WAEC SSCE - Physics - 1999 (Essay)
Tambaya 1 Rahoto
(a) State Heisenberg's uncertainty principle.
(b) Mention two phenomena that can only be explained in terms of the particulate nature of light.
(a) Heisenberg's uncertainty principle: it is impossible to determine simultaneously, with perfect accuracy, both the position and the momentum of a particle. The more accurately one is known, the less accurately the other can be known. In symbols:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]where \(\Delta x\) is the uncertainty in position, \(\Delta p\) the uncertainty in momentum and \(h\) Planck's constant.
(b) Two phenomena explained only by the particulate (quantum) nature of light:
- The photoelectric effect.
- The Compton effect (scattering of X-rays by electrons). Emission of line/black-body spectra is also acceptable.
Bayanin Amsa
(a) Heisenberg's uncertainty principle: it is impossible to determine simultaneously, with perfect accuracy, both the position and the momentum of a particle. The more accurately one is known, the less accurately the other can be known. In symbols:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]where \(\Delta x\) is the uncertainty in position, \(\Delta p\) the uncertainty in momentum and \(h\) Planck's constant.
(b) Two phenomena explained only by the particulate (quantum) nature of light:
- The photoelectric effect.
- The Compton effect (scattering of X-rays by electrons). Emission of line/black-body spectra is also acceptable.
Tambaya 2 Rahoto
(a) State Hooke's law of elasticity
(b) A spiral spring, loaded with a piece of metal, extends by 10.5cm hair. When the metal is fully submerged in water, the spring extends by 6.8cm. Calculate the relative density of the metal. (Assume Hooke's law is obeyed).
(a) Hooke's law: provided the elastic limit is not exceeded, the extension of an elastic material (spring) is directly proportional to the stretching force (load) applied to it, i.e. \( F = ke \).
(b) Because the spring obeys Hooke's law, its extension is directly proportional to the force stretching it.
In air: the extension is caused by the full weight \(W\) of the metal, so \( W \propto 10.5 \).
In water: the metal experiences an upthrust, so the spring is stretched only by the apparent weight \(W'\), where \( W' \propto 6.8 \).
The upthrust equals the loss in weight (weight of water displaced):
\[ U = W - W' \propto (10.5 - 6.8) = 3.7 \]Relative density:
\[ \text{R.D.} = \frac{\text{weight of metal in air}}{\text{upthrust in water}} = \frac{W}{W - W'} = \frac{10.5}{3.7} \] \[ \text{R.D.} = 2.84 \]Answer: the relative density of the metal is \(2.84\).
Bayanin Amsa
(a) Hooke's law: provided the elastic limit is not exceeded, the extension of an elastic material (spring) is directly proportional to the stretching force (load) applied to it, i.e. \( F = ke \).
(b) Because the spring obeys Hooke's law, its extension is directly proportional to the force stretching it.
In air: the extension is caused by the full weight \(W\) of the metal, so \( W \propto 10.5 \).
In water: the metal experiences an upthrust, so the spring is stretched only by the apparent weight \(W'\), where \( W' \propto 6.8 \).
The upthrust equals the loss in weight (weight of water displaced):
\[ U = W - W' \propto (10.5 - 6.8) = 3.7 \]Relative density:
\[ \text{R.D.} = \frac{\text{weight of metal in air}}{\text{upthrust in water}} = \frac{W}{W - W'} = \frac{10.5}{3.7} \] \[ \text{R.D.} = 2.84 \]Answer: the relative density of the metal is \(2.84\).
Tambaya 3 Rahoto
(a) State three properties of waves
(b)(i) Describe, with the aid of a labelled diagram, an experiment to show how the frequency of the note emitted by a vibrating string depends on the tension in the string
(ii) State two precautions necessary to ensure accurate results.
(c) Draw a ray diagram showing how a virtual image of an object is formed by a concave mirror.
(a) Three properties of waves are:
- Reflection
- Refraction
- Diffraction
(b)(i) Experiment to show the relation between frequency and tension
A sonometer is set up as shown below. A uniform wire is stretched over two knife-edge bridges, A and B. One end of the wire passes over a smooth pulley and supports a scale pan. The distance AB is kept constant throughout the experiment.
- Place a small paper rider at the middle of the wire between A and B.
- Put a known total mass, including the scale pan, on the hanger. The tension in the wire is then \(T=mg\).
- Strike a tuning fork of known frequency and hold its stem on the sonometer box. Pluck the wire gently.
- Adjust the load until resonance occurs. At resonance the vibration of the wire is large and the paper rider is thrown off.
- Record the total suspended mass and the frequency of the tuning fork.
- Repeat the procedure with other tuning forks, keeping the vibrating length AB unchanged.
A set of readings is:
| Frequency, \(f\) (Hz) | Total suspended mass, \(m\) (kg) | \(\sqrt{m}\) (kg1/2) |
|---|---|---|
| 256 | 1.00 | 1.00 |
| 320 | 1.5625 | 1.25 |
| 384 | 2.25 | 1.50 |
| 512 | 4.00 | 2.00 |
The graph of \(f\) against \(\sqrt{m}\) is:
The straight line passes through the origin. Its gradient is
\[\frac{512-256}{2.00-1.00}=256\ \text{Hz kg}^{-1/2}.\]
Hence \(f=256\sqrt{m}\). Since \(T=mg\), \(\sqrt{m}=\sqrt{T/g}\), and therefore
\[f\propto \sqrt{T}.\]
(b)(ii) Precautions
- Keep the distance between the knife-edge bridges fixed, and ensure that the wire is taut and firmly held at the bridges.
- Strike the tuning fork gently on a rubber pad and identify resonance only when the paper rider is thrown off, so that the frequency of the fork is not altered or damped.
(c) Virtual image formed by a concave mirror
When the object is between the pole \(P\) and principal focus \(F\), the reflected rays diverge. Their backward extensions meet behind the mirror at a virtual, erect and magnified image.
Bayanin Amsa
(a) Three properties of waves are:
- Reflection
- Refraction
- Diffraction
(b)(i) Experiment to show the relation between frequency and tension
A sonometer is set up as shown below. A uniform wire is stretched over two knife-edge bridges, A and B. One end of the wire passes over a smooth pulley and supports a scale pan. The distance AB is kept constant throughout the experiment.
- Place a small paper rider at the middle of the wire between A and B.
- Put a known total mass, including the scale pan, on the hanger. The tension in the wire is then \(T=mg\).
- Strike a tuning fork of known frequency and hold its stem on the sonometer box. Pluck the wire gently.
- Adjust the load until resonance occurs. At resonance the vibration of the wire is large and the paper rider is thrown off.
- Record the total suspended mass and the frequency of the tuning fork.
- Repeat the procedure with other tuning forks, keeping the vibrating length AB unchanged.
A set of readings is:
| Frequency, \(f\) (Hz) | Total suspended mass, \(m\) (kg) | \(\sqrt{m}\) (kg1/2) |
|---|---|---|
| 256 | 1.00 | 1.00 |
| 320 | 1.5625 | 1.25 |
| 384 | 2.25 | 1.50 |
| 512 | 4.00 | 2.00 |
The graph of \(f\) against \(\sqrt{m}\) is:
The straight line passes through the origin. Its gradient is
\[\frac{512-256}{2.00-1.00}=256\ \text{Hz kg}^{-1/2}.\]
Hence \(f=256\sqrt{m}\). Since \(T=mg\), \(\sqrt{m}=\sqrt{T/g}\), and therefore
\[f\propto \sqrt{T}.\]
(b)(ii) Precautions
- Keep the distance between the knife-edge bridges fixed, and ensure that the wire is taut and firmly held at the bridges.
- Strike the tuning fork gently on a rubber pad and identify resonance only when the paper rider is thrown off, so that the frequency of the fork is not altered or damped.
(c) Virtual image formed by a concave mirror
When the object is between the pole \(P\) and principal focus \(F\), the reflected rays diverge. Their backward extensions meet behind the mirror at a virtual, erect and magnified image.
Tambaya 4 Rahoto
In an electrolysis experiment, the ammeter records a steady current of 1A. The mass of copper deposited is 0.66g in 30 minutes. Calculate the error in the ammeter reading (Electrochemical equivalent of copper = 0.00033g C\(^{-1}\)).
Faraday's first law of electrolysis: \( m = zIt \), where \(z\) is the electrochemical equivalent.
Step 1: Convert the time. \( t = 30 \times 60 = 1800\,\text{s} \).
Step 2: Find the true (actual) current that must have flowed to deposit the copper.
\[ I = \frac{m}{zt} = \frac{0.66}{0.00033 \times 1800} = \frac{0.66}{0.594} = 1.11\,\text{A} \]Step 3: Compare with the ammeter reading. The ammeter shows \(1.0\,\text{A}\), but the mass of copper proves the true current was \(1.11\,\text{A}\).
\[ \text{Error} = 1.11 - 1.00 = 0.11\,\text{A} \]Answer: the ammeter under-reads by \(0.11\,\text{A}\). (As a percentage of the true value this is \(\tfrac{0.11}{1.11}\times100\% \approx 10\%\).)
Bayanin Amsa
Faraday's first law of electrolysis: \( m = zIt \), where \(z\) is the electrochemical equivalent.
Step 1: Convert the time. \( t = 30 \times 60 = 1800\,\text{s} \).
Step 2: Find the true (actual) current that must have flowed to deposit the copper.
\[ I = \frac{m}{zt} = \frac{0.66}{0.00033 \times 1800} = \frac{0.66}{0.594} = 1.11\,\text{A} \]Step 3: Compare with the ammeter reading. The ammeter shows \(1.0\,\text{A}\), but the mass of copper proves the true current was \(1.11\,\text{A}\).
\[ \text{Error} = 1.11 - 1.00 = 0.11\,\text{A} \]Answer: the ammeter under-reads by \(0.11\,\text{A}\). (As a percentage of the true value this is \(\tfrac{0.11}{1.11}\times100\% \approx 10\%\).)
Tambaya 5 Rahoto
(a) Explain electrolysis
(b) Classify the following substances as electrolytes and non-electrolytes: Sugar solution: paraffin, suit solution and grape fruit juice.
(a) Electrolysis is the chemical decomposition of an electrolyte, in its molten state or in aqueous solution, brought about by the passage of a direct electric current through it. During the process the positive ions (cations) move to the cathode and the negative ions (anions) move to the anode, where they are discharged and the substance is broken down into its components.
(b) Classification. A substance is an electrolyte only if, when molten or dissolved, it contains free-moving ions that can carry the current. Sugar dissolves as neutral molecules and paraffin is a covalent non-conductor, so neither provides ions. Common salt dissolves to give \(\text{Na}^{+}\) and \(\text{Cl}^{-}\) ions, and grape fruit juice contains dissolved acids (such as citric and ascorbic acids) that release ions, so both conduct.
| Substance | Classification |
|---|---|
| Sugar solution | Non-electrolyte |
| Paraffin | Non-electrolyte |
| Salt solution | Electrolyte |
| Grape fruit juice | Electrolyte |
Common misconception: a solution is not an electrolyte simply because it is a liquid or tastes of something dissolved. Sugar solution is a liquid and conducts almost no current because sugar stays as whole neutral molecules; it is a non-electrolyte. What matters is whether free ions are present.
Bayanin Amsa
(a) Electrolysis is the chemical decomposition of an electrolyte, in its molten state or in aqueous solution, brought about by the passage of a direct electric current through it. During the process the positive ions (cations) move to the cathode and the negative ions (anions) move to the anode, where they are discharged and the substance is broken down into its components.
(b) Classification. A substance is an electrolyte only if, when molten or dissolved, it contains free-moving ions that can carry the current. Sugar dissolves as neutral molecules and paraffin is a covalent non-conductor, so neither provides ions. Common salt dissolves to give \(\text{Na}^{+}\) and \(\text{Cl}^{-}\) ions, and grape fruit juice contains dissolved acids (such as citric and ascorbic acids) that release ions, so both conduct.
| Substance | Classification |
|---|---|
| Sugar solution | Non-electrolyte |
| Paraffin | Non-electrolyte |
| Salt solution | Electrolyte |
| Grape fruit juice | Electrolyte |
Common misconception: a solution is not an electrolyte simply because it is a liquid or tastes of something dissolved. Sugar solution is a liquid and conducts almost no current because sugar stays as whole neutral molecules; it is a non-electrolyte. What matters is whether free ions are present.
Tambaya 6 Rahoto
(a) Sketch the magnetic flux pattern around a long; straight, current-carrying wire
(b) State two methods by which the sensitivity of a moving-coil galvanometer can be increased.
(c) A series RLC circuit comprises a 100-\(\Omega\) resistor, a 3-H inductor and a 4-\(\mu\)f capacitor. The a.c source of tile circuit has an e.m.f of 100V and a frequency of 160 Hz.
(i) Draw the circuit diagram of the arrangement. Calculate the:
(ii) capacitive reactance;
(iii) inductive reactance:
(iv) impedance of the circuit;
(v) current in the circuit;
(vi) average power dissipated in the circuit.
(a) Magnetic flux pattern around a long straight current-carrying wire
The wire carries current out of the plane of the paper, as indicated by the dot. The magnetic field lines are concentric circles centred on the wire. Their direction is anticlockwise, given by the right-hand grip rule.
(b) The sensitivity of a moving-coil galvanometer can be increased by:
- using a weaker control spring, so that the restoring torque is smaller;
- using a stronger magnet to increase the magnetic flux density.
(c) Given: \(R=100\,\Omega\), \(L=3\,\text{H}\), \(C=4\,\mu\text{F}=4\times10^{-6}\,\text{F}\), \(V=100\,\text{V}\), and \(f=\dfrac{160}{\pi}\,\text{Hz}\).
(i) Series RLC circuit
(ii) Capacitive reactance
\[X_C=\frac{1}{2\pi fC}=\frac{1}{2\pi\left(\dfrac{160}{\pi}\right)\left(4\times10^{-6}\right)}=781.25\,\Omega\]
\[\boxed{X_C\approx781\,\Omega}\]
(iii) Inductive reactance
\[X_L=2\pi fL=2\pi\left(\dfrac{160}{\pi}\right)(3)=960\,\Omega\]
\[\boxed{X_L=960\,\Omega}\]
(iv) Impedance
\[Z=\sqrt{R^2+(X_L-X_C)^2}\]
\[Z=\sqrt{100^2+(960-781.25)^2}=204.8\,\Omega\]
\[\boxed{Z\approx205\,\Omega}\]
(v) Current in the circuit
\[I=\frac{V}{Z}=\frac{100}{204.8}=0.488\,\text{A}\]
\[\boxed{I\approx0.49\,\text{A}}\]
(vi) Average power dissipated
Only the resistor dissipates average power:
\[P=I^2R=(0.488)^2(100)=23.8\,\text{W}\]
\[\boxed{P\approx24\,\text{W}}\]
Bayanin Amsa
(a) Magnetic flux pattern around a long straight current-carrying wire
The wire carries current out of the plane of the paper, as indicated by the dot. The magnetic field lines are concentric circles centred on the wire. Their direction is anticlockwise, given by the right-hand grip rule.
(b) The sensitivity of a moving-coil galvanometer can be increased by:
- using a weaker control spring, so that the restoring torque is smaller;
- using a stronger magnet to increase the magnetic flux density.
(c) Given: \(R=100\,\Omega\), \(L=3\,\text{H}\), \(C=4\,\mu\text{F}=4\times10^{-6}\,\text{F}\), \(V=100\,\text{V}\), and \(f=\dfrac{160}{\pi}\,\text{Hz}\).
(i) Series RLC circuit
(ii) Capacitive reactance
\[X_C=\frac{1}{2\pi fC}=\frac{1}{2\pi\left(\dfrac{160}{\pi}\right)\left(4\times10^{-6}\right)}=781.25\,\Omega\]
\[\boxed{X_C\approx781\,\Omega}\]
(iii) Inductive reactance
\[X_L=2\pi fL=2\pi\left(\dfrac{160}{\pi}\right)(3)=960\,\Omega\]
\[\boxed{X_L=960\,\Omega}\]
(iv) Impedance
\[Z=\sqrt{R^2+(X_L-X_C)^2}\]
\[Z=\sqrt{100^2+(960-781.25)^2}=204.8\,\Omega\]
\[\boxed{Z\approx205\,\Omega}\]
(v) Current in the circuit
\[I=\frac{V}{Z}=\frac{100}{204.8}=0.488\,\text{A}\]
\[\boxed{I\approx0.49\,\text{A}}\]
(vi) Average power dissipated
Only the resistor dissipates average power:
\[P=I^2R=(0.488)^2(100)=23.8\,\text{W}\]
\[\boxed{P\approx24\,\text{W}}\]
Tambaya 7 Rahoto
(a) List two types of waves, apart from light, that can be plane polarised
(b) State two applications of plane polarised light.
(a) Two types of waves, apart from light, that can be plane polarised. Only transverse waves can be polarised. Examples:
- Radio waves (electromagnetic).
- Microwaves (electromagnetic). (Other acceptable answers: X-rays, or transverse mechanical waves on a stretched string.)
(b) Two applications of plane polarised light:
- Polaroid sunglasses and camera filters, used to cut out glare from reflected light.
- Measuring the concentration of optically active solutions such as sugar (polarimetry / saccharimetry); also used in stress analysis of materials (photoelasticity) and in LCD screens.
Bayanin Amsa
(a) Two types of waves, apart from light, that can be plane polarised. Only transverse waves can be polarised. Examples:
- Radio waves (electromagnetic).
- Microwaves (electromagnetic). (Other acceptable answers: X-rays, or transverse mechanical waves on a stretched string.)
(b) Two applications of plane polarised light:
- Polaroid sunglasses and camera filters, used to cut out glare from reflected light.
- Measuring the concentration of optically active solutions such as sugar (polarimetry / saccharimetry); also used in stress analysis of materials (photoelasticity) and in LCD screens.
Tambaya 8 Rahoto
(a) State the conditions for the equilibrium of a rigid body acted upon by parallel forces
(b)(i) Describe an experiment, using the principle of moments, to determine the mass of a metre rule
(ii) State two precautions necessary to ensure accurate results.
(c) A bullet of mass 120g is fired horizontally into a fixed wooden block with a speed of 20ms\(^{-1}\). The bullet is brought to rest in the block in 0.1s by a constant resistance. Calculate the:
(i) magnitude of the resistance;
(ii) distance moved by the bullet in the wood.
(a) Conditions for equilibrium of a rigid body under parallel forces:
- The algebraic sum of the forces is zero, i.e. the total upward force equals the total downward force (no net force).
- The algebraic sum of the moments of the forces about any point is zero, i.e. the total clockwise moment equals the total anticlockwise moment (principle of moments).
(b)(i) Experiment to find the mass of a metre rule (principle of moments):
- Balance the metre rule alone on a knife-edge and note the balance point; this is its centre of gravity \(G\) (near the 50 cm mark).
- Now shift the knife-edge to one side of \(G\) and suspend a known mass \(m\) from the short arm; adjust its position until the rule balances horizontally.
- Measure the distance \(a\) of the known mass from the knife-edge and the distance \(b\) of \(G\) from the knife-edge.
- By the principle of moments, \( m \times a = M \times b \), so the mass of the rule \( M = \dfrac{m\,a}{b} \).
(ii) Two precautions:
- Read the balance positions with the eye directly above the scale to avoid parallax error.
- Ensure the knife-edge is sharp and the rule is horizontal (balanced) before taking readings.
(c) Bullet: \( m = 120\,\text{g} = 0.12\,\text{kg}, \; u = 20\,\text{m s}^{-1}, \; v = 0, \; t = 0.1\,\text{s}. \)
(i) Resistance (magnitude of the constant force):
\[ F = ma = m\frac{(u - v)}{t} = 0.12 \times \frac{20}{0.1} = 0.12 \times 200 = 24\,\text{N} \](ii) Distance moved in the wood:
\[ s = \left(\frac{u + v}{2}\right)t = \left(\frac{20 + 0}{2}\right)(0.1) = 10 \times 0.1 = 1.0\,\text{m} \]Answers: resistance \(= 24\,\text{N}\); distance \(= 1.0\,\text{m}\).
Bayanin Amsa
(a) Conditions for equilibrium of a rigid body under parallel forces:
- The algebraic sum of the forces is zero, i.e. the total upward force equals the total downward force (no net force).
- The algebraic sum of the moments of the forces about any point is zero, i.e. the total clockwise moment equals the total anticlockwise moment (principle of moments).
(b)(i) Experiment to find the mass of a metre rule (principle of moments):
- Balance the metre rule alone on a knife-edge and note the balance point; this is its centre of gravity \(G\) (near the 50 cm mark).
- Now shift the knife-edge to one side of \(G\) and suspend a known mass \(m\) from the short arm; adjust its position until the rule balances horizontally.
- Measure the distance \(a\) of the known mass from the knife-edge and the distance \(b\) of \(G\) from the knife-edge.
- By the principle of moments, \( m \times a = M \times b \), so the mass of the rule \( M = \dfrac{m\,a}{b} \).
(ii) Two precautions:
- Read the balance positions with the eye directly above the scale to avoid parallax error.
- Ensure the knife-edge is sharp and the rule is horizontal (balanced) before taking readings.
(c) Bullet: \( m = 120\,\text{g} = 0.12\,\text{kg}, \; u = 20\,\text{m s}^{-1}, \; v = 0, \; t = 0.1\,\text{s}. \)
(i) Resistance (magnitude of the constant force):
\[ F = ma = m\frac{(u - v)}{t} = 0.12 \times \frac{20}{0.1} = 0.12 \times 200 = 24\,\text{N} \](ii) Distance moved in the wood:
\[ s = \left(\frac{u + v}{2}\right)t = \left(\frac{20 + 0}{2}\right)(0.1) = 10 \times 0.1 = 1.0\,\text{m} \]Answers: resistance \(= 24\,\text{N}\); distance \(= 1.0\,\text{m}\).
Tambaya 9 Rahoto
(a) Explain (i) work: (ii) power
(b) Show that the efficiency E, the force ratio M.A. and the velocity ratio V.R. of a machine are related by the equation: E = M.A. x 100% V.R.
(c) An inclined plane of angle 15° is used raise a load of 4500N through a height of 2m. If the plane is 75% efficient calculate:
(i) velocity ratio of the plane;
(ii) work done on the load
(d) Explain Charles' law using the Kinetic theory of matter.
(a)(i) Work is done when a force moves its point of application in the direction of the force; it is the product of the force and the distance moved in that direction, \( W = F \times d \) (unit: joule).
(ii) Power is the rate of doing work (or the rate of transfer of energy), \( P = \dfrac{W}{t} \) (unit: watt).
(b) Relation between E, M.A. and V.R. For a machine:
\[ E = \frac{\text{work output}}{\text{work input}} = \frac{\text{load} \times \text{distance moved by load}}{\text{effort} \times \text{distance moved by effort}} \] \[ E = \left(\frac{\text{load}}{\text{effort}}\right) \times \left(\frac{\text{distance moved by load}}{\text{distance moved by effort}}\right) = \text{M.A.} \times \frac{1}{\text{V.R.}} \]since \( \text{V.R.} = \dfrac{\text{distance moved by effort}}{\text{distance moved by load}} \). Expressed as a percentage:
\[ E = \frac{\text{M.A.}}{\text{V.R.}} \times 100\% \](c) Inclined plane: angle \(=15^\circ\), load \(=4500\,\text{N}\), height \(=2\,\text{m}\), efficiency \(=75\%\).
(i) Velocity ratio of an inclined plane:
\[ \text{V.R.} = \frac{1}{\sin\theta} = \frac{1}{\sin 15^\circ} = \frac{1}{0.2588} = 3.86 \](ii) Work done on the load (useful work output):
\[ W_{\text{load}} = \text{load} \times \text{height} = 4500 \times 2 = 9000\,\text{J} \](d) Charles' law from the kinetic theory: at constant pressure, raising the temperature increases the average kinetic energy of the gas molecules, so they move faster and strike the walls harder and more often. To keep the pressure constant the gas must expand, so that the number of collisions per unit area per unit time is unchanged. Hence the volume increases in proportion to the absolute temperature, \( V \propto T \).
Bayanin Amsa
(a)(i) Work is done when a force moves its point of application in the direction of the force; it is the product of the force and the distance moved in that direction, \( W = F \times d \) (unit: joule).
(ii) Power is the rate of doing work (or the rate of transfer of energy), \( P = \dfrac{W}{t} \) (unit: watt).
(b) Relation between E, M.A. and V.R. For a machine:
\[ E = \frac{\text{work output}}{\text{work input}} = \frac{\text{load} \times \text{distance moved by load}}{\text{effort} \times \text{distance moved by effort}} \] \[ E = \left(\frac{\text{load}}{\text{effort}}\right) \times \left(\frac{\text{distance moved by load}}{\text{distance moved by effort}}\right) = \text{M.A.} \times \frac{1}{\text{V.R.}} \]since \( \text{V.R.} = \dfrac{\text{distance moved by effort}}{\text{distance moved by load}} \). Expressed as a percentage:
\[ E = \frac{\text{M.A.}}{\text{V.R.}} \times 100\% \](c) Inclined plane: angle \(=15^\circ\), load \(=4500\,\text{N}\), height \(=2\,\text{m}\), efficiency \(=75\%\).
(i) Velocity ratio of an inclined plane:
\[ \text{V.R.} = \frac{1}{\sin\theta} = \frac{1}{\sin 15^\circ} = \frac{1}{0.2588} = 3.86 \](ii) Work done on the load (useful work output):
\[ W_{\text{load}} = \text{load} \times \text{height} = 4500 \times 2 = 9000\,\text{J} \](d) Charles' law from the kinetic theory: at constant pressure, raising the temperature increases the average kinetic energy of the gas molecules, so they move faster and strike the walls harder and more often. To keep the pressure constant the gas must expand, so that the number of collisions per unit area per unit time is unchanged. Hence the volume increases in proportion to the absolute temperature, \( V \propto T \).
Tambaya 10 Rahoto
(a) What is the principle upon which the lighting in fluorescent tubes operate?
(b) State two factors on which the colour of light from a fluorescent tube depends.
(a) Principle of the fluorescent tube. It works on the principle of fluorescence. An electric discharge is passed through mercury vapour in the tube; the excited mercury atoms emit ultraviolet radiation. This ultraviolet light strikes the fluorescent (phosphor) powder coating the inside of the tube, which absorbs the invisible ultraviolet and re-emits the energy as visible light.
(b) Two factors on which the colour of the light depends:
- The nature (chemical composition) of the fluorescent powder (phosphor) coating the tube.
- The nature of the gas or vapour (and the pressure) inside the tube.
Bayanin Amsa
(a) Principle of the fluorescent tube. It works on the principle of fluorescence. An electric discharge is passed through mercury vapour in the tube; the excited mercury atoms emit ultraviolet radiation. This ultraviolet light strikes the fluorescent (phosphor) powder coating the inside of the tube, which absorbs the invisible ultraviolet and re-emits the energy as visible light.
(b) Two factors on which the colour of the light depends:
- The nature (chemical composition) of the fluorescent powder (phosphor) coating the tube.
- The nature of the gas or vapour (and the pressure) inside the tube.
Tambaya 11 Rahoto
(a) Explain: (i) Fusion. (ii) Fission.
(b) State three advantages of fusion over fission in the generation of power
(c) Calculate in joules, the binding energy for
\(\frac{59}{27}\) Co
(Atomic mass of \(^{59}_{2} Co\) = 58.9332 u)
(Mass of proton = 1.00783u)
(Mass of neutron = 1.00867 u)
(Unified atomic mass unit, U = 931 MeV)
(1 eV = 1.6 x 10\(^{-19}\) J)
(a)(i) Fusion: the joining together of two light nuclei to form a single heavier nucleus, with the release of a large amount of energy.
(ii) Fission: the splitting of a heavy nucleus into two lighter nuclei of comparable mass (together with a few neutrons), with the release of a large amount of energy.
(b) Three advantages of fusion over fission for power generation:
- Fusion releases more energy per unit mass of fuel than fission.
- Its fuel (hydrogen isotopes from water) is very cheap and practically unlimited.
- Its products are non-radioactive (or far less radioactive), so there is little dangerous nuclear waste, and there is no risk of an uncontrolled chain reaction.
(c) Binding energy of \( ^{59}_{27}\text{Co} \): it has \(27\) protons and \(59 - 27 = 32\) neutrons.
Total mass of the separate nucleons:
\[ = 27(1.00783) + 32(1.00867) = 27.2114 + 32.2774 = 59.4888\,\text{u} \]Mass defect:
\[ \Delta m = 59.4888 - 58.9332 = 0.5556\,\text{u} \]Binding energy (using \(1\,\text{u} = 931\,\text{MeV}\)):
\[ E = 0.5556 \times 931 = 517.3\,\text{MeV} \]Converting to joules \((1\,\text{MeV} = 10^{6}\times1.6\times10^{-19}\,\text{J}):\)
\[ E = 517.3\times10^{6} \times 1.6\times10^{-19} = 8.28\times10^{-11}\,\text{J} \]Answer: the binding energy is about \(8.28\times10^{-11}\,\text{J}\).
Bayanin Amsa
(a)(i) Fusion: the joining together of two light nuclei to form a single heavier nucleus, with the release of a large amount of energy.
(ii) Fission: the splitting of a heavy nucleus into two lighter nuclei of comparable mass (together with a few neutrons), with the release of a large amount of energy.
(b) Three advantages of fusion over fission for power generation:
- Fusion releases more energy per unit mass of fuel than fission.
- Its fuel (hydrogen isotopes from water) is very cheap and practically unlimited.
- Its products are non-radioactive (or far less radioactive), so there is little dangerous nuclear waste, and there is no risk of an uncontrolled chain reaction.
(c) Binding energy of \( ^{59}_{27}\text{Co} \): it has \(27\) protons and \(59 - 27 = 32\) neutrons.
Total mass of the separate nucleons:
\[ = 27(1.00783) + 32(1.00867) = 27.2114 + 32.2774 = 59.4888\,\text{u} \]Mass defect:
\[ \Delta m = 59.4888 - 58.9332 = 0.5556\,\text{u} \]Binding energy (using \(1\,\text{u} = 931\,\text{MeV}\)):
\[ E = 0.5556 \times 931 = 517.3\,\text{MeV} \]Converting to joules \((1\,\text{MeV} = 10^{6}\times1.6\times10^{-19}\,\text{J}):\)
\[ E = 517.3\times10^{6} \times 1.6\times10^{-19} = 8.28\times10^{-11}\,\text{J} \]Answer: the binding energy is about \(8.28\times10^{-11}\,\text{J}\).
Tambaya 12 Rahoto
Explain plane polarisation of light
Plane polarisation of light. Ordinary (unpolarised) light is a transverse wave whose vibrations of the electric field take place in all directions (all planes) perpendicular to the direction of travel of the wave.
Plane polarisation is the process by which these vibrations are restricted so that they occur in only one plane containing the direction of travel. Light in which the vibrations are confined to a single plane is called plane-polarised light.
When ordinary light passes through a polariser (for example a Polaroid sheet or by reflection at a suitable angle, or through a Nicol prism), only the component of vibration in one particular plane is transmitted; the rest are absorbed or removed. If this plane-polarised light then meets a second polariser (the analyser) whose axis is at right angles to the first, no light is transmitted, confirming that the light is polarised. Only transverse waves can be polarised, so polarisation is proof that light is a transverse wave.
Bayanin Amsa
Plane polarisation of light. Ordinary (unpolarised) light is a transverse wave whose vibrations of the electric field take place in all directions (all planes) perpendicular to the direction of travel of the wave.
Plane polarisation is the process by which these vibrations are restricted so that they occur in only one plane containing the direction of travel. Light in which the vibrations are confined to a single plane is called plane-polarised light.
When ordinary light passes through a polariser (for example a Polaroid sheet or by reflection at a suitable angle, or through a Nicol prism), only the component of vibration in one particular plane is transmitted; the rest are absorbed or removed. If this plane-polarised light then meets a second polariser (the analyser) whose axis is at right angles to the first, no light is transmitted, confirming that the light is polarised. Only transverse waves can be polarised, so polarisation is proof that light is a transverse wave.
Tambaya 13 Rahoto
A stone is projected horizontally from the top of a tower with a speed of 5ms-1. It lands on the ground level at a horizontal distance of 20m from the foot of the tower. Calculate the height of the tower. (g = 10ms\(^{-2}\))
Horizontal projectile. The horizontal and vertical motions are independent. The horizontal velocity stays constant while the vertical motion is free fall from rest.
Step 1: Find the time of flight from the horizontal motion. There is no horizontal acceleration, so
\[ x = ut \implies 20 = 5 \times t \implies t = 4\,\text{s} \]Step 2: Use the time in the vertical motion. Vertically the stone starts with zero vertical velocity and falls a height \(h\):
\[ h = ut + \tfrac{1}{2}gt^2 = 0 + \tfrac{1}{2}(10)(4)^2 = \tfrac{1}{2}(10)(16) \] \[ h = 80\,\text{m} \]Answer: the height of the tower is \(80\,\text{m}\).
Bayanin Amsa
Horizontal projectile. The horizontal and vertical motions are independent. The horizontal velocity stays constant while the vertical motion is free fall from rest.
Step 1: Find the time of flight from the horizontal motion. There is no horizontal acceleration, so
\[ x = ut \implies 20 = 5 \times t \implies t = 4\,\text{s} \]Step 2: Use the time in the vertical motion. Vertically the stone starts with zero vertical velocity and falls a height \(h\):
\[ h = ut + \tfrac{1}{2}gt^2 = 0 + \tfrac{1}{2}(10)(4)^2 = \tfrac{1}{2}(10)(16) \] \[ h = 80\,\text{m} \]Answer: the height of the tower is \(80\,\text{m}\).
Tambaya 14 Rahoto
(a) What is surface tension?
(b) State two methods by which the surface tension of a liquid can be reduced.
(a) Surface tension is the force per unit length acting along (and at right angles to) an imaginary line drawn in the free surface of a liquid, tending to make the surface behave like a stretched elastic skin and to contract to the smallest possible area. Its unit is \(\text{N m}^{-1}\).
(b) Two methods of reducing the surface tension of a liquid:
- Raising the temperature of the liquid.
- Adding impurities such as soap, detergent or oil (surfactants) to the liquid.
Bayanin Amsa
(a) Surface tension is the force per unit length acting along (and at right angles to) an imaginary line drawn in the free surface of a liquid, tending to make the surface behave like a stretched elastic skin and to contract to the smallest possible area. Its unit is \(\text{N m}^{-1}\).
(b) Two methods of reducing the surface tension of a liquid:
- Raising the temperature of the liquid.
- Adding impurities such as soap, detergent or oil (surfactants) to the liquid.
Tambaya 15 Rahoto
(a) Define Young's modulus.
(b) State the physical quantities one has to measure in order to determine the Young's modulus of a wire.
(a) Young's modulus of a material is the ratio of the tensile stress to the tensile strain, provided the elastic limit is not exceeded:
\[ Y = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/L} = \frac{FL}{Ae} \]Its unit is \(\text{N m}^{-2}\) (Pa).
(b) Physical quantities to measure for a wire:
- The stretching force (load) \(F\) applied to the wire.
- The original (unstretched) length \(L\) of the wire.
- The extension \(e\) produced by the load.
- The diameter of the wire (to obtain the cross-sectional area \(A\)).
Bayanin Amsa
(a) Young's modulus of a material is the ratio of the tensile stress to the tensile strain, provided the elastic limit is not exceeded:
\[ Y = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/L} = \frac{FL}{Ae} \]Its unit is \(\text{N m}^{-2}\) (Pa).
(b) Physical quantities to measure for a wire:
- The stretching force (load) \(F\) applied to the wire.
- The original (unstretched) length \(L\) of the wire.
- The extension \(e\) produced by the load.
- The diameter of the wire (to obtain the cross-sectional area \(A\)).
