WAEC SSCE - Physics - 2009 (Essay)
Tambaya 1 Rahoto
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 streams of negatively charged particles (electrons); hence they are deflected by both electric and magnetic fields (towards the positive plate).
(Other valid properties: they possess kinetic energy and momentum, they can produce heat, they cause certain substances to fluoresce, and they can produce X-rays on striking a metal target.)
(b) Increasing the intensity and energy of cathode rays
Intensity: The intensity (number of electrons emitted per second) is increased by raising the temperature of the cathode/filament, that is by increasing the heating (filament) current. A hotter cathode emits more electrons per second by thermionic emission, giving a more intense beam.
Energy: The energy (speed) of the cathode rays is increased by raising the accelerating potential difference (the anode voltage) between the cathode and anode. Each electron gains energy \(eV\), so a larger \(V\) gives faster, more energetic electrons.
Bayanin Amsa
(a) Two properties of cathode rays
- They travel in straight lines and cast sharp shadows of objects placed in their path.
- They are streams of negatively charged particles (electrons); hence they are deflected by both electric and magnetic fields (towards the positive plate).
(Other valid properties: they possess kinetic energy and momentum, they can produce heat, they cause certain substances to fluoresce, and they can produce X-rays on striking a metal target.)
(b) Increasing the intensity and energy of cathode rays
Intensity: The intensity (number of electrons emitted per second) is increased by raising the temperature of the cathode/filament, that is by increasing the heating (filament) current. A hotter cathode emits more electrons per second by thermionic emission, giving a more intense beam.
Energy: The energy (speed) of the cathode rays is increased by raising the accelerating potential difference (the anode voltage) between the cathode and anode. Each electron gains energy \(eV\), so a larger \(V\) gives faster, more energetic electrons.
Tambaya 2 Rahoto
(a) State two differences between a sound wave and a radio wave.
(b) Explain why a vibrating tuning fork sounds louder when its stem is pressed against a table top than when held in air.
(c)State two conditions necessary for the:
- production of stationary wave in a medium;
- formation of interference wave patterns;
- occurrence of total internal reflection of a wave.
(d) A ray of light is incident on one face of an equilateral glass prism.
- Draw a ray diagram to show the path of the ray through the prism.
- Calculate the refractive index of the glass if the angle of minimum deviation is 41\(^o\).
(a)
| Sound wave | Radio wave |
|---|---|
| It is a mechanical wave and requires a material medium for propagation. | It is an electromagnetic wave and can travel through a vacuum. |
| It is longitudinal in air. | It is transverse. |
(b)
The vibrating tuning fork forces the table top to vibrate. The table top has a much larger vibrating surface than the prongs of the fork and therefore sets a larger volume of air into vibration. More sound energy is transmitted to the air per second, so the sound is louder.
(c)
- Production of stationary waves:
- Two waves of the same frequency and equal or nearly equal amplitude must be present.
- The waves must travel in opposite directions along the same line and superpose.
- Formation of interference patterns:
- The sources must be coherent, that is, have the same frequency and a constant phase difference.
- The waves must overlap in the same region with equal or nearly equal amplitudes.
- Occurrence of total internal reflection:
- The wave must travel from an optically denser medium to an optically less dense medium.
- The angle of incidence in the denser medium must be greater than the critical angle.
(d)(i) At minimum deviation, the path through the equilateral prism is symmetrical and the refracted ray is parallel to the base.
(d)(ii)
For an equilateral prism,
\[A=60^\circ, \qquad D_m=41^\circ\]
\[n=\frac{\sin\left(\frac{A+D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\]
\[n=\frac{\sin\left(\frac{60^\circ+41^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)}=\frac{\sin 50.5^\circ}{\sin 30^\circ}\]
\[n=\frac{0.7716}{0.5000}=1.54\]
Therefore, the refractive index of the glass is \(1.54\).
Bayanin Amsa
(a)
| Sound wave | Radio wave |
|---|---|
| It is a mechanical wave and requires a material medium for propagation. | It is an electromagnetic wave and can travel through a vacuum. |
| It is longitudinal in air. | It is transverse. |
(b)
The vibrating tuning fork forces the table top to vibrate. The table top has a much larger vibrating surface than the prongs of the fork and therefore sets a larger volume of air into vibration. More sound energy is transmitted to the air per second, so the sound is louder.
(c)
- Production of stationary waves:
- Two waves of the same frequency and equal or nearly equal amplitude must be present.
- The waves must travel in opposite directions along the same line and superpose.
- Formation of interference patterns:
- The sources must be coherent, that is, have the same frequency and a constant phase difference.
- The waves must overlap in the same region with equal or nearly equal amplitudes.
- Occurrence of total internal reflection:
- The wave must travel from an optically denser medium to an optically less dense medium.
- The angle of incidence in the denser medium must be greater than the critical angle.
(d)(i) At minimum deviation, the path through the equilateral prism is symmetrical and the refracted ray is parallel to the base.
(d)(ii)
For an equilateral prism,
\[A=60^\circ, \qquad D_m=41^\circ\]
\[n=\frac{\sin\left(\frac{A+D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\]
\[n=\frac{\sin\left(\frac{60^\circ+41^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)}=\frac{\sin 50.5^\circ}{\sin 30^\circ}\]
\[n=\frac{0.7716}{0.5000}=1.54\]
Therefore, the refractive index of the glass is \(1.54\).
Tambaya 3 Rahoto
(a) State two essential differences between a moving coil galvanometer and a d.c. generator.
(b) Explain the term eddy currents and state two devices in which the currents are applied.
(c) State the principle on which the potentiometer is based when it is functioning.
(d) A source of e.m.f. 110 V and frequency 60Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 2A, the potential differences across the resistor is 80 V and that across the inductor is 40 V. Draw the vector diagram of the potential differences across the inductor, the capacitor and the resistor.
Calculate the:
(i) potential difference across the capacitor;
(ii) capacitance of the capacitor;
(iii) inductance of the inductor. [π = 3.14]
(a) Differences between a moving-coil galvanometer and a d.c. generator
| Moving-coil galvanometer | D.C. generator |
|---|---|
| It detects or measures small electric currents. | It generates electrical energy. |
| It converts electrical energy to mechanical deflection; its coil turns through a limited angle and is controlled by hair springs. | It converts mechanical energy to electrical energy; its coil rotates continuously and uses a split-ring commutator and carbon brushes. |
(b) Eddy currents
Eddy currents are circulating currents induced in the body of a conductor when the magnetic flux through it changes. They flow in closed paths and oppose the change producing them. They are applied in an induction furnace and an eddy-current brake.
(c) Principle of a potentiometer
When a steady current flows through a uniform wire of constant cross-sectional area, the potential difference across the wire is directly proportional to its length:
\[V \propto l.\]
(d) Vector diagram
Taking the current, and hence \(V_R\), as the horizontal reference, \(V_L\) leads the current by \(90^\circ\), while \(V_C\) lags the current by \(90^\circ\).
\[V^2=V_R^2+(V_L-V_C)^2\]
\[110^2=80^2+(40-V_C)^2\]
\[(40-V_C)^2=12100-6400=5700\]
Since the circuit is net capacitive, \(V_C>V_L\):
\[V_C=40+\sqrt{5700}=40+75.5=115.5\text{ V}\]
(i) Potential difference across the capacitor:
\[\boxed{V_C=115.5\text{ V}}\]
(ii) Capacitive reactance:
\[X_C=\frac{V_C}{I}=\frac{115.5}{2}=57.75\ \Omega\]
\[C=\frac{1}{2\pi fX_C}=\frac{1}{2\times3.14\times60\times57.75}=4.59\times10^{-5}\text{ F}\]
\[\boxed{C\approx46\ \mu\text{F}}\]
(iii) Inductive reactance:
\[X_L=\frac{V_L}{I}=\frac{40}{2}=20\ \Omega\]
\[L=\frac{X_L}{2\pi f}=\frac{20}{2\times3.14\times60}=0.0531\text{ H}\]
\[\boxed{L\approx0.053\text{ H}}\]
Bayanin Amsa
(a) Differences between a moving-coil galvanometer and a d.c. generator
| Moving-coil galvanometer | D.C. generator |
|---|---|
| It detects or measures small electric currents. | It generates electrical energy. |
| It converts electrical energy to mechanical deflection; its coil turns through a limited angle and is controlled by hair springs. | It converts mechanical energy to electrical energy; its coil rotates continuously and uses a split-ring commutator and carbon brushes. |
(b) Eddy currents
Eddy currents are circulating currents induced in the body of a conductor when the magnetic flux through it changes. They flow in closed paths and oppose the change producing them. They are applied in an induction furnace and an eddy-current brake.
(c) Principle of a potentiometer
When a steady current flows through a uniform wire of constant cross-sectional area, the potential difference across the wire is directly proportional to its length:
\[V \propto l.\]
(d) Vector diagram
Taking the current, and hence \(V_R\), as the horizontal reference, \(V_L\) leads the current by \(90^\circ\), while \(V_C\) lags the current by \(90^\circ\).
\[V^2=V_R^2+(V_L-V_C)^2\]
\[110^2=80^2+(40-V_C)^2\]
\[(40-V_C)^2=12100-6400=5700\]
Since the circuit is net capacitive, \(V_C>V_L\):
\[V_C=40+\sqrt{5700}=40+75.5=115.5\text{ V}\]
(i) Potential difference across the capacitor:
\[\boxed{V_C=115.5\text{ V}}\]
(ii) Capacitive reactance:
\[X_C=\frac{V_C}{I}=\frac{115.5}{2}=57.75\ \Omega\]
\[C=\frac{1}{2\pi fX_C}=\frac{1}{2\times3.14\times60\times57.75}=4.59\times10^{-5}\text{ F}\]
\[\boxed{C\approx46\ \mu\text{F}}\]
(iii) Inductive reactance:
\[X_L=\frac{V_L}{I}=\frac{40}{2}=20\ \Omega\]
\[L=\frac{X_L}{2\pi f}=\frac{20}{2\times3.14\times60}=0.0531\text{ H}\]
\[\boxed{L\approx0.053\text{ H}}\]
Tambaya 4 Rahoto
A particle is projected horizontally at 15ms\(^{-1}\) from a height of 20m.
Calculate the horizontal distance covered by the particle just before hitting the ground.
[g = 10 ms\(^{-2}\)]
The particle is projected horizontally, so its vertical motion is free fall from rest and its horizontal velocity stays constant.
Time to reach the ground (vertical fall of \( h = 20\,\text{m} \)): \[ h = \tfrac{1}{2} g t^2 \Rightarrow t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2 \times 20}{10}} = \sqrt{4} = 2\,\text{s}. \]
Horizontal distance (range): \[ x = u \times t = 15 \times 2 = 30\,\text{m}. \]
The horizontal distance covered before hitting the ground is 30 m.
Bayanin Amsa
The particle is projected horizontally, so its vertical motion is free fall from rest and its horizontal velocity stays constant.
Time to reach the ground (vertical fall of \( h = 20\,\text{m} \)): \[ h = \tfrac{1}{2} g t^2 \Rightarrow t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2 \times 20}{10}} = \sqrt{4} = 2\,\text{s}. \]
Horizontal distance (range): \[ x = u \times t = 15 \times 2 = 30\,\text{m}. \]
The horizontal distance covered before hitting the ground is 30 m.
Tambaya 5 Rahoto
Explain the following terms:
(a) tensile stress;
(b)Young’s modulus
(a) Tensile stress
Tensile stress is the stretching force acting per unit cross-sectional area of a material when it is subjected to a pull along its length.
\[ \text{Tensile stress} = \frac{\text{Force (tension)}}{\text{Cross-sectional area}} = \frac{F}{A} \]Its S.I. unit is the pascal (\(\text{Pa}\)) or \(\text{N m}^{-2}\).
(b) Young's modulus
Young's modulus (the modulus of elasticity) is the ratio of tensile stress to tensile strain for a material, within the limit of proportionality (the region where Hooke's law holds).
\[ E = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/l} = \frac{Fl}{Ae} \]where \(F\) is the stretching force, \(A\) the cross-sectional area, \(l\) the original length and \(e\) the extension. Its S.I. unit is the pascal (\(\text{Pa}\)) or \(\text{N m}^{-2}\). A large value of \(E\) indicates a stiff material that is difficult to stretch.
Bayanin Amsa
(a) Tensile stress
Tensile stress is the stretching force acting per unit cross-sectional area of a material when it is subjected to a pull along its length.
\[ \text{Tensile stress} = \frac{\text{Force (tension)}}{\text{Cross-sectional area}} = \frac{F}{A} \]Its S.I. unit is the pascal (\(\text{Pa}\)) or \(\text{N m}^{-2}\).
(b) Young's modulus
Young's modulus (the modulus of elasticity) is the ratio of tensile stress to tensile strain for a material, within the limit of proportionality (the region where Hooke's law holds).
\[ E = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/l} = \frac{Fl}{Ae} \]where \(F\) is the stretching force, \(A\) the cross-sectional area, \(l\) the original length and \(e\) the extension. Its S.I. unit is the pascal (\(\text{Pa}\)) or \(\text{N m}^{-2}\). A large value of \(E\) indicates a stiff material that is difficult to stretch.
Tambaya 6 Rahoto
Give three observations in support of de Broglie's assumption that moving particles behave like waves
De Broglie proposed that a moving particle of momentum \(p = mv\) has an associated wavelength \(\lambda = \dfrac{h}{mv}\). The following observations support this wave nature of moving particles:
- Electron diffraction (Davisson and Germer): a beam of electrons directed at a crystal (nickel) produced a diffraction pattern of maxima and minima, just as X-rays do. Diffraction is a wave property, so the electrons behaved as waves.
- Thomson's experiment: when a beam of fast electrons was passed through a thin metal foil, it produced concentric diffraction rings on a photographic plate, similar to those given by X-rays passing through the same foil.
- The electron microscope: its successful operation relies on the very short wavelength associated with fast-moving electrons; the electron beam can be focused and diffracted like a wave, which is why it resolves far finer detail than a light microscope.
(A further supporting observation is the diffraction of other particles such as neutrons and protons by crystals.)
Bayanin Amsa
De Broglie proposed that a moving particle of momentum \(p = mv\) has an associated wavelength \(\lambda = \dfrac{h}{mv}\). The following observations support this wave nature of moving particles:
- Electron diffraction (Davisson and Germer): a beam of electrons directed at a crystal (nickel) produced a diffraction pattern of maxima and minima, just as X-rays do. Diffraction is a wave property, so the electrons behaved as waves.
- Thomson's experiment: when a beam of fast electrons was passed through a thin metal foil, it produced concentric diffraction rings on a photographic plate, similar to those given by X-rays passing through the same foil.
- The electron microscope: its successful operation relies on the very short wavelength associated with fast-moving electrons; the electron beam can be focused and diffracted like a wave, which is why it resolves far finer detail than a light microscope.
(A further supporting observation is the diffraction of other particles such as neutrons and protons by crystals.)
Tambaya 7 Rahoto
(a) Explain why it is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure.
(b) State the effect of an increase in pressure on the (i) boiling point; and (ii) melting point of water.
(c) The graph below shows the saturated vapour pressure (s.v.p.) of water plotted against temperature.
Pure water is known to boil at 100°C and at an atmospheric pressure of 760 mmHg. What general conclusion can be drawn from the information given above?
(d) A thread of mercury of length 20 cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end. With the tube vertical and the open end uppermost, the length of the trapped air column is 15cm. Calculate the length of the air column when the tube is held: (i) horizontally; (ii) vertically with the open end underneath. [Atmospheric pressure = 76 cmHg]
(a) A clinical thermometer has a small temperature range, approximately \(35\,^{\circ}\mathrm{C}\) to \(43\,^{\circ}\mathrm{C}\). In boiling water, the mercury expands excessively and may produce enough pressure to crack or burst the glass thermometer.
(b)
- An increase in pressure increases the boiling point of water.
- An increase in pressure decreases the melting point of ice.
(c) A liquid boils when its saturated vapour pressure is equal to the external atmospheric pressure. Thus, at \(100\,^{\circ}\mathrm{C}\), the saturated vapour pressure of pure water is \(760\ \mathrm{mmHg}\).
(d) The arrangements of the trapped air and mercury thread are represented below.
Since the capillary tube has uniform cross-sectional area, volume is proportional to length. Hence, Boyle's law may be written as:
\[P_1L_1=P_2L_2\]
Initially, with the open end uppermost, the pressure of the trapped air is
\[P_1=76+20=96\ \mathrm{cmHg}\]
and \(L_1=15\ \mathrm{cm}\). Therefore,
\[P_1L_1=96\times15=1440\ \mathrm{cmHg\,cm}.\]
(i) Tube held horizontally
The pressure of the trapped air is atmospheric pressure:
\[P_2=76\ \mathrm{cmHg}\]
\[96\times15=76L_2\]
\[L_2=\frac{1440}{76}=18.95\ \mathrm{cm}\approx\boxed{18.9\ \mathrm{cm}}\]
(ii) Tube held vertically with the open end underneath
The pressure of the trapped air is
\[P_3=76-20=56\ \mathrm{cmHg}\]
\[96\times15=56L_3\]
\[L_3=\frac{1440}{56}=25.7\ \mathrm{cm}\]
\[\boxed{L_3=25.7\ \mathrm{cm}}\]
Bayanin Amsa
(a) A clinical thermometer has a small temperature range, approximately \(35\,^{\circ}\mathrm{C}\) to \(43\,^{\circ}\mathrm{C}\). In boiling water, the mercury expands excessively and may produce enough pressure to crack or burst the glass thermometer.
(b)
- An increase in pressure increases the boiling point of water.
- An increase in pressure decreases the melting point of ice.
(c) A liquid boils when its saturated vapour pressure is equal to the external atmospheric pressure. Thus, at \(100\,^{\circ}\mathrm{C}\), the saturated vapour pressure of pure water is \(760\ \mathrm{mmHg}\).
(d) The arrangements of the trapped air and mercury thread are represented below.
Since the capillary tube has uniform cross-sectional area, volume is proportional to length. Hence, Boyle's law may be written as:
\[P_1L_1=P_2L_2\]
Initially, with the open end uppermost, the pressure of the trapped air is
\[P_1=76+20=96\ \mathrm{cmHg}\]
and \(L_1=15\ \mathrm{cm}\). Therefore,
\[P_1L_1=96\times15=1440\ \mathrm{cmHg\,cm}.\]
(i) Tube held horizontally
The pressure of the trapped air is atmospheric pressure:
\[P_2=76\ \mathrm{cmHg}\]
\[96\times15=76L_2\]
\[L_2=\frac{1440}{76}=18.95\ \mathrm{cm}\approx\boxed{18.9\ \mathrm{cm}}\]
(ii) Tube held vertically with the open end underneath
The pressure of the trapped air is
\[P_3=76-20=56\ \mathrm{cmHg}\]
\[96\times15=56L_3\]
\[L_3=\frac{1440}{56}=25.7\ \mathrm{cm}\]
\[\boxed{L_3=25.7\ \mathrm{cm}}\]
Tambaya 8 Rahoto
a) Define diffusion.
(b) State two applications of electrical conduction through gases.
(a) Diffusion
Diffusion is the gradual movement or spreading of the molecules of a substance from a region of higher concentration to a region of lower concentration until the molecules are uniformly distributed. It occurs because the molecules are in continuous random motion, and it is faster in gases than in liquids because gas molecules move faster and are farther apart.
(b) Two applications of electrical conduction through gases
- Fluorescent tubes and neon (discharge) lamps - light is produced when an electric current is passed through a gas at low pressure.
- Advertising / sodium and mercury vapour street lamps - conduction through the vapour produces the characteristic coloured light used in signs and lighting.
(Other acceptable examples: the cathode-ray tube / X-ray tube, and the lightning discharge, which is conduction through air.)
Bayanin Amsa
(a) Diffusion
Diffusion is the gradual movement or spreading of the molecules of a substance from a region of higher concentration to a region of lower concentration until the molecules are uniformly distributed. It occurs because the molecules are in continuous random motion, and it is faster in gases than in liquids because gas molecules move faster and are farther apart.
(b) Two applications of electrical conduction through gases
- Fluorescent tubes and neon (discharge) lamps - light is produced when an electric current is passed through a gas at low pressure.
- Advertising / sodium and mercury vapour street lamps - conduction through the vapour produces the characteristic coloured light used in signs and lighting.
(Other acceptable examples: the cathode-ray tube / X-ray tube, and the lightning discharge, which is conduction through air.)
Tambaya 9 Rahoto
a) Given a retort stand and clamp, a stout pin, a simple pendulum and a pencil, describe how you would use these apparatus to determine the centre of gravity of an irregularly shaped piece of cardboard of a moderate size.
(b) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph:
(i) acceleration;
(ii) total distance covered.
(c ) A body at rest is given an initial uniform acceleration of \(6.0\ \mathrm{ms}^{-2}\) for \(20\mathrm{s}\) after which the acceleration is reduced to \(4.0\ \mathrm{ms}^{-2}\) for the next \(10\mathrm{s}\).
The body maintains the speed attained for \(30\mathrm{s}\).
Draw the velocity-time graph of the motion using the information given above. From the graph, calculate the:
- maximum speed attained during the motion;
- total distance traveled during the first \(30\ \mathrm{s}\);
- average speed during the same time interval as in (ii) above.
(a) Determination of the centre of gravity of the cardboard
- Make three small, well-spaced holes near the edge of the irregular cardboard.
- Clamp the stout pin horizontally in the retort stand. Suspend the cardboard freely from the pin through one hole.
- Hang the simple pendulum from the same pin, with its string very close to the surface of the cardboard.
- Allow the cardboard and pendulum to come to rest. Trace the vertical line indicated by the pendulum string on the cardboard with the pencil.
- Suspend the cardboard successively from each of the other two holes and trace the vertical line in each case.
- The point of intersection of the three vertical lines is the centre of gravity of the cardboard.
Precautions: The pin must be firmly clamped; the cardboard must swing freely; and the pendulum must be at rest before each line is traced.
(b) Velocity-time graph
(i) Acceleration
Acceleration is the gradient of the velocity-time graph. For the straight-line section joining points A and B,
\[a=\frac{\text{change in velocity}}{\text{time taken}}=\frac{v_B-v_A}{t_B-t_A}.\]
(ii) Total distance covered
The total distance covered is the area between the velocity-time graph and the time axis. Thus, for the graph shown, the distance is the area under the graph from the starting time to the final time.
(c) Velocity-time graph of the motion
For the first 20 s:
\[v_1=0+(6.0\times20)=120\ \text{m s}^{-1}.\]
For the next 10 s:
\[v_2=120+(4.0\times10)=160\ \text{m s}^{-1}.\]
The body then continues at \(160\ \text{m s}^{-1}\) for a further 30 s.
The graph consists of straight-line sections through \((0,0)\), \((20,120)\), \((30,160)\), and \((60,160)\), where time is in seconds and velocity is in \(\text{m s}^{-1}\).
(i) Maximum speed
\[\boxed{160\ \text{m s}^{-1}}\]
(ii) Total distance travelled during the first 30 s
This is the area under the graph from 0 s to 30 s:
\[\begin{aligned}s_1&=\frac12\times20\times120=1200\ \text{m},\\s_2&=\frac12(120+160)\times10=1400\ \text{m}.\end{aligned}\]
\[s=1200+1400=\boxed{2600\ \text{m}}.\]
(iii) Average speed during the first 30 s
\[\text{Average speed}=\frac{\text{total distance}}{\text{total time}}=\frac{2600}{30}=86.7\ \text{m s}^{-1}.\]
\[\boxed{86.7\ \text{m s}^{-1}}\]
Bayanin Amsa
(a) Determination of the centre of gravity of the cardboard
- Make three small, well-spaced holes near the edge of the irregular cardboard.
- Clamp the stout pin horizontally in the retort stand. Suspend the cardboard freely from the pin through one hole.
- Hang the simple pendulum from the same pin, with its string very close to the surface of the cardboard.
- Allow the cardboard and pendulum to come to rest. Trace the vertical line indicated by the pendulum string on the cardboard with the pencil.
- Suspend the cardboard successively from each of the other two holes and trace the vertical line in each case.
- The point of intersection of the three vertical lines is the centre of gravity of the cardboard.
Precautions: The pin must be firmly clamped; the cardboard must swing freely; and the pendulum must be at rest before each line is traced.
(b) Velocity-time graph
(i) Acceleration
Acceleration is the gradient of the velocity-time graph. For the straight-line section joining points A and B,
\[a=\frac{\text{change in velocity}}{\text{time taken}}=\frac{v_B-v_A}{t_B-t_A}.\]
(ii) Total distance covered
The total distance covered is the area between the velocity-time graph and the time axis. Thus, for the graph shown, the distance is the area under the graph from the starting time to the final time.
(c) Velocity-time graph of the motion
For the first 20 s:
\[v_1=0+(6.0\times20)=120\ \text{m s}^{-1}.\]
For the next 10 s:
\[v_2=120+(4.0\times10)=160\ \text{m s}^{-1}.\]
The body then continues at \(160\ \text{m s}^{-1}\) for a further 30 s.
The graph consists of straight-line sections through \((0,0)\), \((20,120)\), \((30,160)\), and \((60,160)\), where time is in seconds and velocity is in \(\text{m s}^{-1}\).
(i) Maximum speed
\[\boxed{160\ \text{m s}^{-1}}\]
(ii) Total distance travelled during the first 30 s
This is the area under the graph from 0 s to 30 s:
\[\begin{aligned}s_1&=\frac12\times20\times120=1200\ \text{m},\\s_2&=\frac12(120+160)\times10=1400\ \text{m}.\end{aligned}\]
\[s=1200+1400=\boxed{2600\ \text{m}}.\]
(iii) Average speed during the first 30 s
\[\text{Average speed}=\frac{\text{total distance}}{\text{total time}}=\frac{2600}{30}=86.7\ \text{m s}^{-1}.\]
\[\boxed{86.7\ \text{m s}^{-1}}\]
Tambaya 10 Rahoto
Explain why mercury does not wet glass while water does.
Whether a liquid wets a solid surface depends on the relative sizes of two types of intermolecular force:
- Cohesion - the force of attraction between molecules of the same substance (liquid to liquid).
- Adhesion - the force of attraction between molecules of different substances (liquid to solid).
Water on glass: The adhesive force between water molecules and glass molecules is greater than the cohesive force between the water molecules themselves. Water molecules are therefore pulled towards the glass, spread over it, and the water forms a concave meniscus with an acute angle of contact. Because it clings to and spreads on the surface, water is said to wet glass.
Mercury on glass: The cohesive force between mercury molecules is greater than the adhesive force between mercury and glass. The mercury molecules are pulled more strongly towards one another than towards the glass, so the mercury draws itself together into rounded drops, forms a convex meniscus with an obtuse angle of contact, and does not spread. Hence mercury does not wet glass.
In short: a liquid wets a solid when adhesion exceeds cohesion (water and glass) and does not wet it when cohesion exceeds adhesion (mercury and glass).
Bayanin Amsa
Whether a liquid wets a solid surface depends on the relative sizes of two types of intermolecular force:
- Cohesion - the force of attraction between molecules of the same substance (liquid to liquid).
- Adhesion - the force of attraction between molecules of different substances (liquid to solid).
Water on glass: The adhesive force between water molecules and glass molecules is greater than the cohesive force between the water molecules themselves. Water molecules are therefore pulled towards the glass, spread over it, and the water forms a concave meniscus with an acute angle of contact. Because it clings to and spreads on the surface, water is said to wet glass.
Mercury on glass: The cohesive force between mercury molecules is greater than the adhesive force between mercury and glass. The mercury molecules are pulled more strongly towards one another than towards the glass, so the mercury draws itself together into rounded drops, forms a convex meniscus with an obtuse angle of contact, and does not spread. Hence mercury does not wet glass.
In short: a liquid wets a solid when adhesion exceeds cohesion (water and glass) and does not wet it when cohesion exceeds adhesion (mercury and glass).
Tambaya 11 Rahoto
(a) Explain what is meant by cations
(b) Draw and label an electrolytic cell
(a) Cations are positively charged ions. During electrolysis, they are attracted to and discharged at the cathode.
(b) A labelled electrolytic cell is shown below.
The anode is connected to the positive terminal of the d.c. supply, while the cathode is connected to the negative terminal. Cations move towards the cathode and anions move towards the anode through the electrolyte.
Bayanin Amsa
(a) Cations are positively charged ions. During electrolysis, they are attracted to and discharged at the cathode.
(b) A labelled electrolytic cell is shown below.
The anode is connected to the positive terminal of the d.c. supply, while the cathode is connected to the negative terminal. Cations move towards the cathode and anions move towards the anode through the electrolyte.
Tambaya 12 Rahoto
(a) State Faraday’s second law of electrolysis.
(b) An electric charge of 9.6 x 10\(^4\) C liberates 1 mole of substance containing 6.0 x 10\(^{23}\) atoms. Determine the value of the electronic charge
(a) Faraday's second law of electrolysis
When the same quantity of electricity is passed through different electrolytes, the masses of the different substances liberated (or deposited) are directly proportional to their chemical equivalents (that is, to the ratio of their relative atomic mass to valency).
(b) Value of the electronic charge
The substance is monovalent, so each atom is discharged by one electron. One mole contains \(N = 6.0\times10^{23}\) atoms and is liberated by a charge \(Q = 9.6\times10^{4}\ \text{C}\).
The charge carried by one electron (one atom) is:
\[ e = \frac{Q}{N} = \frac{9.6\times10^{4}}{6.0\times10^{23}} \] \[ e = 1.6\times10^{-19}\ \text{C} \]The value of the electronic charge is \(1.6\times10^{-19}\ \text{C}\).
Bayanin Amsa
(a) Faraday's second law of electrolysis
When the same quantity of electricity is passed through different electrolytes, the masses of the different substances liberated (or deposited) are directly proportional to their chemical equivalents (that is, to the ratio of their relative atomic mass to valency).
(b) Value of the electronic charge
The substance is monovalent, so each atom is discharged by one electron. One mole contains \(N = 6.0\times10^{23}\) atoms and is liberated by a charge \(Q = 9.6\times10^{4}\ \text{C}\).
The charge carried by one electron (one atom) is:
\[ e = \frac{Q}{N} = \frac{9.6\times10^{4}}{6.0\times10^{23}} \] \[ e = 1.6\times10^{-19}\ \text{C} \]The value of the electronic charge is \(1.6\times10^{-19}\ \text{C}\).
Tambaya 13 Rahoto
State three methods of polirizing an unpolarized light
Unpolarized light (in which the vibrations occur in all planes perpendicular to the direction of travel) can be polarized by the following methods:
- Selective absorption (using a Polaroid): passing the light through a Polaroid sheet, which transmits vibrations in only one plane and absorbs those in all other planes.
- Reflection: light reflected from a smooth non-metallic surface (such as glass or water) at the polarizing (Brewster) angle becomes plane polarized.
- Double refraction (using a crystal): passing the light through a doubly refracting crystal such as calcite or tourmaline (or a Nicol prism), which splits it into two plane-polarized rays.
(Scattering, for example by dust or air molecules, also produces partially polarized light.)
Bayanin Amsa
Unpolarized light (in which the vibrations occur in all planes perpendicular to the direction of travel) can be polarized by the following methods:
- Selective absorption (using a Polaroid): passing the light through a Polaroid sheet, which transmits vibrations in only one plane and absorbs those in all other planes.
- Reflection: light reflected from a smooth non-metallic surface (such as glass or water) at the polarizing (Brewster) angle becomes plane polarized.
- Double refraction (using a crystal): passing the light through a doubly refracting crystal such as calcite or tourmaline (or a Nicol prism), which splits it into two plane-polarized rays.
(Scattering, for example by dust or air molecules, also produces partially polarized light.)
Tambaya 14 Rahoto
(a) Briefly explain the following terms:
(i) emission line spectra;
(ii) line absorption spectra.
(b) Draw a labeled diagram showing the structure of a simple type of photocell and explain its mode of operation.
(c) State two
(i) reasons to show that x-rays are waves;
(ii) uses of x-rays other than in medicine.
(d) An electron jumps from an energy level of \(-1.6\ \text{eV}\) to one of \(-1.4\ \text{eV}\) in an atom. Calculate the energy and wavelength of the emitted radiation. [ \(h = 6.6 \times 10^{-34}\ \text{Js}\); \(c = 3.00 \times 10^8\ \text{ms}^{-1}\); \(\text{eV} = 1.6 \times 10^{-19}\ \text{J}\) ]
(a)(i) Emission line spectrum
An emission line spectrum consists of separate bright lines on a dark background. Each line has a definite wavelength. It is produced when excited atoms emit photons as electrons fall from higher to lower energy levels.
(a)(ii) Line absorption spectrum
A line absorption spectrum consists of dark lines in an otherwise continuous spectrum. It is produced when light passes through a cooler gas: atoms in the gas absorb photons of particular wavelengths to raise electrons to higher energy levels.
(b) Simple photoemissive photocell
Light must have a frequency at least equal to the threshold frequency of the photosensitive cathode. The light then causes electrons to be emitted from the cathode by the photoelectric effect. Since the anode is positive relative to the cathode, it attracts and collects these electrons. Their movement through the external circuit produces a photocurrent, detected by the galvanometer. Increasing light intensity increases the number of emitted electrons per second, so the photocurrent increases.
(c)(i) Evidence that X-rays are waves
- X-rays undergo diffraction, for example when diffracted by the regularly spaced planes of atoms in a crystal.
- X-rays produce interference patterns. Diffraction and interference are characteristic wave behaviours.
(c)(ii) Uses of X-rays other than in medicine
- Industrial radiography: detecting internal cracks or flaws in welded joints and metal castings.
- Determining crystal structure by X-ray diffraction.
(d) Energy change and wavelength
There is an inconsistency in the wording: an electron moving from \( -1.6\ \text{eV} \) to \( -1.4\ \text{eV} \) moves to a higher energy level, because \( -1.4\ \text{eV} \) is greater than \( -1.6\ \text{eV} \). Therefore, the electron must absorb radiation; it cannot emit radiation during this transition.
The magnitude of the energy absorbed is:
\[ \Delta E=(-1.4)-(-1.6)=0.2\ \text{eV} \] \[ \Delta E=0.2(1.6\times10^{-19})=3.2\times10^{-20}\ \text{J} \]Using \(E=\dfrac{hc}{\lambda}\):
\[ \lambda=\frac{hc}{E} =\frac{(6.6\times10^{-34})(3.00\times10^8)}{3.2\times10^{-20}} \] \[ \lambda=6.19\times10^{-6}\ \text{m}\approx6.2\times10^{-6}\ \text{m} \]Thus, for the levels printed in the question, radiation of energy \(0.20\ \text{eV}\), or \(3.2\times10^{-20}\ \text{J}\), and wavelength \(6.2\times10^{-6}\ \text{m}\) is absorbed. Emission would occur only for the reverse transition, from \( -1.4\ \text{eV} \) to \( -1.6\ \text{eV} \).
Bayanin Amsa
(a)(i) Emission line spectrum
An emission line spectrum consists of separate bright lines on a dark background. Each line has a definite wavelength. It is produced when excited atoms emit photons as electrons fall from higher to lower energy levels.
(a)(ii) Line absorption spectrum
A line absorption spectrum consists of dark lines in an otherwise continuous spectrum. It is produced when light passes through a cooler gas: atoms in the gas absorb photons of particular wavelengths to raise electrons to higher energy levels.
(b) Simple photoemissive photocell
Light must have a frequency at least equal to the threshold frequency of the photosensitive cathode. The light then causes electrons to be emitted from the cathode by the photoelectric effect. Since the anode is positive relative to the cathode, it attracts and collects these electrons. Their movement through the external circuit produces a photocurrent, detected by the galvanometer. Increasing light intensity increases the number of emitted electrons per second, so the photocurrent increases.
(c)(i) Evidence that X-rays are waves
- X-rays undergo diffraction, for example when diffracted by the regularly spaced planes of atoms in a crystal.
- X-rays produce interference patterns. Diffraction and interference are characteristic wave behaviours.
(c)(ii) Uses of X-rays other than in medicine
- Industrial radiography: detecting internal cracks or flaws in welded joints and metal castings.
- Determining crystal structure by X-ray diffraction.
(d) Energy change and wavelength
There is an inconsistency in the wording: an electron moving from \( -1.6\ \text{eV} \) to \( -1.4\ \text{eV} \) moves to a higher energy level, because \( -1.4\ \text{eV} \) is greater than \( -1.6\ \text{eV} \). Therefore, the electron must absorb radiation; it cannot emit radiation during this transition.
The magnitude of the energy absorbed is:
\[ \Delta E=(-1.4)-(-1.6)=0.2\ \text{eV} \] \[ \Delta E=0.2(1.6\times10^{-19})=3.2\times10^{-20}\ \text{J} \]Using \(E=\dfrac{hc}{\lambda}\):
\[ \lambda=\frac{hc}{E} =\frac{(6.6\times10^{-34})(3.00\times10^8)}{3.2\times10^{-20}} \] \[ \lambda=6.19\times10^{-6}\ \text{m}\approx6.2\times10^{-6}\ \text{m} \]Thus, for the levels printed in the question, radiation of energy \(0.20\ \text{eV}\), or \(3.2\times10^{-20}\ \text{J}\), and wavelength \(6.2\times10^{-6}\ \text{m}\) is absorbed. Emission would occur only for the reverse transition, from \( -1.4\ \text{eV} \) to \( -1.6\ \text{eV} \).
Tambaya 15 Rahoto
A particle is dropped from a vertical height \(h\) and falls freely for a time \(t\). With the aid of a sketch, explain how \(h\) varies with
(a) \(t\);
(b) \(t^2\).
For a particle dropped from rest, the initial velocity, \(u=0\). Hence,
\[h=ut+\frac{1}{2}gt^2=\frac{1}{2}gt^2.\]
Taking \(g=9.8\ \text{m s}^{-2}\),
\[h=4.9t^2.\]
(a) Graph of \(h\) against \(t\)
The graph is a parabola passing through the origin. Thus, \(h\) varies parabolically with \(t\), since \(h\propto t^2\). Its gradient increases as time increases.
(b) Graph of \(h\) against \(t^2\)
This is a straight line through the origin. Therefore,
\[h\propto t^2.\]
The gradient of the graph is
\[\frac{h}{t^2}=4.9=\frac{g}{2}.\]
Hence, \(g=2\times4.9=9.8\ \text{m s}^{-2}\).
Bayanin Amsa
For a particle dropped from rest, the initial velocity, \(u=0\). Hence,
\[h=ut+\frac{1}{2}gt^2=\frac{1}{2}gt^2.\]
Taking \(g=9.8\ \text{m s}^{-2}\),
\[h=4.9t^2.\]
(a) Graph of \(h\) against \(t\)
The graph is a parabola passing through the origin. Thus, \(h\) varies parabolically with \(t\), since \(h\propto t^2\). Its gradient increases as time increases.
(b) Graph of \(h\) against \(t^2\)
This is a straight line through the origin. Therefore,
\[h\propto t^2.\]
The gradient of the graph is
\[\frac{h}{t^2}=4.9=\frac{g}{2}.\]
Hence, \(g=2\times4.9=9.8\ \text{m s}^{-2}\).
