WAEC SSCE - Physics - 2018
Pregunta 1 Informe
In which of the following activities is an induction coil not useful?
Detalles de la respuesta
An induction coil is not useful in A.C. power transmission. An induction coil is a device that can produce high voltages by inducing a high voltage in a secondary coil through electromagnetic induction. It works on the principle of changing the magnetic field around a conductor to induce an electromotive force (EMF) in a nearby conductor. In the investigation of high voltages, the induction coil is useful as it can produce very high voltages that are needed for some experiments. In the study of electric discharge, induction coils are used to create a spark in a gas-filled tube or bulb. In the operation of X-ray tubes, induction coils are used to generate a high voltage potential difference between the cathode and anode of the tube. This voltage accelerates electrons towards the anode, causing X-rays to be produced. However, in A.C. power transmission, the voltage is already alternating, and the transformer is used instead of an induction coil to increase or decrease the voltage level. A transformer works on the same principle of electromagnetic induction but can handle higher power levels and has more efficient energy transfer characteristics. Thus, an induction coil is not useful in A.C. power transmission. Therefore, the correct answer is A.C. power transmission.
Pregunta 2 Informe
Two plane mirrors are inclined to each other such that an object placed between them has 11 images. Determine the angle of inclination
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Pregunta 3 Informe
A student uses two compass needles to investigate the magnetic field around a bar magnet. Which of the following diagrams show the correct directions of the needles?
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Pregunta 5 Informe
According to Newton's law of cooling, the heat energy lost per unit area of a body depends on the
I. Surface area
II. Nature of the surface
III. Temperature of the surface
IV. Excess temperature over the surroundings
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Pregunta 6 Informe
The general definition of elastic modulus is
Detalles de la respuesta
The definition of elastic modulus is the ratio of stress to strain in a material. Stress is defined as the force applied per unit area, and strain is defined as the deformation or change in length per unit length of a material under that stress. The elastic modulus is a measure of a material's stiffness and its resistance to deformation. The formula for elastic modulus is: \(\frac{\text{stress}}{\text{strain}}\) In simpler terms, elastic modulus is a number that tells us how much a material will stretch or compress under a given amount of force. A material with a high elastic modulus will be stiffer and less likely to deform, while a material with a low elastic modulus will be more flexible and more likely to deform.
Pregunta 7 Informe
The diagram above represents a transverse travelling wave. Which two points are 180° out of phase?
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Pregunta 8 Informe
When two flat plate conductors are arranged in such a way that they possess equal and opposite charges seperated by a dielectric, they form a/an
Detalles de la respuesta
When two flat plate conductors are arranged in such a way that they possess equal and opposite charges separated by a dielectric, they form a capacitor. A capacitor is an electrical component that stores electric charge and energy in an electric field. It consists of two conductive plates separated by a non-conductive material, which is called a dielectric. When a potential difference, or voltage, is applied across the plates, electric charge accumulates on each plate, creating an electric field between them. The amount of charge that a capacitor can store, or its capacitance, depends on the size and distance between the plates, as well as the properties of the dielectric material. In the given scenario, two flat plate conductors are arranged in such a way that they possess equal and opposite charges separated by a dielectric. This arrangement creates a capacitor, where the two conductive plates act as the two electrodes of the capacitor, and the dielectric material between them acts as the insulator. When a potential difference is applied across the plates, charge accumulates on the plates, and the capacitor stores energy in the electric field. Therefore, the correct answer is capacitor.
Pregunta 9 Informe
A wire of length 400cm is stretched between two fixed points. When plucked, its fundamental frequency is 150Hz. Calculate the speed of the wave produced.
Detalles de la respuesta
The speed of a wave can be calculated using the formula: v = fλ where v is the wave speed, f is the frequency of the wave, and λ is the wavelength of the wave. In this problem, we are given the length of the wire, but we need to find the wavelength of the wave. The fundamental frequency of a stretched wire fixed at both ends is given by: f = (1/2L) * √(T/μ) where L is the length of the wire, T is the tension in the wire, and μ is the linear mass density of the wire. We are given the length of the wire (L) and the fundamental frequency (f), so we can solve for T/μ: T/μ = (2Lf)^2 Now we can use the wavelength formula to find the wave speed: v = fλ = f * 2L We know f and L, so we can plug in those values and solve for v: v = 150Hz * 2 * 400cm v = 120000cm/s We need to convert centimeters per second to meters per second: v = 1200m/s Therefore, the speed of the wave produced is 1200 m/s. The answer is option (C).
Pregunta 10 Informe
The diagram above illustrates a simple barometer. Which distance measures the atmospheric pressure?
Detalles de la respuesta
The distance that measures atmospheric pressure in the diagram of the barometer is QR. This is because atmospheric pressure is the force exerted by the weight of air on the surface below it. In a simple barometer like this one, the atmospheric pressure pushes down on the surface of the mercury in the tube, causing it to rise up the tube. The height of the mercury column in the tube, measured from the level of the mercury in the bowl, is directly proportional to the atmospheric pressure pushing down on the surface of the mercury in the bowl. Therefore, the distance QR, which represents the height of the mercury column in the tube, is the measure of atmospheric pressure.
Pregunta 11 Informe
Which of the following radiations has the longest wavelength?
Detalles de la respuesta
Electromagnetic radiation is characterized by its wavelength and frequency. The wavelength is the distance between consecutive peaks of the wave, and the frequency is the number of waves that pass a given point in a certain amount of time. In general, the longer the wavelength, the lower the frequency and energy of the radiation. (Gamma rays) Gamma rays have the shortest wavelength and highest frequency and energy of the four types of radiation listed. (Radio waves) Radio waves have the longest wavelength and lowest frequency and energy of the four types of radiation listed. (Infrared rays) Infrared rays have a longer wavelength than X-rays and gamma rays, but shorter than radio waves. (X-rays) X-rays have a shorter wavelength than infrared rays and radio waves, but longer than gamma rays. So, the answer is radio waves have the longest wavelength.
Pregunta 12 Informe
The pitch of a note is not affected by
Detalles de la respuesta
The pitch of a note refers to how high or low a sound seems to a listener. It is determined by the frequency of the sound wave, or the number of waves that pass by a point in a given amount of time. The tension in a vibrating string, the mass of a vibrating string, and the amplitude of sound wave do not directly affect the pitch of a note. However, they can affect the volume (loudness) and quality (timbre) of the sound, which can influence our perception of the pitch.
Pregunta 13 Informe
A body is projected with an initial velocity u at angle \(\theta\) to the horizontal. If \(R_{max}\) is the maximum range of the projectile, what does the relation \(\frac{u^{2}}{R_{max}}\) represent?
Detalles de la respuesta
The relation \(\frac{u^{2}}{R_{max}}\) represents the horizontal distance covered by the projectile. When a body is projected at an angle \(\theta\) to the horizontal with an initial velocity u, it follows a curved path called a projectile motion. The path of the projectile can be divided into two components: the horizontal component and the vertical component. The horizontal component of the projectile's velocity remains constant throughout the motion because there is no acceleration acting on it in that direction. On the other hand, the vertical component of the velocity changes due to the acceleration due to gravity. The maximum range \(R_{max}\) of the projectile is the horizontal distance covered by it before hitting the ground. It can be shown that the maximum range is given by the formula: \(R_{max} = \frac{u^{2}\sin 2\theta}{g}\) where g is the acceleration due to gravity. Rearranging this equation, we get: \(\frac{u^{2}}{R_{max}} = \frac{g}{\sin 2\theta}\) Therefore, the relation \(\frac{u^{2}}{R_{max}}\) represents the inverse of the sine of twice the angle of projection multiplied by the acceleration due to gravity. This shows that the relation is a constant value for a given angle of projection and acceleration due to gravity. It represents the horizontal distance covered by the projectile, which is the maximum range.
Pregunta 14 Informe
In the diagram illustrated a block of mass M is at rest on a plane inclined at an angle \(\theta\) to the horizontal. As \(\theta\) increases, the velocity ratio of the plane
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Pregunta 15 Informe
Which of the following factors does not increase the rate of evaporation of water in a lake?
Detalles de la respuesta
An increase in the pressure of the atmosphere, rise in temperature, increase in the surface area of the lake, and increase in the speed of the wind all contribute to an increase in the rate of evaporation of water in a lake. Therefore, the factor that does not increase the rate of evaporation of water in a lake is an increase in the pressure of the atmosphere. The reason for this is that an increase in atmospheric pressure means that there is more air above the surface of the lake, which makes it more difficult for water molecules to escape into the air. This results in a decrease in the rate of evaporation of water in the lake.
Pregunta 16 Informe
The diagram above illustrates a waveform. Determine the number of wavelengths in it
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Pregunta 17 Informe
The graph of density against temperature for water is correctly illustrated by
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Pregunta 18 Informe
A body of mass m moving around a circle of radius r with a uniform speed v experiences a centripetal force F. The work done by the centripetal force on it is
Detalles de la respuesta
The centripetal force is the force that keeps an object moving in a circular path. It always acts towards the center of the circle and is perpendicular to the object's velocity. Since the velocity of the object is always perpendicular to the centripetal force, no work is done by the centripetal force on the object. This means that the work done by the centripetal force is zero. Therefore, the answer to the question is "zero".
Pregunta 19 Informe
Which of the following statements about magnetic lines of force is not correct? They
Detalles de la respuesta
Magnetic lines of force are imaginary lines used to visualize the magnetic field around a magnet. They help to understand the behavior of magnetic fields. Out of the given options, the statement that "magnetic lines of force intersect one another" is not correct. This is because magnetic lines of force never intersect each other. If they were to intersect, it would mean that there are two magnetic poles at the same location, which is not possible. The magnetic field lines always run from the north pole of a magnet to its south pole, and they never cross each other. Moreover, magnetic lines of force pass through all materials, whether magnetic or non-magnetic. However, the strength of the magnetic field decreases as it passes through a non-magnetic material. Additionally, the magnetic lines of force are in a state of tension and tend to be shorted. They always try to minimize their energy and move in such a way as to shorten their length.
Pregunta 20 Informe
Which of the following statements about parallel plate capacitors is/are correct?
I. Capacitance of the capacitor is inversely proportional to the distance between the plates
II. Capacitors can filter a.c signals from d.c
III. Capacitors can store charges
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Pregunta 22 Informe
During a training session, two footballers pass a ball repeatedly between each other. The 'to and fro' motion of the ball is not simple harmonic because the
I. Acceleration of the ball is not directed towards a fixed point
II. Restoring force is not directed towards the centre
III. Acceleration of the ball is not directly proportional to the displacement from a fixed point
Which of the statements above are correct?
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Pregunta 23 Informe
The diagram above illustrates the mercury-in-glass thermometer used for determining the temperature of a room. Use the data on it to determine the room temperature.
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Pregunta 24 Informe
A bar AB is balanced horizontally on two knife edges as illustrated as in the diagram above. Determine the weight of the bar.
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Pregunta 25 Informe
A resistor with fixed resistance value is called a
Detalles de la respuesta
A resistor with a fixed resistance value is called a standard resistor. A standard resistor is an electrical component that provides a precise amount of resistance to an electrical circuit. The resistance value of a standard resistor is determined by its physical properties, such as its length, cross-sectional area, and the type of material used to make it. Standard resistors are often used in electrical and electronic circuits as a reference or comparison standard for measuring other resistors or components. They are also used in voltage divider circuits and as current-limiting resistors. Resistance box, shunt resistor, and multiplier resistor are different types of resistors with specific functions. A resistance box is a device that contains multiple resistors that can be connected in different combinations to produce a specific resistance value. A shunt resistor is used to measure current in high-current circuits by providing a low-resistance path for the current to flow. A multiplier resistor is used in conjunction with other components to increase the output voltage of a circuit.
Pregunta 26 Informe
A football is kicked at an angle of 45° to the horizontal over a defense line up with a velocity of 15\(ms^{-1}\). Calculate the magnitudeof horizontal velocity of the ball at its highest point [Neglect friction, g = 10\(ms^{-2}\)].
Detalles de la respuesta
When a projectile is launched at an angle to the horizontal, its initial velocity can be resolved into two components: a horizontal component and a vertical component. The horizontal component remains constant throughout the projectile's motion, while the vertical component is affected by gravity. At the highest point of the projectile's motion, its vertical velocity is zero, but its horizontal velocity remains the same as its initial horizontal velocity. Therefore, the magnitude of the horizontal velocity of the ball at its highest point is 15\(ms^{-1}\)cos(45°) which is approximately equal to 10.6\(ms^{-1}\). Therefore, the correct option is (c) 10.6\(ms^{-1}\).
Pregunta 27 Informe
Two objects P and Q of masses 100kg and 20kg respectively are 1.2m apart on a horizontal surface. Calculate the net gravitational force acting on Q. \([G= 6.67 \times 10^{-11} Nm^{2}kg^{-2}]\)
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Pregunta 28 Informe
An object at rest is said to possess
Detalles de la respuesta
An object at rest is said to possess potential energy. Potential energy is stored energy that an object has because of its position or arrangement. For example, an object placed on a high shelf has potential energy due to its position above the ground. If the object were to fall, the potential energy would be converted into kinetic energy, which is the energy of motion. Chemical energy, electrical energy, and other forms of energy can also be converted into potential energy and kinetic energy depending on the situation.
Pregunta 29 Informe
In which of the following fields is radioisotopes not used?
Detalles de la respuesta
Radioisotopes are not used in weather forecast. Radioisotopes are isotopes (variants of chemical elements) that have unstable nuclei, which means they spontaneously decay and emit radiation in the form of particles or energy. These radioactive properties make them useful in various fields. In medicine, radioisotopes are used in a variety of ways, including for imaging and diagnosis of diseases and in radiation therapy for cancer treatment. In the oil industry, radioisotopes are used to determine the porosity and permeability of rocks and to locate oil reserves. In agriculture, radioisotopes are used to study plant metabolism and nutrient uptake, and to control pests and diseases. However, radioisotopes are not used in weather forecast. Weather forecasting relies on various types of data, including atmospheric pressure, temperature, humidity, and wind speed and direction, which are collected using weather instruments and analyzed using computer models. Radioisotopes do not play a role in this process.
Pregunta 30 Informe
A simple barometer is constructed using a liquid \(Q_{1}\) of density 16.2\(g/cm^{3}\). Calculate the least lenth of the tube to be used to measure pressure of 1.5atm. [Density of mercury = 13.6\(g/cm^{3}\), 1atm = 76.0 cm of Mercury]
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Pregunta 31 Informe
The S.I unit for sound energy is
Detalles de la respuesta
The SI unit for sound energy is J, which stands for Joule. Sound is a form of energy that is produced by the vibration of an object. When an object vibrates, it creates a disturbance in the air molecules around it, which causes the molecules to vibrate and carry the sound energy from the object to our ears. The amount of sound energy produced by an object is related to the amplitude (height) of the sound wave, which determines how loud the sound is, and the frequency (number of vibrations per second), which determines the pitch of the sound. The Joule is a unit of energy that is commonly used in physics to measure different forms of energy, including sound energy. When sound waves travel through a medium, they transfer energy to the medium, which can be measured in Joules. So, the SI unit for sound energy is J.
Pregunta 32 Informe
The property of a body to remain at rest or to continue in uniform motion in a straight line is called
Detalles de la respuesta
The property being described here is called "inertia." In simple terms, it means that an object will keep doing whatever it's already doing (whether that's sitting still, or moving at a constant speed in a straight line) unless a force acts on it to change that state. For example, if you're sitting in a chair, you won't start moving on your own - it takes a force (like someone pushing you) to get you moving. And once you're moving, you'll keep moving at the same speed and direction unless another force (like friction or air resistance) slows you down or changes your direction. This idea of inertia is one of the fundamental principles of physics, and it's closely related to the concept of momentum (which describes how hard it is to stop a moving object) and energy (which is related to how much work is required to change the motion of an object). However, for this particular question, the correct answer is "inertia."
Pregunta 33 Informe
To obtain a virtual, erect and magnified image from a thin converging lens, an object must be placed
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Pregunta 34 Informe
Which of the following is emitted in a process of natural radioactivity?
Detalles de la respuesta
Among the given options, natural radioactivity emits two types of particles: alpha particles and beta particles. Alpha particles are made up of two protons and two neutrons and are essentially helium nuclei. When an unstable nucleus emits an alpha particle, it loses two protons and two neutrons, decreasing its atomic number by two and its atomic mass by four. Beta particles, on the other hand, are high-energy electrons or positrons emitted by a nucleus. Beta decay occurs when a neutron in an unstable nucleus is converted into a proton, and an electron or a positron is emitted. X-rays and lambda rays (gamma rays) are not emitted in natural radioactivity. X-rays are usually produced artificially, while gamma rays are often produced during nuclear reactions or decay processes.
Pregunta 35 Informe
The primary of a transformer has 100 turns and its secondary has 200 turns. If the voltage at the primary is 100V, determine the voltage at the secondary.
Detalles de la respuesta
The voltage at the secondary of a transformer can be found using the equation: V_secondary = (N_secondary / N_primary) * V_primary Where N_secondary is the number of turns on the secondary, N_primary is the number of turns on the primary, and V_primary is the voltage on the primary. Plugging in the values we have: V_secondary = (200 turns / 100 turns) * 100V = 200V So the voltage at the secondary of the transformer is 200V.
Pregunta 36 Informe
Which of the following electromagnetic waves is most energetic?
Detalles de la respuesta
Among the given options, gamma rays are the most energetic form of electromagnetic radiation. Gamma rays have the shortest wavelength and highest frequency among all the electromagnetic waves. This high frequency implies that each photon of gamma rays carries a lot of energy. In comparison, radio waves have the longest wavelength and lowest frequency, while X-rays and ultraviolet rays have intermediate wavelengths and frequencies. Thus, gamma rays are more energetic than X-rays, ultraviolet rays, and radio waves.
Pregunta 37 Informe
The diagram above illustrates the wire connections to a three-pin plug. The wires P, Q and R respectively are:
Detalles de la respuesta
The wires in the diagram are connected to a three-pin plug, which is a type of electrical plug used in many countries for electrical safety. The three wires are typically labeled as live, neutral, and earth. The live wire carries the electric current that powers your device. It is usually colored brown or red. The neutral wire returns the electric current back to the source after it has flowed through your device. It is usually colored blue or black. The earth wire is there for safety. In case of a fault in the device, it provides a path for the electric current to safely flow to the ground. It is usually colored green or green/yellow. So, the correct order of the wires is earth, neutral, and live.
Pregunta 38 Informe
In a series R-L-C circuit, R = 10\(\Omega\), \(X_{c}\)= 4\(\Omega\) and \(X_{L}\) = 9\(\Omega\). The impedance of the circuit is
Detalles de la respuesta
The impedance of a circuit is the total opposition to the flow of an alternating current, consisting of both resistance and reactance. The formula for impedance in a series R-L-C circuit is: Z = \(\sqrt{R^2 + (X_L - X_C)^2}\) where R is the resistance, \(X_L\) is the inductive reactance, and \(X_C\) is the capacitive reactance. Substituting the given values, we get: Z = \(\sqrt{10^2 + (9 - 4)^2}\) = \(\sqrt{100 + 25}\) = \(\sqrt{125}\) = 11.2\(\Omega\) Therefore, the impedance of the given series R-L-C circuit is 11.2\(\Omega\).
Pregunta 39 Informe
The bob of a simple pendulum takes 8.0s to complete 10 oscillations. Determine the frequency of oscillation of the bob.
Detalles de la respuesta
The frequency of oscillation of a pendulum can be calculated using the formula: f = (number of oscillations) / (time taken for the oscillations) Plugging in the values we have: f = (10 oscillations) / (8.0s) = 1.25Hz So the frequency of oscillation of the bob is 1.25Hz. This means the bob completes 1.25 oscillations in one second.
Pregunta 40 Informe
Rainbow formation is a natural phenomenon illustrating that white light
Detalles de la respuesta
Rainbow formation is a natural phenomenon that occurs when white light is dispersed into its individual colours after passing through raindrops in the air. White light is made up of many different colours of light, each with its own wavelength. When the light enters a raindrop, it slows down and bends. This bending causes the light to separate into its individual colours, just like a glass prism would. The different colours then exit the raindrop at slightly different angles, creating a circular arc of coloured light in the sky. So, the correct answer is that rainbow formation illustrates that white light is dispersed into its constituent colours after rainfall.
Pregunta 41 Informe
Frequency is measured in
Detalles de la respuesta
Frequency is measured in hertz (Hz). Hertz is the unit of measurement for frequency, and it represents the number of cycles per second of a periodic signal, such as a sound wave or an electromagnetic wave. One hertz is equivalent to one cycle per second. For example, if a sound wave completes 440 cycles in one second, its frequency is 440 Hz. Hertz can also be used to measure the frequency of other periodic events, such as the rotation of a motor or the vibration of a string. Meter per second is the unit of measurement for speed or velocity. Seconds are units of time, and farad is the unit of measurement for capacitance.
Pregunta 42 Informe
A dam is able to hold a large quantity of water because its wall is
Detalles de la respuesta
The dam is able to hold a large quantity of water because its wall is thickest at the bottom. A dam is a large structure built across a river to store water. It is designed to hold back a large volume of water and withstand the pressure of the water it holds. The force exerted by the water on the dam increases with depth, so the wall of the dam needs to be thicker at the bottom to support the weight of the water. The wall of the dam is usually shaped like a triangle, with the base of the triangle at the bottom of the wall. This is known as a trapezoidal profile. The thickness of the wall gradually decreases towards the top of the dam. By making the wall thicker at the bottom, the dam can support the weight of the water and prevent it from flowing over the top of the dam. The thicker bottom part of the dam also provides a stable foundation for the structure, making it more resistant to damage from the pressure of the water. Therefore, the correct answer is "thickest at the bottom".
Pregunta 44 Informe
A block is acted upon by TWO horizontal forces as illustrated in the diagram above. The block accelerates at 1.5\(ms^{-2}\). Calculate the mass of the block.
Detalles de la respuesta
The acceleration of the block is given as 1.5\(ms^{-2}\). Let's denote the mass of the block as m. From Newton's second law of motion, we know that the force applied on an object is equal to its mass multiplied by its acceleration. In this case, there are two forces acting on the block, both in the horizontal direction. Let's denote these forces as F1 and F2. The net force acting on the block is the vector sum of these two forces: Net Force = F1 + F2 Using the formula for force and substituting the given values, we get: Net Force = ma = 1.5m We know that the acceleration is 1.5\(ms^{-2}\), so we can write: ma = 1.5m Simplifying this equation, we get: m(F1 + F2) = 1.5m Dividing both sides by (F1 + F2), we get: m = 1.5 / (F1 + F2) We don't have the values of F1 and F2, but we can still find the mass of the block using the fact that the answer choices are given in integers. If we assume that the masses are integers, we can try out each of the answer choices by substituting them for m and calculating the values of F1 and F2. If the sum of the forces equals the mass multiplied by the acceleration (1.5\(ms^{-2}\)), then that answer choice is correct. Let's try out the first answer choice of 6kg: m = 6kg ma = 1.5m a = 1.5m / m = 1.5\(ms^{-2}\) We don't have enough information to calculate the forces directly, but we can use the fact that the net force must equal the mass multiplied by the acceleration. If F1 and F2 are the forces acting on the block, then we have: F1 + F2 = ma = 1.5m = 9N If we assume that F1 and F2 are integers, then the only possible combination that adds up to 9N is F1 = 3N and F2 = 6N. Checking that the net force equals the mass multiplied by the acceleration, we get: Net Force = F1 + F2 = 3N + 6N = 9N ma = 6kg x 1.5\(ms^{-2}\) = 9N Since the net force equals the mass multiplied by the acceleration, the answer choice of 6kg is correct. Therefore, the mass of the block is 6kg.
Pregunta 45 Informe
In which of the following situations is photoelectric effect employed?
I. Automatic devices for switching on light at dusk
II. Automatic doors in departmental stores
III. Automatic devices for switching on air conditioners
IV. Automatic voltage stabilizers
Detalles de la respuesta
The photoelectric effect is employed in option I only, which is automatic devices for switching on light at dusk. The photoelectric effect is a phenomenon in which electrons are emitted from a material when it absorbs electromagnetic radiation, such as light. This effect is used in devices such as photoelectric cells or photodiodes, which convert light energy into electrical energy. In option I, the use of photoelectric cells is common in automatic devices for switching on lights at dusk, such as streetlights or security lights. These devices use a photoelectric cell to detect the level of ambient light and automatically switch on the lights when it gets dark. Option II, automatic doors in departmental stores, do not use the photoelectric effect. Instead, they use motion sensors or infrared sensors to detect the presence of a person and trigger the door to open. Option III, automatic devices for switching on air conditioners, do not use the photoelectric effect. They use thermostats or temperature sensors to detect the temperature and trigger the air conditioner to turn on or off. Option IV, automatic voltage stabilizers, do not use the photoelectric effect. They use a voltage regulator or stabilizer circuit to maintain a constant voltage output regardless of fluctuations in the input voltage. Therefore, the correct answer is option I, II and IV only.
Pregunta 46 Informe
Which of the following quantities are dimensionally independent?
I. Dielectric constant II. Momentum III. Efficiency IV. Refractive index
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Pregunta 47 Informe
Which of the following statements is not an advantage of an alkaline accumulator?
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Pregunta 48 Informe
The circuit contains three ammeters A1, A2 and A3 with readings of A2 and A3 shown as indicated in the diagram. Determine the reading of A1.
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Pregunta 49 Informe
When the speed of a car is halved, its kinetic energy is
Detalles de la respuesta
When the speed of a car is halved, its kinetic energy is quartered. Kinetic energy is the energy an object possesses due to its motion. It is dependent on the object's mass and velocity. The formula for kinetic energy is 1/2 x mass x velocity squared. When the speed of a car is halved, its velocity is reduced by a factor of 2. Using the formula for kinetic energy, we can see that the kinetic energy is proportional to the square of the velocity. Therefore, if the velocity is halved, the kinetic energy is reduced to 1/4th of its original value (0.5 x m x (v/2)^2 = 0.25 x m x v^2). So, if the speed of a car is halved, its kinetic energy is quartered.
Pregunta 50 Informe
A nuclide is represented by \(_{32}^{70}B\). Determine its neutron-proton ratio
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Pregunta 51 Informe
a) What does the acronym LASER stand for?
b) What is a laser?
(a) LASER stands for Light Amplification by Stimulated Emission of Radiation.
(b) A laser is a device that produces an intense, narrow, highly directional beam of light that is monochromatic (single wavelength) and coherent (the waves are in phase), by the process of stimulated emission of radiation.
Detalles de la respuesta
(a) LASER stands for Light Amplification by Stimulated Emission of Radiation.
(b) A laser is a device that produces an intense, narrow, highly directional beam of light that is monochromatic (single wavelength) and coherent (the waves are in phase), by the process of stimulated emission of radiation.
Pregunta 52 Informe
Name three classes of magnetic materials.
Three classes of magnetic materials
- Ferromagnetic materials (e.g. iron, nickel, cobalt) which are strongly attracted by a magnet.
- Paramagnetic materials (e.g. aluminium, platinum) which are weakly attracted by a magnet.
- Diamagnetic materials (e.g. copper, bismuth) which are weakly repelled by a magnet.
Detalles de la respuesta
Three classes of magnetic materials
- Ferromagnetic materials (e.g. iron, nickel, cobalt) which are strongly attracted by a magnet.
- Paramagnetic materials (e.g. aluminium, platinum) which are weakly attracted by a magnet.
- Diamagnetic materials (e.g. copper, bismuth) which are weakly repelled by a magnet.
Pregunta 53 Informe
(a) What is an intrinsic semiconductor?
(b) Distinguish between the p-type and n-type semi-conductors.
(a) Intrinsic semiconductor
An intrinsic semiconductor is a pure semiconductor (such as pure silicon or germanium) with no added impurity, in which the number of free electrons equals the number of holes, both being produced only by thermal breaking of covalent bonds.
(b) p-type versus n-type semiconductors
| p-type | n-type |
|---|---|
| Formed by doping with a trivalent impurity (e.g. boron, aluminium). | Formed by doping with a pentavalent impurity (e.g. phosphorus, arsenic). |
| Majority charge carriers are holes (positive). | Majority charge carriers are free electrons (negative). |
| The impurity is called an acceptor. | The impurity is called a donor. |
Detalles de la respuesta
(a) Intrinsic semiconductor
An intrinsic semiconductor is a pure semiconductor (such as pure silicon or germanium) with no added impurity, in which the number of free electrons equals the number of holes, both being produced only by thermal breaking of covalent bonds.
(b) p-type versus n-type semiconductors
| p-type | n-type |
|---|---|
| Formed by doping with a trivalent impurity (e.g. boron, aluminium). | Formed by doping with a pentavalent impurity (e.g. phosphorus, arsenic). |
| Majority charge carriers are holes (positive). | Majority charge carriers are free electrons (negative). |
| The impurity is called an acceptor. | The impurity is called a donor. |
Pregunta 54 Informe
(a) Define strain.
(b) A rubber band is stretched to twice its original length. Calculate the strain on the rubber band.
(a)
Strain is a measure of the deformation or elongation of a material when subjected to an external force or stress. It is defined as the ratio of the change in length or shape of an object to its original length or shape.
(b)
In this case, the rubber band has been stretched to twice its original length. Therefore, the change in length is equal to twice the original length (2L), where L is the original length.
To calculate the strain, we use the formula:
Strain = (Change in length) / (Original length)
Substituting the values, we get:
Strain = (2L - L) / L = L / L = 1
So, the strain on the rubber band is 1, which means that the rubber band has elongated by 100% of its original length. This is because the change in length is equal to the original length, so the strain is equal to 1 (or 100%).
Detalles de la respuesta
(a)
Strain is a measure of the deformation or elongation of a material when subjected to an external force or stress. It is defined as the ratio of the change in length or shape of an object to its original length or shape.
(b)
In this case, the rubber band has been stretched to twice its original length. Therefore, the change in length is equal to twice the original length (2L), where L is the original length.
To calculate the strain, we use the formula:
Strain = (Change in length) / (Original length)
Substituting the values, we get:
Strain = (2L - L) / L = L / L = 1
So, the strain on the rubber band is 1, which means that the rubber band has elongated by 100% of its original length. This is because the change in length is equal to the original length, so the strain is equal to 1 (or 100%).
Pregunta 55 Informe
(a) List two factors each that affect heat loss by:
(i) radiation;
(ii) convection.
(b) State two factors that determine the quantity of heat in a body.
(c) Explain the statement: The vecilic latent heat of vaporization of mercury is 2.72 x 10\(^5\) Jkg\(^{-1}\).
(d)A jug of heat capacity 250 Jkg\(^{-1}\) contains water at 28°C. An electric heater of resistance 35\(\Omega\) connected to a 220 V source is used to raise the temperature of the water until it boils at 100°C in 4 minutes. After. another 5 minutes, 300 g of water has evaporated. Assuming no heat is lost to the surroundings, calculate the:
(i) mass of water in the jug before heating;
(ii) specific latent heat of vaporization of steam. [Specific heat capacity of water = 4200 kg\(^{-1}\)K\(^{-1}\)]
(a) Factors affecting heat loss
(i) By radiation:
- The temperature of the body (and of its surroundings).
- The nature and colour of the surface (dull/black surfaces radiate more than shiny/white ones) and its surface area.
(ii) By convection:
- The temperature difference between the body and the surrounding fluid.
- The exposed surface area and the freedom of the fluid to circulate (draught/movement of the fluid).
(b) Factors determining the quantity of heat in a body
- The mass of the body.
- The temperature (and the nature/specific heat capacity of the material).
(c) The statement "the specific latent heat of vaporization of mercury is \(2.72\times10^{5}\ \text{J kg}^{-1}\)" means that \(2.72\times10^{5}\) joules of heat is required to change 1 kilogram of mercury from liquid to vapour at its boiling point without any change in temperature.
(d) The jug of water
Power of heater: \(P = \dfrac{V^{2}}{R} = \dfrac{220^{2}}{35} = \dfrac{48400}{35} = 1382.9\ \text{W}\).
(i) Mass of water before heating. In the first \(t_{1} = 4\text{ min} = 240\text{ s}\) the temperature rises from \(28^{\circ}\text{C}\) to \(100^{\circ}\text{C}\), so \(\Delta\theta = 72\ \text{K}\).
Heat supplied \(= P t_{1} = 1382.9\times240 = 3.319\times10^{5}\ \text{J}\).
Taking heat capacity of jug \(C = 250\ \text{J K}^{-1}\) and \(c_{water} = 4200\ \text{J kg}^{-1}\text{K}^{-1}\):
\((m c_{water} + C)\Delta\theta = P t_{1}\)
\((4200m + 250)\times 72 = 331886\)
\(4200m + 250 = 4609.5 \Rightarrow 4200m = 4359.5 \Rightarrow m = 1.04\ \text{kg}\).
The mass of water before heating is about 1.04 kg.
(ii) Specific latent heat of vaporization of steam. In the next \(t_{2} = 5\text{ min} = 300\text{ s}\) the water boils at constant temperature and \(0.3\ \text{kg}\) evaporates.
Heat supplied \(= P t_{2} = 1382.9\times300 = 4.149\times10^{5}\ \text{J}\).
\(L = \dfrac{P t_{2}}{m_{evap}} = \dfrac{414857}{0.3} = 1.38\times10^{6}\ \text{J kg}^{-1}\).
Detalles de la respuesta
(a) Factors affecting heat loss
(i) By radiation:
- The temperature of the body (and of its surroundings).
- The nature and colour of the surface (dull/black surfaces radiate more than shiny/white ones) and its surface area.
(ii) By convection:
- The temperature difference between the body and the surrounding fluid.
- The exposed surface area and the freedom of the fluid to circulate (draught/movement of the fluid).
(b) Factors determining the quantity of heat in a body
- The mass of the body.
- The temperature (and the nature/specific heat capacity of the material).
(c) The statement "the specific latent heat of vaporization of mercury is \(2.72\times10^{5}\ \text{J kg}^{-1}\)" means that \(2.72\times10^{5}\) joules of heat is required to change 1 kilogram of mercury from liquid to vapour at its boiling point without any change in temperature.
(d) The jug of water
Power of heater: \(P = \dfrac{V^{2}}{R} = \dfrac{220^{2}}{35} = \dfrac{48400}{35} = 1382.9\ \text{W}\).
(i) Mass of water before heating. In the first \(t_{1} = 4\text{ min} = 240\text{ s}\) the temperature rises from \(28^{\circ}\text{C}\) to \(100^{\circ}\text{C}\), so \(\Delta\theta = 72\ \text{K}\).
Heat supplied \(= P t_{1} = 1382.9\times240 = 3.319\times10^{5}\ \text{J}\).
Taking heat capacity of jug \(C = 250\ \text{J K}^{-1}\) and \(c_{water} = 4200\ \text{J kg}^{-1}\text{K}^{-1}\):
\((m c_{water} + C)\Delta\theta = P t_{1}\)
\((4200m + 250)\times 72 = 331886\)
\(4200m + 250 = 4609.5 \Rightarrow 4200m = 4359.5 \Rightarrow m = 1.04\ \text{kg}\).
The mass of water before heating is about 1.04 kg.
(ii) Specific latent heat of vaporization of steam. In the next \(t_{2} = 5\text{ min} = 300\text{ s}\) the water boils at constant temperature and \(0.3\ \text{kg}\) evaporates.
Heat supplied \(= P t_{2} = 1382.9\times300 = 4.149\times10^{5}\ \text{J}\).
\(L = \dfrac{P t_{2}}{m_{evap}} = \dfrac{414857}{0.3} = 1.38\times10^{6}\ \text{J kg}^{-1}\).
Pregunta 56 Informe
A black body radiates maximum energy when its surface temperature T and the corresponding wavelength \(\lambda\)max are related by the equation \(\lambda\)max T = constant. Given the values of the constant and surface temperature as 2.9 x 10\(^{-3}\) mK and 57°C respectively; Calculate the frequency of the energy radiated.
Wien's displacement law
Given \(\lambda_{\max}T = 2.9\times10^{-3}\ \text{m K}\), and \(T = 57^{\circ}\text{C} = 57 + 273 = 330\ \text{K}\).
\(\lambda_{\max} = \dfrac{2.9\times10^{-3}}{330} = 8.79\times10^{-6}\ \text{m}\).
Frequency: \(f = \dfrac{c}{\lambda_{\max}} = \dfrac{3.0\times10^{8}}{8.79\times10^{-6}}\).
\(f = 3.41\times10^{13}\ \text{Hz}\).
The frequency of the radiated energy is \(3.4\times10^{13}\ \text{Hz}\).
Detalles de la respuesta
Wien's displacement law
Given \(\lambda_{\max}T = 2.9\times10^{-3}\ \text{m K}\), and \(T = 57^{\circ}\text{C} = 57 + 273 = 330\ \text{K}\).
\(\lambda_{\max} = \dfrac{2.9\times10^{-3}}{330} = 8.79\times10^{-6}\ \text{m}\).
Frequency: \(f = \dfrac{c}{\lambda_{\max}} = \dfrac{3.0\times10^{8}}{8.79\times10^{-6}}\).
\(f = 3.41\times10^{13}\ \text{Hz}\).
The frequency of the radiated energy is \(3.4\times10^{13}\ \text{Hz}\).
Pregunta 57 Informe
A missile is projected so as to attain its maximum range. Calculate the maximum height attained if the initial velocity of projection is 200 ms\(^{-1}\). [g = 10ms\(^{-2}\)]
Pregunta 58 Informe
(a) Define uniform acceleration.
(b) Forces act on a car in motion. List the
(i) horizontal forces and their directions;
(ii) vertical forces and their directions
(c) A car starts from rest and accelerate uniformly for 20s to attain a speed of 25 ms\(^{-1}\). It maintains this speed for 30s before decelerating uniformly to rest. The total time for the journey is 60s.
(i) Sketch a velocity-tune graph for the motion.
(ii) Use the graph to determine the (\(\alpha\)) total distance travelled by the car (\(\beta\)) deceleration of the car.
The figure here illustrates force-extension graph for a stretched spiral spring. Determine the work done on the spring.
(a) Uniform acceleration. Uniform acceleration is a constant rate of change of velocity with time, that is, the velocity of the body changes by equal amounts in equal intervals of time.
(b) Forces acting on a moving car.
(i) Horizontal forces and their directions:
- The driving (tractive) force from the engine through the wheels, acting forward in the direction of motion.
- Friction between the tyres and road together with air resistance (drag), acting backward, opposite to the direction of motion.
(ii) Vertical forces and their directions:
- The weight of the car, acting vertically downward.
- The normal reaction from the road surface, acting vertically upward.
(c)(i) Velocity-time graph. The car speeds up uniformly from rest to \(25\text{ m s}^{-1}\) in the first \(20\text{ s}\) (line \(OA\)), holds \(25\text{ m s}^{-1}\) for the next \(30\text{ s}\), that is until \(t=50\text{ s}\) (line \(AB\)), then slows uniformly to rest by \(t=60\text{ s}\) (line \(BC\)).
(c)(ii)(\(\alpha\)) Total distance travelled. The distance is the area under the graph, which is the trapezium \(OABC\). The parallel sides are the whole base \(OC=60\text{ s}\) and the top \(AB=30\text{ s}\); the height is the speed \(25\text{ m s}^{-1}\):
\[ s=\tfrac{1}{2}(OC+AB)\times v=\tfrac{1}{2}(60+30)\times 25=\tfrac{1}{2}\times 90\times 25=1125\text{ m.} \]
(c)(ii)(\(\beta\)) Deceleration. The deceleration is the magnitude of the slope of the final line \(BC\), where the speed falls from \(25\text{ m s}^{-1}\) to \(0\) in the last \(10\text{ s}\) (from \(t=50\text{ s}\) to \(t=60\text{ s}\)):
\[ a=\frac{\text{change in velocity}}{\text{time}}=\frac{0-25}{60-50}=\frac{-25}{10}=-2.5\text{ m s}^{-2}. \]
So the car decelerates at \(2.5\text{ m s}^{-2}\). Note the value is \(2.5\text{ m s}^{-2}\), not \(25\text{ m s}^{-2}\): the change of \(25\text{ m s}^{-1}\) is divided by the \(10\text{ s}\) taken.
Work done on the spring (force-extension graph). For a spring obeying Hooke's law the work done, which is the elastic potential energy stored, equals the area under the force-extension line. From the graph the line runs from the origin to \(F=12\text{ N}\) at an extension \(e=0.5\text{ cm}=0.5\times10^{-2}\text{ m}=0.005\text{ m}\), so the area is a triangle:
\[ W=\tfrac{1}{2}\,F\,e=\tfrac{1}{2}\times 12\times 0.005=0.03\text{ J.} \]
The work done on the spring is \(0.03\text{ J}\), that is \(3.0\times10^{-2}\text{ J}\). Remember to convert the extension from centimetres to metres before multiplying, otherwise the energy comes out one hundred times too large.
Detalles de la respuesta
(a) Uniform acceleration. Uniform acceleration is a constant rate of change of velocity with time, that is, the velocity of the body changes by equal amounts in equal intervals of time.
(b) Forces acting on a moving car.
(i) Horizontal forces and their directions:
- The driving (tractive) force from the engine through the wheels, acting forward in the direction of motion.
- Friction between the tyres and road together with air resistance (drag), acting backward, opposite to the direction of motion.
(ii) Vertical forces and their directions:
- The weight of the car, acting vertically downward.
- The normal reaction from the road surface, acting vertically upward.
(c)(i) Velocity-time graph. The car speeds up uniformly from rest to \(25\text{ m s}^{-1}\) in the first \(20\text{ s}\) (line \(OA\)), holds \(25\text{ m s}^{-1}\) for the next \(30\text{ s}\), that is until \(t=50\text{ s}\) (line \(AB\)), then slows uniformly to rest by \(t=60\text{ s}\) (line \(BC\)).
(c)(ii)(\(\alpha\)) Total distance travelled. The distance is the area under the graph, which is the trapezium \(OABC\). The parallel sides are the whole base \(OC=60\text{ s}\) and the top \(AB=30\text{ s}\); the height is the speed \(25\text{ m s}^{-1}\):
\[ s=\tfrac{1}{2}(OC+AB)\times v=\tfrac{1}{2}(60+30)\times 25=\tfrac{1}{2}\times 90\times 25=1125\text{ m.} \]
(c)(ii)(\(\beta\)) Deceleration. The deceleration is the magnitude of the slope of the final line \(BC\), where the speed falls from \(25\text{ m s}^{-1}\) to \(0\) in the last \(10\text{ s}\) (from \(t=50\text{ s}\) to \(t=60\text{ s}\)):
\[ a=\frac{\text{change in velocity}}{\text{time}}=\frac{0-25}{60-50}=\frac{-25}{10}=-2.5\text{ m s}^{-2}. \]
So the car decelerates at \(2.5\text{ m s}^{-2}\). Note the value is \(2.5\text{ m s}^{-2}\), not \(25\text{ m s}^{-2}\): the change of \(25\text{ m s}^{-1}\) is divided by the \(10\text{ s}\) taken.
Work done on the spring (force-extension graph). For a spring obeying Hooke's law the work done, which is the elastic potential energy stored, equals the area under the force-extension line. From the graph the line runs from the origin to \(F=12\text{ N}\) at an extension \(e=0.5\text{ cm}=0.5\times10^{-2}\text{ m}=0.005\text{ m}\), so the area is a triangle:
\[ W=\tfrac{1}{2}\,F\,e=\tfrac{1}{2}\times 12\times 0.005=0.03\text{ J.} \]
The work done on the spring is \(0.03\text{ J}\), that is \(3.0\times10^{-2}\text{ J}\). Remember to convert the extension from centimetres to metres before multiplying, otherwise the energy comes out one hundred times too large.
Pregunta 59 Informe
State three materials used for making optical fibres.
Materials used for making optical fibres
- Silica (fused quartz / glass).
- Plastic (transparent polymer, e.g. acrylic/PMMA).
- Doped glass such as germanium-doped or fluoride glass used for the core and cladding.
Detalles de la respuesta
Materials used for making optical fibres
- Silica (fused quartz / glass).
- Plastic (transparent polymer, e.g. acrylic/PMMA).
- Doped glass such as germanium-doped or fluoride glass used for the core and cladding.
Pregunta 60 Informe
(a) Define:
(i) reactance;
(ii) impedance in an a.c.
The diagram here illustrates an a.c. generator. When the coil is rotated, an e.m.f is induced in the coil.
(i) Explain why an e.m.f. is induced.
(ii) State the purpose of the slip-rings.
(iii) Name and state the law used to determine the direction of the induced current.
(iv) State two ways to increase the induced e.m.f.
(c) A lamp is rated 12 V. 6 W. Calculate the amount of energy transformed by the lamp in 5 minutes.
Detalles de la respuesta
None
Pregunta 61 Informe
(a) Define binding energy in an atom.
(b) List three evidence to support the claim that X-rays are electromagnetic waves.
(c) List three peaceful uses of nuclear energy.
(d) Light of wavelength 4.5 x 10\(^{-7}\) in is incident on a metal resulting in the emission of photo electrons. If the work function of the metal is 3.0 x 10\(^{-9}\) J, calculate the:
(i) frequency of the incident light;
(ii) energy of the incident light;
(iii) energy of the photoelectrons. [Speed of light = 3.0 x 10\(^8\) ms\(^{-1}\), h = 6.6 x 10\(^{-34}\) Js]
(a) Binding energy
The binding energy of a nucleus is the energy required to completely separate the nucleus into its individual constituent protons and neutrons (equivalently, the energy released when the separate nucleons combine to form the nucleus). It equals the mass defect times \(c^{2}\).
(b) Three evidences that X-rays are electromagnetic waves
- They travel in straight lines with the speed of light and are undeflected by electric or magnetic fields (so they are uncharged).
- They undergo diffraction and interference when passed through crystals.
- They can be polarised.
(c) Three peaceful uses of nuclear energy
- Generation of electricity in nuclear power stations.
- Medical uses: radiotherapy to treat cancer and use of radioisotopes as tracers in diagnosis.
- Industrial and agricultural uses: sterilisation of equipment, food preservation, and thickness/level gauging.
(d) Photoelectric calculation
Data: \(\lambda = 4.5\times10^{-7}\ \text{m}\), work function \(W_{0} = 3.0\times10^{-19}\ \text{J}\) (taking the stated value as \(3.0\times10^{-19}\ \text{J}\)), \(c = 3.0\times10^{8}\ \text{m s}^{-1}\), \(h = 6.6\times10^{-34}\ \text{J s}\).
(i) Frequency: \(f = \dfrac{c}{\lambda} = \dfrac{3.0\times10^{8}}{4.5\times10^{-7}} = 6.67\times10^{14}\ \text{Hz}\).
(ii) Energy of the incident light: \(E = hf = 6.6\times10^{-34}\times6.67\times10^{14} = 4.4\times10^{-19}\ \text{J}\).
(iii) Energy (max KE) of the photoelectrons: \(E_{k} = E - W_{0} = 4.4\times10^{-19} - 3.0\times10^{-19} = 1.4\times10^{-19}\ \text{J}\).
Detalles de la respuesta
(a) Binding energy
The binding energy of a nucleus is the energy required to completely separate the nucleus into its individual constituent protons and neutrons (equivalently, the energy released when the separate nucleons combine to form the nucleus). It equals the mass defect times \(c^{2}\).
(b) Three evidences that X-rays are electromagnetic waves
- They travel in straight lines with the speed of light and are undeflected by electric or magnetic fields (so they are uncharged).
- They undergo diffraction and interference when passed through crystals.
- They can be polarised.
(c) Three peaceful uses of nuclear energy
- Generation of electricity in nuclear power stations.
- Medical uses: radiotherapy to treat cancer and use of radioisotopes as tracers in diagnosis.
- Industrial and agricultural uses: sterilisation of equipment, food preservation, and thickness/level gauging.
(d) Photoelectric calculation
Data: \(\lambda = 4.5\times10^{-7}\ \text{m}\), work function \(W_{0} = 3.0\times10^{-19}\ \text{J}\) (taking the stated value as \(3.0\times10^{-19}\ \text{J}\)), \(c = 3.0\times10^{8}\ \text{m s}^{-1}\), \(h = 6.6\times10^{-34}\ \text{J s}\).
(i) Frequency: \(f = \dfrac{c}{\lambda} = \dfrac{3.0\times10^{8}}{4.5\times10^{-7}} = 6.67\times10^{14}\ \text{Hz}\).
(ii) Energy of the incident light: \(E = hf = 6.6\times10^{-34}\times6.67\times10^{14} = 4.4\times10^{-19}\ \text{J}\).
(iii) Energy (max KE) of the photoelectrons: \(E_{k} = E - W_{0} = 4.4\times10^{-19} - 3.0\times10^{-19} = 1.4\times10^{-19}\ \text{J}\).
Pregunta 62 Informe
(a) Define dffraction.
(b)(i) Explain critical angle. The diagram here illustrates a ray of light passing through a rectangular transparent plastic block \(\alpha\) Determine the value of the critical angle. \(\beta\) Calculate the refractive index of the block.
(c) A pipe closed at one end has fundamental frequency of 200Hz. The frequency of the first overtone of the closed pipe is equal to the frequency of the first overtone of an open pipe. Calculate the:
(i) fundamental frequency of the open pipe;
(ii) length of the closed pipe;
(iii) length of the open pipe. [Speed of sound in air = 330 ms\(^{-1}\)]
(a) Diffraction
Diffraction is the spreading of a wave into the region behind an obstacle, or the bending of a wave round the edges of an obstacle or through an aperture, as the wave passes them. It is most noticeable when the width of the aperture or obstacle is comparable with the wavelength of the wave.
(b)(i) Critical angle
The critical angle is the angle of incidence in the optically denser medium for which the angle of refraction in the less dense medium is exactly \(90^\circ\). For any angle of incidence greater than the critical angle the light undergoes total internal reflection.
(\(\alpha\)) Value of the critical angle
From the diagram the ray travels along (grazes) the top surface of the block, so its angle of incidence in air is \(90^\circ\). It is refracted into the block, and the refracted ray makes \(44^\circ\) with the surface. Its angle measured from the normal is therefore
\[ r = 90^\circ - 44^\circ = 46^\circ. \]At grazing incidence the angle of refraction is equal to the critical angle of the block, hence
\[ c = 46^\circ. \](\(\beta\)) Refractive index of the block
\[ n = \frac{1}{\sin c} = \frac{1}{\sin 46^\circ} = \frac{1}{0.7193} = 1.39. \](c) Pipes
A pipe closed at one end sounds only the odd harmonics. Its fundamental is \(f_c = 200\,\text{Hz}\), so its first overtone is the third harmonic:
\[ 3 \times 200 = 600\,\text{Hz}. \](i) Fundamental frequency of the open pipe
An open pipe sounds all harmonics, so its first overtone is the second harmonic, \(2f_o\). Given that the two first overtones are equal:
\[ 2f_o = 600 \Rightarrow f_o = 300\,\text{Hz}. \](ii) Length of the closed pipe
For a closed pipe \(f_c = \dfrac{v}{4L_c}\), so
\[ L_c = \frac{v}{4f_c} = \frac{330}{4 \times 200} = 0.41\,\text{m}. \](iii) Length of the open pipe
For an open pipe \(f_o = \dfrac{v}{2L_o}\), so
\[ L_o = \frac{v}{2f_o} = \frac{330}{2 \times 300} = 0.55\,\text{m}. \]Detalles de la respuesta
(a) Diffraction
Diffraction is the spreading of a wave into the region behind an obstacle, or the bending of a wave round the edges of an obstacle or through an aperture, as the wave passes them. It is most noticeable when the width of the aperture or obstacle is comparable with the wavelength of the wave.
(b)(i) Critical angle
The critical angle is the angle of incidence in the optically denser medium for which the angle of refraction in the less dense medium is exactly \(90^\circ\). For any angle of incidence greater than the critical angle the light undergoes total internal reflection.
(\(\alpha\)) Value of the critical angle
From the diagram the ray travels along (grazes) the top surface of the block, so its angle of incidence in air is \(90^\circ\). It is refracted into the block, and the refracted ray makes \(44^\circ\) with the surface. Its angle measured from the normal is therefore
\[ r = 90^\circ - 44^\circ = 46^\circ. \]At grazing incidence the angle of refraction is equal to the critical angle of the block, hence
\[ c = 46^\circ. \](\(\beta\)) Refractive index of the block
\[ n = \frac{1}{\sin c} = \frac{1}{\sin 46^\circ} = \frac{1}{0.7193} = 1.39. \](c) Pipes
A pipe closed at one end sounds only the odd harmonics. Its fundamental is \(f_c = 200\,\text{Hz}\), so its first overtone is the third harmonic:
\[ 3 \times 200 = 600\,\text{Hz}. \](i) Fundamental frequency of the open pipe
An open pipe sounds all harmonics, so its first overtone is the second harmonic, \(2f_o\). Given that the two first overtones are equal:
\[ 2f_o = 600 \Rightarrow f_o = 300\,\text{Hz}. \](ii) Length of the closed pipe
For a closed pipe \(f_c = \dfrac{v}{4L_c}\), so
\[ L_c = \frac{v}{4f_c} = \frac{330}{4 \times 200} = 0.41\,\text{m}. \](iii) Length of the open pipe
For an open pipe \(f_o = \dfrac{v}{2L_o}\), so
\[ L_o = \frac{v}{2f_o} = \frac{330}{2 \times 300} = 0.55\,\text{m}. \]
