WAEC SSCE - Physics - 2020
Tambaya 1 Rahoto
A diver steps off a diving platform that is 10m above the water. If there is no air resistance during the fall, there will be a decrease in the diver's
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When the diver steps off the diving platform, they start to fall towards the water due to the force of gravity. As the diver falls, their height above the water decreases, and therefore their gravitational potential energy decreases. At the same time, the diver's speed increases as they fall, which means their kinetic energy increases. Since there is no air resistance during the fall, the total mechanical energy of the diver (which is the sum of their kinetic energy and gravitational potential energy) remains constant. This is known as the conservation of energy. However, momentum is not directly affected by the decrease in height or the increase in speed. Momentum is defined as the product of an object's mass and velocity, and it only changes if there is a net force acting on the object. In the absence of external forces, the momentum of the diver remains constant. Therefore, the correct answer is: the diver's gravitational potential energy decreases, while their kinetic energy increases, and their momentum remains constant.
Tambaya 2 Rahoto
Which of the following statements about the motion of a simple pendulum is true?
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Tambaya 3 Rahoto
What factors determine the frequency of a note emitted by a vibrating string?
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The frequency of a note emitted by a vibrating string is determined by the tension in the string, the mass per unit length of the string, and the length of the string. When a string is plucked or struck, it vibrates at a certain frequency, producing a sound wave with a particular pitch. The tension in the string determines how tightly the string is stretched and how fast it vibrates. The mass per unit length of the string affects how easily the string vibrates and the frequency of the sound wave. The longer the string, the lower the frequency of the sound wave, while the shorter the string, the higher the frequency of the sound wave. Therefore, the correct option is: "Mass per unit length of string, tension in string and length of string".
Tambaya 4 Rahoto
The temperature of an object is raised by 120°C. The resulting increase in its absolute temperature is
Tambaya 5 Rahoto
The average distance moved by a molecule between collisions is called
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The correct term for the average distance moved by a molecule between collisions is "mean free path". This is the average distance that a molecule travels before it collides with another molecule or object in its surroundings. Think of it like a crowded room where people are moving around. If you're a person in the room, you're going to bump into other people as you move around. The average distance you travel before bumping into someone else is like the mean free path for a molecule in a gas. The mean free path is affected by several factors, including the density of the gas, the temperature, and the size of the molecules. In general, if the gas is denser or the molecules are larger, the mean free path will be shorter because there are more collisions happening. Similarly, if the temperature is higher, the molecules will be moving faster and the mean free path will be longer. Understanding the concept of mean free path is important in fields like physics and chemistry, where scientists study the behavior of gases and other materials at the molecular level.
Tambaya 6 Rahoto
Which of the current-voltage characteristics shown below is that exhibited an ohmic conductor?
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Tambaya 7 Rahoto
A parallel plate capacitor is charged and the charging battery is subsequently disconnected. lf the plates of the capacitor are moved farther apart by means of insulating handles, the
Tambaya 8 Rahoto
Which of the following statements about the process of melting of a solid are true? The temperature of the solid will
I. remain steady until melting starts.
ll. keep rising until melting starts,
Ill remain steady as melting proceeds.
V, keep rising as melting proceeds.
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Tambaya 9 Rahoto
An a.c. generator can be converted to a d.c. electric motor by replacing the
Bayanin Amsa
An AC generator and a DC motor are different in their basic operating principles and construction. To convert an AC generator to a DC motor, certain modifications must be made. The main difference between an AC generator and a DC motor is that an AC generator produces alternating current (AC) while a DC motor runs on direct current (DC). In order to convert an AC generator to a DC motor, the AC output of the generator must be converted to DC. This can be achieved by replacing the slip rings of the AC generator with a split ring, and connecting a DC source (such as a battery) to the split ring. The split ring acts as a rectifier, converting the AC output of the generator to DC. Therefore, to convert an AC generator to a DC motor, you would replace the slip rings with a split ring and connect a DC source.
Tambaya 10 Rahoto
The velocity of sound in air at 15°C is 35400ms\(^{-1}\). Calculate the velocity at 47\(^o\)C
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Tambaya 11 Rahoto
In collision between two objects, kinetic energy is conserved only if
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In a collision between two objects, kinetic energy is conserved only if the collision is elastic. An elastic collision is a type of collision in which the total kinetic energy of the two objects before and after the collision remains the same. In other words, no kinetic energy is lost or gained during the collision. In an inelastic collision, on the other hand, some or all of the kinetic energy is converted into other forms of energy, such as heat or sound. This means that kinetic energy is not conserved in an inelastic collision. So, the correct option is "the collision is elastic". If the collision is inelastic, then the kinetic energy is not conserved. The other options are not correct because the conservation of kinetic energy does not depend on whether one of the objects was initially at rest or whether potential energy is converted to work.
Tambaya 12 Rahoto
Two identical cups containing the same volume of water at 45°C and 5°C, are left in a room at 25°C. Which of the following graphs correctly illustrates the variation of temperature with time?
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Tambaya 13 Rahoto
When two cells of negligible internal resistances and equal e.m.f. denoted by E\(_1\) and E\(_2\) are connected in parallel, combined e.m.f. E is given by
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When two cells of negligible internal resistances are connected in parallel, their positive terminals are connected together and their negative terminals are connected together. This forms a circuit with two paths for current flow. Since the cells have the same electromotive force (emf) E1 and E2, they are trying to push the same amount of charge through each path. Thus, the total emf of the combination E must be the same as E1 and E2. Therefore, the correct option is E = E1 = E2. is incorrect because emfs in parallel do not add up. is incorrect because it represents the formula for calculating the combined resistance of two resistors in parallel, not emfs. is incorrect because it represents the formula for calculating the ratio of emfs in a series circuit.
Tambaya 14 Rahoto
In photoelectric effect, the number of electrode emitted per second from a metallic surface is proportional to the
Tambaya 16 Rahoto
How would the capacitance of a parallel plate capacitor be affected if the distance of separation of its plates is decreased? It will
Bayanin Amsa
The capacitance of a parallel plate capacitor is directly proportional to the surface area of the plates and inversely proportional to the distance between the plates. This relationship can be expressed as C = εA/d, where C is the capacitance, ε is the permittivity of free space, A is the surface area of the plates, and d is the distance between the plates. When the distance of separation between the plates is decreased, the denominator of this equation decreases, causing the overall value of the capacitance to increase. So, the answer is that the capacitance will increase in value.
Tambaya 17 Rahoto
Which of the following statements explain why hot soapy water is more effective in cleaning oil-stained dishes?
Bayanin Amsa
The statement that explains why hot soapy water is more effective in cleaning oil-stained dishes is that "soap and heat decrease the surface tension of oil." When oil gets stuck on dishes, it forms a layer on top of the surface, making it difficult for water to penetrate and remove it. However, soap molecules are designed to break down the surface tension of oil, allowing water to mix with it and wash it away. The addition of heat to the soapy water further helps to reduce the surface tension of the oil, making it easier to clean. Therefore, using hot soapy water is more effective in cleaning oil-stained dishes.
Tambaya 18 Rahoto
In the hydraulic press, the force F applied is related to the diameter d of the cylinder by
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Tambaya 19 Rahoto
In an electric circuit, an inductor of inductance 0.5 H and resistance 50\(\Omega\) is connected to an alternating current source of frequency 60 Hz. Calculate the impedance of the circuit.
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The impedance of the circuit can be calculated using the formula: Z = √(R^2 + X^2) Where R is the resistance of the inductor and X is the reactance of the inductor, which is given by: X = 2πfL Where f is the frequency of the AC source and L is the inductance of the inductor. Substituting the given values, we get: X = 2π(60)(0.5) = 188.5 Ω R = 50 Ω Z = √(50^2 + 188.5^2) = 195.0 Ω Therefore, the impedance of the circuit is 195.0 Ω. To explain it in simpler terms, the impedance of the circuit is like the total resistance that the AC source "sees" when it is connected to the circuit. It takes into account both the resistance and the reactance (which is like a resistance but specific to AC circuits) of the circuit. In this case, the inductor has a reactance of 188.5 Ω which, when combined with the resistance of 50 Ω, gives an impedance of 195.0 Ω.
Tambaya 20 Rahoto
Which of the following statement about viscosity is not true? it
Tambaya 21 Rahoto
The diagram illustrates a ball of mass m sliding down a plank inclined at an angle of 0 to the horizontal. The kinetic friction between the ball and the plank is F and acceleration of free fall is g. The normal force on the ball is?
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Tambaya 22 Rahoto
Water waves have a wavelength of 3.6cm and speed of 18cms\(^{-1 }\) in deep water. If the waves enter shallow water with wavelength of 2.0cm, calculate the speed of the waves in the shallow water.
Tambaya 23 Rahoto
The vacuum between the double walls of a thermos flask reduces heat loss through
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The vacuum between the double walls of a thermos flask helps to reduce heat loss in two ways: conduction and radiation. Conduction is the transfer of heat through a material, such as the metal walls of the thermos flask. When there is a vacuum between the walls, there are no particles to transfer heat through conduction, so heat cannot escape or enter the flask easily. Radiation is the transfer of heat through electromagnetic waves, such as the infrared radiation that is emitted by hot objects. The vacuum between the walls of the thermos flask helps to reduce heat loss through radiation by reflecting the infrared radiation back into the flask, rather than allowing it to escape through the walls. Therefore, the vacuum between the double walls of a thermos flask reduces heat loss through both conduction and radiation, making it an effective insulator for keeping hot liquids hot and cold liquids cold.
Tambaya 24 Rahoto
The diagram above illustrates a ball of mass m sliding down a plank inclined at an angle of 0 to the horizontal. The kinetic friction between the ball and the plank is F and the acceleration of free fall is g. The net force on the ball along the plank is?
Bayanin Amsa
The net force acting on the ball is the vector sum of all the forces acting on it along the plank. The forces acting on the ball are the gravitational force and the kinetic frictional force. The gravitational force acting on the ball can be resolved into two components, one perpendicular to the plank and the other parallel to the plank. The component perpendicular to the plank is balanced by the normal force of the plank on the ball. The component parallel to the plank is responsible for the acceleration of the ball down the plank and is given by mg sin θ, where θ is the angle of inclination of the plank. The kinetic frictional force acts in the opposite direction to the motion of the ball and is given by F. Therefore, the net force acting on the ball along the plank is the vector sum of the gravitational force and the kinetic frictional force: Net force = mg sin θ - F Therefore, the correct answer is option (C): mgsin θ - F.
Tambaya 25 Rahoto
An electrical device has 50 turns in its primary coil and 20 turns in the secondary coil. The device can be a\an?
Bayanin Amsa
The electrical device described is a step-down transformer. A transformer is a device that can change the voltage level of an alternating current (AC) electrical signal. It has two coils, a primary coil and a secondary coil, which are wrapped around a common iron core. In this case, the primary coil has more turns than the secondary coil, which means that the voltage in the secondary coil will be lower than the voltage in the primary coil. This is because transformers work by transferring energy from one coil to the other through the magnetic field created by the alternating current. A step-up transformer would have more turns in the secondary coil than the primary coil, resulting in a higher voltage in the secondary coil. A DC generator produces direct current, which is different from the alternating current used in transformers. An AC generator produces alternating current, but it does not have the two coils of a transformer, so it cannot be a transformer.
Tambaya 26 Rahoto
A ray of light travels obliquely from a less dense medium to a denser medium. Which of the following statements is true about the light?
Bayanin Amsa
When light travels from a less dense medium (such as air) to a denser medium (such as glass), it slows down. According to Snell's Law of Refraction, the ratio of the speed of light in the two media is proportional to the ratio of their refractive indices. This means that the speed of the light decreases in the denser medium. As the speed of light is proportional to its frequency, a decrease in the speed of light results in a decrease in its frequency. This means that there is a change in the frequency of the light when it travels from the less dense medium to the denser medium. Additionally, when light travels obliquely from a less dense medium to a denser medium, it bends towards the normal (perpendicular) to the surface separating the two media. This is known as refraction and is responsible for the bending of light at the surface of a lens or the surface of a swimming pool. So, the answer is that the light refracts towards the normal and there is a change in the frequency of the light.
Tambaya 27 Rahoto
The material used to slow down the neutrons in a nuclear reactor is
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The material used to slow down the neutrons in a nuclear reactor is graphite. In a nuclear reactor, neutrons are released from the fission of uranium atoms, and these neutrons need to be slowed down or moderated so that they can be absorbed by other uranium atoms to sustain the chain reaction. Graphite is an excellent moderator because it contains many carbon atoms that are arranged in a crystalline lattice structure. The neutrons can easily collide with the carbon atoms in the graphite lattice and transfer some of their kinetic energy to the graphite atoms, thereby slowing down the neutrons. Slower neutrons are more likely to be absorbed by uranium atoms and initiate fission reactions. Boron and copper are not typically used as moderators in nuclear reactors. Boron is often used as a neutron absorber to control the rate of the nuclear reaction, while copper is a good conductor of heat and electricity but is not an effective moderator. Uranium is the fuel used in the reactor to sustain the nuclear reaction, but it is not used as a moderator because it is more likely to absorb neutrons than to slow them down.
Tambaya 28 Rahoto
A motorcycle starting from rest is uniformly accelerated such that its velocity in 10s is 72kmhr\(^{-1}\). What is its acceleration?
Bayanin Amsa
The formula for acceleration is given by: acceleration = change in velocity / change in time. In this case, the change in velocity is 72 km/hr and the change in time is 10s. To use the formula, we first need to convert the velocity from km/hr to m/s. 72 km/hr = 72 x 1000/3600 m/s = 20 m/s So, acceleration = 20 m/s / 10s = 2 m/s\(^2\) Therefore, the acceleration of the motorcycle is 2 m/s\(^2\).
Tambaya 29 Rahoto
The length of an iron bar is 100cm at 20°C. At what temperature will its length increase by 0.01%? [Linear expansivity of iron = 1.2 x 10\(^{-5}\)°C]
Bayanin Amsa
The problem requires us to find the temperature at which the length of the iron bar increases by 0.01%. We can use the formula for linear expansion of a solid: ΔL = α L ΔT where ΔL is the change in length, L is the original length, ΔT is the change in temperature, and α is the coefficient of linear expansion. Rearranging this equation, we get: ΔT = ΔL / (α L) We can substitute the given values into this equation to obtain the temperature change required to produce the given change in length: ΔT = (0.01/100) / (1.2 x 10^(-5) x 100) = 8.33°C Since we are asked for the temperature at which the length of the iron bar will increase by 0.01%, we need to add this temperature change to the initial temperature of 20°C: T_final = T_initial + ΔT = 20 + 8.33 = 28.3°C Therefore, the correct answer is option (C), 28.3°C.
Tambaya 30 Rahoto
A beam consisting of \(\alpha\)-particles. \(\beta\)-particles and \(\gamma\)-rays pass through a magnetic field at right angles to the direction of the field. Which of the following observations would be made about the \(\alpha\)-particles, \(\beta\)-particles, and \(\gamma\)-rays respectively?
Bayanin Amsa
The correct observation would be "Deflected, not deflected, not deflected." \(\alpha\)-particles are positively charged particles and are therefore deflected when they pass through a magnetic field. The direction of deflection is determined by the right-hand rule and depends on the direction of the magnetic field and the direction of the velocity of the particle. \(\beta\)-particles are also charged particles, but they are negatively charged, so they are not deflected by a magnetic field. \(\gamma\)-rays are not charged particles and therefore do not experience a force when they pass through a magnetic field. Thus, they are not deflected. In summary, the deflection of charged particles in a magnetic field depends on the charge and velocity of the particle, while uncharged particles are not deflected.
Tambaya 31 Rahoto
Which of the following actions would increase the electric force between two positively charged particles?
Bayanin Amsa
The electric force between two charged particles is determined by two factors: the magnitude of the charges and the distance between the particles. Like charges, such as two positively charged particles, repel each other, so an increase in the electric force would mean an increase in the repulsion between the particles. To increase the electric force between two positively charged particles, we need to either increase the magnitude of the charges or decrease the distance between the particles. Decreasing the mass of the particles or increasing the mass of the particles has no effect on the electric force. Similarly, "increasing the assistance" between the particles is not a meaningful term in this context, so it is not a relevant option. Therefore, the correct answer is: decreasing the distance between the particles. This is because the closer the particles are to each other, the stronger the electric force of repulsion between them will be, according to Coulomb's law.
Tambaya 32 Rahoto
A moving-coil galvanometer which gives a full-scale on deflection with 0.005A is converted to a voltmeter reading up to 5V using an external resistance of 975\(\Omega\). What is the resistance of the metre?
Bayanin Amsa
To convert a moving-coil galvanometer into a voltmeter, an external resistance is added in series with the galvanometer. The value of the external resistance is chosen so that the full-scale deflection of the galvanometer corresponds to the maximum voltage that the voltmeter is intended to measure. In this case, the moving-coil galvanometer gives a full-scale deflection with a current of 0.005A. To convert it into a voltmeter that reads up to 5V, an external resistance of 975\(\Omega\) is used. Using Ohm's Law, we can calculate the resistance of the galvanometer (meter resistance) as follows: Meter resistance = (maximum voltage / full-scale current) - external resistance Meter resistance = (5 V / 0.005 A) - 975\(\Omega\) Meter resistance = 1000\(\Omega\) - 975\(\Omega\) Meter resistance = 25\(\Omega\) Therefore, the resistance of the meter is 25.00\(\Omega\). Option C is the correct answer.
Tambaya 33 Rahoto
The diagram above illustrates a beam of parallel rays from a distant object O, incident on one side of the total reflecting prism. Which diagram does not represent the correct path of the beam when it emerges from the prism?
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Tambaya 34 Rahoto
Which of the following units is not fundamental?
Tambaya 35 Rahoto
Which of the following statements about photoelectrons is correct?
Tambaya 36 Rahoto
The volume of a fixed mass of a gas varies inversely as the pressure on it provided the temperature is constant. This statement is
Bayanin Amsa
The statement "The volume of a fixed mass of a gas varies inversely as the pressure on it provided the temperature is constant" is describing Boyle's law. Boyle's law states that the volume of a fixed amount of gas at a constant temperature is inversely proportional to the pressure on it. In other words, if the pressure on a gas increases, its volume will decrease, and vice versa. So, the statement is Boyle's law.
Tambaya 37 Rahoto
An electron of mass m and charge e moves in a circular path in a magnetic field of flux density B. How long does it take to complete one orbit?
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Tambaya 38 Rahoto
A ray of light traveling from glass into ethyl alcohol is incident at the boundary at an angle of incidence 30\(^o\). Calculate the angle of refraction.
[Refractive index of glass =1.5; refractive index of ethyl alcohol = 1.36]
Bayanin Amsa
The angle of refraction can be calculated using Snell's Law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two materials. Mathematically, this can be written as: sin(angle of incidence) / sin(angle of refraction) = refractive index of material 1 / refractive index of material 2 For this problem, we can substitute the given values and solve for the angle of refraction: sin(30\(^o\)) / sin(angle of refraction) = 1.5 / 1.36 Taking the inverse sine of both sides: angle of refraction = sin^-1(sin(30\(^o\)) * 1.36 / 1.5) Using a calculator, the angle of refraction can be calculated to be approximately 33.5\(^o\). So, the angle of refraction is approximately 33.5\(^o\).
Tambaya 39 Rahoto
A luminous object is one that
Bayanin Amsa
A luminous object is one that gives off light of its own. This means it does not need any other source of light to shine. It produces light on its own, making it visible even in the dark. Examples of luminous objects include the sun, stars, fireflies, and light bulbs.
Tambaya 40 Rahoto
A 500N box rests on a horizontal floor. A constant horizontal force is exerted on the box so that it moves through 8m. If the coefficient of kinetic friction between the floor and the box is 0.22, calculate the workdone on the box
Bayanin Amsa
The work done on the box is equal to the force exerted on the box multiplied by the distance it moves in the direction of the force. In this case, the force exerted on the box is the force required to overcome friction, which is equal to the normal force multiplied by the coefficient of kinetic friction. First, we need to calculate the normal force, which is the force exerted on the box by the floor in a direction perpendicular to the surface. Since the box is at rest, the normal force is equal to the weight of the box, which is 500N. Next, we can calculate the force required to overcome friction, which is equal to the normal force multiplied by the coefficient of kinetic friction. force of friction = 500N x 0.22 = 110N Finally, we can calculate the work done on the box by multiplying the force required to overcome friction by the distance moved. work done = force of friction x distance moved = 110N x 8m = 880J Therefore, the work done on the box is 880J, and the correct option is (a) 880J.
Tambaya 41 Rahoto
A 60kg man stands on a weighing balance in an elevator. If the elevator accelerates upwards at 5ms\(^{-2}\), determine the reading of the scale. [g = 10ms\(^{-2}\)]
Bayanin Amsa
The reading of the scale in this situation is equal to the weight of the man plus the force due to the acceleration of the elevator. The weight of the man can be calculated as follows: Weight = mass x gravity Weight = 60kg x 10ms\(^{-2}\) Weight = 600N The force due to the acceleration of the elevator can be calculated as follows: Force = mass x acceleration Force = 60kg x 5ms\(^{-2}\) Force = 300N Adding these two forces together, we get the total force acting on the man: Total force = Weight + Force due to acceleration Total force = 600N + 300N Total force = 900N So the reading of the scale would be 900N.
Tambaya 42 Rahoto
An image formed on a screen is always?
Tambaya 43 Rahoto
The depolarising agent in a Leclanche cell is?
Bayanin Amsa
The depolarizing agent in a Leclanche cell is manganese (IV) oxide. A Leclanche cell is a type of battery that uses a chemical reaction to generate electricity. Inside the cell, there are two electrodes, a zinc plate and a carbon rod, immersed in an electrolyte solution of ammonium chloride. When the cell is in use, the zinc plate undergoes a chemical reaction with the ammonium chloride, releasing electrons and forming zinc chloride and hydrogen gas. This process creates a buildup of positive charge on the zinc plate, which can eventually inhibit the flow of electrons and reduce the cell's efficiency. To counteract this buildup of positive charge, the manganese (IV) oxide acts as a depolarizing agent. It reacts with the hydrogen gas produced by the reaction on the zinc plate, converting it to water and preventing the buildup of positive charge. This allows the cell to continue generating electricity for a longer period of time.
Tambaya 44 Rahoto
Which of the following waves requires a material medium for its propagation?
Bayanin Amsa
The wave that requires a material medium for its propagation is sound waves. Sound waves are a type of mechanical wave, which means they need a material medium, such as air, water, or a solid, to travel through. This is because sound waves involve the compression and rarefaction of the medium, which causes the wave to propagate. Radio waves and light waves are types of electromagnetic waves, which do not require a material medium to travel through. They can travel through a vacuum, such as outer space, and do not involve the compression and rarefaction of a medium. X-rays are also a type of electromagnetic wave, and like other electromagnetic waves, they do not require a material medium to propagate. They can travel through a vacuum and do not involve the compression and rarefaction of a medium.
Tambaya 45 Rahoto
The sketched graph above illustrates the heating curve of a 0.02kg of water. Determine the approximate value of the specific latent heat of vapourization of water
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Tambaya 46 Rahoto
The earpiece of a telephone handset converts energy from
Bayanin Amsa
The earpiece of a telephone handset converts energy from one form to another. In this case, the earpiece converts electrical energy into sound energy that can be heard by the person using the telephone. When a person speaks into the telephone, their voice is converted into an electrical signal which is sent along the telephone wire to the other person's telephone. This electrical signal is received by the earpiece, which then converts it back into sound energy that can be heard by the person holding the telephone. Therefore, the correct answer is option (A), "electrical to sound".
Tambaya 47 Rahoto
The quantity of motion of a body is its
Bayanin Amsa
The quantity of motion of a body is its momentum. Momentum is a measure of how much "oomph" an object has when it is moving. It is calculated by multiplying the object's mass by its velocity, or speed in a particular direction. Think of a heavy truck moving slowly and a small car moving quickly. Even though the car is lighter than the truck, it can still have a greater momentum because it is moving faster. Momentum is important in physics because it helps us understand how objects interact with each other during collisions and other events. Acceleration refers to how quickly an object's velocity changes over time, while displacement is a measure of how far an object has moved from its original position. Velocity, on the other hand, is a measure of how fast an object is moving in a particular direction.
Tambaya 49 Rahoto
An electron of mass 9.1x 10\(^{31}\) kg is travelling at a speed of 2.0 x 10\(^6\)ms\(^{-1}\). Calculate the associate wavelength of the electron. [h = 6.6. x 10\(^{34}\) Js]
Bayanin Amsa
The wavelength of a moving particle, such as an electron, is given by the de Broglie equation: λ = h/mv where λ is the wavelength, h is Planck's constant, m is the mass of the particle, and v is its velocity. Substituting the given values, we get: λ = (6.6 x 10^-34 J s) / (9.1 x 10^-31 kg) x (2.0 x 10^6 m/s) Simplifying this expression, we get: λ = 3.63 x 10^-10 m Therefore, the associated wavelength of the electron is 3.63 x 10^-10 m Explanation: This question requires the application of the de Broglie equation, which relates the wavelength of a moving particle to its momentum. In this case, we are given the mass and velocity of an electron and asked to calculate its associated wavelength. We can do this by simply plugging in the given values into the de Broglie equation and solving for λ. The answer we obtain is the wavelength of the electron.
Tambaya 50 Rahoto
The magnitude of the force experienced by a charge of 1.6 x 10\(^{-8}\)C in a uniform electric field of intensity 5 x 105NC\(^{-1}\) is
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Tambaya 51 Rahoto
(a) In the design of an optical fibre, what type of material is most suitable for the design of the core?
(b) State one condition necessary to confine signals to the core of an optical fibre.
a) In the design of an optical fiber, a material with a higher refractive index than the cladding material is most suitable for the design of the core. This allows for the light to be confined within the core and transmitted along the fiber. Common materials used for the core include glass or plastic with a high refractive index.
b) One condition necessary to confine signals to the core of an optical fiber is that the refractive index of the core must be higher than that of the cladding material surrounding it. This difference in refractive index creates a phenomenon known as total internal reflection, which causes the light to be trapped within the core and transmitted along the fiber.
Bayanin Amsa
a) In the design of an optical fiber, a material with a higher refractive index than the cladding material is most suitable for the design of the core. This allows for the light to be confined within the core and transmitted along the fiber. Common materials used for the core include glass or plastic with a high refractive index.
b) One condition necessary to confine signals to the core of an optical fiber is that the refractive index of the core must be higher than that of the cladding material surrounding it. This difference in refractive index creates a phenomenon known as total internal reflection, which causes the light to be trapped within the core and transmitted along the fiber.
Tambaya 52 Rahoto
(a) (i) Define Optical angle.
(ii) Explain two conditions necessary for total internal reflection to occur.
(iii) List three practical applications of total internal reflection.
(b) State two effects of refraction.
(c)(i) Define progressive waves.
(ii) A plane progressive wave is represented by the equation y = 0.5 sin(1000\(\pi\)r = \(\frac{100 \pi \lambda}{17}\)) where y is in millimetres, t in seconds and x in metres. Calculate the: (\(\alpha\)) frequency of the wave; (\(\beta\))of the wave; (\(\gamma\)) speed of the wave
(a)
(i) Optical angle is the angle between the incident ray and the refracted ray at the point of incidence on a surface.
(ii) The two conditions necessary for total internal reflection to occur are: (1) the light must be travelling from a denser medium to a less dense medium, and (2) the angle of incidence must be greater than the critical angle for the two media.
(iii) Three practical applications of total internal reflection are: (1) optical fibers used for communication systems, (2) periscopes used in submarines and binoculars, and (3) diamond cutting tools.
(b) The two effects of refraction are: (1) bending of light when it passes from one medium to another with a different refractive index, and (2) change in the speed of light as it passes through different media.
(c)
(i) Progressive waves are waves that transfer energy from one point to another without any permanent displacement of the medium.
(ii) (a) The frequency of the wave is 500 Hz. (b) The wavelength of the wave is 0.017 m. (c) The speed of the wave is 8.5 m/s.
To find the frequency, we use the formula: frequency = wave speed / wavelength. Substituting the given values, we get: frequency = (1000 * pi * 0.017) / 17 = 500 Hz.
To find the wavelength, we rearrange the equation to get: wavelength = wave speed / frequency. Substituting the given values, we get: wavelength = 8.5 / 500 = 0.017 m.
To find the speed of the wave, we rearrange the equation to get: wave speed = frequency * wavelength. Substituting the given values, we get: wave speed = 500 * 0.017 = 8.5 m/s.
Bayanin Amsa
(a)
(i) Optical angle is the angle between the incident ray and the refracted ray at the point of incidence on a surface.
(ii) The two conditions necessary for total internal reflection to occur are: (1) the light must be travelling from a denser medium to a less dense medium, and (2) the angle of incidence must be greater than the critical angle for the two media.
(iii) Three practical applications of total internal reflection are: (1) optical fibers used for communication systems, (2) periscopes used in submarines and binoculars, and (3) diamond cutting tools.
(b) The two effects of refraction are: (1) bending of light when it passes from one medium to another with a different refractive index, and (2) change in the speed of light as it passes through different media.
(c)
(i) Progressive waves are waves that transfer energy from one point to another without any permanent displacement of the medium.
(ii) (a) The frequency of the wave is 500 Hz. (b) The wavelength of the wave is 0.017 m. (c) The speed of the wave is 8.5 m/s.
To find the frequency, we use the formula: frequency = wave speed / wavelength. Substituting the given values, we get: frequency = (1000 * pi * 0.017) / 17 = 500 Hz.
To find the wavelength, we rearrange the equation to get: wavelength = wave speed / frequency. Substituting the given values, we get: wavelength = 8.5 / 500 = 0.017 m.
To find the speed of the wave, we rearrange the equation to get: wave speed = frequency * wavelength. Substituting the given values, we get: wave speed = 500 * 0.017 = 8.5 m/s.
Tambaya 53 Rahoto
TEST OF PRACTICAL KNOWLEDGE QUESTION
You are provided with a resistance box R, voltmeter, key, cell of e.m.f. E, constantan wire, standard resistor, RX, an ammeter, and other necessary apparatus.
(i) Measure and record the e.m.f. E of the cell provided.
(ii) Set up the circuit as shown in the diagram above.
(iii) Set R to 1\(\Omega\). Close the key, read and record the current I and the corresponding voltage V.
(iv) Repeat the procedure for four other values of R= 2\(\Omega\), 4\(\Omega, 6\(\Omega\), and 8\(\Omega\).
(v) In each case, read and record I and V.
(vi) Tabulate the reading.
(vii) Plot a graph of V on the vertical axis and I on the horizontal axis.
(viii) Determine the slope s, of the graph.
(ix) State two precautions are taken to ensure accurate results
(b) )State two advantages of connecting identical cells in parallel.
(ii) State two factors to consider in choosing the material for the design of a resistor
Experiment with Resistors and EMF
(a): Measuring EMF and Plotting I-V Characteristics
(i) Measuring EMF (E):
- Connect the voltmeter directly across the cell's terminals (positive to positive, negative to negative) with the key open (circuit off).
- Read the voltmeter value and record it as the cell's EMF (E).
(ii) Setting Up the Circuit:
- Refer to the provided diagram to ensure correct connection of the following components:
- Cell
- Ammeter (in series)
- Resistor box (R)
- Standard resistor (RX) (likely in series with R)
- Voltmeter (in parallel with RX)
- Key
(iii - v) Measuring Current (I) and Voltage (V):
- Set the resistance box (R) to 1 Ω.
- Close the key (circuit on).
- Read the ammeter value (current, I) and voltmeter value (voltage, V) across the standard resistor (RX). Record both values.
- Repeat steps 1-3 for four other resistance values: R = 2 Ω, 4 Ω, 6 Ω, and 8 Ω.
(vi) Tabulating Readings:
- Create a table with columns for Resistance (R), Current (I), and Voltage (V).
- Record the corresponding values obtained in steps (iii) - (v).
(vii) Plotting the I-V Graph:
- Plot a graph with Voltage (V) on the vertical axis and Current (I) on the horizontal axis.
- Use the data points from your table to plot the graph.
(viii) Determining the Slope (s):
- The slope (s) of the graph represents the reciprocal of the total resistance in the circuit (including the cell's internal resistance).
- Use the linear regression method or graphical methods to calculate the slope (s) of the I-V graph.
(ix) Precautions for Accurate Results:
- Minimize contact resistance: Ensure good connections between components to avoid introducing unwanted resistance. Tighten all screws in binding posts.
- Use appropriate measuring ranges: Select ammeter and voltmeter ranges suitable for the expected current and voltage values in the circuit. Using incorrect ranges can lead to inaccurate readings.
(b) Connecting Cells in Parallel
(i) Advantages of Parallel Connection:
- Increased Current Output: When cells are connected in parallel, the total voltage remains the same as that of a single cell, but the current output increases. This is because each cell provides a parallel path for the current to flow. This is useful in applications requiring high current, such as powering electric motors.
- Redundancy: If one cell in a parallel connection fails, the circuit can still function using the remaining cells. This redundancy provides a backup system and ensures continued operation in some cases.
(ii) Factors for Resistor Material Selection:
- Resistivity: This is a material's inherent property that reflects its opposition to electrical current flow. Lower resistivity is desirable for low-resistance applications, while high resistivity is preferred for high-resistance applications.
- Temperature Coefficient of Resistance (TCR): This indicates how much a material's resistance changes with temperature. Materials with low TCR are preferred for applications where resistance needs to remain stable across temperature variations.
- Power Rating: The maximum power a resistor can handle without overheating.
- Cost and Availability
- Mechanical Properties: Such as size, shape, and ability to withstand physical stress.
Bayanin Amsa
Experiment with Resistors and EMF
(a): Measuring EMF and Plotting I-V Characteristics
(i) Measuring EMF (E):
- Connect the voltmeter directly across the cell's terminals (positive to positive, negative to negative) with the key open (circuit off).
- Read the voltmeter value and record it as the cell's EMF (E).
(ii) Setting Up the Circuit:
- Refer to the provided diagram to ensure correct connection of the following components:
- Cell
- Ammeter (in series)
- Resistor box (R)
- Standard resistor (RX) (likely in series with R)
- Voltmeter (in parallel with RX)
- Key
(iii - v) Measuring Current (I) and Voltage (V):
- Set the resistance box (R) to 1 Ω.
- Close the key (circuit on).
- Read the ammeter value (current, I) and voltmeter value (voltage, V) across the standard resistor (RX). Record both values.
- Repeat steps 1-3 for four other resistance values: R = 2 Ω, 4 Ω, 6 Ω, and 8 Ω.
(vi) Tabulating Readings:
- Create a table with columns for Resistance (R), Current (I), and Voltage (V).
- Record the corresponding values obtained in steps (iii) - (v).
(vii) Plotting the I-V Graph:
- Plot a graph with Voltage (V) on the vertical axis and Current (I) on the horizontal axis.
- Use the data points from your table to plot the graph.
(viii) Determining the Slope (s):
- The slope (s) of the graph represents the reciprocal of the total resistance in the circuit (including the cell's internal resistance).
- Use the linear regression method or graphical methods to calculate the slope (s) of the I-V graph.
(ix) Precautions for Accurate Results:
- Minimize contact resistance: Ensure good connections between components to avoid introducing unwanted resistance. Tighten all screws in binding posts.
- Use appropriate measuring ranges: Select ammeter and voltmeter ranges suitable for the expected current and voltage values in the circuit. Using incorrect ranges can lead to inaccurate readings.
(b) Connecting Cells in Parallel
(i) Advantages of Parallel Connection:
- Increased Current Output: When cells are connected in parallel, the total voltage remains the same as that of a single cell, but the current output increases. This is because each cell provides a parallel path for the current to flow. This is useful in applications requiring high current, such as powering electric motors.
- Redundancy: If one cell in a parallel connection fails, the circuit can still function using the remaining cells. This redundancy provides a backup system and ensures continued operation in some cases.
(ii) Factors for Resistor Material Selection:
- Resistivity: This is a material's inherent property that reflects its opposition to electrical current flow. Lower resistivity is desirable for low-resistance applications, while high resistivity is preferred for high-resistance applications.
- Temperature Coefficient of Resistance (TCR): This indicates how much a material's resistance changes with temperature. Materials with low TCR are preferred for applications where resistance needs to remain stable across temperature variations.
- Power Rating: The maximum power a resistor can handle without overheating.
- Cost and Availability
- Mechanical Properties: Such as size, shape, and ability to withstand physical stress.
Tambaya 54 Rahoto
State three properties of lasers that make them preferable to ordinary light beam.
Lasers have several properties that make them preferable to ordinary light beams:
- Coherence: Lasers produce light that is highly coherent, meaning that the light waves are all in phase and of the same frequency. This results in a very focused and intense beam of light.
- Monochromaticity: Lasers produce light that is monochromatic, meaning that it consists of a single color or wavelength. This makes lasers useful for applications where a specific color or wavelength is required, such as spectroscopy or laser printing.
- Directionality: Lasers produce light that is highly directional, meaning that the beam is focused in a specific direction and does not spread out as much as an ordinary light beam. This makes lasers useful for applications where a focused beam of light is required, such as cutting or welding materials.
These properties make lasers highly useful for a wide range of applications and are why they are preferred over ordinary light beams in many cases.
Bayanin Amsa
Lasers have several properties that make them preferable to ordinary light beams:
- Coherence: Lasers produce light that is highly coherent, meaning that the light waves are all in phase and of the same frequency. This results in a very focused and intense beam of light.
- Monochromaticity: Lasers produce light that is monochromatic, meaning that it consists of a single color or wavelength. This makes lasers useful for applications where a specific color or wavelength is required, such as spectroscopy or laser printing.
- Directionality: Lasers produce light that is highly directional, meaning that the beam is focused in a specific direction and does not spread out as much as an ordinary light beam. This makes lasers useful for applications where a focused beam of light is required, such as cutting or welding materials.
These properties make lasers highly useful for a wide range of applications and are why they are preferred over ordinary light beams in many cases.
Tambaya 55 Rahoto
(a)i) Define a torque.
(ii)State three factors that determine a torque
(b)(i) Define free fall.
(ii) A body is thrown vertically upwards from the top of a tower 40.0m high with a velocity of 10.0ms\(^{-1}\). Calculate the time taken for the body to reach the ground. [g.= 10.0ms\(^{-2}\)]
c) A cube of wood of side 8.0cm, floats at the interface between oil and water with 2.0cm of its surface below the interface as shown in the diagram below. Given that the relative densities of oil and water are 0.72 and 1.00 respectively, calculate the mass of the wood
(a)
i) Torque is the twisting force that causes rotational motion in an object.
ii) Three factors that determine torque are the magnitude of the force applied, the distance between the force and the axis of rotation, and the angle between the force and the lever arm.
(b)
i) Free fall is the motion of an object falling under the influence of gravity without any resistance from the air or other medium.
ii) To solve this problem, we can use the equation:
h = ut + 0.5gt\(^2\)
where h = height of the tower = 40.0m, u = initial velocity = 10.0ms\(^{-1}\), g = acceleration due to gravity = 10.0ms\(^{-2}\), and t = time taken to reach the ground.
Rearranging the equation, we get:
t = sqrt(2h/g) = sqrt(2 x 40.0/10.0) = 4.0s
Therefore, the time taken for the body to reach the ground is 4.0 seconds.
(c)
To calculate the mass of the wood, we can use the principle of buoyancy. The buoyant force acting on the wood is equal to the weight of the water displaced by the wood.
The volume of water displaced by the wood is equal to the volume of the wood that is submerged, which is given by:
V = l x b x h
where l, b, and h are the length, breadth, and height of the submerged portion of the wood, respectively.
From the diagram, we can see that h = 8.0 - 2.0 = 6.0cm, and the length and breadth are also equal to 8.0cm.
Therefore, V = l x b x h = 8.0 x 8.0 x 6.0 = 384cm\(^3\)
The weight of water displaced by the wood is given by:
W = V x density of water x g
where g is the acceleration due to gravity and the density of water is 1.00g/cm\(^3\).
Substituting the values, we get:
W = 384 x 1.00 x 10.0 = 3840g
The weight of the wood is equal to the weight of the water displaced, so:
mass of wood = 3840/0.72 = 5333.33g
Therefore, the mass of the wood is approximately 5333g.
Bayanin Amsa
(a)
i) Torque is the twisting force that causes rotational motion in an object.
ii) Three factors that determine torque are the magnitude of the force applied, the distance between the force and the axis of rotation, and the angle between the force and the lever arm.
(b)
i) Free fall is the motion of an object falling under the influence of gravity without any resistance from the air or other medium.
ii) To solve this problem, we can use the equation:
h = ut + 0.5gt\(^2\)
where h = height of the tower = 40.0m, u = initial velocity = 10.0ms\(^{-1}\), g = acceleration due to gravity = 10.0ms\(^{-2}\), and t = time taken to reach the ground.
Rearranging the equation, we get:
t = sqrt(2h/g) = sqrt(2 x 40.0/10.0) = 4.0s
Therefore, the time taken for the body to reach the ground is 4.0 seconds.
(c)
To calculate the mass of the wood, we can use the principle of buoyancy. The buoyant force acting on the wood is equal to the weight of the water displaced by the wood.
The volume of water displaced by the wood is equal to the volume of the wood that is submerged, which is given by:
V = l x b x h
where l, b, and h are the length, breadth, and height of the submerged portion of the wood, respectively.
From the diagram, we can see that h = 8.0 - 2.0 = 6.0cm, and the length and breadth are also equal to 8.0cm.
Therefore, V = l x b x h = 8.0 x 8.0 x 6.0 = 384cm\(^3\)
The weight of water displaced by the wood is given by:
W = V x density of water x g
where g is the acceleration due to gravity and the density of water is 1.00g/cm\(^3\).
Substituting the values, we get:
W = 384 x 1.00 x 10.0 = 3840g
The weight of the wood is equal to the weight of the water displaced, so:
mass of wood = 3840/0.72 = 5333.33g
Therefore, the mass of the wood is approximately 5333g.
Tambaya 56 Rahoto
TEST OF PRACTICAL KNOWLEDGE QUESTION
You are provided with a glass block, plane mirror, and optical pins.
(i) Place the glass block on a drawing sheet and trace its outline ABCD as shown in the diagram above.
(ii) Remove the block, measure and record the width W of the block.
(iii) Draw a normal ON to DC at a point about one-quarter the length of DC.
(iv) Draw a line making an angle i = 10° with the normal.
(v) Replace the block on its outline and mount the plane mirror vertically behind the block such that it makes good contact with the face AB.
(vi) Stick two pins P\(_{1}\) and P\(_{2}\) on the line MO.
(vii) Looking through the face CD, stick two other pins P\(_{3}\) and P\(_{4}\) such that they appear to be in a straight line with the images of pins P\(_{1}\) and P\(_{2}\) seen through the block.
(viii) Join P\(_{3}\) and P\(_{4}\) with a straight line and extend it to touch the face CD at O\(^{1}\).
(ix)Draw a perpendicular line from the midpoint of OO\(^{1}\) to meet AB at Q.
(x) Draw lines OQ, O\(^{1}\)Q and normal O\(^{1}\)N\(^{1}\) produced.
(xi) Measure and record \cos\theta, e, and d.
(xii) Evaluate m = sin e, and n cos\(\frac{\theta}{2}\)
(xii)Repeat the procedure for i = 20°, 30°, 40° and 50.
(xiv) Tabulate your readings.
(xv) Plot a graph with m on the vertical axis and n on the horizontal axis.
(xvi) Determine the slope, s, of the graph and evaluate \cos\theta = 2Ws.
(xvii) State two precautions are taken to ensure accurate results.
(xviii) Sketch a diagram to show the path of the ray through the glass block when the angle of incidence i = 90° in the experiment above.
(xix) A coin lies at the bottom of a tank containing water to a depth of 130cm. If the refractive index of water is 1.3, calculate the apparent displacement of the coin when viewed vertically from above.
(i) The width W of the glass block measured and recorded in em to at least 1 decimal place.
(ii) 5 complete traces attached, showing at least incident and emergent rays, the normals ON and 01 N1 , and the rays OQ and Q01 as illustrated in the diagram.
(iii) 5 values of the emergent angles e measured and recorded in Degrees.
(iv) 5 values of angle 6 measured and recorded in degrees
(v) 5 values of d measured and recorded in em to at least 1 decimal place
(vi)5 values of m = sine e) correctly evaluated to at least 3 decimal places
(vii) 5 values of n = cos(θ/2)correctly evaluated to at least 3 decimal places
(viii) Composite table showing i, θ, e, m and n
(xvii) Precautions for Accurate Results:
- Ensure accurate measurements: Use a protractor or ruler to carefully measure angles (incident and refraction) and the width (W) of the block.
- Precise pin placement: Use straight pins and position them precisely for better alignment when looking through the block.
(xviii) Ray Path at 90° Incidence:
When the angle of incidence (i) is 90 degrees (light strikes the block perpendicular to its surface), the light ray will travel straight through the block without bending. There will be no refraction in this case. You can sketch a ray entering the block at 90 degrees, going straight through the block parallel to the sides, and exiting the block at the opposite face, again perpendicular to the surface.
(xix) Apparent Displacement of Coin in Water:
This question requires applying the concept of refractive index but to a different scenario (light traveling from air to water). Here's how to solve it:
- Angle of Incidence: Since you're viewing the coin vertically from above, the angle of incidence (i) can be considered 0 degrees.
- Refractive Index and Formula: We know the refractive index of water (n = 1.3). Use the formula: n = 1.3 = sin(0°) / sin(angle of refraction in water).
- Solve for Angle of Refraction: Solve the equation to find the angle of refraction in water.
- Apparent Depth and Displacement: Use trigonometry based on the actual depth (130 cm) and the angle of refraction in water to calculate the apparent depth. The difference between the actual depth and the apparent depth is the apparent displacement of the coin.
Bayanin Amsa
(i) The width W of the glass block measured and recorded in em to at least 1 decimal place.
(ii) 5 complete traces attached, showing at least incident and emergent rays, the normals ON and 01 N1 , and the rays OQ and Q01 as illustrated in the diagram.
(iii) 5 values of the emergent angles e measured and recorded in Degrees.
(iv) 5 values of angle 6 measured and recorded in degrees
(v) 5 values of d measured and recorded in em to at least 1 decimal place
(vi)5 values of m = sine e) correctly evaluated to at least 3 decimal places
(vii) 5 values of n = cos(θ/2)correctly evaluated to at least 3 decimal places
(viii) Composite table showing i, θ, e, m and n
(xvii) Precautions for Accurate Results:
- Ensure accurate measurements: Use a protractor or ruler to carefully measure angles (incident and refraction) and the width (W) of the block.
- Precise pin placement: Use straight pins and position them precisely for better alignment when looking through the block.
(xviii) Ray Path at 90° Incidence:
When the angle of incidence (i) is 90 degrees (light strikes the block perpendicular to its surface), the light ray will travel straight through the block without bending. There will be no refraction in this case. You can sketch a ray entering the block at 90 degrees, going straight through the block parallel to the sides, and exiting the block at the opposite face, again perpendicular to the surface.
(xix) Apparent Displacement of Coin in Water:
This question requires applying the concept of refractive index but to a different scenario (light traveling from air to water). Here's how to solve it:
- Angle of Incidence: Since you're viewing the coin vertically from above, the angle of incidence (i) can be considered 0 degrees.
- Refractive Index and Formula: We know the refractive index of water (n = 1.3). Use the formula: n = 1.3 = sin(0°) / sin(angle of refraction in water).
- Solve for Angle of Refraction: Solve the equation to find the angle of refraction in water.
- Apparent Depth and Displacement: Use trigonometry based on the actual depth (130 cm) and the angle of refraction in water to calculate the apparent depth. The difference between the actual depth and the apparent depth is the apparent displacement of the coin.
Tambaya 57 Rahoto
(a) In an experiment to measure the specific latent heat of vapourisation of water, a student places a heater in a beaker containing water. The beaker stands on an electronic balance so that the mass of the beaker and water could be measured. The heater is switched on and readings were taken every 100s when the water starts boiling.
The table below shows the readings.
| Time/s | 0 | 100 | 200 | 300 | 400 |
| Reading on balance/g | 203.22 | 201.62 | 199.79 | 198.26 | 196.50 |
| Mass of water evaporated/g | 0 | ||||
| Energy supplied by heated/J | 0 |
(I) Fill in the mass of water evaporated.
(ii) Given that the heater supplies energy at the rate of 38J/s, fill in the values of the energy supplied by the heater in 100s, 200s, 300s, and 400s.
(ii) Plot a graph of energy supplied on the vertical axis and mass of water evaporated on the horizontal axis, starting both axes from the origin(0,0).
(iv) Determine the slope of the graph.
(v) what does the value of the slope mean?
(i)
(a)(i)
| Time/s | 0 | 100 | 200 | 300 | 400 |
| Reading on balance/g | 203.22 | 201.62 | 199.79 | 198.26 | 196.50 |
| Mass of water evaporated/g | 0 | 1.60 (12 )
|
3.43 (12
)
|
4.96 (12 ) | 6.72 (12 ) |
| Energy supplied by heated/J | 0 | 3800 (12 ) | 7600 (12 ) | 11400 (12 ) | 15200 (12 ) |
To fill in the mass of water evaporated, the difference between the initial mass and the final mass of the water can be calculated. For example, at 100s, the mass of water evaporated would be 203.22g - 201.62g = 1.60g.
(ii) To fill in the values of the energy supplied by the heater, the energy can be calculated by multiplying the heater's power by the time it has been on. For example, at 100s, the energy supplied would be 38 J/s x 100s = 3800 J.
(iii) A graph of energy supplied on the vertical axis and mass of water evaporated on the horizontal axis can be plotted as follows:
(iv) The slope of the graph can be determined by finding the ratio of the change in energy supplied to the change in mass of water evaporated. The slope represents the specific latent heat of vaporization of water.
Determination of slope of graph
Slope = ΔQΔM
= 14.0−2.0)×1036.1−0.8
= 2.26 x 103 J/g
(v) The value of the slope represents the specific latent heat of vaporization of water, which is the amount of energy required to change 1 g of a substance from a liquid to a gas at a constant temperature. In this experiment, the slope represents the specific latent heat of vaporization of water, which is the amount of energy required to evaporate 1 g of water at the boiling point.
2.26 x 102 J of heat energy is needed to evaporate 1g of water at constant temperature.
Bayanin Amsa
(i)
(a)(i)
| Time/s | 0 | 100 | 200 | 300 | 400 |
| Reading on balance/g | 203.22 | 201.62 | 199.79 | 198.26 | 196.50 |
| Mass of water evaporated/g | 0 | 1.60 (12 )
|
3.43 (12
)
|
4.96 (12 ) | 6.72 (12 ) |
| Energy supplied by heated/J | 0 | 3800 (12 ) | 7600 (12 ) | 11400 (12 ) | 15200 (12 ) |
To fill in the mass of water evaporated, the difference between the initial mass and the final mass of the water can be calculated. For example, at 100s, the mass of water evaporated would be 203.22g - 201.62g = 1.60g.
(ii) To fill in the values of the energy supplied by the heater, the energy can be calculated by multiplying the heater's power by the time it has been on. For example, at 100s, the energy supplied would be 38 J/s x 100s = 3800 J.
(iii) A graph of energy supplied on the vertical axis and mass of water evaporated on the horizontal axis can be plotted as follows:
(iv) The slope of the graph can be determined by finding the ratio of the change in energy supplied to the change in mass of water evaporated. The slope represents the specific latent heat of vaporization of water.
Determination of slope of graph
Slope = ΔQΔM
= 14.0−2.0)×1036.1−0.8
= 2.26 x 103 J/g
(v) The value of the slope represents the specific latent heat of vaporization of water, which is the amount of energy required to change 1 g of a substance from a liquid to a gas at a constant temperature. In this experiment, the slope represents the specific latent heat of vaporization of water, which is the amount of energy required to evaporate 1 g of water at the boiling point.
2.26 x 102 J of heat energy is needed to evaporate 1g of water at constant temperature.
Tambaya 58 Rahoto
(a)Explain resonance frequency as applied in RLC series Circuit.
(ii) Sketch a diagram to illustrate the variation of frequency, f, with the resistance, R, the capacitive reactance, X\(_c\) and the inductive reactance X\(_L\), in RLC series circuit.
(iii) Using the diagram drawn in (a)(ii) state whether the current in the circuit leads, lags or is in phase with the supply voltage when: (\(\alpha\)) f = f\(_o\); (\(\beta\)) f < f\(_o\) ; (\(\gamma\))f\(_o\); when f\(_o\) is the resonant frequency.
b)(i) Define mutual inductance.
(ii) The coil of an electric generator has 500 turns and 8.0cm diameter. If it rotates in a magnetic field of density 0.25T, calculate the angular speed when its peak voltage is 480V. [\(\pi\) = 3.142].
(a) Resonance frequency refers to the frequency at which the capacitive reactance (Xc) and inductive reactance (XL) in an RLC series circuit are equal, resulting in the impedance of the circuit being at a minimum. This results in a maximum current flow through the circuit and is useful in applications such as tuning circuits for radio communication.
(ii) The diagram illustrates that as resistance increases, the resonant frequency decreases, while as the capacitive reactance or inductive reactance increases, the resonant frequency increases.
(iii) When f = fo, the current is in phase with the supply voltage, meaning that they reach their maximum and minimum values at the same time. When f < fo, the current lags behind the voltage, meaning that the current reaches its maximum and minimum values after the voltage. When f > fo, the current leads the voltage, meaning that the current reaches its maximum and minimum values before the voltage.
(b) Mutual inductance is a measure of the ability of two coils to induce a voltage in each other. It is defined as the ratio of the induced electromotive force (emf) in one coil to the rate of change of current in the other coil.
(ii) To solve the problem, we need to use the formula for the peak voltage of a generator: Vp = NBA?, where Vp is the peak voltage, N is the number of turns in the coil, B is the magnetic field density, A is the area of the coil, and ? is the angular speed of the coil. We can solve for ? by rearranging the formula: ? = Vp / NBA. Plugging in the given values, we get:
? = 480 / (500 * ? * (0.08/2)\(^2\) * 0.25)
? ? 150 radians/second
Therefore, the angular speed of the coil is approximately 150 radians per second.
Bayanin Amsa
(a) Resonance frequency refers to the frequency at which the capacitive reactance (Xc) and inductive reactance (XL) in an RLC series circuit are equal, resulting in the impedance of the circuit being at a minimum. This results in a maximum current flow through the circuit and is useful in applications such as tuning circuits for radio communication.
(ii) The diagram illustrates that as resistance increases, the resonant frequency decreases, while as the capacitive reactance or inductive reactance increases, the resonant frequency increases.
(iii) When f = fo, the current is in phase with the supply voltage, meaning that they reach their maximum and minimum values at the same time. When f < fo, the current lags behind the voltage, meaning that the current reaches its maximum and minimum values after the voltage. When f > fo, the current leads the voltage, meaning that the current reaches its maximum and minimum values before the voltage.
(b) Mutual inductance is a measure of the ability of two coils to induce a voltage in each other. It is defined as the ratio of the induced electromotive force (emf) in one coil to the rate of change of current in the other coil.
(ii) To solve the problem, we need to use the formula for the peak voltage of a generator: Vp = NBA?, where Vp is the peak voltage, N is the number of turns in the coil, B is the magnetic field density, A is the area of the coil, and ? is the angular speed of the coil. We can solve for ? by rearranging the formula: ? = Vp / NBA. Plugging in the given values, we get:
? = 480 / (500 * ? * (0.08/2)\(^2\) * 0.25)
? ? 150 radians/second
Therefore, the angular speed of the coil is approximately 150 radians per second.
Tambaya 59 Rahoto
A satellite launched with velocity V\(_E\) just escapes the earth's gravitational attraction. Given that the radius of the earth is R, show that V\(_E\) = \(\sqrt{20R}\) [g = 10ms\(^{-2}\)
The gravitational potential energy, U, of an object of mass m at a distance r from the center of the Earth is given by:
U = - (GMm) / r
where G is the gravitational constant and M is the mass of the Earth. The gravitational potential energy is negative because it takes energy to move an object away from the Earth's gravitational pull.
The kinetic energy of the object is given by:
K = (1/2)mv2
where v is the velocity of the object and m is its mass.
When the object just escapes the Earth's gravitational pull, its kinetic energy is equal to its gravitational potential energy:
(1/2)mv2 = (GMm) / r
Simplifying and solving for v, we get:
v = sqrt[(2GM) / r]
Substituting G = 6.67 × 10-11 Nm2/kg2, M = 5.97 × 1024 kg, and r = R + h (where h is the height of the satellite above the Earth's surface), we get:
v = sqrt[(2 × 6.67 × 10-11 Nm2/kg2 × 5.97 × 1024 kg) / (R + h)]
When the satellite just escapes the Earth's gravitational pull, its potential energy is zero, so we can set the kinetic energy equal to zero and solve for the maximum height, h, that the satellite can reach:
(1/2)mv2 = 0
v = sqrt[(2GM) / r] = sqrt[(2GM) / (R + h)]
Solving for h, we get:
h = R[(v2 / VE2) - 1]
where VE is the escape velocity from the Earth's gravitational pull. When h is at its maximum value, the satellite has just escaped the Earth's gravitational pull, so its escape velocity is equal to v. Substituting this into the equation for h and simplifying
Bayanin Amsa
The gravitational potential energy, U, of an object of mass m at a distance r from the center of the Earth is given by:
U = - (GMm) / r
where G is the gravitational constant and M is the mass of the Earth. The gravitational potential energy is negative because it takes energy to move an object away from the Earth's gravitational pull.
The kinetic energy of the object is given by:
K = (1/2)mv2
where v is the velocity of the object and m is its mass.
When the object just escapes the Earth's gravitational pull, its kinetic energy is equal to its gravitational potential energy:
(1/2)mv2 = (GMm) / r
Simplifying and solving for v, we get:
v = sqrt[(2GM) / r]
Substituting G = 6.67 × 10-11 Nm2/kg2, M = 5.97 × 1024 kg, and r = R + h (where h is the height of the satellite above the Earth's surface), we get:
v = sqrt[(2 × 6.67 × 10-11 Nm2/kg2 × 5.97 × 1024 kg) / (R + h)]
When the satellite just escapes the Earth's gravitational pull, its potential energy is zero, so we can set the kinetic energy equal to zero and solve for the maximum height, h, that the satellite can reach:
(1/2)mv2 = 0
v = sqrt[(2GM) / r] = sqrt[(2GM) / (R + h)]
Solving for h, we get:
h = R[(v2 / VE2) - 1]
where VE is the escape velocity from the Earth's gravitational pull. When h is at its maximum value, the satellite has just escaped the Earth's gravitational pull, so its escape velocity is equal to v. Substituting this into the equation for h and simplifying
Tambaya 60 Rahoto
In a solar panel for heat supply, state the function of each of the following parts: (a) metal flat plate; (b) thermal insulator: (c) tubes.
(a) The metal flat plate is the absorbing surface of the solar panel. Its main function is to absorb sunlight and convert it into heat energy. The plate is typically coated with a dark, heat-absorbing material that maximizes the amount of solar energy that can be absorbed.
(b) The thermal insulator is a layer of material that is placed behind the metal flat plate to reduce heat loss from the back of the panel. Its main function is to prevent heat energy from escaping through the back of the panel and ensuring that the absorbed heat is retained in the panel.
(c) The tubes are the part of the solar panel where a heat transfer fluid, usually water or a mixture of water and antifreeze, circulates. The tubes are in contact with the metal flat plate and absorb the heat energy that is transferred from the plate. The heated fluid then circulates through the tubes and is used to heat water or air for domestic or industrial use.
Bayanin Amsa
(a) The metal flat plate is the absorbing surface of the solar panel. Its main function is to absorb sunlight and convert it into heat energy. The plate is typically coated with a dark, heat-absorbing material that maximizes the amount of solar energy that can be absorbed.
(b) The thermal insulator is a layer of material that is placed behind the metal flat plate to reduce heat loss from the back of the panel. Its main function is to prevent heat energy from escaping through the back of the panel and ensuring that the absorbed heat is retained in the panel.
(c) The tubes are the part of the solar panel where a heat transfer fluid, usually water or a mixture of water and antifreeze, circulates. The tubes are in contact with the metal flat plate and absorb the heat energy that is transferred from the plate. The heated fluid then circulates through the tubes and is used to heat water or air for domestic or industrial use.
Tambaya 61 Rahoto
An elastic material of length 3m is to be stretched to reduce an extension three times its original length. Calculate the force required to produce the extension
To calculate the force required to produce the extension, the formula for Hooke's law can be used. Hooke's law states that the force applied to an elastic material is proportional to the extension produced, provided that the extension is within the elastic limit. The formula for Hooke's law is given by:
F = kx
where F is the force, k is the spring constant, and x is the extension.
In this case, the original length of the material is 3m, and the extension is three times this length, or 3 x 3m = 9m. The force required to produce this extension can be calculated as follows:
F = k x 9m
The spring constant, k, is a property of the material and can be determined experimentally. If the spring constant is known, the force can be calculated.
Note that this calculation assumes that the extension is within the elastic limit of the material. If the extension exceeds the elastic limit, the material will not return to its original length when the force is removed and will have permanent deformation.
Bayanin Amsa
To calculate the force required to produce the extension, the formula for Hooke's law can be used. Hooke's law states that the force applied to an elastic material is proportional to the extension produced, provided that the extension is within the elastic limit. The formula for Hooke's law is given by:
F = kx
where F is the force, k is the spring constant, and x is the extension.
In this case, the original length of the material is 3m, and the extension is three times this length, or 3 x 3m = 9m. The force required to produce this extension can be calculated as follows:
F = k x 9m
The spring constant, k, is a property of the material and can be determined experimentally. If the spring constant is known, the force can be calculated.
Note that this calculation assumes that the extension is within the elastic limit of the material. If the extension exceeds the elastic limit, the material will not return to its original length when the force is removed and will have permanent deformation.
Tambaya 62 Rahoto
TEST OF PRACTICAL KNOWLEDGE QUESTION
(a)
You are provided with a metre rule, a knife-edge, set of masses, inextensible string, retort support and other necessary apparatus.
i. Place the metre rule on the knife edge. Read and record the point G where the metre rule balances horizontally, as shown in Fig (a).
ii. Suspend the metre rule at G with the aid of the string provided and attach the string to the retort support as shown in Fig 1(b). Keep the string attached to this point throughout the experiment.
iii. Attach the mass M\(_{0}\) at the 80cm mark of the metre rule. Determine the distance of y from G. Keep M\(_{0}\) at this
position throughout the experiment.
iv. Suspend a mass M= 40g on the side AG and adjust its position until the metre rule balances horizontally.
v. Measure and record the distance of x of M from G. Evaluate x\(^{-1}\)
vi. Repeat the procedure for four other values of M= 60g, 80g, 100g and 120g. Measure and record x and evaluate x\(^{-1}\) in each case.
vii. Tabulate the readings.
viii. Plot a graph of M on the vertical axis and x on the horizontal axis, starting both axes from the origin (0,0).
ix. Determine the slope s of the graph:
x. Given that s = yM\(_{0}\), determine M\(_{0}\).
xi. State two precautions taken to obtain accurate results.
(b) i. Define the moment of a force about a point.
ii. A uniform metre rule is suspended by an inextensible string at its centre of gravity. If a mass of 60g is placed at the 25cm mark, what mass should be placed at the 80cm mark of the metre rule to balance it horizontally?
(a)
- Place the metre rule on the knife-edge and adjust it until it balances horizontally. Read and record the point G where the metre rule balances horizontally.
- Suspend the metre rule at G with the aid of the inextensible string provided and attach the string to the retort support. Keep the string attached to this point throughout the experiment.
- Attach the mass M0 at the 80cm mark of the metre rule. Determine the distance of y from G. Keep M0 at this position throughout the experiment.
- Suspend a mass M = 40g on the side AG and adjust its position until the metre rule balances horizontally.
- Measure and record the distance of x of M from G. Evaluate x-1.
- Repeat the procedure for four other values of M: 60g, 80g, 100g, and 120g. Measure and record x and evaluate x-1 in each case.
- Tabulate the readings.
- Plot a graph of M on the vertical axis and x on the horizontal axis, starting both axes from the origin (0,0).
- Determine the slope s of the graph.
- Given that s = yM0, determine M0.
- State two precautions taken to obtain accurate results.
(b)
- The moment of a force about a point is defined as the product of the magnitude of the force and the perpendicular distance of its line of action from the point.
- To balance the uniform metre rule horizontally, the moment of the force due to the mass at 80cm mark, M80, must be equal and opposite to the moment of the force due to the 60g mass at 25cm mark, M25, i.e., M80 = M25. Since the metre rule is uniform, the centre of gravity is at its midpoint (50cm mark). Therefore, the distance from the 80cm mark to the midpoint is 30cm, and the distance from the midpoint to the 25cm mark is 25cm. Using the principle of moments, we have:
M80 = M25
40g x x40 = 60g x 25cm
x40 = (60g x 25cm)/(40g) - 80cm
x40 = 25cm - 80cm = -55cm
Thus, a mass of 40g should be placed at the 80cm mark, on the same side as the 60g mass at the 25cm mark, to balance the metre rule horizontally.
Bayanin Amsa
(a)
- Place the metre rule on the knife-edge and adjust it until it balances horizontally. Read and record the point G where the metre rule balances horizontally.
- Suspend the metre rule at G with the aid of the inextensible string provided and attach the string to the retort support. Keep the string attached to this point throughout the experiment.
- Attach the mass M0 at the 80cm mark of the metre rule. Determine the distance of y from G. Keep M0 at this position throughout the experiment.
- Suspend a mass M = 40g on the side AG and adjust its position until the metre rule balances horizontally.
- Measure and record the distance of x of M from G. Evaluate x-1.
- Repeat the procedure for four other values of M: 60g, 80g, 100g, and 120g. Measure and record x and evaluate x-1 in each case.
- Tabulate the readings.
- Plot a graph of M on the vertical axis and x on the horizontal axis, starting both axes from the origin (0,0).
- Determine the slope s of the graph.
- Given that s = yM0, determine M0.
- State two precautions taken to obtain accurate results.
(b)
- The moment of a force about a point is defined as the product of the magnitude of the force and the perpendicular distance of its line of action from the point.
- To balance the uniform metre rule horizontally, the moment of the force due to the mass at 80cm mark, M80, must be equal and opposite to the moment of the force due to the 60g mass at 25cm mark, M25, i.e., M80 = M25. Since the metre rule is uniform, the centre of gravity is at its midpoint (50cm mark). Therefore, the distance from the 80cm mark to the midpoint is 30cm, and the distance from the midpoint to the 25cm mark is 25cm. Using the principle of moments, we have:
M80 = M25
40g x x40 = 60g x 25cm
x40 = (60g x 25cm)/(40g) - 80cm
x40 = 25cm - 80cm = -55cm
Thus, a mass of 40g should be placed at the 80cm mark, on the same side as the 60g mass at the 25cm mark, to balance the metre rule horizontally.
