement (or distance), \(s\)
- Velocity/speed (initial \(u\) and final \(v\))
- Acceleration, \(a\)
- Time, \(t\)
(c) Variation of potential energy and kinetic energy with time in S.H.M.
In simple harmonic motion the total mechanical energy \(E\) stays constant, so the potential energy (P.E.) and kinetic energy (K.E.) continuously interchange. When the body is at the equilibrium position the K.E. is maximum and the P.E. is zero; at the extreme positions the P.E. is maximum and the K.E. is zero. Both curves are \(\sin^2\)/\(\cos^2\) shapes that always add up to the constant total energy \(E\).
Key features of the sketch: both curves lie between \(0\) and \(E\); K.E. starts at its maximum \(E\) (body passing through equilibrium) while P.E. starts at zero; wherever one curve peaks the other is zero, and for every instant \(\text{K.E.} + \text{P.E.} = E\).
(d) The spring-and-block problem
Given: compression \(e = 5\,\text{cm} = 0.05\,\text{m}\); energy stored \(W = 10\,\text{J}\); mass of P, \(m_P = 2.0\,\text{kg}\); mass of Q, \(m_Q = 4.0\,\text{kg}\); time after release \(t = 0.25\,\text{s}\).
(i) Value of the force constant \(k\)
The energy stored in a compressed spring is \(W = \tfrac{1}{2}k e^{2}\), so
\[ k = \frac{2W}{e^{2}} = \frac{2 \times 10}{(0.05)^{2}} = \frac{20}{0.0025} = 8000\ \text{N m}^{-1}. \](ii) Force used to compress the spring
\[ F = k e = 8000 \times 0.05 = 400\ \text{N}. \](iii) Acceleration of block P after release
Using Newton's second law with the spring force acting on P:
\[ a = \frac{F}{m_P} = \frac{400}{2.0} = 200\ \text{m s}^{-2}. \](iv) Common speed after the inelastic collision
First find the velocity of P just before impact. Taking the acceleration to act over \(t = 0.25\,\text{s}\):
\[ u_P = a t = 200 \times 0.25 = 50\ \text{m s}^{-1}. \]The collision is inelastic, so the blocks move together. Applying conservation of linear momentum:
\[ m_P u_P + m_Q u_Q = (m_P + m_Q)v \] \[ (2.0 \times 50) + (4.0 \times 0) = (2.0 + 4.0)\,v \] \[ 100 = 6v \quad\Rightarrow\quad v = \frac{100}{6} = 16.7\ \text{m s}^{-1}. \]The common speed of the two blocks after collision is \(16.7\ \text{m s}^{-1}\).
Question 13 Report
Name one use of 'LASER' in each of the following areas:
(a) communication;
(b) medicine;
(c) security
Uses of the LASER
- (a) Communication: transmission of signals (voice, data, TV) through optical fibre cables using laser light.
- (b) Medicine: laser surgery, for example bloodless cutting of tissue, welding a detached retina, or removing tumours.
- (c) Security: laser beams used in burglar-alarm/intruder-detection systems (and in barcode/scanner and printing security).
Answer Details
Uses of the LASER
- (a) Communication: transmission of signals (voice, data, TV) through optical fibre cables using laser light.
- (b) Medicine: laser surgery, for example bloodless cutting of tissue, welding a detached retina, or removing tumours.
- (c) Security: laser beams used in burglar-alarm/intruder-detection systems (and in barcode/scanner and printing security).
Question 14 Report
A mass of 11.0 kg is suspended from a rigid support by an aluminum wire of length 2.0 m, diameter 2.0 mm and Young's modulus 7.0 x 10\(^{11}\) Nm\(^{-2}\). Determine the extension produced. [g = 10 ms\(^{-2}\); \(\pi\) = 3.142]
Question 15 Report
Write down the name of:
(a) two particles used in explaining the wave nature of matter;
(b) one device whose invention is based on the wave nature of matter.
(a) Two particles used in explaining the wave nature of matter:
- The electron.
- The neutron (a proton is also acceptable).
(b) One device based on the wave nature of matter:
- The electron microscope.
Answer Details
(a) Two particles used in explaining the wave nature of matter:
- The electron.
- The neutron (a proton is also acceptable).
(b) One device based on the wave nature of matter:
- The electron microscope.
