WAEC SSCE - Physics - 1999 (Essay)

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(a) Two types of waves, apart from light, that can be plane polarised. Only transverse waves can be polarised. Examples:

1. Radio waves (electromagnetic).
2. Microwaves (electromagnetic). (Other acceptable answers: X-rays, or transverse mechanical waves on a stretched string.)

(b) Two applications of plane polarised light:

1. Polaroid sunglasses and camera filters, used to cut out glare from reflected light.
2. Measuring the concentration of optically active solutions such as sugar (polarimetry / saccharimetry); also used in stress analysis of materials (photoelasticity) and in LCD screens.

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(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 \(2.84\).

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(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:

1. Raising the temperature of the liquid.
2. Adding impurities such as soap, detergent or oil (surfactants) to the liquid.

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(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:

1. Fusion releases more energy per unit mass of fuel than fission.
2. Its fuel (hydrogen isotopes from water) is very cheap and practically unlimited.
3. 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 \(8.28\times10^{-11}\,\text{J}\).

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(a) Conditions for equilibrium of a rigid body under parallel forces:

1. The algebraic sum of the forces is zero, i.e. the total upward force equals the total downward force (no net force).
2. 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:

1. Read the balance positions with the eye directly above the scale to avoid parallax error.
2. 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 \(= 1.0\,\text{m}\).

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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 \(80\,\text{m}\).

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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 \(\tfrac{0.11}{1.11}\times100\% \approx 10\%\).)

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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.

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(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.

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(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:

1. The photoelectric effect.
2. The Compton effect (scattering of X-rays by electrons). Emission of line/black-body spectra is also acceptable.

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(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:

1. The nature (chemical composition) of the fluorescent powder (phosphor) coating the tube.
2. The nature of the gas or vapour (and the pressure) inside the tube.

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(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, \( V \propto T \).

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(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:

1. The stretching force (load) \(F\) applied to the wire.
2. The original (unstretched) length \(L\) of the wire.
3. The extension \(e\) produced by the load.
4. The diameter of the wire (to obtain the cross-sectional area \(A\)).