WAEC SSCE - Physics - 1992

Akọwa Nkọwa

The force applied parallel to the surface of the table is the maximum force of friction between the block and the table that is just sufficient to make the block move on the table. The frictional force (f) acting on the block is given by the product of the coefficient of friction (μ) and the normal force (N) acting on the block. Therefore, we can write: f = μN where N = mg is the normal force on the block due to the gravitational force acting on it. Since the block is just about to move, the force applied (20N) is equal to the maximum force of friction, i.e., f = 20N Substituting the values of N and f in the equation f = μN, we get: 20 = μ × 4 × 10 Solving for μ, we get: μ = 20/40 μ = 0.5 Therefore, the coefficient of friction between the block and the table is 0.5. Hence, the correct option is (c) 0.50.

Akọwa Nkọwa

Akọwa Nkọwa

When an object is immersed in a fluid, it experiences an upthrust equal to the weight of the fluid it displaces. Let V be the volume of the solid. The weight of the solid in air is its actual weight, W_a = 0.040 N. The weight of the solid in the liquid is reduced by the upthrust, which is equal to the weight of the displaced liquid. The weight of the displaced liquid is equal to its volume times its density times the acceleration due to gravity, or Vρg. Thus, the weight of the solid in the liquid is W_l = 0.040 - Vρg. Since the solid is fully immersed in the liquid, it displaces a volume of liquid equal to its own volume. Therefore, W_l = Vρg - Vρ_lg, where ρ_l is the density of the liquid. Setting the two expressions for W equal, we have: 0.040 - Vρg = Vρg - Vρ_lg Simplifying and solving for V, we get: V = (0.040)/(2ρg - ρ_lg) Substituting the given values, we get: V = (0.040)/(2 x 800 x 9.81 - 9.81) V ≈ 2.0 x 10^-6 m³ Therefore, the volume of the solid is approximately 2.0 x 10^-6 m³. The correct option is (i) 2.0 x 10^-6m³.

Akọwa Nkọwa

The longitudinal waves are waves in which the particles of the medium vibrate parallel to the direction of propagation of the wave. Ripples on the surface of water are not longitudinal waves because the particles of water move in circular motion, which is perpendicular to the direction of propagation of the wave. Waves produced by a tuning fork vibrating in air are longitudinal waves because the vibrations of the tuning fork cause the particles of air to vibrate back and forth in the same direction as the propagation of the wave. Light waves are not longitudinal waves because they are transverse waves. In a transverse wave, the particles of the medium vibrate perpendicular to the direction of propagation of the wave. Waves produced by a flute are not longitudinal waves because the vibrations of the air column in the flute are perpendicular to the direction of propagation of the wave. Therefore, the correct answer is II and IV only.

Akọwa Nkọwa

Akọwa Nkọwa

The pressure at any point in a fluid is given by the formula P = hρg, where P is the pressure at the point, h is the height of the fluid above the point, ρ is the density of the fluid and g is the acceleration due to gravity. In this problem, the height of the mercury column above the membrane is (76 - 10) = 66 cm = 0.66 m. The density of mercury is 13600 kg/m³ and the acceleration due to gravity is 10 m/s². Substituting these values into the formula gives: P = hρg = 0.66 x 13600 x 10 = 8976 N/m² = 8.98 x 10⁴ Nm⁻² (to 3 significant figures) Therefore, the correct option is: 1.17 x 10⁵ Nm^-2.

Akọwa Nkọwa

The expression that is correct is F = EAe/l. Here's why: Young's modulus (E) is a measure of the stiffness of a material. The larger the value of E, the harder it is to stretch the material. The cross-sectional area (A) of the string is a measure of how much material there is in the string. A larger area means more material, which means a greater force is required to stretch the string. The length of the string (l) is the distance between the two points that the string is attached to. The length extension (e) is the amount by which the length of the string increases when a force F is applied. The formula for Young's modulus is E = stress/strain, where stress is the force applied per unit area and strain is the extension per unit length. In this problem, we can use Hooke's law to relate the force applied to the extension of the string: F = kx, where k is the spring constant and x is the displacement of the string. Since the string is elastically stretched, we can assume that Hooke's law applies. The strain in the string is e/l, and the stress is F/A. Therefore, we can write: E = (F/A)/(e/l) Rearranging this equation gives: F = EAe/l So the correct expression is F = EAe/l.

Akọwa Nkọwa

We can use the formula: distance = (initial velocity × time) + (1/2 × acceleration × time²) In this case, the body starts from rest, so the initial velocity is 0. The acceleration is given as 3 ms⁻¹ and the time is 8 seconds. Therefore, distance = (0 × 8) + (1/2 × 3 × 8²) = 0 + (1/2 × 3 × 64) = 96m Therefore, the distance covered by the body during the acceleration is 96m. So the correct option is (E) 96m.

Akọwa Nkọwa

The second statement "Magnetic fields are scalar quantities" is not correct. Explanation: Scalar quantities are those that have only magnitude but no direction, such as temperature or mass. On the other hand, magnetic fields are vector quantities, which have both magnitude and direction. The direction of the magnetic field is given by the direction of the force that a north pole of a compass would experience if placed in the field. Therefore, magnetic fields cannot be described as scalar quantities.

Akọwa Nkọwa

Charles' Law states that, for a given mass of gas at constant pressure, the volume of the gas is directly proportional to its temperature measured in Kelvin. So we can use the formula V1/T1 = V2/T2, where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature. Converting the temperatures to Kelvin, we have T1 = 283K and T2 = 303K. We know that the gas obeys Charles' law exactly, so we can use the formula to find the final volume: V2 = (V1/T1) x T2 V2 = (283/283) x 303 V2 = 303 Therefore, the final volume of the gas at 30°C is 303cm^3. So the correct option is 303cm^3.

Akọwa Nkọwa

Akọwa Nkọwa

Electrical appliances and lamps are arranged in parallel across the mains so as to enable the voltage across the appliances not to be affected when the lamps are switched on and off. When devices are connected in parallel, they receive the same voltage as each other. This is because each device is connected to the same two points in the circuit, and the voltage difference between those points is the same for each device. In contrast, if the devices were connected in series, the voltage would be divided between them, and the devices would not receive the same voltage. Therefore, in a domestic circuit, electrical appliances and lamps are arranged in parallel across the mains to ensure that the voltage across each device remains constant regardless of whether other devices in the circuit are switched on or off.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

The power P dissipated by an electric heater of resistance R when connected to a voltage V is given by the formula: P = V²/R where P is in watts, V is in volts and R is in ohms. In this case, the electric heater has a resistance of 1.8 x 10³ Ω and is connected to a 240 V mains supply. So, the power dissipated by the electric heater is: P = V²/R = (240²)/(1.8 x 10³) = 32 W To calculate the time t in which 4.8 kJ of energy would be expended, we need to use the formula: E = P x t where E is the energy expended in joules, P is the power in watts, and t is the time in seconds. In this case, E = 4.8 kJ = 4,800 J and P = 32 W. Substituting these values into the formula, we get: 4,800 = 32 x t Solving for t, we get: t = 4,800 / 32 = 150 seconds Therefore, the time in which 4.8 kJ of energy would be expended when the electric heater is used on a 240 V mains supply is 150 seconds. Answer: 150.0s.

Akọwa Nkọwa

When a material is heated, it expands in length. The linear expansion ΔL of a material can be calculated using the formula: ΔL = αLΔT where α is the coefficient of linear expansion, L is the original length of the material, and ΔT is the change in temperature. In this case, we need to calculate the amount of linear expansion of the steel bars between 28°C and 40°C. The change in temperature is: ΔT = 40°C - 28°C = 12°C The original length of each bar is: L = 3.0 m = 300 cm The coefficient of linear expansion for steel is given as 1.0 x 10^-5°C. Therefore, the linear expansion of each bar is: ΔL = αLΔT = (1.0 x 10^-5 °C)^-1 (300 cm) (12°C) = 0.036 cm To ensure that the bars do not expand into each other, a safety gap must be left between them. This safety gap is equal to the amount of linear expansion of each bar. Therefore, the safety gap is: 0.036 cm = 3.6 x 10^-2cm Therefore, the correct answer is: 3.6 x 10^-2cm.

Akọwa Nkọwa

Images formed by a convex mirror are always erect, virtual, and diminished. A convex mirror is a mirror that bulges outward and reflects light outwards. When an object is placed in front of a convex mirror, the light rays from the object are reflected by the mirror and appear to come from a point behind the mirror. This point is known as the virtual image of the object. The image formed by a convex mirror is always erect, meaning it is upright and not inverted. It is also virtual, meaning that the light rays do not actually converge at the point where the image appears to be located. Instead, they only appear to be coming from that point. Finally, the image is diminished, meaning that it is smaller in size than the actual object. Convex mirrors are commonly used as side-view mirrors in cars because they provide a wider field of view and allow the driver to see more of the surrounding area. The fact that the image formed by a convex mirror is always smaller than the actual object also helps to minimize the distortion of the image and prevent objects from appearing closer than they actually are. In summary, images formed by a convex mirror are always erect, virtual, and diminished. They are commonly used as side-view mirrors in cars due to their wider field of view and ability to minimize distortion.

Akọwa Nkọwa

Akọwa Nkọwa

Black surfaces will radiate heat energy best. All objects radiate heat energy in the form of infrared radiation. The rate of radiation depends on several factors, including the temperature and color of the object's surface. Darker surfaces, such as black, absorb more light and radiation compared to lighter surfaces, such as white. Because of this, they also emit more heat energy. When a surface absorbs light or heat, the energy is either reflected or absorbed by the surface. If the surface is a dark color, more of the energy will be absorbed, causing the surface to become hotter. As the temperature of the surface increases, the amount of heat energy that is radiated back into the environment also increases. Therefore, in the given options, black surfaces will radiate heat energy best because they absorb the most amount of heat energy and also emit the most amount of heat energy.

Akọwa Nkọwa

The potential energy of an object is given by mgh, where m is the mass of the object, g is the acceleration due to gravity and h is the height of the object. As the object falls from a height of 10m, its potential energy is converted to kinetic energy, which is given by (1/2)mv², where m is the mass of the object and v is its velocity. At the instant just before the object strikes the ground, all its potential energy has been converted to kinetic energy. Therefore, the kinetic energy just before the object strikes the ground is given by: KE = (1/2)mv² where m = 5kg and g = 10ms^-2 The velocity of the object just before it strikes the ground can be found using the equation: v² - u² = 2gh where u is the initial velocity of the object (which is zero, as the object is dropped from rest). Substituting the given values, we get: v² = 2gh = 2 x 10 x 10 = 200 v = sqrt(200) = 14.14ms^-1 Substituting the value of v in the equation for kinetic energy, we get: KE = (1/2)mv² = (1/2) x 5 x 14.14² = 500J Therefore, the kinetic energy of the body just before it strikes the ground is 500J. Answer: 500J

Akọwa Nkọwa

The correct option that explains why a person suffers a more severe burn when his skin is exposed to steam than when boiling water pours on his skin is: - Steam possesses greater heat energy per unit mass than boiling water. When a person's skin is exposed to steam, the steam condenses on the skin and releases its latent heat of vaporization, which is much higher than the specific heat capacity of boiling water. This means that steam possesses greater heat energy per unit mass than boiling water. Therefore, when a person's skin is exposed to steam, more heat energy is transferred to the skin in a shorter time, causing more severe burns than boiling water. Additionally, steam has the ability to spread more easily over a wider area of the skin than boiling water, increasing the likelihood of more severe burns.

Akọwa Nkọwa

Akọwa Nkọwa

The electric potential at a point distance r from a proton of charge q placed in a medium of permittivity ε_o is given by the formula: V = q/4πε_or Here, V represents the electric potential (also known as voltage) at the point, q represents the charge of the proton, r represents the distance from the proton, and ε_o represents the permittivity of the medium. This formula states that the electric potential at a point is directly proportional to the charge of the proton and inversely proportional to the distance from the proton. It also depends on the permittivity of the medium, which is a measure of how much the medium can resist the formation of an electric field. Therefore, the correct answer is: q/4πε_or².

Akọwa Nkọwa

When the nucleus of a uranium atom is split into two fragments of nearly equal mass, it releases a significant amount of energy. This energy is known as nuclear energy. According to Einstein's famous equation E=mc^2, mass can be converted into energy. So, when the uranium nucleus is split, the sum of the masses of the two fragments produced is less than the mass of the original nucleus. This mass difference is converted into nuclear energy, which can be harnessed for various purposes, such as generating electricity. Therefore, the correct option is "nuclear energy released."

Akọwa Nkọwa

Akọwa Nkọwa

To find the time taken to reach the maximum height, we need to use the fact that the vertical component of the velocity of the projectile at the maximum height is zero. This occurs when the projectile reaches the top of its trajectory. The initial vertical component of the velocity is Usin?. The vertical component of the velocity at the top of the trajectory is zero. Using the equation of motion for vertical motion, we can find the time taken to reach the maximum height. The equation of motion for vertical motion is given by: v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. At the maximum height, the final velocity is zero, the initial velocity is Usin?, and the acceleration is the acceleration due to gravity, g. Therefore, we can write: 0 = Usin? - gt Rearranging this equation, we get: t = Usin? / g Therefore, the time taken to reach the maximum height is Usin? / g. The correct option is Usin? / g.

Akọwa Nkọwa

The correct option is "Change of an electron from a higher to a lower energy level in the atom". Line spectra are the unique set of wavelengths of light that are emitted when electrons in an atom transition from a higher energy level to a lower energy level. When an electron absorbs a photon, it jumps to a higher energy level, and when it releases a photon, it jumps to a lower energy level. Since electrons can only exist in specific energy levels in an atom, each transition between these levels results in the emission of a photon of a specific wavelength. This gives rise to the characteristic line spectra that can be observed in experiments. The other options listed do not accurately describe the phenomenon of line spectra.

Akọwa Nkọwa

A scalar quantity is a physical quantity that has only magnitude (size) and no direction. Therefore, the answer to the question is a physical quantity that only has a size. Momentum and force are vector quantities because they have both magnitude and direction. Acceleration and displacement are vector quantities when the direction is considered, but can also be scalar quantities when only the magnitude is considered. Distance, on the other hand, is a scalar quantity because it only has magnitude and no direction. So the answer to the question is Distance.

Akọwa Nkọwa

The time taken for the sound to travel from the boy to the cliff and back to the boy (i.e., the time for the echo) is 0.9s. Since the sound has to travel twice the distance between the boy and the cliff, we can use the formula: distance = (speed of sound × time taken) ÷ 2 Plugging in the given values, we get: distance = (330 × 0.9) ÷ 2 = 148.5m Therefore, the boy should stand at a distance of 148.5m from the cliff to hear the echo of his clap 0.9s later. The correct option is (c) 148.50m.

Akọwa Nkọwa

The resistance of a wire is given by the formula: R = (ρ * L)/A Where R is resistance, ρ is resistivity, L is the length of the wire and A is the cross-sectional area of the wire. To find the length of the constantan wire required to construct a standard resistor of resistance 21Ω, we rearrange the formula to get: L = (RA)/ρ Substituting the given values, we have: L = (21 * 4π x 10^-8)/ (1.1 x 10^-6) L = 0.76 meters Since the value of π was given as 22/7, we use the approximate value of π = 3.14 in the calculation. Therefore, the length of the constantan wire required to construct a standard resistor of resistance 21 Ω is 0.76 meters. So, the correct option is (b) 2.40 m.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

If an element has a half-life of 3 years, it means that after 3 years, half of the atoms in a sample of that element will have decayed. After another 3 years (a total of 6 years), half of the remaining atoms will have decayed. After another 3 years (a total of 9 years), half of the remaining atoms will have decayed again. So, after 9 years, the number of atoms remaining would be 1/2 x 1/2 x 1/2 times the original number of atoms, or (1/2)^3 times N. Simplifying this expression, we get (1/8) times N. Therefore, the number of atoms that would have decayed after 9 years would be the original number of atoms minus (1/8) times the original number of atoms, or 7/8 times N. So, the answer is 7 N atoms8.

Akọwa Nkọwa

The iris is used for controlling the amount of light entering the eye. It is a thin, circular structure located behind the cornea and in front of the lens. The iris contains muscles that control the size of the pupil, the black circular opening in the center of the iris. In bright light, the iris muscles contract and the pupil becomes smaller, allowing less light to enter the eye. In dim light, the iris muscles relax and the pupil becomes larger, allowing more light to enter the eye. This process of adjusting the size of the pupil to control the amount of light entering the eye is called pupillary reflex.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

The correct option is electrons. Cathode rays are streams of negatively charged particles called electrons. These electrons are accelerated to high velocities and are deflected by electric and magnetic fields. The speed of cathode rays is relatively high, but not as fast as the speed of light.

Akọwa Nkọwa

Akọwa Nkọwa

The eye defect illustrated in the diagram is myopia, also known as nearsightedness. In myopia, the image is formed in front of the retina instead of on the retina, resulting in a blurred image of distant objects. To correct this defect, a diverging lens, also known as a concave lens, is used. The concave lens is thinner at the center than at the edges, which causes the light rays to diverge or spread out. When the diverging lens is placed in front of the myopic eye, the lens causes the light rays to diverge even more before they enter the eye. As a result, the image is shifted back towards the retina, allowing the myopic eye to focus on the object and produce a clear image.

Akọwa Nkọwa

Akọwa Nkọwa

When light passes through a prism, it is refracted at both its entry and exit faces. The amount of refraction depends on the wavelength of the light, with shorter wavelengths (violet and blue) being refracted more than longer wavelengths (red and orange). As a result, white light, which is a mixture of all colors, is separated into its constituent colors, forming a spectrum of colors. The angle of deviation, which is the angle between the incident ray and the emergent ray, decreases in the order of violet, blue, green, yellow, orange, and red. Therefore, the correct option is: blue, red and orange.

Akọwa Nkọwa

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas. The combined gas law is given by: (P1 x V1) / T1 = (P2 x V2) / T2 where P1, V1, and T1 are the initial pressure, volume, and temperature of the gas, and P2, V2, and T2 are the final pressure, volume, and temperature of the gas. In this problem, we know that the initial temperature is 30°C, or 303 K (since temperature must be in Kelvin for gas laws). We also know that the initial volume is V, and the final volume is two-thirds of the initial volume, or (2/3)V. Finally, we know that the final pressure is twice the initial pressure, or 2P. Substituting these values into the combined gas law, we get: (P x V) / 303 = (2P x 2V/3) / T2 Simplifying this equation, we get: T2 = (2P x 2V/3 x 303) / P x V Canceling out the P and V terms, we get: T2 = (2 x 2/3 x 303) / 1 Simplifying this equation, we get: T2 = 404 K Converting this temperature back to Celsius, we get: T2 = 404 - 273 = 131°C Therefore, the temperature of the gas when the volume is reduced to two-thirds of its original value by applying a pressure twice the original value is 131°C. The correct option is 131^oC.

Akọwa Nkọwa

A barometer is an instrument used to measure atmospheric pressure. The weight of the air above the barometer exerts a pressure on the mercury in the barometer. This pressure is measured and used to determine the atmospheric pressure. Knowing the atmospheric pressure at a particular location, we can estimate the height of the location relative to sea level. As we go higher, the atmospheric pressure decreases due to a decrease in the weight of air above us. Therefore, a barometer can be used to determine the height of a mountain. The barometer can't be used to determine the depth of a mine because the atmospheric pressure doesn't change significantly with depth. Dew point is the temperature at which the air becomes saturated with water vapor and dew starts to form. Barometers are not used to determine the dew point. Therefore, the correct option is: I and II only.

Akọwa Nkọwa

Akọwa Nkọwa

The fundamental frequency of a closed pipe is given by: f = v/4L where f is the frequency, v is the velocity of sound and L is the length of the pipe. Rearranging this equation to solve for v, we get: v = 4Lf Substituting the values given in the question, we get: v = 4 x 1.0m x 85Hz v = 340.0ms^-1 Therefore, the velocity of sound in air is 340.0ms^-1. Answer: 340.0ms^-1.

Akọwa Nkọwa

The process by which a metal, heated to a high temperature, gives off electrons from its surface is known as thermionic emission. When a metal is heated to a high temperature, some of its electrons gain enough energy to overcome the attractive forces of the metal atoms and are emitted from the surface of the metal. This process is known as thermionic emission, and the emitted electrons are called thermions. Thermionic emission is commonly used in vacuum tubes, such as cathode ray tubes and vacuum tubes used in radio and television equipment. In these devices, a metal filament is heated to a high temperature, causing the emission of electrons, which are then accelerated towards a positively charged electrode, producing an electric current. In summary, thermionic emission is the process by which a metal emits electrons from its surface when heated to a high temperature. It is commonly used in vacuum tubes for various electronic applications.

Akọwa Nkọwa

The moment of a force has the same unit as work, which is joule (J). The moment of a force is the product of the force and the perpendicular distance from the line of action of the force to the pivot point. It is also known as torque. The unit of force is newton (N), and the unit of distance is meter (m). Therefore, the unit of moment of a force is Nm. Work is defined as the product of the force and the distance moved in the direction of the force. The unit of work is also joule (J). Therefore, the moment of a force and work have the same unit, which is joule (J). The other options do not have the same unit as the moment of a force. Force is measured in newtons (N), power is measured in watts (W), momentum is measured in kilogram meters per second (kg m/s), and impulse is measured in newton seconds (N s).

Akọwa Nkọwa

Akọwa Nkọwa

Lenz’s law of electromagnetic induction states that "the induced current in a conductor is in such a direction as to oppose the change producing it". This means that when a magnetic field is changing near a conductor, an electromotive force (EMF) is induced in the conductor, which in turn generates a current. This induced current will always flow in a direction that creates a magnetic field that opposes the change in the original magnetic field. In simple terms, the law states that nature always tries to resist any change in the magnetic field through a conductor.

Akọwa Nkọwa

The instrument used to measure relative humidity is called a hygrometer. A hygrometer measures the amount of water vapor in the air, which is used to determine the relative humidity. The other instruments listed are used to measure other properties such as the density and pressure of gases and liquids, but not specifically for measuring relative humidity. Therefore, the correct answer is hygrometer.

Akọwa Nkọwa

Akọwa Nkọwa

The change of the direction of a wavefront as it moves from one medium to another due to a change in its velocity is called refraction. When a wave encounters a boundary between two media with different densities or refractive indices, its velocity changes, and this causes the wave to change direction. The bending of light as it passes through a lens is a common example of refraction. The amount of bending that occurs depends on the angle of incidence and the refractive indices of the two media.

Akọwa Nkọwa

Akọwa Nkọwa

The stream of alpha particles projected into an electric field between two plates X and Y will be deflected towards plate Y. This is because alpha particles are positively charged and therefore attracted to the negatively charged plate Y. The direction of deflection depends on the direction of the electric field and the initial velocity of the alpha particles. The other options are incorrect because the particles are not accelerated in a straight line, directionally reversed at the end of the plates, or attracted by both plates. The particles are only deflected towards one plate due to the electric field.

Akọwa Nkọwa

Increasing the length of the string will reduce the frequency of oscillation of a simple pendulum. The frequency of oscillation of a simple pendulum is given by the formula: f = 1/(2π) √(g/L) where f is the frequency, g is the acceleration due to gravity, and L is the length of the pendulum. As we can see from this equation, the frequency of oscillation is directly proportional to the square root of the acceleration due to gravity and inversely proportional to the square root of the length of the pendulum. Therefore, increasing the length of the pendulum will decrease the frequency of oscillation. The mass of the bob and the amplitude of oscillation do not affect the frequency of oscillation, as long as they remain small. The mass of the bob and the amplitude of oscillation affect the period of oscillation, which is the time taken for one complete oscillation.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

(a) Photoelectric emission

Photoelectric emission is the emission of electrons from the surface of a metal when electromagnetic radiation of sufficiently high frequency falls on it.

(b) Simple photocell

The photocell consists of an evacuated glass or quartz bulb containing a curved photosensitive cathode, usually coated with caesium, and a collecting anode. When light of frequency equal to or greater than the threshold frequency falls on the caesium-coated cathode, photoelectrons are emitted. The positively charged anode attracts and collects these electrons. Electrons then flow through the external circuit and a current is registered by the microammeter. Increasing the intensity of light increases the number of electrons emitted per second and hence increases the photocurrent.

(c) Applications of photoelectric emission

1. Automatic switching on of street lights at dusk.
2. Burglar and security alarm systems.
3. Television camera tubes.
4. Industrial counting and control devices, for example counting objects moving on a conveyor belt.

(d)

Before photoelectric emission begins, the light energy is absorbed by electrons in the metal. For light below the threshold frequency, the energy supplied to each electron is less than the work function, so the electron cannot escape from the metal surface; the absorbed energy is dissipated mainly as heat. At the threshold frequency, each photon supplies just enough energy to overcome the binding energy of an electron, and emission begins.

One factor affecting the number of emitted electrons is the intensity of the incident light, provided that its frequency is at least the threshold frequency. Greater intensity produces more emitted electrons per second.

Akọwa Nkọwa

(a) Step-down transformer

A step-down transformer has more turns in the primary coil than in the secondary coil, that is, \(N_p > N_s\). When an alternating e.m.f. is applied to the primary coil, an alternating current flows in it and produces a changing magnetic flux in the laminated soft-iron core. This changing flux links the secondary coil and induces an alternating e.m.f. in it by electromagnetic induction.

Since the induced e.m.f. is proportional to the number of turns,

\[\frac{E_s}{E_p}=\frac{N_s}{N_p}\]

As \(N_s < N_p\), \(E_s < E_p\). Hence, the output voltage is lower than the input voltage.

(b) Energy losses and their reduction

1. Copper loss: Heat is produced in the windings because of their resistance. This is minimized by using thick, low-resistance copper wire.
2. Eddy-current loss: Eddy currents are induced in the iron core and produce heating. This is minimized by laminating the core into thin insulated sheets.
3. Hysteresis loss: Energy is dissipated during repeated magnetisation and demagnetisation of the core. This is minimized by using soft iron or a soft magnetic material with low hysteresis loss.

(c)

Given: \(V_p=4400\text{ V}\), \(V_s=220\text{ V}\), lamp power \(P_{out}=60\text{ W}\), and efficiency \(\eta=95\%=0.95\).

(i) Turns ratio

\[\frac{N_p}{N_s}=\frac{V_p}{V_s}=\frac{4400}{220}=20\]

Therefore, the ratio of primary turns to secondary turns is:

\[\boxed{N_p:N_s=20:1}\]

(ii) Current taken from the mains

\[\eta=\frac{P_{out}}{P_{in}}\]

\[0.95=\frac{60}{P_{in}}\]

\[P_{in}=\frac{60}{0.95}=63.16\text{ W}\]

\[I_p=\frac{P_{in}}{V_p}=\frac{63.16}{4400}=0.0144\text{ A}\]

\[\boxed{I_p=0.0144\text{ A}=14.4\text{ mA}}\]

Akọwa Nkọwa

(a)(i) Resonance occurs when a body is made to vibrate at its own natural frequency by a periodic force having the same frequency; the amplitude of vibration then becomes a maximum.

(ii) Experiment to illustrate resonance (resonance tube):

• Hold a vibrating tuning fork of known frequency just above the open end of a tube whose air-column length can be varied (e.g. a tube dipping in water, raised or lowered).
• Slowly change the length of the air column.
• At a particular length the sound suddenly becomes very loud; here the natural frequency of the air column equals the frequency of the fork, i.e. resonance occurs.
• Measure that resonating length.

(iii) When the stem of the vibrating fork is pressed on a table top, it forces the large table surface to vibrate at the same frequency (forced vibration / resonance). The large surface sets a much greater volume of air in motion than the thin prongs alone, so more sound energy reaches the ear and the note sounds louder. In air, only the small prongs move the air, so the sound is faint.

(b) Concave-mirror uses.

(i) To ignite carbon paper: the mirror faces the sun, whose rays are parallel to the axis; the mirror converges them to its principal focus \(F\). The paper placed at \(F\) receives concentrated heat and burns.

(ii) To produce an exact copy (same size) of a picture on a screen: place the object at the centre of curvature \(C\) (i.e. at \(2f\)). The mirror then forms a real, inverted image, of exactly the same size, also at \(C\), which can be received on a screen.

Akọwa Nkọwa

(a) The statement means that \(3.4\times10^{5}\,\text{J}\) of heat is required to change \(1\,\text{kg}\) of ice at \(0^\circ\text{C}\) completely into water at \(0^\circ\text{C}\), without any change in temperature.

(b) Experiment (method of mixtures):

• Half-fill a lagged calorimeter (of known mass \(m_c\) and specific heat \(c_c\)) with warm water; measure the mass of water \(m_w\) and the initial temperature \(\theta_1\).
• Dry some small pieces of pure melting ice at \(0^\circ\text{C}\) with blotting paper and drop them into the water; stir until all the ice just melts.
• Record the lowest (final) temperature \(\theta_2\), and find the mass of ice added \(m_i\) by reweighing.
• Heat lost by water and calorimeter = heat used to melt the ice + heat to warm the melted ice from \(0^\circ\text{C}\) to \(\theta_2\):

from which the specific latent heat of fusion \(L\) is calculated.

Two precautions: dry the ice thoroughly and use ice already at \(0^\circ\text{C}\); lag the calorimeter and stir gently to reduce heat exchange with the surroundings.

(c) Kinetic-theory explanation. When ice melts at \(0^\circ\text{C}\), the heat supplied is used to break down the rigid bonds of the ice lattice, increasing the potential energy of the molecules and letting them move past one another as a liquid. It does not increase their average kinetic energy. Since temperature depends on the average kinetic energy of the molecules, the temperature stays constant at \(0^\circ\text{C}\) throughout melting.