WAEC SSCE - Physics - 1996

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In a nuclear reactor, a chain reaction is sustained by the collision of neutrons with atomic nuclei, which leads to the release of more neutrons and a large amount of energy. However, not all neutrons are capable of causing this chain reaction, as only neutrons with sufficient energy (known as fast neutrons) can interact with the nuclei to cause the chain reaction. To increase the efficiency of the chain reaction, it is desirable to slow down these fast-moving neutrons so that they are more likely to interact with atomic nuclei and cause the chain reaction to continue. This is achieved by using a material known as a moderator, which is placed in the reactor core to slow down the fast neutrons. Of the options given, the material commonly used as a moderator in nuclear reactors is graphite rods. Graphite is made of carbon atoms and has a layered structure that allows it to efficiently slow down fast neutrons. Other materials such as heavy water and beryllium can also be used as moderators, but graphite is the most commonly used. Therefore, the correct answer to the question is "Graphite rods".

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When a body of mass is suspended by a string, the tension in the string is equal to the weight of the body. The weight of the body is given by the formula: Weight = mass x gravity where "mass" is the mass of the body and "gravity" is the acceleration due to gravity. In this case, the mass of the body is 2 kg and the acceleration due to gravity is 10 m/s^2. However, the lift is accelerating upwards with an acceleration of 2 m/s^2. This means that the net acceleration of the body (relative to the lift) is the difference between the acceleration due to gravity and the acceleration of the lift: Net acceleration = acceleration due to gravity - acceleration of the lift Net acceleration = 10 m/s^2 - 2 m/s^2 Net acceleration = 8 m/s^2 Therefore, the net force acting on the body (relative to the lift) is given by the formula: Net force = mass x net acceleration Substituting the given values, we have: Net force = 2 kg x 8 m/s^2 Net force = 16 N Since the tension in the string is equal to the net force acting on the body, the tension in the string is 16 N. Therefore, the correct answer to the question is "16N".

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Water boils when its vapor pressure equals the external pressure acting on its surface. At sea level, the atmospheric pressure is high, and water requires a higher temperature to achieve the vapor pressure necessary to boil. However, at the top of a mountain, the atmospheric pressure is lower due to the decrease in the weight of the air column above it. As a result, the vapor pressure required for water to boil is lower, and water boils at a lower temperature. This is because the atmospheric pressure is not sufficient to hold the liquid water molecules together, and they escape into the atmosphere as water vapor, which is the gaseous state of water. Therefore, the correct option that explains why water boils at a lower temperature when heated at the top of a mountain than at sea-level is: "pressure of the atmosphere is lower".

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In a series R-C circuit, the total impedance is given by: Z = √(R^2 + Xc^2) where R is the resistance and Xc is the reactance of the capacitor. Since the impedance of the circuit is 13 Ω and the resistance is 5 Ω, we have: 13 = √(5^2 + Xc^2) Squaring both sides, we get: 169 = 25 + Xc^2 Simplifying, we get: Xc^2 = 144 Taking the square root of both sides, we get: Xc = ±12 Since we are looking for the magnitude of the reactance, we take the positive value: Xc = 12 Ω Therefore, the reactance of the capacitor is 12 Ω.

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The distance between two successive crests of a wave is called its wavelength, denoted by λ. In this question, the wavelength is given as 1m. The number of complete oscillations (or cycles) of a particle on the surface of the water in 1 second is called the frequency, denoted by f. In this question, the frequency is given as 2 cycles per second (or 2 Hz). The speed of a wave is given by the formula: v = f × λ Substituting the given values, we get: v = 2 Hz × 1m v = 2 m/s Therefore, the speed of the wave is 2 m/s. The correct answer is option (C) 2.0 ms^-1.

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The count rate of an alpha-particle source refers to the number of alpha particles emitted by the source in one minute. The half-life of a radioactive source is the time it takes for half of the radioactive atoms to decay. If the count rate of the alpha-particle source is 400 per minute and the half-life of the source is 5 days, this means that after 5 days, the count rate will decrease to 200 per minute (half of 400). After another 5 days (a total of 10 days), the count rate will decrease to 100 per minute (half of 200). After another 5 days (a total of 15 days), the count rate will decrease to 50 per minute (half of 100). Therefore, the count rate per minute after 15 days will be 50, and the correct answer is option (C) 50.

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The velocity of a sound wave depends on the properties of the medium through which it is traveling. The formula for the velocity of a sound wave is: v = λf where v is the velocity, λ is the wavelength, and f is the frequency of the wave. In this question, we are given that the velocity of sound in air is 350 ms^-1 and the ratio of the wavelength of sound in water to that in air is 425:100. Let's assume that the frequency of the wave remains constant as it passes from air into water. If the wavelength of the sound wave in air is λ_air, then the wavelength of the same wave in water is: λ_water = (425/100)λ_air = 4.25λ_air Now we can use the formula for the velocity of a sound wave to find the velocity of the wave in water: v_water = λ_waterf Since the frequency of the wave remains constant, we can write: v_water = (4.25λ_air)f We know that the velocity of the wave in air is 350 ms^-1, and we can write: v_air = λ_airf Solving for f, we get: f = v_air/λ_air Substituting this expression for f into the previous equation, we get: v_water = (4.25λ_air)v_air/λ_air Simplifying, we get: v_water = 4.25v_air Substituting the given value for the velocity of sound in air, we get: v_water = 4.25 x 350 ms^-1 Simplifying, we get: v_water = 1487.5 ms^-1 Therefore, the velocity of the wave in water is 1487.5 ms^-1. Option (D) is the correct answer.

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The critical angle is the angle of incidence in the medium for which the angle of refraction is 90 degrees (the refracted ray lies along the boundary). The critical angle is given by the formula `sin c = 1/n`, where `c` is the critical angle and `n` is the refractive index of the medium relative to air. Substituting `n = 1.8` into the formula, we get `sin c = 1/1.8 = 0.556`. Taking the inverse sine of both sides, we get `c = sin^{-1}(0.556)`. Evaluating this on a calculator, we get `c ≈ 34°`. Therefore, the critical angle for the medium is approximately 34 degrees. Answer is the correct answer.

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The number of oscillations of a pendulum in a given time depends on its period of oscillation. The period is the time taken for one complete oscillation. The relationship between the period and the number of oscillations is direct; that is, as the period of a pendulum decreases, the number of oscillations it makes in a given time increases. In this question, we are given that the period of oscillation of pendulum x is 1.5 seconds, and it makes 400 oscillations in equal time. We are also given that pendulum y makes 500 oscillations in equal time. Since both pendula are making oscillations in equal time, we can assume that the time taken for each pendulum to make one oscillation is the same. To find the period of oscillation of pendulum y, we can use the relationship between the period and the number of oscillations. If pendulum x with a period of 1.5 seconds makes 400 oscillations in equal time, then the total time taken for 400 oscillations is: Time taken by pendulum x = 400 x 1.5 seconds = 600 seconds Since pendulum y makes 500 oscillations in the same time, the period of oscillation of y can be calculated as follows: Time taken by pendulum y for 500 oscillations = 600 seconds Period of pendulum y = Total time taken / Number of oscillations Period of pendulum y = 600 seconds / 500 oscillations Period of pendulum y = 1.2 seconds Therefore, the period of oscillation of pendulum y is 1.2 seconds.

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The magnitude of the force required to make an object of mass M move with speed V in a circular path of radius R is given by the expression mv²/r. This is known as the centripetal force. When an object moves in a circle, it experiences a change in direction which means it is accelerating. Since there is acceleration, there must be a force acting on the object. The centripetal force is the force that acts on an object moving in a circular path, directed towards the center of the circle, and is necessary to keep the object moving in a circular path. The magnitude of this force is directly proportional to the mass of the object, the square of its velocity and inversely proportional to the radius of the circular path. Thus, the correct option is (C) mv²/r.

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Lenz's law of electromagnetic induction is a fundamental law in physics that describes the relationship between the change in magnetic flux and the induced electromotive force (EMF) and current. The law states that when there is a change in the magnetic flux through a closed loop, an EMF is induced in the loop that is proportional to the rate of change of the flux. Furthermore, the direction of the induced EMF is such that it opposes the change that produced it. In other words, the induced current flows in a direction that creates a magnetic field that opposes the change in the original magnetic field. This means that if the magnetic flux through a circuit increases, the induced current will create a magnetic field that opposes the increase in flux. Similarly, if the flux through the circuit decreases, the induced current will create a magnetic field that opposes the decrease in flux. Therefore, among the given options, the correct answer is (B) the induced current in a conductor is in such a direction as to oppose the change producing it.

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The decay constant (λ) is a measure of the probability that a radioactive nucleus will decay per unit time. It is related to the half-life (t_1/2) of the radioactive element by the equation: λ = ln(2) / t_1/2 where ln(2) is the natural logarithm of 2, which is approximately 0.693. Substituting the given half-life of 3 seconds into the equation, we have: λ = ln(2) / 3 s λ ≈ 0.231s^-1 Therefore, the decay constant of the radioactive element is approximately 0.231s^-1. The option closest to this value is 0.23s^-1.

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The saturated vapor pressure of a liquid increases as the temperature of the liquid increases. When a liquid is heated, its molecules gain kinetic energy and move faster, causing more molecules to escape into the gas phase. This increases the concentration of vapor molecules in the air above the liquid, leading to an increase in the vapor pressure. Conversely, if the temperature of the liquid decreases, the vapor pressure will decrease as fewer molecules have enough energy to escape into the gas phase. The volume and mass of the liquid do not directly affect the saturated vapor pressure.

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Soft iron is used in making the armature of an electric bell because it loses its magnetism readily. This is due to its low coercivity, which means that it is easily magnetized and demagnetized. In an electric bell, the armature needs to be magnetized and demagnetized rapidly in order to vibrate back and forth, striking the bell and producing the sound. If the armature were made of a material that retained its magnetism for a long time, it would not vibrate back and forth as required, and the bell would not ring properly. Therefore, soft iron is an ideal material for the armature of an electric bell because it can be easily magnetized and demagnetized, allowing the bell to ring properly.

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The impulse experienced by the ball can be calculated using the impulse-momentum theorem, which states that the impulse on an object is equal to the change in momentum of the object. In this case, the ball has an initial momentum of p = mv, where m is its mass and v is its velocity, and it rebounds with the same speed, but in the opposite direction. This means that the final momentum of the ball is -mv. The change in momentum, therefore, is: Δp = -mv - mv = -2mv The impulse experienced by the ball is equal to the change in momentum, so: impulse = Δp = -2mv Substituting the given values, we get: impulse = -2(5.0 kg)(2 m/s) = -20.0 kg m/s Since impulse is a vector quantity, its magnitude is always positive. Therefore, we take the absolute value of the calculated impulse to get: impulse = 20.0 kg m/s So the correct answer to the question is 20.0 kg m/s.

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The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of 1 kg of the substance by 1 Kelvin or 1 degree Celsius. In this problem, we have an immersion heater that provides 400 W of power to a liquid with a mass of 0.5 kg. The temperature of the liquid increases by 2.5^oC in one second. We can use the formula: Heat energy supplied = mass x specific heat capacity x change in temperature to calculate the specific heat capacity of the liquid. First, we need to calculate the heat energy supplied by the immersion heater in one second. We can use the formula: Power = Heat energy supplied / time Rearranging this formula, we get: Heat energy supplied = Power x time Substituting the given values, we get: Heat energy supplied = 400 W x 1 s = 400 J Next, we can use the formula for specific heat capacity to solve for it: specific heat capacity = Heat energy supplied / (mass x change in temperature) Substituting the given values, we get: specific heat capacity = 400 J / (0.5 kg x 2.5^oC) = 320 Jkg^-1K^-1 Therefore, the specific heat capacity of the liquid is 320 Jkg^-1K^-1. The answer is.

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The main reason why rice cooks faster in a pressure cooker than in a cooking pot is that the boiling point of water in the cooker is raised. In a pressure cooker, water boils at a higher temperature due to the increased pressure inside the cooker. This increase in temperature means that the rice cooks faster as it is exposed to higher temperatures than it would be in a cooking pot. Additionally, less heat escapes from the cooker, and the vapour pressure in the cooker is constant, contributing to the rice cooking faster.

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When the loaded test-tube is depressed and then released, it experiences a restoring force that brings it back to its original position. This restoring force arises from the displaced water that exerts an upward buoyant force on the test-tube. Since the restoring force acts in the opposite direction to the displacement, the motion of the test-tube is oscillatory, with the test-tube moving back and forth around its equilibrium position. Therefore, the correct answer to the question is "Oscillatory".

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The correct answer is "Alpha particle". Alpha particles are not part of the electromagnetic spectrum. Alpha particles are actually helium-4 nuclei, consisting of two protons and two neutrons, and they are emitted by some types of radioactive decay. The other options are all part of the electromagnetic spectrum. The electromagnetic spectrum includes radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. They are all forms of electromagnetic radiation characterized by different wavelengths and frequencies.

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The instrument that produces sound by the vibration of an air column is a flute. The flute is a wind instrument that produces sound by blowing air across the mouthpiece. The air column inside the flute vibrates and produces sound waves, which then travel through the instrument and out of the end, producing the sound that we hear. In contrast, a piano produces sound through the vibration of strings, a guitar produces sound through the vibration of strings, a handbell produces sound through the vibration of metal, and a talking drum produces sound through the vibration of a stretched membrane.

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A platinum-resistance thermometer has a resistance of 5 Ω at 0°C and 9 Ω at 100°C, and it is assumed that resistance changes uniformly with temperature. We need to calculate the resistance of the thermometer when the temperature is 45°C. We can start by finding the change in resistance as the temperature changes from 0°C to 100°C: ΔR = 9 Ω - 5 Ω = 4 Ω This change occurs over a temperature range of 100°C - 0°C = 100°C, so the resistance changes by 4 Ω per 100°C. To find the resistance at 45°C, we can use the following proportion: (ΔR / ΔT) = (R₂ - R₁) / (T₂ - T₁) Where: ΔR = change in resistance ΔT = change in temperature R₁ = initial resistance (at 0°C) R₂ = final resistance (at 45°C) T₁ = initial temperature (0°C) T₂ = final temperature (45°C) Substituting the known values, we get: (4 Ω / 100°C) = (R₂ - 5 Ω) / (45°C - 0°C) Simplifying the equation: R₂ - 5 Ω = (4 Ω / 100°C) x 45°C R₂ - 5 Ω = 1.8 Ω R₂ = 6.8 Ω Therefore, the resistance of the platinum-resistance thermometer at 45°C is 6.8 Ω. Answer: 6.8 Ω.

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The main function of the mouth of a telephone is to convert sound energy into electrical energy. When a person speaks into the mouthpiece of a telephone, the sound waves from their voice cause a diaphragm inside the microphone to vibrate. This vibration is then converted into an electrical signal, which is transmitted through the telephone wires to the receiver at the other end. The receiver then converts the electrical signal back into sound waves, which can be heard by the person on the other end of the line. Therefore, the mouth of a telephone does not convert sound energy into heat, light, chemical, or mechanical energy, but rather into electrical energy.

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The components of a DC electric motor include an armature, a commutator, field magnets, carbon brushes, and sometimes, brushes or slip rings. The armature is the rotating part of the motor that converts electrical energy into mechanical energy. The commutator is a device that changes the direction of the current flowing through the armature, ensuring that it continues to rotate in the same direction. The field magnets produce a magnetic field that interacts with the current flowing through the armature, causing it to rotate. Carbon brushes are used to transfer electrical current between stationary wires and moving parts of the motor, such as the commutator or slip rings. Slip rings are used in some types of motors as an alternative to the commutator to transfer electrical current between the stationary wires and the rotating parts of the motor. Therefore, the component that is not a component of a DC electric motor is slip rings. Although slip rings can be used in some types of electric motors, they are not commonly used in DC electric motors. Instead, DC motors typically use a commutator to transfer electrical current between the stationary wires and the rotating armature. So the correct answer to the question is "Slip rings".

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The speed of a wave can be calculated using the formula: speed = frequency x wavelength In a stationary wave, the distance between two consecutive nodes (or antinodes) represents half of the wavelength. Therefore, the wavelength of the wave in the given diagram can be calculated as: wavelength = 2 x distance(N1N2) = 2 x 1.5m = 3m Given that the frequency of the wave is 256Hz, we can now calculate its speed: speed = frequency x wavelength = 256Hz x 3m = 768ms^-1 Therefore, the speed of the wave is 768ms^-1.

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According to Boyle's law, at a constant temperature, the pressure and volume of a fixed mass of gas are inversely proportional. This means that as the volume of the gas is halved, its pressure will double. In this case, the initial pressure of the gas is 2.0 x 10⁵Nm^-2. When the volume is halved, the pressure will double to 4.0 x 10⁵Nm^-2. Therefore, the answer is 4.0 x 10⁵Nm^-2.

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In this problem, we can use the principle of conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it. The initial momentum of the system before the collision is given by: p1 = m1 * v1 where m1 is the mass of the first body (4.2 kg) and v1 is its velocity (10 m/s). p1 = 4.2 kg * 10 m/s p1 = 42 kg m/s The second body is stationary, so its initial momentum is zero: p2 = 0 kg m/s The total initial momentum of the system is therefore: p_initial = p1 + p2 p_initial = 42 kg m/s After the collision, the two bodies stick together and move with a common velocity V. The total mass of the system is the sum of the masses of the two bodies: m_total = m1 + m2 m_total = 4.2 kg + 2.8 kg m_total = 7 kg The final momentum of the system is: p_final = m_total * V According to the principle of conservation of momentum, the total initial momentum of the system is equal to the total final momentum of the system: p_initial = p_final Substituting the values we have found, we get: 42 kg m/s = 7 kg * V Solving for V, we get: V = 42 kg m/s / 7 kg V = 6 m/s Therefore, the value of V after the collision is 6 m/s due east. Thus, the correct answer is "6ms^-1".

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Electrical conductivity is a measure of how easily a material can conduct electricity. It is the reciprocal of electrical resistivity. The formula for electrical conductivity is given by: σ = 1/ρ Where σ is electrical conductivity and ρ is electrical resistivity. We are given the length (L) of the wire as 100 cm, the cross-sectional area (A) as 2.0 x 10^-3 cm^2, and the resistance (R) as 0.10 Ω. We can use Ohm's law to calculate the resistivity (ρ): R = ρ*L/A ρ = RA/L Substituting the given values, we get: ρ = (0.10 Ω)*(2.0 x 10^-3 cm^2)/100 cm ρ = 2.0 x 10^-6 Ω cm Now, we can calculate the electrical conductivity using the formula: σ = 1/ρ σ = 1/(2.0 x 10^-6 Ω cm) σ = 5.0 x 10^5 Ω^-1 cm^-1 Therefore, the electrical conductivity of the wire is 5.0 x 10^5 Ω^-1 cm^-1. Answer: 5.0 x 10^5 Ω^-1 cm^-1

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The circuit shown is a shunt or parallel circuit. We know that the full-scale deflection current of the galvanometer, I_G = 10mA = 10 x 10^-3A. Let R be the resistance of the shunt resistor. Since the galvanometer and shunt resistor are in parallel, the total current passing through the circuit is I_T = I_G + I. Here, I is the current passing through the shunt resistor. We also know that the current I should be such that the ammeter should read a full-scale deflection when the current I_T = 1A. According to the question, the resistance of the galvanometer is 29Ω. We can calculate the value of the current I_T when a potential difference of 29V is applied across the galvanometer: I_T = V / (R_G + R) where V is the applied potential difference, R_G is the resistance of the galvanometer, and R is the resistance of the shunt resistor. Since we know that I_T should be 1A when the ammeter reads full-scale deflection, we can set up the following equation: 1 = 29 / (29 + R) Solving for R gives: R = 29Ω x (1 - 1/1) = 0Ω This is not a practical value for a shunt resistor. It means that the ammeter will read zero current regardless of the actual current passing through the circuit. Therefore, we need to select an option which states that the value of R is very small. The correct option is therefore: 2.0 x 10^-3 Ω.

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The electric field intensity at point P due to the isolated charged sphere is given by Coulomb's law, which states that the magnitude of the electric field intensity is directly proportional to the charge on the sphere and inversely proportional to the square of the distance between the sphere and point P. Mathematically, it is expressed as: E = Q/(4πε_or²) where E is the electric field intensity, Q is the charge on the sphere, ε_o is the permittivity of the medium, and r is the distance between the sphere and point P. Therefore, the correct option is (b) Q/(4πε_or²).

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The inside of a vacuum flask is usually coated with silver to reduce heat lost by radiation. Heat can transfer from one object to another through convection, conduction, and radiation. In a vacuum flask, there is a vacuum between the two walls of the flask which minimizes heat loss by convection and conduction. However, heat can still be transferred by radiation, which is the emission of electromagnetic waves from a body. The silver coating on the inner wall of the vacuum flask reflects the heat radiation back into the flask, reducing the amount of heat lost to the surroundings. Therefore, silver is used to minimize heat loss by radiation in a vacuum flask.

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Let the original length of metal P be x and the original length of metal Q be y. We know that the increase in length is given by: ΔL = L₀αΔθ where ΔL is the increase in length, L₀ is the original length, α is the coefficient of linear expansion, and Δθ is the change in temperature. Since the increase in length is the same for both metals, we can write: ΔL_P = ΔL_Q L_Pα_PΔθ = L_Qα_QΔθ L_P/L_Q = α_Q/α_P Since we are told that the linear expansivity of P is twice that of Q, we can write: α_P = 2α_Q Substituting into the equation above, we get: L_P/L_Q = α_Q/2α_Q = 1/2 Therefore, the ratio of the original length of P to that of Q is 1:2. Answer: 1:2

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The perforated pads on the walls and ceilings of standard auditoria are there to reduce the effect of reverberations of sound waves. When sound waves are produced in a room, they bounce off the walls, floor, and ceiling, creating an echo-like effect that can make it difficult to hear and understand speech or music clearly. The perforations in the pads allow sound waves to pass through them and get absorbed by a layer of sound-absorbing material behind them. This reduces the amount of sound waves bouncing around the room, leading to clearer and more intelligible sound. The pads do not increase the intensity or loudness of sound waves, nor do they increase the interference effect or decrease the frequency of sound waves.

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When a ray of light is incident normally on an air-glass interface, its angle of refraction is 0 degrees. This is because when light travels from one medium to another, it changes direction, a phenomenon known as refraction. The amount of refraction depends on the angle at which the light enters the second medium, as well as the properties of the two media, such as their refractive indices. In the case of an air-glass interface, since the angle of incidence is normal, that is, the ray of light is perpendicular to the surface, the angle of refraction is also zero. This means that the light continues to travel in a straight line without changing its direction. So the correct answer to the question is 0 degrees.

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Out of the given options, palm oil is the most viscous at room temperature. Viscosity is a measure of a fluid's resistance to flow, and is affected by factors such as temperature and molecular structure of the fluid. Palm oil has a higher molecular weight and longer hydrocarbon chains compared to the other options, which leads to stronger intermolecular forces and greater resistance to flow. This makes it more viscous than water, alcohol, petrol, and kerosene at room temperature.

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The heating element in an electric kettle is usually located near the bottom of the kettle because of convection currents. When the heating element is turned on, it heats up the water in contact with it. Hot water is less dense than cold water, so it rises to the top of the kettle while the colder water sinks to the bottom. This creates a convection current, where the hot water at the top of the kettle is replaced by the colder water at the bottom, which in turn is heated by the heating element. This process continues until all the water in the kettle is heated to the desired temperature. If the heating element were located near the top of the kettle, the convection current would be less efficient because the hot water at the top would not mix as easily with the colder water at the bottom. By placing the heating element near the bottom of the kettle, the convection currents are more effective at carrying the heat to all parts of the water, resulting in faster heating and more efficient energy use. Therefore, the correct answer is option (D) - the convectional currents which are set up can carry heat to all parts of the water.

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(a) The statement means that the capacitor stores a charge of 2 microcoulombs for every volt of potential difference between its plates:

\[C=2\,\mu\text{F}=2\times10^{-6}\ \text{C V}^{-1}.\]

Thus, \(Q=CV\).

(b)(i) The capacitance of a parallel-plate capacitor depends on:

1. the overlapping area of the plates;
2. the separation between the plates;
3. the nature or permittivity of the dielectric material between the plates.

(b)(ii) Uses of capacitors include:

1. storing electric charge and energy;
2. smoothing the output of rectifier power supplies;
3. coupling alternating signals while blocking direct current in electronic circuits.

(c) When a capacitor is being charged, the potential difference is not constant. If the instantaneous charge on it is \(q\), then

\[V=\frac{q}{C}.\]

The work done in adding a small charge \(dq\) is

\[dW=Vdq=\frac{q}{C}dq.\]

Hence the energy stored when the charge increases from \(0\) to \(Q\) is

\[W=\int_0^Q\frac{q}{C}\,dq=\frac{1}{C}\left[\frac{q^2}{2}\right]_0^Q.\]

Therefore,

\[\boxed{W=\frac{Q^2}{2C}}\]

Since \(Q=CV\), equivalent forms are \(W=\frac12QV=\frac12CV^2\).

(d) The required parallel arrangement is shown below.

For capacitors in parallel, the potential difference across each capacitor equals the source voltage:

\[V_1=V_2=50\ \text{V}.\]

(i) Charge on either plate

For the \(2\,\mu\text{F}\) capacitor:

\[Q_1=C_1V=(2\times10^{-6})(50)=100\times10^{-6}\ \text{C}=\boxed{100\,\mu\text{C}}.\]

For the \(3\,\mu\text{F}\) capacitor:

\[Q_2=C_2V=(3\times10^{-6})(50)=150\times10^{-6}\ \text{C}=\boxed{150\,\mu\text{C}}.\]

Each capacitor has equal and opposite charges of these magnitudes on its two plates.

(ii) Potential difference across each capacitor

\[\boxed{V_1=V_2=50\ \text{V}}\]

(iii) Energy stored in the combination

\[C_{\text{total}}=C_1+C_2=2\,\mu\text{F}+3\,\mu\text{F}=5\times10^{-6}\ \text{F}.\]

\[W=\frac12C_{\text{total}}V^2=\frac12(5\times10^{-6})(50)^2.\]

\[\boxed{W=6.25\times10^{-3}\ \text{J}}\]

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(a) Dispersion of white light is the separation of white light into its constituent colours when it passes through a refracting medium such as a glass prism.

(b) In ascending order of wavelength:

Violet, Indigo, Blue, Green, Yellow, Orange, Red (VIBGYOR).

(c)

1. The colour deviated least is red.
2. The colour deviated most is violet.

(d) White light is made up of colours of different wavelengths. The refractive index of glass is different for the different colours. Consequently, the colours travel at different speeds in the prism and are refracted through different angles. Violet light is refracted most, while red light is refracted least; hence the colours spread out to form a spectrum.

(e) Production of a pure spectrum of white light

An illuminated narrow slit is placed at the principal focus of a converging lens, L₁. The lens produces a parallel beam of white light which falls on a glass prism. The prism disperses the light into separate parallel beams of different colours. A second converging lens, L₂, brings each coloured beam to a separate focus on a white screen placed in its focal plane. Thus, sharp non-overlapping images of the slit form a pure spectrum.

Akọwa Nkọwa

(a) The apparent cubic expansivity of a liquid is the apparent increase in volume per unit original volume of the liquid per unit rise in temperature. It is the expansion observed when the expansion of the containing vessel is not allowed for.

(b)(i) Determination of apparent cubic expansivity

The apparatus is arranged as shown below.

A clean, dry density bottle fitted with a stopper having a fine capillary hole is weighed empty. Let its mass be \(M_1\). It is filled completely with the liquid at the initial temperature \(T_1\), the stopper is inserted, and the outside is wiped dry. The mass of the bottle and liquid is found as \(M_2\).

The bottle is suspended in a water bath, with the stopper and capillary orifice above the water level. A thermometer is placed in the bath. The water is heated slowly to a final temperature \(T_2\). As the liquid expands, some of it escapes through the capillary hole. Heating is continued until no further liquid escapes at \(T_2\).

The bottle is removed, allowed to cool, wiped dry and weighed again. Let its final mass be \(M_3\).

Thus,

\[\text{mass of liquid expelled}=M_2-M_3\]

\[\text{mass of liquid remaining}=M_3-M_1\]

\[\Delta T=T_2-T_1\]

Therefore, the apparent cubic expansivity is

\[\gamma_a=\frac{M_2-M_3}{(M_3-M_1)(T_2-T_1)}.\]

(ii) Precautions

1. Heat the water bath slowly and stir it so that the liquid and the bath are at a uniform temperature.
2. Read the thermometer at eye level to avoid parallax error.
3. Keep the capillary orifice above the water level and wipe the bottle completely dry before each weighing.

(c) Calculation of the real cubic expansivity

For a temperature rise of \(50\,\mathrm{K}\):

\[\gamma_a=\frac{44.25-42.02}{42.02\times 50}\]

\[\gamma_a=\frac{2.23}{2101}=1.06\times10^{-3}\,\mathrm{K^{-1}}.\]

The cubic expansivity of the glass bottle is

\[\gamma_g=3\alpha_g=3(1.0\times10^{-5})=3.0\times10^{-5}\,\mathrm{K^{-1}}.\]

Since \(\gamma_r=\gamma_a+\gamma_g\),

\[\gamma_r=1.06\times10^{-3}+3.0\times10^{-5}\]

\[\boxed{\gamma_r=1.09\times10^{-3}\,\mathrm{K^{-1}}}.\]

Akọwa Nkọwa

(a)(i) Viscosity is the property of a fluid by which it opposes relative motion between its adjacent layers; it is a kind of internal friction in the fluid.

(ii) Terminal velocity is the constant (maximum) velocity attained by a body falling through a fluid when the net force on it is zero, i.e. when its weight is balanced by the upthrust plus the viscous drag.

(b)(i) Experiment (terminal velocity of a steel ball in glycerine):

• Fill a tall glass jar with glycerine and fix two rubber bands (marks) round it, well below the surface and a measured distance \(d\) apart.
• Gently release a small steel ball into the centre of the glycerine and let it fall.
• Using a stopwatch, time the ball \((t)\) as it travels between the two marks. In that lower region the velocity is constant.
• Terminal velocity \( v = \dfrac{d}{t} \). Repeat and average.

(ii) Two precautions: release the ball centrally so it falls far from the walls of the jar; take timings only in the lower region where the velocity has become steady, and avoid parallax when reading the marks.

(c)(i) Two effects of viscosity: it opposes/retards the motion of bodies moving through a fluid (drag); it reduces the rate of flow of the fluid and generates heat.

(ii) Two applications of viscosity: lubrication of machine parts by oils; use of viscous fluids in shock absorbers (dashpots) and hydraulic brakes.