WAEC SSCE - Physics - 1989

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Temperature is NOT a vector quantity. Vector quantities have both magnitude and direction, whereas scalar quantities have only magnitude. Momentum, force, velocity, and displacement are all vector quantities because they have both magnitude and direction. Temperature, on the other hand, only has a magnitude (e.g. 25 degrees Celsius), and does not have a direction. Therefore, temperature is a scalar quantity.

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In a wave, the maximum displacement of particles from their equilibrium positions is called the amplitude. The amplitude represents the maximum height of a wave or the distance between the crest (highest point) and the trough (lowest point) of a wave. It is a measure of the wave's energy, and the greater the amplitude, the more energy the wave carries. For example, in a sound wave, the amplitude determines the loudness of the sound, and in a light wave, it determines the brightness of the light.

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Water in an open container boils when the vapor pressure of the water equals the atmospheric pressure. At higher altitudes, the atmospheric pressure is lower than at sea-level because the column of air above is shorter. As a result, water boils at a lower temperature at higher altitudes because the lower pressure means that water molecules need less thermal energy to break the intermolecular bonds and evaporate into the air. Therefore, the correct option is that at the top of a mountain, the pressure is lower than that at sea-level, leading to a lower boiling point of water.

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The tangential velocity of a circular motion is given by the product of the radius and the angular velocity. When the string is cut, the centripetal force acting on the stone is removed, and the stone moves off the circle with the same tangential velocity. So, the tangential velocity of the stone when the string is cut is: v = rω = 4m × 2 rad/s = 8 m/s Therefore, the answer is 8.0 ms^-1.

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The property that is not associated with longitudinal waves is polarization. Longitudinal waves are waves in which the oscillations of the particles of the medium are in the same direction as the direction of wave propagation. In other words, the wave moves parallel to the oscillations of the particles. Examples of longitudinal waves include sound waves and seismic waves. Compression is a property of longitudinal waves in which the particles of the medium are pushed together, while rarefaction is when the particles are spread apart. Reflection is the property where a wave bounces back after hitting a boundary. Refraction is the property where a wave bends as it passes from one medium to another. Diffraction is the property where a wave bends around a barrier or through an opening.

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The image in a pinhole camera is always inverted. This is because the light passing through the small hole in the camera travels in straight lines and forms an inverted image on the opposite side of the camera. The size of the image depends on the size of the hole, the distance between the hole and the image plane, and the distance between the object and the hole. However, the image is always inverted regardless of these factors.

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A hydrometer is used to determine the relative density of the acid in a car battery. A hydrometer is a device that measures the specific gravity or relative density of a liquid. The acid in a car battery is a mixture of sulfuric acid and water, and its density changes with the state of charge of the battery. A hydrometer can be used to measure the relative density of the acid, which provides an indication of the state of charge of the battery. When the battery is fully charged, the acid has a higher density than when it is discharged. Therefore, by measuring the relative density of the acid, it is possible to determine the state of charge of the battery.

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Black surfaces will radiate heat energy best. This is because black surfaces absorb all wavelengths of light and reflect very little, which means they absorb more heat and therefore radiate more heat energy. White and light-colored surfaces reflect most of the incoming light, and absorb less heat energy. On the other hand, dark colors, especially black, absorb more light and heat energy, which makes them better radiators of heat. Therefore, black surfaces are considered to be the best radiators of heat energy among the given options.

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The instrument that can be used to compare the relative magnitudes of charges on two given bodies is the gold-leaf electroscope. The gold-leaf electroscope works by using two thin gold leaves suspended from a metal rod within a glass container. When a charged object is brought near the metal rod, the charges induce a separation of charges in the gold leaves, causing them to repel each other and move apart. The amount of separation of the leaves is proportional to the magnitude of the charge on the object. To compare the charges on two bodies, you would first bring one body near the metal rod and observe the amount of separation of the leaves. Then, you would discharge the electroscope by touching the metal rod with your hand, and repeat the process with the other body. By comparing the amount of separation of the leaves in each case, you can determine which body has the greater charge. The other options listed are not directly used to compare charges on two bodies. The electrophorous is used to generate a static charge on an object, the ebonite rod can be charged by rubbing it with fur or silk, the proof plane is used to transfer charge to the gold-leaf electroscope, and the capacitor is used to store electrical charge.

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To calculate the heat energy required to vaporize 50g of water initially at 80^oC, we need to first calculate the amount of heat energy required to raise the temperature of the water from 80^oC to its boiling point, which is 100^oC. Using the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, we have: Q₁ = (50g)(4.2Jg^-1K)(100^oC - 80^oC) = 4200J The amount of heat energy required to vaporize the water at its boiling point can be calculated using the formula Q = mL, where Q is the heat energy, m is the mass, and L is the specific latent heat of vaporization. We have: Q₂ = (50g)(2260Jg^-1) = 113000J Therefore, the total amount of heat energy required to vaporize 50g of water initially at 80^oC is: Q_total = Q₁ + Q₂ = 4200J + 113000J = 117200J Thus, the answer is 117200J.

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A stringed instrument produces sound through the vibration of its strings, which are usually stretched between two points. The sound is amplified through a resonating chamber. Therefore, the stringed instrument among the options listed is the piano. The piano has strings that are struck by hammers when the keys are pressed, producing different notes and sounds.

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The velocity of a wave, v is given by v = fλ, where f is the frequency of the wave and λ is the wavelength. For a standing wave in a string, the wavelength λ is equal to twice the length of the string, since it is fixed at both ends. Therefore, λ = 2 × 50 cm = 100 cm = 1 m The velocity of the wave is given as 300 m/s. So, 300 m/s = f × 1 m Hence, frequency f = 300 Hz The number of vibrations (cycles) per second is equal to the frequency of the wave, therefore the number of vibrations made by the string in 1s is 300. Therefore, the answer is 300 vibrations per second.

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The heat generated by the current passing through the resistor is given by the formula H = I²RT, where H is the heat energy, I is the current, R is the resistance, and T is the time for which the current flows. In this case, I = 2A, R = 6Ω, and T = 2.5s. So, the heat generated is: H = (2A)² x 6Ω x 2.5s = 60J This heat is used to evaporate 5g of a liquid at its boiling point. The specific latent heat of vaporization (L) of the liquid is given by the formula: L = H/m where m is the mass of the liquid. In this case, m = 5g, so: L = 60J / 5g = 12J/g Therefore, the specific latent heat of the liquid is 12J/g. The option closest to this value is 120 Jg^-1, so that would be the answer.

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The motion of the prongs of a sounding tuning fork is vibratory. When the prongs are struck or set into vibration, they oscillate back and forth in opposite directions, creating a series of compressions and rarefactions in the surrounding air molecules. This causes the air molecules to vibrate and propagate sound waves through the air, which we perceive as sound. The vibratory motion of the prongs is what sets up the sound wave, which is why tuning forks are often used as a reference pitch for tuning musical instruments.

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Glass is not a good conductor of electricity. This is because glass is an insulator and has very few free electrons that can move and conduct electricity. In contrast, metals such as aluminium and copper are good conductors of electricity because they have many free electrons that can move freely and carry an electric current. The human body and the Earth can also conduct electricity, although their conductivity is lower than that of metals.

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When the sound pulse is sent vertically downwards into the earth, it is reflected from the two layers and returns back to the surface. The time interval between the two echoes is the time taken by the pulse to travel from the first layer to the second layer and back. Let's assume that the distance from the surface to the first layer is 'd' meters and the distance between the first and second layer is 'x' meters. The total distance covered by the pulse would be equal to the sum of the distance travelled to reach the first layer, the distance travelled between the two layers, and the distance travelled to return to the surface. Total distance = d + 2x Using the formula for speed, distance and time, we can express the time taken for the pulse to travel to the first layer and back as: 2d/2000 Similarly, the time taken for the pulse to travel to the second layer and back can be expressed as: 2(d + x)/2000 Given that the time interval between the two echoes is 0.2s, we can set up the following equation: 2(d + x)/2000 - 2d/2000 = 0.2 Simplifying this equation, we get: d + x - d = 200 x = 200m Therefore, the distance between the two layers is 200m. Thus, option A is the correct answer.

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The correct option is: charges generated by friction in the tanker can be conducted to the earth. The short chain is usually made of metal and is attached to the back of a petrol tanker trailing behind it to prevent the buildup of static electricity caused by friction between the tanker and the road surface. As the tanker moves, friction between the wheels and the road generates static electricity, which can accumulate on the metal body of the tanker. If the static charge buildup is not dissipated, it can cause a spark, which can ignite the fuel vapor, leading to an explosion. The short chain attached to the tanker helps to conduct the static charge to the earth, reducing the risk of explosion.

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In a transverse wave, nodes are the points that do not move while the anti-nodes are the points of maximum displacement. The distance between a node and an adjacent anti-node is equal to one-fourth of the wavelength. This is because, in one complete cycle of the wave, there is one node and one anti-node, and the distance between them is half a wavelength. Thus, the distance between a node and an anti-node is one-fourth of a wavelength. Therefore, the correct option is "one-fourth of the wavelength."

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The power of an engine is the rate at which it does work. We can use the formula P = W/t, where P is the power, W is the work done and t is the time taken. In this question, the work done is the product of the force applied and the distance moved in the direction of the force. Since the engine is raising a mass of 1000kg, the force applied is equal to the weight of the water, which is given by F = m*g, where m is the mass and g is the acceleration due to gravity. Thus, F = 1000kg * 10m/s^2 = 10,000N. The distance moved by the water is the height through which it is raised, which is 60m. Therefore, the work done by the engine is given by W = F*d = 10,000N * 60m = 600,000J. Finally, we can substitute these values into the power formula to obtain P = W/t = 600,000J / 20s = 30,000W. Therefore, the power of the engine is 30,000W. Answer: 30,000 W.

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A derived unit is a unit of measurement that is derived from the seven base units of the International System of Units (SI). The base units are the units for length, mass, time, electric current, temperature, amount of substance, and luminous intensity. The units of metre, kilogram, second and ampere are base units, whereas the unit of coulomb is a derived unit, defined as the quantity of electric charge transported by a constant current of one ampere in one second. Therefore, the correct answer is Coulomb.

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The gravitational potential due to a molecule of mass, m and at a distance, r from it is given by the formula: Gm/r. The gravitational potential is a measure of the work done per unit mass in moving an object from infinity to a point in the gravitational field. Gravitational potential energy is the energy associated with an object due to its position in a gravitational field. The higher the gravitational potential energy, the more work is required to move the object against the force of gravity. The formula for gravitational potential due to a point mass is a simple expression of the strength of the gravitational field created by a mass at a distance r. This formula is widely used in physics to calculate the gravitational potential at a point due to a body of mass m.

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(a)

Wave motion is the propagation of a disturbance from one point to another, transmitting energy without any net transfer of matter.

For the wave equation

\[y=A\sin\left[\frac{2\pi}{\lambda}(Vt-X)\right],\]

0 is the amplitude, that is, the maximum displacement of a particle of the medium from its equilibrium position.

\(\lambda\) is the wavelength, that is, the distance between two successive points vibrating in the same phase, such as two successive crests.

(b)(i)

Set up a resonance tube containing water and a movable vertical tube. Strike the sound source, such as a tuning fork, gently and hold it just above the open end of the tube. Raise the tube slowly so that the length \(l\) of the enclosed air column increases.

The sound becomes loudest at resonance. At the first resonance, the air column has a node at the water surface and an antinode at the open end; hence, neglecting end correction,

\[l=\frac{\lambda}{4}.\]

Therefore,

\[\lambda=4l\]

and, if \(v\) is the velocity of sound in air, the frequency \(f\) of the source is

\[f=\frac{v}{\lambda}=\frac{v}{4l}.\]

Alternatively, locate two successive resonance lengths \(l_1\) and \(l_2\). Then \(l_2-l_1=\lambda/2\), so that

\[f=\frac{v}{2(l_2-l_1)}.\]

(b)(ii)

For the fundamental note of a pipe closed at one end,

\[L=\frac{\lambda}{4}.\]

Thus,

\[\lambda=4L=4(1)=4\ \text{m}.\]

\[f=\frac{v}{\lambda}=\frac{340}{4}=85\ \text{Hz}.\]

Frequency of the note = \(85\ \text{Hz}\).

(c)

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.

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(a) Laws of electromagnetic induction

• Faraday's law: whenever there is a change in the magnetic flux linking a circuit, an e.m.f. is induced, and the magnitude of the induced e.m.f. is directly proportional to the rate of change of the magnetic flux linkage.
• Lenz's law: the direction of the induced current is always such that it opposes the change producing it.

(b)(i) Simple experiment

Connect a coil of wire to a sensitive galvanometer. When a bar magnet is pushed into (or pulled out of) the coil, the galvanometer deflects, showing that an e.m.f. (and current) is induced. When the magnet is held still there is no deflection.

(b)(ii) Two factors on which the induced e.m.f. depends

• The rate at which the magnetic flux changes (the speed of the magnet).
• The number of turns on the coil (and the strength of the magnet).

(c) r.m.s. value of an alternating current

The root-mean-square value of an alternating current is the value of the steady direct current that would dissipate heat in a given resistor at the same rate as the alternating current does.

(d)(i) Meaning of the symbols in \( I = I_o \sin \omega t \)

• \( I \) is the instantaneous value of the current at time t.
• \( I_o \) is the peak (maximum) value of the current.
• \( \omega \) is the angular frequency, \( \omega = 2\pi f \).
• \( \omega t \) is the phase angle of the current at time t.

(d)(ii) Instantaneous value

Peak value: \( I_o = I_{rms}\sqrt{2} = 15.0 \times 1.414 = 21.2\,\text{A} \).

At phase angle \( \omega t = 30^\circ \):

\[ I = I_o \sin 30^\circ = 21.2 \times 0.5 = 10.6\,\text{A} \]

The instantaneous current is about 10.6 A.

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(a)(i) Photoelectric emission and threshold frequency

• Photoelectric emission is the ejection of electrons from the surface of a metal when electromagnetic radiation (light) of sufficiently high frequency falls on it.
• Threshold frequency is the minimum frequency of the incident radiation below which no electrons are emitted, no matter how intense the light.

(a)(ii) Symbols in \( E = hf - hf_o \)

• \( E \) is the maximum kinetic energy of the emitted (photo)electron.
• \( hf \) is the energy of the incident photon (h = Planck's constant, f = its frequency).
• \( hf_o \) is the work function of the metal (\( f_o \) is the threshold frequency).

(b) Frequency of the incident radiation

By Einstein's equation, the photon energy equals the work function plus the kinetic energy given to the electron:

\[ hf = W + KE = (1.33 \times 10^{-16}) + (1.97 \times 10^{-16}) = 3.30 \times 10^{-16}\,\text{J} \]

\[ f = \frac{hf}{h} = \frac{3.30 \times 10^{-16}}{6.60 \times 10^{-34}} = 5.0 \times 10^{17}\,\text{Hz} \]

The frequency of the incident radiation is \( 5.0 \times 10^{17}\,\text{Hz} \) (treating the given energies in joules).

(c) Below the threshold frequency, each photon does not carry enough energy to free an electron; the light energy absorbed by the surface electrons is simply re-radiated or converted into heat, so no electron gains sufficient energy to overcome the work function and escape. The number of electrons emitted (once above threshold) depends on the intensity (brightness) of the incident light.

(d) Duality of matter

The duality of matter means that matter (such as electrons) can behave both as particles and as waves. It behaves as a particle in phenomena such as the photoelectric effect and collisions, and as a wave in phenomena such as electron diffraction and interference, where a beam of electrons produces a diffraction pattern like that of light. The de Broglie relation \( \lambda = \dfrac{h}{mv} \) links the two aspects.

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(a) Meaning of a linear expansivity of \( 1.0 \times 10^{-5}\,\text{K}^{-1} \)

It means that each unit length of the solid increases by \( 1.0 \times 10^{-5} \) of its original length for every 1 K (1 °C) rise in temperature. For example, a 1 m length would lengthen by \( 1.0 \times 10^{-5}\,\text{m} \) for each degree rise.

(b)(i) Experiment to determine the linear expansivity of a steel rod

Measure the original length \( l_0 \) of the steel rod and its initial temperature \( \theta_1 \). Place the rod in a jacket through which steam is passed, with one end fixed and the other end pressing against a micrometer/spherometer (or an optical lever). Record the increase in length \( \Delta l \) as the rod is heated to the steam temperature \( \theta_2 \). The linear expansivity is

\[ \alpha = \frac{\Delta l}{l_0(\theta_2 - \theta_1)} \]

(b)(ii) Safety gap between rails

Expansion of one bar: \( \Delta l = l_0 \, \alpha \, \Delta\theta \)

\[ \Delta l = 3 \times (1.0 \times 10^{-5}) \times (41 - 29) \]

\[ \Delta l = 3 \times 1.0 \times 10^{-5} \times 12 = 3.6 \times 10^{-4}\,\text{m} = 0.36\,\text{mm} \]

A gap of about 0.36 mm must be left between successive bars.

(c) Advantages and disadvantages of thermal expansion of solids

Three advantages:

• Fitting a metal tyre or ring onto a wheel (heat to expand, then cools and grips tightly).
• Hot riveting of metal plates (rivet contracts on cooling and clamps the plates firmly).
• Use in bimetallic strips for thermostats, thermometers and fire alarms.

Two disadvantages:

• Railway lines and bridges may buckle or bend if no expansion gaps are provided.
• Overhead telephone/electric cables sag in hot weather and may snap when they contract in cold weather; tight-fitting parts may jam or crack.