WAEC SSCE - Physics - 1997
Pergunta 2 Relatório
In a photocell, light energy is converted to
Detalhes da Resposta
In a photocell, light energy is converted to electrical energy. A photocell is a device that uses the photoelectric effect to produce an electric current. When light falls on a photocell, the energy of the photons is transferred to the electrons in the material of the photocell. This energy is sufficient to cause some of the electrons to break free from their atoms, creating a flow of electrons or an electric current. Thus, the light energy is converted to electrical energy by the photocell.
Pergunta 3 Relatório
At what distance from a simple microscope must an object be placed so that the image 5 times the size of the object is produced 20cm from the lens?
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Pergunta 4 Relatório
An alternating current with a peak value of 5A passes through a resistor of resistance 10Ω. Calculate the rate at which energy is dissipated in the resistor
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Pergunta 5 Relatório
The magnification produced with a converging lens is 5. If the object is a square wire gauze of side 2cm, calculate the area of the image
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Pergunta 6 Relatório
A capacitor of capacitance 25µF is connected to an a.c power source of frequency 200/πHz. Calculate the reactance of the capacitor
Detalhes da Resposta
The reactance of a capacitor can be calculated using the formula X = 1/(2πfC), where X is the reactance, f is the frequency, and C is the capacitance. Given that the capacitance of the capacitor is 25µF and the frequency of the a.c power source is 200/πHz, we can substitute these values into the formula: X = 1/(2πfC) X = 1/(2π*(200/π)*25 x 10^-6) X = 1/(2*(200)*25 x 10^-6) X = 1/(10 x 10^-3) X = 100Ω Therefore, the reactance of the capacitor is 100Ω. Answer: 100Ω.
Pergunta 7 Relatório
When a paint brush is removed from clean water, the bristles of the brush are pulled together because of
Detalhes da Resposta
The correct answer is surface tension forces. Surface tension is a phenomenon that occurs because of the cohesive forces between the molecules of a liquid. When the paintbrush is immersed in water, the water molecules adhere to the brush bristles. When the brush is removed from water, the cohesive forces between the water molecules cause them to pull together, which in turn pulls the bristles of the brush together. This is due to the surface tension of the water, which acts like a thin film on the surface and creates the surface tension forces. The viscosity of the surrounding air, the low density of water, the weight or mass of the brush bristles do not play a significant role in this phenomenon.
Pergunta 8 Relatório
Which of the following explains why a thick glass cup cracks when boiling water is poured into it?
Detalhes da Resposta
The correct option that explains why a thick glass cup cracks when boiling water is poured into it is "Unequal expansion of the exterior and interior walls of the cup." When boiling water is poured into a glass cup, the temperature of the cup increases rapidly. The outer layer of the cup expands due to the sudden increase in temperature while the inner layer remains cooler. Since the outer layer of the cup is hotter than the inner layer, it expands more than the inner layer. This uneven expansion causes stress on the glass, which may exceed its strength and cause it to crack or break. Therefore, the unequal expansion of the exterior and interior walls of the cup is the reason why it cracks when boiling water is poured into it.
Pergunta 9 Relatório
A progressive wave has a wavelength of 50cm. calculate the phase difference between two points at a distance of 20cm apart.
Detalhes da Resposta
The phase difference between two points on a wave is given by the equation: Δφ = 2π(Δx / λ) where Δx is the distance between the two points and λ is the wavelength of the wave. In this question, the wavelength is given as 50 cm and the distance between the two points is 20 cm. Therefore, substituting these values into the equation gives: Δφ = 2π(20 / 50) = π/5 Therefore, the phase difference between the two points is π/5. So the correct answer is: 4/5 π
Pergunta 10 Relatório
A fixed mass of gas occupies a volume of 20\(cm^3\) at a fixed pressure of 700mmHg. Assuming that the temperature remains constant, what will be the volume of the gas at 750mmHg?
Detalhes da Resposta
The relationship between pressure and volume of a fixed mass of gas at a constant temperature is described by Boyle's law which states that the product of pressure and volume of a fixed mass of gas is constant as long as the temperature remains constant. That is: P1V1 = P2V2 Where P1 and V1 are the initial pressure and volume of the gas, respectively, and P2 and V2 are the final pressure and volume of the gas, respectively. Using the values given in the question, we have: P1 = 700mmHg, V1 = 20cm³, P2 = 750mmHg, and V2 = ? Substituting these values into the equation, we get: 700mmHg x 20cm³ = 750mmHg x V2 Solving for V2 gives: V2 = (700mmHg x 20cm³)/750mmHg V2 = 18.7cm³ Therefore, the volume of the gas at 750mmHg is 18.7cm³. So, the answer is 18.7\(cm^3\).
Pergunta 11 Relatório
A solid material of volume 100\(cm^{-3}\) is heated through a temperature difference of 40°C. Calculate the increase in the volume of the material if its linear expansitivity is 2 x \(10^{-6}K{-1}\)
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Pergunta 12 Relatório
The half-life of a radioactive material is 6hrs. What quantity of 1kg of the material would decay in 24 hours?
Detalhes da Resposta
The half-life of a radioactive material is the time taken for half of the initial quantity of the material to decay. Therefore, if the half-life of a material is 6hrs, after 6 hours half of the initial quantity of the material would have decayed, and after 12 hours, a quarter of the initial quantity would have decayed. After 18 hours, 1/8 of the initial quantity would have decayed, and after 24 hours, 4 half-lives would have passed, meaning that 1/16 of the initial quantity would have decayed. Therefore, the quantity of 1kg of the material that would decay in 24 hours is 1/16kg. So the correct option is: 1/16kg.
Pergunta 13 Relatório
The relationship between the directions of magnetic field, the current-carrying wire in the field is easily remembered using
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Pergunta 14 Relatório
Lenz’s law of electromagnetic induction is essentially a statement of the
Detalhes da Resposta
Lenz's law of electromagnetic induction is a statement of the law of conservation of energy. It states that when there is a change in magnetic flux through a closed loop, an induced emf (electromotive force) is generated in the loop which creates a current that opposes the change in magnetic flux. This means that energy is conserved because the induced current opposes the change in magnetic field, thereby slowing it down and preventing a sudden change in energy.
Pergunta 15 Relatório
The odor of a leaking gas is perceived at a distance from the source. This is made possible by the process of
Detalhes da Resposta
The process that allows the odor of a leaking gas to be perceived at a distance from the source is diffusion. Diffusion is the movement of particles from an area of higher concentration to an area of lower concentration, resulting in the mixing of substances. In this case, the gas particles are moving from the source of the leak to the surrounding air, where they become more spread out and less concentrated. As a result, the odor can be detected even at some distance from the source due to the diffusion of the gas particles in the air.
Pergunta 16 Relatório
Which of the following is not true of the similarities between a camera and the human eye?
Detalhes da Resposta
The option that is not true of the similarities between a camera and the human eye is "The distance between the lens and the screen is fixed in both." In a camera, the distance between the lens and the screen can be adjusted to focus on objects at different distances. In contrast, in the human eye, the distance between the lens (cornea) and the screen (retina) is fixed and cannot be adjusted. Instead, the eye changes focus by altering the shape of the lens through a process called accommodation.
Pergunta 17 Relatório
A piece of copper of mass 30g loses 60J of heat energy. If the specific heat capacity of copper is 400J\(kg^{-1}K^{-1}\), Calculate the change in the temperature of the copper
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Pergunta 18 Relatório
Calculate the quantity of heat released when 100g of steam at 100°C condenses to water(Take the specific latent heat of vaporization of water as 2.3x \(10^6Jkg^{-1}\))
Detalhes da Resposta
When steam at 100°C condenses to water, it releases energy in the form of heat. The heat released is equal to the product of the mass of the steam and the specific latent heat of vaporization of water. Given, mass of steam = 100g = 0.1 kg Specific latent heat of vaporization of water = 2.3 x 10^6 J/kg Therefore, the quantity of heat released when 100g of steam at 100°C condenses to water can be calculated as: Q = m × L = 0.1 kg × 2.3 × 10^6 J/kg = 2.3 × 10^5 J So, the correct option is: 2.3x\(10^{5}J\)
Pergunta 19 Relatório
An object of mass 0.40 kg attached to the end of a string is whirled around in a horizontal circle of radius 2.0m with a constant speed of 8\(ms^{-1}\). Calculate the angular velocity of the object
Detalhes da Resposta
The angular velocity of the object can be found using the formula: ω = v/r where ω is the angular velocity, v is the velocity of the object and r is the radius of the circle. In this case, the velocity of the object is given as 8\(ms^{-1}\) and the radius of the circle is given as 2.0m. So, ω = 8\(ms^{-1}\)/2.0m = 4 rad\(s^{-1}\) Therefore, the angular velocity of the object is 4 rad\(s^{-1}\). Hence, the correct option is (c) 4 rad\(s^{-1}\).
Pergunta 20 Relatório
Whenever light waves are restricted to a specific plane they are said to be
Detalhes da Resposta
Whenever light waves are restricted to a specific plane they are said to be "plane polarized". In other words, when the vibrations of light waves occur only in a single plane perpendicular to the direction of propagation, they are said to be plane polarized. This can be achieved by using polarizing filters that block light waves vibrating in all but one direction. Plane polarized light waves have several important applications in science and technology, including in the study of crystal structures, 3D movie technology, and LCD screens.
Pergunta 21 Relatório
Two blocks of the same dimension, one steel and the other wooden are dropped simultaneously from the same height. If they fall freely neglecting air resistance the
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Pergunta 22 Relatório
When a resistor of resistance R is connected across a cell, the terminal p.d of the cell is reduced to three quarters of its e.m.f. The internal resistance of the cell is expressed as
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Pergunta 23 Relatório
A converging lens of focal length 15cm is used to obtain a real image magnified \(1\frac{1}{2}\) times. Calculate the distance of the image from the lens?
Detalhes da Resposta
The formula for magnification is given as: magnification = - (image distance) / (object distance) where magnification is negative for real images. For a converging lens, the focal length is positive. Using the lens formula: 1/f = 1/u + 1/v where f is the focal length, u is the object distance, and v is the image distance. Given that the focal length of the lens is 15cm, the object distance is assumed to be greater than the focal length, since a real image is formed. Also, the magnification is given as 1.5. Using the formula for magnification, we can rewrite the image distance as: image distance = - (1.5) x (object distance) Substituting this into the lens formula: 1/15 = 1/u - 2/3u Solving for u, we get: u = 22.5cm Finally, using the lens formula to find the image distance: 1/15 = 1/22.5 + 1/v Solving for v, we get: v = 37.5cm Therefore, the distance of the image from the lens is 37.5cm. The correct option is: 37.5cm.
Pergunta 24 Relatório
What is the frequency of a radio wave of wavelength 150m if the velocity of radio waves in free space is 3 x \(10^8ms^{-1}\)?
Detalhes da Resposta
The formula relating wavelength, frequency, and velocity is: velocity = wavelength x frequency Rearranging the equation, we have: frequency = velocity / wavelength Substituting the values given in the question, we get: frequency = (3 x 10^8) / 150 frequency = 2 x 10^6 Hz Therefore, the frequency of the radio wave is 2 x 10^6 Hz.
Pergunta 25 Relatório
Steel bars each of length 3m at 28°C are to be used for constructing a rail line. If the linear expansitivity of the steel is 1 x \(10^{-5}K^{-1}\),what is the safety gap that must be left between the successive bars if the highest temperature expected is 40°C?
Detalhes da Resposta
The steel bars will undergo thermal expansion as the temperature increases from 28°C to 40°C. To prevent the bars from buckling or bending, a safety gap must be left between successive bars. The change in length of a steel bar can be calculated using the formula: ΔL = αLΔT Where: α = linear expansitivity of steel = 1 x 10^-5 K^-1 L = original length of steel bar = 3m ΔT = change in temperature = (40°C - 28°C) = 12°C Substituting the given values: ΔL = (1 x 10^-5 K^-1)(3m)(12°C) = 3.6 x 10^-4 m The safety gap to be left between the successive bars would be equal to the change in length of a single bar, which is 3.6 x 10^-4 m or 0.36 mm (since 1 m = 100 cm). Converting this to cm, we get: 0.36 mm = 0.036 cm Therefore, the safety gap that must be left between successive bars is 0.036 cm. Hence, the correct option is: - 3.6x10^-1 cm
Pergunta 26 Relatório
An object falls freely from a height of 25m onto the roof a building 5m high. Calculate the velocity with which the object strikes the roof (g=10\(ms^{-2}\)
Detalhes da Resposta
To solve this problem, we can use the formula for the final velocity of an object undergoing free fall: v^2 = u^2 + 2gh Where v is the final velocity, u is the initial velocity (which is zero for an object dropped from rest), g is the acceleration due to gravity (given as 10 m/s^2), and h is the height from which the object is dropped. In this case, the object is dropped from a height of 25m, so h = 25m. It falls until it reaches the roof of the building, which is 5m high. So the distance it falls is 25 - 5 = 20m. Substituting these values into the formula, we get: v^2 = 0 + 2 * 10 * 20 v^2 = 400 v = sqrt(400) v = 20 m/s Therefore, the velocity with which the object strikes the roof is 20 m/s. So, the correct option is (b) 20\(ms^{-2}\).
Pergunta 27 Relatório
A solid plastic cube of side 0.2m is submerged in a liquid density of 0.8kg\(m^{-3}\). Calculate the upthrust of the liquid on the cube (g=10\(ms^{-2}\)
Detalhes da Resposta
The upthrust on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. The volume of the cube is given by: V = (0.2m) x (0.2m) x (0.2m) = 0.008m³ Since the cube is submerged in a liquid density of 0.8kg\(m^{-3}\), the mass of the fluid displaced by the cube is: m = ρV = 0.8kg\(m^{-3}\) x 0.008m³ = 0.0064kg The weight of the fluid displaced is: W = mg = (0.0064kg) x (10\(ms^{-2}\)) = 0.064N Therefore, the upthrust on the cube is 0.064N. So the correct option is: - 0.064N
Pergunta 28 Relatório
When white light is incident on a glass prism the spectrum produced on a screen placed beyond the prism is due to?
Detalhes da Resposta
When white light is incident on a glass prism, the spectrum produced on a screen placed beyond the prism is due to refraction. White light consists of seven colors of different wavelengths. When white light enters the glass prism, it gets refracted at different angles depending on the wavelength of each color. The different colors are bent at different angles due to the different refractive indices of the prism material for each wavelength. This causes the light to separate into its constituent colors, creating a spectrum. The spectrum ranges from violet with the shortest wavelength, to red with the longest wavelength, with the other colors arranged in between.
Pergunta 29 Relatório
A missile weighing 400N on the earth surface is shot into the atmosphere to an altitude of 6.4 x \(10^{-6}\)m. Taking the earth as a sphere of radius 6.4 x \(10^{-6}\)m and assuming the inverse-square law of universal gravitation, what would be the weight of the missile at that altitude?
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Pergunta 30 Relatório
A cup containing 100g of pure water at 20°C is placed in a refrigerator. If the refrigerator extracts heat at a rate of 840J per minute, calculate the time taken for the water to freeze. Neglect the heat capacity of the material of the cup. (Specific heat capacity of water =4.2J\(g^{-1}K^{-2}\))((Specific latent heat fusion of water =336J\(g^{-1}\)
Detalhes da Resposta
To freeze water, it must be cooled from its initial temperature of 20°C to 0°C and then undergo a phase change from liquid to solid at 0°C. The amount of heat extracted from the water during cooling is given by: Q1 = mcΔT Where Q1 is the amount of heat, m is the mass of the water, c is its specific heat capacity, and ΔT is the change in temperature. Substituting the given values: Q1 = (100g) x (4.2J/gK) x (20°C-0°C) = 8,400J The amount of heat required for the water to freeze is given by: Q2 = mL Where Q2 is the amount of heat, m is the mass of the water, and L is the specific latent heat of fusion. Substituting the given values: Q2 = (100g) x (336J/g) = 33,600J The total amount of heat that must be extracted from the water to freeze it is the sum of Q1 and Q2: Qtot = Q1 + Q2 = 8,400J + 33,600J = 42,000J The rate at which heat is extracted from the water is given as 840J/min. Therefore, the time required to extract the total amount of heat required for freezing is: t = Qtot / heat extraction rate = 42,000J / 840J/min = 50 minutes Therefore, the time taken for the water to freeze is 50 minutes. Answer: 4
Pergunta 31 Relatório
Which of the following statements is not true of nuclear fission?
Detalhes da Resposta
The statement "Two light nuclei are made to combine to form a heavy nucleus" is not true of nuclear fission. This statement actually describes nuclear fusion, which involves the combination of light nuclei to form a heavier nucleus, releasing a large amount of energy in the process. In nuclear fission, a heavy nucleus is split into two smaller nuclei, which also releases a large amount of energy. So, this option is the answer to the question.
Pergunta 32 Relatório
During the electrolysis of copper (II) tetraoxosulphate (VI) solution, an ammeter shows a steady reading of 1A for 30minutes while 6.6 x \(10^{-4}\)kg of copper is liberated. Calculate the error in the ammeter The electrochemical equivalent of copper is 3.30 x \(10^{-7}kgC{-1}\)
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Pergunta 33 Relatório
In a nuclear reaction, the mass defect is 2 x \(10^{-6}\)g. Calculate the energy released given that the velocity of light is 3 x \(10^{8}ms^{-1}\)
Detalhes da Resposta
When a nuclear reaction takes place, the total mass of the reactants may not be equal to the total mass of the products due to the conversion of some mass into energy. This is because mass and energy are interchangeable according to the famous equation E=mc², where E is energy, m is mass, and c is the speed of light. The mass defect is the difference between the total mass of the reactants and the total mass of the products, and it represents the mass that has been converted into energy. To calculate the energy released, we can use the equation E=mc², where m is the mass defect. We can convert the mass defect from grams to kilograms by dividing by 1000: mass defect = 2 x \(10^{-6}\) g = 2 x \(10^{-9}\) kg The speed of light is 3 x \(10^{8}ms^{-1}\). Substituting these values into the equation, we get: E = (2 x \(10^{-9}\) kg) x (3 x \(10^{8}ms^{-1}\))² = 1.8 x \(10^{8}\) J Therefore, the energy released in the nuclear reaction is 1.8 x \(10^{8}\) J. The correct option is: 1.8 x \(10^{8}\) J
Pergunta 34 Relatório
A pipe open at both ends, produces a fundamental note. If the velocity of sound in air is V and L the length of the pipe, which of the following expresses the frequency of the note? Neglect end correction
Detalhes da Resposta
When a pipe open at both ends vibrates with its fundamental frequency, it produces a wavelength equal to twice the length of the pipe. This means that the wavelength of the fundamental frequency is equal to 2L, where L is the length of the pipe. The velocity of sound in air is given by V. Therefore, we can use the formula for the speed of sound, which is equal to the product of frequency and wavelength: Speed of sound = frequency x wavelength V = f(2L) Solving for f gives us: f = V/2L Therefore, the correct option is: V/2L.
Pergunta 35 Relatório
A ball of mass 0.1kg approaching a tennis player with a velocity of 10\(ms^{-1}\),is hit back in the opposite direction with a velocity of 15\(ms^{-1}\). If the time of impact between the racket and the ball is 0.01s, calculate the magnitude of the force with which the ball is hit
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Pergunta 36 Relatório
The ratio of tensile stress to tensile strain is known as
Detalhes da Resposta
The ratio of tensile stress to tensile strain is known as the "Young’s modulus". It is a measure of the stiffness of a given material. In other words, it represents the ability of a material to resist deformation under tension or compression. Young's modulus is usually represented by the letter E and is expressed in pascals (Pa) or newtons per square meter (N/m²).
Pergunta 37 Relatório
A charge of 1 x \(10^{-5}\)C experiences a force of 40N at a certain point in space. What is the magnitude of the electric field intensity at the point in Newton and couloumb
Detalhes da Resposta
The electric field intensity at a point is defined as the electric force per unit charge experienced by a charge placed at that point. Given, a charge of 1 x \(10^{-5}\)C experiences a force of 40N at a certain point in space. So, electric field intensity (E) at that point can be calculated as follows: E = F/q where F is the force experienced by the charge q. Substituting the given values, we get: E = 40 N / (1 x \(10^{-5}\)C) E = 4 x \(10^{6}\) N/C Therefore, the magnitude of the electric field intensity at the point is 4 x \(10^{6}\) N/C. Note: Coulomb is the unit of charge, not electric field intensity.
Pergunta 38 Relatório
A magnet is being inserted into a coil of wire. On which of the following factors does the induced e.m.f in the coil depend? I. Number of turns in the coil II. Strength of the coil III. Speed with which magnet is inserted into the coil.
Detalhes da Resposta
The induced electromotive force (e.m.f) in a coil of wire depends on the rate of change of magnetic flux through the coil. The magnetic flux is directly proportional to the strength of the magnet and the speed at which it is inserted into the coil. Therefore, option III is correct. The induced e.m.f also depends on the number of turns in the coil, which determines the amount of magnetic flux linking the coil. Additionally, the strength of the coil is also a factor since a stronger coil can produce a stronger magnetic field and, hence, a larger magnetic flux. Therefore, options I and II are also correct. Thus, the answer is option V - I, II, and III only.
Pergunta 39 Relatório
The invisible part of the spectrum of a white light consists of the following colors
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Pergunta 40 Relatório
An atom radiates 1.5 x \(10^{-19}\)J of energy when an electron jumps from one energy level to another. What is the wavelength of the emitted radiation? (Planck’s constant =6.6 x \(10^{-34}\)Js. Speed of light in air=3 x \(10^{8}ms^{-1}\)
Detalhes da Resposta
The formula for calculating the wavelength of emitted radiation is given by: \begin{equation*} \lambda =\frac{hc}{\Delta E} \end{equation*} where $\lambda$ is the wavelength, $h$ is Planck's constant, $c$ is the speed of light and $\Delta E$ is the energy difference between the two energy levels. Given that the energy radiated is 1.5 x $10^{-19}$ J and the other constants are: Planck’s constant = 6.6 x $10^{-34}$ Js and speed of light in air = 3 x $10^{8}$ ms$^{-1}$, we can substitute these values into the formula and solve for the wavelength: \begin{align*} \lambda &=\frac{hc}{\Delta E}\\ &=\frac{(6.6\times 10^{-34}\text{Js})(3\times 10^8\text{ms}^{-1})}{1.5\times 10^{-19}\text{J}}\\ &=1.32\times 10^{-6}\text{m} \end{align*} Therefore, the wavelength of the emitted radiation is 1.32 x $10^{-6}$ m. Thus, the correct option is (a) 1.32 x $10^{-6}$ m.
Pergunta 41 Relatório
Which of the following statements about kinetic theory of matter is not correct?
Detalhes da Resposta
The statement "When a body is heated the average kinetic energy of its molecules decreases" is not correct in the context of kinetic theory of matter. According to the theory, when a body is heated, the average kinetic energy of its molecules increases because heat is the energy of molecular motion. Therefore, the correct option is (B). Explanation: The kinetic theory of matter explains the behavior of matter in terms of the motion of its constituent particles. The theory assumes that matter is made up of very tiny particles called molecules, which are always in motion. When a body is heated, the kinetic energy of its molecules increases, and the temperature of the body increases as a result. Similarly, when a body is cooled, the kinetic energy of its molecules decreases, and the temperature of the body decreases as a result. In addition, the theory explains that the molecules of a gas move more freely than those of a liquid because the intermolecular forces between gas molecules are weak. The molecules of a liquid move more freely than those of a solid, but less freely than those of a gas because the intermolecular forces in liquids are stronger than those in gases but weaker than those in solids. Therefore, option (C) is correct.
Pergunta 42 Relatório
Two forces 3N and 4N act on a body in directions due north and due east respectively. Calculate their equilibrant
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Pergunta 43 Relatório
The current in a series R-L-C circuit attains its maximum value when the
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Pergunta 44 Relatório
The current in the primary coil of a transformer is 2.5A. If the coil has 50 turns and the secondary250 turns, calculate the current in the secondary coil. Neglect energy losses in the transformer
Detalhes da Resposta
A transformer is an electrical device that can increase or decrease the voltage of an alternating current. It works on the principle of mutual induction. The current in the primary coil produces a magnetic field that induces a voltage in the secondary coil. According to the law of conservation of energy, the power (P) in the primary coil is equal to the power in the secondary coil. Power is equal to the product of current and voltage (P=IV). Since the transformer is ideal and there are no energy losses, the power in the primary coil is equal to the power in the secondary coil. Therefore, I1V1 = I2V2 where I1 and I2 are the currents in the primary and secondary coils, respectively, and V1 and V2 are the voltages in the primary and secondary coils, respectively. We know that the current in the primary coil is 2.5A, the number of turns in the primary coil is 50 and the number of turns in the secondary coil is 250. According to the transformer equation, the voltage in the secondary coil is equal to the voltage in the primary coil multiplied by the ratio of the number of turns in the secondary coil to the number of turns in the primary coil. Therefore, V2 = V1 x (N2/N1) = V1 x (250/50) = 5V Now we can calculate the current in the secondary coil using the formula for power: P = IV Since the power in the primary coil is equal to the power in the secondary coil, I1V1 = I2V2 2.5A x V1 = I2 x 5V I2 = 2.5A x V1/5V = 0.5A Therefore, the current in the secondary coil is 0.5A.
Pergunta 45 Relatório
Which of the following is not correct about the Rutherford’s model of an atom?
Detalhes da Resposta
The statement that is not correct about the Rutherford’s model of an atom is "The model is applicable to atoms with only one electron in the outer shell". Rutherford's model is applicable to all atoms, regardless of the number of electrons in their outer shells. The model describes an atom as having a small, dense, positively charged nucleus at its center, which contains most of the mass of the atom. The electrons are believed to be present in the space surrounding the nucleus.
Pergunta 46 Relatório
Two identical waves traveling in the same direction are superimposed. What should be the phase difference between the waves for maximum destructive interference to occur?
Detalhes da Resposta
When two identical waves traveling in the same direction superimpose, they interfere constructively when their crests coincide and destructively when their crests coincide with troughs. For maximum destructive interference, the phase difference between the two waves should be half of a wavelength, or 180 degrees. This is because when the crest of one wave meets the trough of the other wave, they cancel each other out and the resultant amplitude is zero. Therefore, the correct answer is 180°.
Pergunta 47 Relatório
(a) Explain the term work
(b) Draw a diagram of a pulley system with a velocity ratio of 5.
(c) A man pulls up a box of mass 70kg using an inclined plane of effective length 5m onto a platform 2.5m high at uniform speed. If the frictional force between the box and the plane is 100N, draw the diagram of all the forces acting on the box when in motion and calculate the
(i) minimum effort applied in pulling up the box
(ii) velocity ratio of the plane
(iii) mechanical advantage of the plane
(iv) efficiency of the plane
(v) energy lost in the system
(vi) work output of the man
(vii) total power developed by the man given that the time taken to raise the box onto the platform is 50s. (g = 10ms\(^{-2}\)).
(a) Work
Work is done when a force moves a body through a distance in the direction of the force. Thus,
\[W=Fs\]
where \(F\) is the force and \(s\) is the displacement in its direction. Work is a scalar quantity and its SI unit is the joule (J).
(b) Pulley system of velocity ratio 5
There are five rope sections supporting the movable block. Hence,
\[V.R.=5\]
(c) Inclined plane
The weight of the box is
\[W=mg=70\times10=700\ \text{N}.\]
The plane makes an angle \(\theta\) with the horizontal such that
\[\sin\theta=\frac{h}{L}=\frac{2.5}{5}=0.5,\qquad \theta=30^\circ.\]
The force diagram for the box moving up the plane is shown below.
(i) Minimum effort applied
Since the box moves at uniform speed, the effort balances the component of weight down the plane and the frictional force:
\[E=mg\sin\theta+F=700(0.5)+100=450\ \text{N}.\]
(ii) Velocity ratio of the plane
\[V.R.=\frac{\text{distance moved by effort}}{\text{distance moved by load}}=\frac{L}{h}=\frac{5}{2.5}=2.\]
(iii) Mechanical advantage of the plane
\[M.A.=\frac{\text{Load}}{\text{Effort}}=\frac{700}{450}=1.56.\]
(iv) Efficiency of the plane
\[\eta=\frac{M.A.}{V.R.}\times100\%=\frac{1.56}{2}\times100\%=77.8\%\approx78\%.\]
(v) Energy lost in the system
Energy is lost in overcoming friction:
\[E_{\text{lost}}=F\times L=100\times5=500\ \text{J}.\]
(vi) Work output
\[W_{\text{output}}=mgh=70\times10\times2.5=1750\ \text{J}.\]
(vii) Total power developed by the man
The work input is
\[W_{\text{input}}=E\times L=450\times5=2250\ \text{J}.\]
Therefore,
\[P=\frac{W_{\text{input}}}{t}=\frac{2250}{50}=45\ \text{W}.\]
Detalhes da Resposta
(a) Work
Work is done when a force moves a body through a distance in the direction of the force. Thus,
\[W=Fs\]
where \(F\) is the force and \(s\) is the displacement in its direction. Work is a scalar quantity and its SI unit is the joule (J).
(b) Pulley system of velocity ratio 5
There are five rope sections supporting the movable block. Hence,
\[V.R.=5\]
(c) Inclined plane
The weight of the box is
\[W=mg=70\times10=700\ \text{N}.\]
The plane makes an angle \(\theta\) with the horizontal such that
\[\sin\theta=\frac{h}{L}=\frac{2.5}{5}=0.5,\qquad \theta=30^\circ.\]
The force diagram for the box moving up the plane is shown below.
(i) Minimum effort applied
Since the box moves at uniform speed, the effort balances the component of weight down the plane and the frictional force:
\[E=mg\sin\theta+F=700(0.5)+100=450\ \text{N}.\]
(ii) Velocity ratio of the plane
\[V.R.=\frac{\text{distance moved by effort}}{\text{distance moved by load}}=\frac{L}{h}=\frac{5}{2.5}=2.\]
(iii) Mechanical advantage of the plane
\[M.A.=\frac{\text{Load}}{\text{Effort}}=\frac{700}{450}=1.56.\]
(iv) Efficiency of the plane
\[\eta=\frac{M.A.}{V.R.}\times100\%=\frac{1.56}{2}\times100\%=77.8\%\approx78\%.\]
(v) Energy lost in the system
Energy is lost in overcoming friction:
\[E_{\text{lost}}=F\times L=100\times5=500\ \text{J}.\]
(vi) Work output
\[W_{\text{output}}=mgh=70\times10\times2.5=1750\ \text{J}.\]
(vii) Total power developed by the man
The work input is
\[W_{\text{input}}=E\times L=450\times5=2250\ \text{J}.\]
Therefore,
\[P=\frac{W_{\text{input}}}{t}=\frac{2250}{50}=45\ \text{W}.\]
Pergunta 48 Relatório
(a) Explain the term resonance and give two examples
(b)(i) Describe, with the aid of a labelled diagram, an experiment to show how the frequency of the note emitted by a vibrating string depends on the length of the string.
(ii) State two precautions necessary to obtain an accurate result.
(c) A sonometer wire is plucked and it vibrates emitting a fundamental note. State the effect on the frequency of the note if the
(i) tension in the wire were made nine times as large with no change in the length of the wire;
(ii) length of the wire were doubled with no change in the tension.
(a) Resonance
Resonance is the phenomenon in which a body is forced to vibrate at its natural frequency by a periodic force of the same frequency, producing vibrations of maximum amplitude.
Examples are:
- tuning a radio receiver to a particular station;
- a diver's board vibrating strongly when it is periodically forced at its natural frequency.
(b)(i) Experiment to investigate the effect of length on the frequency of a vibrating string
A sonometer wire is stretched over two bridges, A and B, on a hollow wooden box. The wire passes over a smooth pulley and is kept taut by a constant load, W. Bridge B is movable, so that the vibrating length, L, between the bridges can be altered. A light paper rider is placed at the middle of the vibrating length.
A tuning fork of known frequency is struck gently with a rubber bung and its stem is placed on the sonometer box. The movable bridge is adjusted until resonance occurs. Resonance is indicated when the paper rider is thrown off the wire or when the sound becomes loud. The resonating length, L, is measured.
The procedure is repeated with tuning forks of different known frequencies while the load, and hence the tension, is kept constant. The readings may be recorded as follows:
| Frequency, f (Hz) | Resonating length, L (m) | 1/L (m−1) |
|---|---|---|
| 100 | 1.600 | 0.625 |
| 128 | 1.250 | 0.800 |
| 160 | 1.000 | 1.000 |
| 200 | 0.800 | 1.250 |
| 256 | 0.625 | 1.600 |
A graph of frequency, f, against reciprocal length, 1/L, is plotted.
The straight line through the origin shows that, for constant tension and the same wire,
\[f \propto \frac{1}{L}.\]
(b)(ii) Precautions
- Keep the tension constant throughout by using the same load and a smooth pulley.
- Strike each tuning fork gently on a rubber bung and measure the distance between the inner edges of the bridges without parallax.
(c) For the fundamental mode of a stretched string,
\[f=\frac{1}{2L}\sqrt{\frac{T}{\mu}},\]
where \(T\) is the tension and \(\mu\) is the mass per unit length of the wire.
(i) If \(T\) becomes \(9T\),
\[f'\propto\sqrt{9T}=3\sqrt{T}.\]
Therefore, the fundamental frequency is tripled.
(ii) If the length becomes \(2L\),
\[f'=\frac{1}{2L}f.\]
Therefore, the fundamental frequency is halved.
Detalhes da Resposta
(a) Resonance
Resonance is the phenomenon in which a body is forced to vibrate at its natural frequency by a periodic force of the same frequency, producing vibrations of maximum amplitude.
Examples are:
- tuning a radio receiver to a particular station;
- a diver's board vibrating strongly when it is periodically forced at its natural frequency.
(b)(i) Experiment to investigate the effect of length on the frequency of a vibrating string
A sonometer wire is stretched over two bridges, A and B, on a hollow wooden box. The wire passes over a smooth pulley and is kept taut by a constant load, W. Bridge B is movable, so that the vibrating length, L, between the bridges can be altered. A light paper rider is placed at the middle of the vibrating length.
A tuning fork of known frequency is struck gently with a rubber bung and its stem is placed on the sonometer box. The movable bridge is adjusted until resonance occurs. Resonance is indicated when the paper rider is thrown off the wire or when the sound becomes loud. The resonating length, L, is measured.
The procedure is repeated with tuning forks of different known frequencies while the load, and hence the tension, is kept constant. The readings may be recorded as follows:
| Frequency, f (Hz) | Resonating length, L (m) | 1/L (m−1) |
|---|---|---|
| 100 | 1.600 | 0.625 |
| 128 | 1.250 | 0.800 |
| 160 | 1.000 | 1.000 |
| 200 | 0.800 | 1.250 |
| 256 | 0.625 | 1.600 |
A graph of frequency, f, against reciprocal length, 1/L, is plotted.
The straight line through the origin shows that, for constant tension and the same wire,
\[f \propto \frac{1}{L}.\]
(b)(ii) Precautions
- Keep the tension constant throughout by using the same load and a smooth pulley.
- Strike each tuning fork gently on a rubber bung and measure the distance between the inner edges of the bridges without parallax.
(c) For the fundamental mode of a stretched string,
\[f=\frac{1}{2L}\sqrt{\frac{T}{\mu}},\]
where \(T\) is the tension and \(\mu\) is the mass per unit length of the wire.
(i) If \(T\) becomes \(9T\),
\[f'\propto\sqrt{9T}=3\sqrt{T}.\]
Therefore, the fundamental frequency is tripled.
(ii) If the length becomes \(2L\),
\[f'=\frac{1}{2L}f.\]
Therefore, the fundamental frequency is halved.
Pergunta 49 Relatório
(a) Briefly explain what would happen to a stable element if it is bombarded by \(\alpha\)-particles
(b) Explain how the bombardment of Uranium with neutrons could lead to nuclear fission chain reaction and hence nuclear explosion.
(c) State three characteristics of nuclear activity
(d) State three applications of atomic energy
(e) State two postulates of Bohr's model of the atom and two limitations of such a model
(a) A stable element bombarded by alpha-particles
The alpha-particle (a helium nucleus) may penetrate the nucleus of the stable atom and be captured, forming a new, heavier and usually unstable (radioactive) nucleus, often with the emission of another particle such as a proton or a neutron. This is artificial transmutation: the target element is changed into a different element (a new nuclide), which may then be radioactive. (For example, alpha bombardment of nitrogen produces oxygen and a proton.)
(b) Nuclear fission chain reaction
When a slow (thermal) neutron strikes a nucleus of uranium-235, the nucleus captures it, becomes highly unstable and splits (fissions) into two lighter nuclei, releasing a large amount of energy. Crucially, each fission also releases two or three fresh neutrons. If enough fissile material is present (a critical mass), these new neutrons go on to split further uranium nuclei, each releasing yet more neutrons, so the number of fissions multiplies rapidly. This self-sustaining, ever-growing sequence is a chain reaction. If it is uncontrolled, an enormous amount of energy is released in a very short time, producing a nuclear explosion. (In a reactor the reaction is controlled by absorbing surplus neutrons.)
(c) Three characteristics of nuclear activity
- It is spontaneous and random, and unaffected by temperature, pressure or chemical combination.
- It releases very large amounts of energy compared with chemical reactions.
- It involves changes within the nucleus, producing new elements and emitting radiation (\(\alpha\), \(\beta\), \(\gamma\)).
(d) Three applications of atomic (nuclear) energy
- Generation of electricity in nuclear power stations.
- Use of radioisotopes in medicine (diagnosis, treatment of cancer, sterilization).
- Use as tracers in agriculture/industry, in carbon dating, and in propulsion (nuclear submarines).
(e) Bohr's model
Two postulates:
- An electron moves round the nucleus only in certain allowed circular orbits (stationary states) in which it does not radiate energy; each orbit has a fixed energy.
- Energy is emitted or absorbed only when an electron jumps from one allowed orbit to another, the photon energy being \(hf = E_2 - E_1\).
Two limitations:
- It successfully explains only the hydrogen (one-electron) atom and fails for atoms with more than one electron.
- It cannot account for the fine structure (splitting) of spectral lines or their relative intensities, and it violates later quantum ideas (the uncertainty principle).
Detalhes da Resposta
(a) A stable element bombarded by alpha-particles
The alpha-particle (a helium nucleus) may penetrate the nucleus of the stable atom and be captured, forming a new, heavier and usually unstable (radioactive) nucleus, often with the emission of another particle such as a proton or a neutron. This is artificial transmutation: the target element is changed into a different element (a new nuclide), which may then be radioactive. (For example, alpha bombardment of nitrogen produces oxygen and a proton.)
(b) Nuclear fission chain reaction
When a slow (thermal) neutron strikes a nucleus of uranium-235, the nucleus captures it, becomes highly unstable and splits (fissions) into two lighter nuclei, releasing a large amount of energy. Crucially, each fission also releases two or three fresh neutrons. If enough fissile material is present (a critical mass), these new neutrons go on to split further uranium nuclei, each releasing yet more neutrons, so the number of fissions multiplies rapidly. This self-sustaining, ever-growing sequence is a chain reaction. If it is uncontrolled, an enormous amount of energy is released in a very short time, producing a nuclear explosion. (In a reactor the reaction is controlled by absorbing surplus neutrons.)
(c) Three characteristics of nuclear activity
- It is spontaneous and random, and unaffected by temperature, pressure or chemical combination.
- It releases very large amounts of energy compared with chemical reactions.
- It involves changes within the nucleus, producing new elements and emitting radiation (\(\alpha\), \(\beta\), \(\gamma\)).
(d) Three applications of atomic (nuclear) energy
- Generation of electricity in nuclear power stations.
- Use of radioisotopes in medicine (diagnosis, treatment of cancer, sterilization).
- Use as tracers in agriculture/industry, in carbon dating, and in propulsion (nuclear submarines).
(e) Bohr's model
Two postulates:
- An electron moves round the nucleus only in certain allowed circular orbits (stationary states) in which it does not radiate energy; each orbit has a fixed energy.
- Energy is emitted or absorbed only when an electron jumps from one allowed orbit to another, the photon energy being \(hf = E_2 - E_1\).
Two limitations:
- It successfully explains only the hydrogen (one-electron) atom and fails for atoms with more than one electron.
- It cannot account for the fine structure (splitting) of spectral lines or their relative intensities, and it violates later quantum ideas (the uncertainty principle).
Pergunta 50 Relatório
(a) Sketch the form of the magnetic flux pattern due to a current flowing
(i) in a long solenoid
(ii) through two long straight parallel wires when the directions of the current are opposite. (Neglect the earth's magnetic field).
(b) Draw a labelled diagram of an electric bell and explain how it works
(c) An electric bell takes a current of 0.2A groin a battery of two dry cells connected in series. Each cell has an e.m.f. of 1.5V and an internal resistance of 1.0\(\Omega\).
(i) Calculate the effective resistance of the bell
(ii) What current would the bells take in the cells were arranged in parallel?
(a) Magnetic flux patterns
(i) Long solenoid
The magnetic field is strong and nearly uniform inside the solenoid. Its pattern is like that of a bar magnet: field lines emerge from the north pole and return to the south pole outside the solenoid.
(ii) Two long parallel wires carrying currents in opposite directions
The magnetic fields are circular about each wire. In the region between the wires, the fields are in the same direction and reinforce; outside the wires, they oppose.
(b) Electric bell
When the switch is closed, current flows through the coils and magnetises the soft-iron cores of the electromagnet. The electromagnet attracts the soft-iron armature, causing the hammer to strike the gong.
As the armature moves towards the electromagnet, it moves away from the contact screw and breaks the circuit. The electromagnet then loses its magnetism. The spring pulls the armature back, restoring contact with the screw and completing the circuit again. This rapid make-and-break action continues while the switch is pressed, making the hammer strike the gong repeatedly.
(c) Calculations
For two cells in series:
\[E=2(1.5)=3.0\text{ V},\qquad r=2(1.0)=2.0\ \Omega\]
(i) If \(R\) is the effective resistance of the bell,
\[I=\frac{E}{R+r}\]
\[0.2=\frac{3.0}{R+2.0}\]
\[R+2.0=\frac{3.0}{0.2}=15\]
\[\boxed{R=13\ \Omega}\]
(ii) With the cells in parallel,
\[E=1.5\text{ V},\qquad r=\frac{1.0}{2}=0.5\ \Omega\]
\[I=\frac{E}{R+r}=\frac{1.5}{13+0.5}=\frac{1.5}{13.5}=0.111\ldots\text{ A}\]
\[\boxed{I\approx 0.11\text{ A}}\]
Detalhes da Resposta
(a) Magnetic flux patterns
(i) Long solenoid
The magnetic field is strong and nearly uniform inside the solenoid. Its pattern is like that of a bar magnet: field lines emerge from the north pole and return to the south pole outside the solenoid.
(ii) Two long parallel wires carrying currents in opposite directions
The magnetic fields are circular about each wire. In the region between the wires, the fields are in the same direction and reinforce; outside the wires, they oppose.
(b) Electric bell
When the switch is closed, current flows through the coils and magnetises the soft-iron cores of the electromagnet. The electromagnet attracts the soft-iron armature, causing the hammer to strike the gong.
As the armature moves towards the electromagnet, it moves away from the contact screw and breaks the circuit. The electromagnet then loses its magnetism. The spring pulls the armature back, restoring contact with the screw and completing the circuit again. This rapid make-and-break action continues while the switch is pressed, making the hammer strike the gong repeatedly.
(c) Calculations
For two cells in series:
\[E=2(1.5)=3.0\text{ V},\qquad r=2(1.0)=2.0\ \Omega\]
(i) If \(R\) is the effective resistance of the bell,
\[I=\frac{E}{R+r}\]
\[0.2=\frac{3.0}{R+2.0}\]
\[R+2.0=\frac{3.0}{0.2}=15\]
\[\boxed{R=13\ \Omega}\]
(ii) With the cells in parallel,
\[E=1.5\text{ V},\qquad r=\frac{1.0}{2}=0.5\ \Omega\]
\[I=\frac{E}{R+r}=\frac{1.5}{13+0.5}=\frac{1.5}{13.5}=0.111\ldots\text{ A}\]
\[\boxed{I\approx 0.11\text{ A}}\]
