WAEC SSCE - Physics - 2011
Question 1 Rapport
two sound waves have frequencies of 12Hz and 10Hz. Calculate their beats period
Détails de la réponse
When two sound waves of different frequencies overlap, they interfere with each other to produce a resultant wave. If the frequencies of the two waves are close, a phenomenon called beats occurs, where the amplitude of the resultant wave varies periodically. The period of beats is given by T = 1/(f1 - f2), where f1 and f2 are the frequencies of the two waves. Using the formula, we can calculate the beats period for the given frequencies: T = 1/(12 Hz - 10 Hz) = 1/2 Hz = 0.5 s Therefore, the answer is 0.5s.
Question 2 Rapport
Which of the following substances does not conduct electricity?
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Question 3 Rapport
Ether produces a cooling effect when it is poured on the hand because it has
Détails de la réponse
Ether produces a cooling effect when it is poured on the hand because it has a high latent heat of vaporization. Latent heat is the energy absorbed or released during a phase change. When ether is poured on the hand, it quickly evaporates due to its low boiling point. As it evaporates, it absorbs heat energy from the hand to overcome the intermolecular forces holding the liquid together, causing a cooling effect on the skin. The higher the latent heat of vaporization, the greater the cooling effect. Therefore, "high latent heat of vaporization" is the correct answer.
Question 4 Rapport
The unit of capacitance is
Détails de la réponse
The unit of capacitance is farad, which is represented by the symbol F. Capacitance is the ability of a capacitor to store electric charge per unit voltage applied. One farad of capacitance is defined as the capacitance of a capacitor that stores one coulomb of charge when a potential difference of one volt is applied across it. In simpler terms, it is a measure of the amount of electrical charge that can be stored in a capacitor for a given voltage.
Question 5 Rapport
Which of the following properties is not applicable to X-rays? They
Détails de la réponse
X-rays are electromagnetic waves that are produced when high-speed electrons are stopped by a metal target. They have many useful properties such as traveling in straight lines, traveling at the speed of light, and affecting photographic films. X-rays, however, are not affected by magnetic fields, meaning they cannot be deflected by a magnetic field. Therefore, the property that is not applicable to X-rays is that they are deflected by magnetic fields.
Question 6 Rapport
The change in mass of fuel during a nuclear reaction is 2.0 x 10-27 kg. Calculate the amount of energy released. (Speed of light = 2.0 x 108ms-1)
Détails de la réponse
The equation that relates mass and energy is E=mc2. Here, the change in mass is given as 2.0 x 10-27 kg. The speed of light is given as 2.0 x 108 ms-1. Substituting these values in the equation, we get: E = (2.0 x 10-27 kg) x (2.0 x 108 ms-1)2 E = 1.8 x 10-10 J Therefore, the amount of energy released is 1.8 x 10-10 J.
Question 7 Rapport
Which of the following statements about anomalous expansion of water are correct? i. There is contraction between 0oC and 4oC. ii. There is expansion between 0oC and 100oC. iii. The volume is minimum at 4oC. iv. The volume is maximum at 4oC
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Question 8 Rapport
Two simple pendulum A and B of equal lengths and of masses 5g and 20g respectively are located in the same environment. The periods TA and TB of their respective oscillations are related by the equations
Détails de la réponse
Question 9 Rapport
Which of the following statements correctly defines a simple machine? A device
Détails de la réponse
The correct statement that defines a simple machine is: "a device with which work can be done easily". A simple machine is any mechanical device that can change the magnitude or direction of an applied force, making it easier to move or lift an object. Examples of simple machines include levers, pulleys, inclined planes, screws, and wheel and axle systems. These machines work by reducing the amount of force required to do a certain amount of work, allowing humans to accomplish tasks that would otherwise be difficult or impossible to perform. Therefore, option C is the correct answer.
Question 10 Rapport
The main difference between echo and reverberation is that
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Question 11 Rapport
A wire of length 30cm is moved with a speed of 2ms-1 at right angles to a magnetic field of flux density 0.4T. calculate the emf induced in the wire
Détails de la réponse
The emf induced in a wire moving at right angles to a magnetic field is given by the equation: emf = B × l × v where B is the magnetic flux density, l is the length of the wire, and v is the velocity of the wire. Plugging in the given values, we have: emf = 0.4T × 0.3m × 2ms-1 = 0.24V Therefore, the emf induced in the wire is 0.24V. The correct option is: 0.24V
Question 12 Rapport
A metal of mass 200g at a temperature of 100oC is placed in 100g of water at 25oC in a container of negligible heat capacity. If the final steady temperature is 30oC, calculate the specific heat capacity of the metal. [specific heat capacity of the water is 4200 Jkg-1 K-1]
Détails de la réponse
Question 13 Rapport
A stone of mass 300g is released from rest from the top of a bulding of height 100m. Determine the kinetic energy gained by the stone when it is a quarter way down from the point of release.[g = 10 ms-2]
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Question 14 Rapport
Using the circuit diagram above. Calculate the total charge in the circuit
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Question 15 Rapport
Which of the following statements about the image formed by a plane mirror is not correct?
Détails de la réponse
The correct option is: "it is formed by actual intersection of rays." Explanation: When an object is placed in front of a plane mirror, its image is formed behind the mirror. The image formed is virtual, which means that it cannot be projected onto a screen. It is laterally inverted, which means that left appears as right and right appears as left. The size of the image is the same as that of the object, and therefore, it has unit magnification. The distance of the image from the mirror is the same as the distance of the object from the mirror, and therefore, the last statement is also true. However, the image is not formed by actual intersection of rays as the rays of light do not actually converge at the position of the image. The image is formed due to the reflection of light from the mirror's surface.
Question 16 Rapport
A body of mass 120g placed at the 10cm mark on a uniform metre rule makes the rule settle horizontally on a fulcrum places at 35cm mark. Calculate the mass of the rule
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Question 18 Rapport
A ray undergoes a minimum deviation at 40o when it is incident on equilateral triangular glass prism. Calculate the refractive index of the prism
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Question 19 Rapport
The property of a fluid that is used to determine its suitability as a lubricant is its
Détails de la réponse
The property of a fluid that is used to determine its suitability as a lubricant is its viscosity. Viscosity is a measure of a fluid's resistance to flow or deformation, and it is directly related to its thickness or consistency. The thicker the fluid, the higher its viscosity, and the more effective it is at providing lubrication and reducing friction between two surfaces. Therefore, high viscosity fluids are preferred as lubricants since they provide better lubrication, while low viscosity fluids are preferred as coolants.
Question 20 Rapport
Which of the following assumptions is not valid for an ideal gas ?
Détails de la réponse
The assumption that is not valid for an ideal gas is that the "force of attraction between molecules is negligible." This assumption is not valid for real gases since molecules have intermolecular forces of attraction. In an ideal gas, it is assumed that the gas particles have no intermolecular forces of attraction, meaning that they do not attract or repel each other, and they move independently. However, in reality, all gases have some degree of intermolecular forces of attraction.
Question 21 Rapport
The image formed by a convex mirror is always
Détails de la réponse
The image formed by a convex mirror is always virtual, erect and diminished. A convex mirror is a diverging mirror, meaning that it spreads out the light rays that enter it. The image formed by a convex mirror is virtual, which means that it cannot be projected onto a screen. It is always upright or erect, which means that it is not inverted like the image formed by a concave mirror. The image formed by a convex mirror is always smaller than the object, hence diminished.
Question 22 Rapport
For photoelectric emission, the threshold frequency is the
Détails de la réponse
The threshold frequency in photoelectric emission is the minimum frequency of incident radiation required to eject electrons. This means that if the frequency of the incident radiation is below the threshold frequency, no electrons will be emitted, regardless of the intensity or duration of the radiation. However, if the frequency of the incident radiation is equal to or greater than the threshold frequency, electrons will be emitted, and their kinetic energy will increase with increasing frequency. Therefore, the correct option is "minimum frequency of incident radiation required to eject electrons."
Question 23 Rapport
The correct relationship between "G" and "g" in a gravitational field is given by the equation (Where the symbols have their usual meanings)
Détails de la réponse
The correct relationship between "G" and "g" in a gravitational field is given by the equation g = \(\frac{GM}{R^2}\), where g represents the acceleration due to gravity, G is the gravitational constant, M is the mass of the attracting body, and R is the distance between the centers of mass of the two objects. This equation is known as Newton's law of gravitation and states that the force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The acceleration due to gravity at any point is the gravitational force experienced by a unit mass at that point. The gravitational constant G is a fundamental constant of nature and is considered to be one of the most accurately measured physical constants.
Question 24 Rapport
Which of the following statements about the electromagnetic waves is correct?
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Question 26 Rapport
When the pointed end of an uncharged optical pin is brought near the cap cf a positively charged electroscope, it is observed that the gold
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Question 27 Rapport
Which of the following statements is not correct?
Détails de la réponse
The statement that is not correct is: "molecules of ice, water and water vapour have equal intermolecular distances." Explanation: - Ice, water, and water vapour are different states of water. They have different intermolecular distances because the intermolecular forces between their molecules are different. - In ice, the intermolecular forces are the strongest due to hydrogen bonding, which results in a rigid and orderly arrangement of water molecules with a fixed intermolecular distance. - In liquid water, the intermolecular forces are weaker than in ice, and the intermolecular distance is more variable. - In water vapour, the intermolecular forces are weakest among the three, and the intermolecular distance is the largest. Therefore, the statement that molecules of ice, water and water vapour have equal intermolecular distances is not correct.
Question 29 Rapport
The radius of a wheel is 30.0 cm and that of its axle is 6.0cm. Calculate the effort required to lift a load of 120.0N using this machine. Assuming 100% efficiency
Détails de la réponse
This is a problem involving the mechanical advantage of a simple machine. The machine in question is a wheel and axle. The effort required to lift a load is dependent on the mechanical advantage of the machine. The mechanical advantage of a wheel and axle is the ratio of the radius of the wheel to the radius of the axle. In this case, the mechanical advantage is: MA = radius of wheel / radius of axle = 30 cm / 6 cm = 5 This means that for every unit of effort applied to the machine, it can lift 5 units of load. Using this information, we can determine the effort required to lift a load of 120.0 N by setting up the equation: MA = load / effort Solving for effort, we get: effort = load / MA = 120.0 N / 5 = 24.0 N Therefore, the effort required to lift a load of 120.0 N using this machine with 100% efficiency is 24.0 N. The correct answer is (c) 24.0N.
Question 30 Rapport
Total internal reflection occurs in an optical prism when light rays
Détails de la réponse
Total internal reflection occurs in an optical prism when light rays are incident in the prism at an angle greater than the critical angle. When a light ray travels from an optically denser medium (like a glass prism) to a less dense medium (like air), it bends away from the normal line. As the angle of incidence increases, the angle of refraction also increases until it reaches a certain angle called the critical angle. If the angle of incidence is greater than the critical angle, the light ray is totally reflected back into the denser medium, which is known as total internal reflection. This phenomenon is used in various optical devices such as prisms, optical fibers, and reflecting telescopes.
Question 31 Rapport
Uranium-234 disintegrated to form Thorium-230 by emitting
Détails de la réponse
Uranium-234 and Thorium-230 are both radioactive isotopes, meaning they undergo spontaneous nuclear decay. In this case, Uranium-234 disintegrates to form Thorium-230 by emitting an alpha particle. An alpha particle is a helium nucleus, consisting of two protons and two neutrons, which is ejected from the nucleus during alpha decay. This results in the atomic number of the daughter nucleus being reduced by 2 and the mass number being reduced by 4. So, is correct.
Question 32 Rapport
An object of mass 5.0 kg moves with a velocity of 10ms-1. Calculate its momentum.
Détails de la réponse
The momentum of an object is defined as the product of its mass and velocity. Mathematically, this can be expressed as p = mv, where p is momentum, m is mass and v is velocity. Using the given values, we can calculate the momentum of the object as follows: p = mv = 5.0 kg x 10ms^-1 = 50.0 kgms^-1 Therefore, the momentum of the object is 50.0 kgms^-1. Answer: 50.0 kgms^-1
Question 33 Rapport
Which of the following practices will lead to losses in electrical power transmission?
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Question 34 Rapport
The resistivity, \(\mathit{P}\), of a material of length l, cross-sectional area A, resistance R, is given by the equation
Détails de la réponse
The resistivity, \(\mathit{P}\), of a material is a measure of its ability to resist the flow of electric current. The resistance, R, of a material is directly proportional to its length, l, and inversely proportional to its cross-sectional area, A. Therefore, we can say that \(\mathit{P} = \frac{RA}{l}\), as this equation combines the resistance, length, and cross-sectional area to give the resistivity of the material.
Question 35 Rapport
A coin is pushed from the edge of a laboratory bench with a horizontal velocity of 15.0 ms-1. If the height of the bench from the floor is 1.5m, calculate the distance from the foot of the bench to the point of impact with the floor [g = 10ms-2]
Détails de la réponse
We can solve this problem using kinematic equations of motion. Since the motion is projectile motion, we can break it into two components, horizontal and vertical. In the horizontal direction, there is no acceleration, so the velocity remains constant throughout the motion. In the vertical direction, the coin is moving upward against the force of gravity, which causes it to slow down, then come to rest at the highest point before falling downward. The time taken to reach the highest point can be calculated using the vertical component of the initial velocity and acceleration due to gravity. Using the equation: v = u + at where: v = final velocity = 0 (at the highest point) u = initial velocity = 0 a = acceleration due to gravity = 10 ms-2 t = time taken to reach the highest point Solving for t, we get: t = u/a = 15.0/10 = 1.5s The maximum height reached by the coin can be calculated using the equation: s = ut + 0.5at^2 where: s = maximum height reached u = initial vertical velocity = 0 a = acceleration due to gravity = 10 ms-2 t = time taken to reach the highest point = 1.5s Solving for s, we get: s = 0 + 0.5 x 10 x (1.5)^2 = 11.25m Now, the coin falls from the maximum height and hits the floor. The time taken to fall from the maximum height can be calculated using the equation: s = ut + 0.5at^2 where: s = height of the bench = 1.5m u = initial vertical velocity = 0 a = acceleration due to gravity = 10 ms-2 t = time taken to fall from maximum height Solving for t, we get: 1.5 = 0 + 0.5 x 10 x t^2 t^2 = 0.3 t = sqrt(0.3) = 0.55s (ignoring the negative root) Now, we can calculate the horizontal distance traveled by the coin during the time it was in the air using the equation: s = ut + 0.5at^2 where: s = horizontal distance traveled u = initial horizontal velocity = 15.0ms-1 a = acceleration in horizontal direction = 0 (no acceleration) t = time taken to fall from maximum height = 0.55s Solving for s, we get: s = 15.0 x 0.55 = 8.25m Therefore, the distance from the foot of the bench to the point of impact with the floor is approximately 8.25m. Answer: 8.22m (rounding off to two decimal places)
Question 36 Rapport
In the absence of gravitational force, the weight of the body is
Détails de la réponse
In the absence of gravitational force, the weight of the body is zero. This is because weight is the force with which a body is attracted towards the center of the Earth or any other massive body. In the absence of such a force, the body will not experience any weight. However, the mass of the body remains the same, as it is a measure of the amount of matter contained in the body. Therefore, - zero - is the correct answer.
Question 37 Rapport
The capacitance of a parallel-plate capacitor
Détails de la réponse
The capacitance of a parallel-plate capacitor is the ratio of the magnitude of the charge on either plate to the potential difference between them. The capacitance is directly proportional to the area of the plates and inversely proportional to the distance between them. Thus, as the separation between the plates decreases, the capacitance increases. The capacitance is also directly proportional to the permittivity of the dielectric material between the plates. A dielectric is an insulating material that can be placed between the plates of a capacitor to increase its capacitance. The permittivity of the dielectric material is a measure of its ability to store electrical energy in an electric field. Thus, adding a dielectric between the plates of a parallel-plate capacitor increases the capacitance. Therefore, the correct answer is: the capacitance of a parallel-plate capacitor is greater with a dielectric material between the plates than without a dielectric.
Question 38 Rapport
Which of the following equations shows how how the linear expansivity (\(\alpha\)) and the cubic expansivity (\(\gamma\)) of a material are related?
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Question 39 Rapport
The density P of a spherical ball of diameter d and mass m is given by
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Question 40 Rapport
Which of the following propeties is not considered in choosing mercury as a thermomemetric liquid?
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Question 41 Rapport
The area under a velocity-time graph represents
Détails de la réponse
The area under a velocity-time graph represents the distance covered by an object. Velocity is the rate of change of displacement with time. Therefore, the area under the velocity-time graph is equal to the displacement covered by the object during the given time interval. This is because the area under the graph is the product of the time interval and the average velocity during that time interval, which is equal to the distance traveled. So, the correct option is "distance covered".
Question 42 Rapport
Using the circuit diagram above. Calculate the potential difference across the capacitors in parallel
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Question 43 Rapport
A metal rod of density 7000 kgm-3 and cross-sectional area of 0.001257m2 has a mass of 1.76kg. Calculate its length.
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Question 44 Rapport
The focal length of the eye-piece of an astronomical telescope is f1, while the focal length of the object is f2. Fro normal adjustment, the angular magnification is given by
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Question 45 Rapport
The speed of light in a certain medium is V while its speed in a vacuum is C. The absolute refractive index of the medium is
Détails de la réponse
The absolute refractive index of a medium is the ratio of the speed of light in a vacuum to the speed of light in that medium. Therefore, the formula for the absolute refractive index of the medium is: \[\text{Absolute refractive index} = \frac{\text{Speed of light in vacuum}}{\text{Speed of light in the medium}} = \frac{C}{V}\] Therefore, the answer is \(\frac{C}{V}\).
Question 46 Rapport
A body is pulled over a distance of 500m by a force of 20N. If the power developed is o.4 kW, calculate the time interval during which the force is applied
Détails de la réponse
Power, P = work done (W) / time interval (t) We know that, work done (W) = force (F) x distance (d) Therefore, power (P) = (F x d) / t Rearranging this equation, we get: t = (F x d) / P Substituting the given values, we get: t = (20 N x 500 m) / 0.4 kW t = 25,000 s Therefore, the time interval during which the force is applied is 25,000 seconds. Answer: 25.0s
Question 48 Rapport
An object is placed 20.0cm in front of a converging lens of focal length 15.0cm. Calculate its image distance
Détails de la réponse
The formula to calculate the image distance for a converging lens is: 1/f = 1/do + 1/di where f is the focal length of the lens, do is the distance of the object from the lens, and di is the distance of the image from the lens. Substituting the given values: 1/15 = 1/20 + 1/di Simplifying and solving for di: di = 60 cm Therefore, the image distance is 60.0cm. Answer: Option D.
Question 49 Rapport
An ammeter with a full scale deflection of 10mA has an internal resistance 0.5\(\Omega\). calculate the resistance of the shunt required to adapt it to read up to 3A
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Question 50 Rapport
(a)(i) State two advantages of alcohol over mercury as a thermometric liquid.
(ii) When the bulb of a thermometer is placed in a beaker of hot water, the level of the mercury first falls and then rises gradually. Explain this observation.
(b) List two factors, other than temperature, that affect the rate of evaporation of a liquid.
(b) A block of lead of mass 100 kg in a crucible and at a temperature of 40 °C was placed in an electric furnace rated 10 kW. If the melting point of lead is 320 °C, calculate the:
(i) quantity of heat required to heat the lead to its melting point;
(ii) additional heat energy required to melt the lead;
(iii) time taken to supply this additional energy. (Specific heat capacity of lead = 120 Jkg\(^{-1}\) K\(^{-1}\) (Specific Idtent heat of fusion lead = 2.5 x 10\(^{4} JK^{-1}\))
(d) State two precautions necessary in an experiment to determine the specific latent heat of vaporization of water.
(a)(i) Two advantages of alcohol over mercury as a thermometric liquid: (1) Alcohol has a much lower freezing point (about -115 °C), so it can measure very low temperatures where mercury would freeze; (2) Alcohol expands more (larger expansivity) for a given temperature rise, making the thermometer more sensitive. (It is also cheaper and can be coloured for easy reading.)
(a)(ii) Mercury level first falls then rises: When the bulb is placed in hot water, the glass bulb is heated first and expands before the heat reaches the mercury. The expansion of the bulb increases the space available, so the mercury level first falls. Shortly afterwards the heat reaches the mercury, which expands much more than the glass, and the level then rises steadily.
(b) Two factors (other than temperature) affecting the rate of evaporation: (1) Surface area exposed; (2) Draught/movement of air over the surface. (Humidity of the surrounding air and nature of the liquid are also acceptable.)
(c) Heating the lead:
(i) Heat to raise the lead to its melting point:
\[ Q = mc\,\Delta\theta = 100\times120\times(320-40) = 100\times120\times280 = 3.36\times10^{6}\,\text{J}. \]
(ii) Additional heat to melt the lead:
\[ Q = mL = 100\times2.5\times10^{4} = 2.5\times10^{6}\,\text{J}. \]
(iii) Time to supply this additional energy (furnace power \(P = 10\,\text{kW} = 10000\,\text{W}\)):
\[ t = \frac{Q}{P} = \frac{2.5\times10^{6}}{10000} = 250\,\text{s}. \]
(d) Two precautions in determining specific latent heat of vaporization of water: (1) Lag (insulate) the apparatus to minimise heat loss to the surroundings; (2) Ensure steady conditions and collect the condensed water only after the water has been boiling steadily (take readings at constant rate). (Correcting for heat losses by repeating at two heating rates is also acceptable.)
Détails de la réponse
(a)(i) Two advantages of alcohol over mercury as a thermometric liquid: (1) Alcohol has a much lower freezing point (about -115 °C), so it can measure very low temperatures where mercury would freeze; (2) Alcohol expands more (larger expansivity) for a given temperature rise, making the thermometer more sensitive. (It is also cheaper and can be coloured for easy reading.)
(a)(ii) Mercury level first falls then rises: When the bulb is placed in hot water, the glass bulb is heated first and expands before the heat reaches the mercury. The expansion of the bulb increases the space available, so the mercury level first falls. Shortly afterwards the heat reaches the mercury, which expands much more than the glass, and the level then rises steadily.
(b) Two factors (other than temperature) affecting the rate of evaporation: (1) Surface area exposed; (2) Draught/movement of air over the surface. (Humidity of the surrounding air and nature of the liquid are also acceptable.)
(c) Heating the lead:
(i) Heat to raise the lead to its melting point:
\[ Q = mc\,\Delta\theta = 100\times120\times(320-40) = 100\times120\times280 = 3.36\times10^{6}\,\text{J}. \]
(ii) Additional heat to melt the lead:
\[ Q = mL = 100\times2.5\times10^{4} = 2.5\times10^{6}\,\text{J}. \]
(iii) Time to supply this additional energy (furnace power \(P = 10\,\text{kW} = 10000\,\text{W}\)):
\[ t = \frac{Q}{P} = \frac{2.5\times10^{6}}{10000} = 250\,\text{s}. \]
(d) Two precautions in determining specific latent heat of vaporization of water: (1) Lag (insulate) the apparatus to minimise heat loss to the surroundings; (2) Ensure steady conditions and collect the condensed water only after the water has been boiling steadily (take readings at constant rate). (Correcting for heat losses by repeating at two heating rates is also acceptable.)
Question 51 Rapport
State three properties of cathode rays which suggest the particle nature of matter
Three properties of cathode rays which suggest the particle (corpuscular) nature of matter are:
- They travel in straight lines and cast sharp shadows of objects placed in their path, as would a stream of moving particles.
- They are deflected by electric and magnetic fields, showing that they consist of charged particles (electrons carrying negative charge) that possess mass and charge.
- They possess momentum and kinetic energy - they can exert a mechanical force, for example turning a small paddle wheel placed in their path, which shows the moving stream carries the properties of particles.
(They also produce heat on striking a target and can be deflected/behave as though made of discrete massive particles.)
Détails de la réponse
Three properties of cathode rays which suggest the particle (corpuscular) nature of matter are:
- They travel in straight lines and cast sharp shadows of objects placed in their path, as would a stream of moving particles.
- They are deflected by electric and magnetic fields, showing that they consist of charged particles (electrons carrying negative charge) that possess mass and charge.
- They possess momentum and kinetic energy - they can exert a mechanical force, for example turning a small paddle wheel placed in their path, which shows the moving stream carries the properties of particles.
(They also produce heat on striking a target and can be deflected/behave as though made of discrete massive particles.)
Question 52 Rapport
In fig. la and fig.1 b above, Ida and Wb represent the respective loads on a spring placedtiear a 30 cm rule, when in air and when in water:
(a) Identify the force causing a shrink in the spring in fig.(b).
(b) Given that the force constant of the spring is 2.0 x 10\(^{11}\) Nm\(^{-1}\), calculate the work done by the force in causing the shrink.
(a) The force causing the spring to shrink in fig.(b)
When the load is lowered into the water it experiences an upthrust (buoyant force) from the water, acting vertically upwards. This upthrust partly supports the weight of the load, so the tension pulling on the spring is reduced and the spring contracts (shrinks). The force responsible for the shrink is therefore the upthrust of the water on the load.
(b) Work done by the force in causing the shrink
Reading the contraction from the metre rule. The pointer attached to the load marks the position of the lower end of the spring against the fixed rule:
- In air (fig. a) the pointer reads \(20\ \text{cm}\) (the spring is fully stretched by the whole weight \(W_a\)).
- In water (fig. b) the upthrust lifts the load, so the pointer rises to \(18\ \text{cm}\).
The contraction (shrink) of the spring is the difference of the two readings:
\[x = 20\ \text{cm}-18\ \text{cm}=2\ \text{cm}=2\times10^{-2}\ \text{m}.\]
Work done. For a spring obeying Hooke's law, the work done in changing its length by \(x\) against a force constant \(k\) is
\[W=\tfrac{1}{2}kx^{2}.\]
With \(k=2.0\times10^{11}\ \text{N m}^{-1}\) and \(x=2\times10^{-2}\ \text{m}\):
\[W=\tfrac{1}{2}\times(2.0\times10^{11})\times(2\times10^{-2})^{2}\]
\[W=\tfrac{1}{2}\times(2.0\times10^{11})\times(4\times10^{-4})\]
\[W=\mathbf{4.0\times10^{7}\ \text{J}}.\]
The key step is to read \(x\) as the difference of the two rule readings (\(20\ \text{cm}\) in air and \(18\ \text{cm}\) in water), giving \(x = 2\ \text{cm}\), and only then substitute into \(W=\tfrac12 kx^{2}\).
Détails de la réponse
(a) The force causing the spring to shrink in fig.(b)
When the load is lowered into the water it experiences an upthrust (buoyant force) from the water, acting vertically upwards. This upthrust partly supports the weight of the load, so the tension pulling on the spring is reduced and the spring contracts (shrinks). The force responsible for the shrink is therefore the upthrust of the water on the load.
(b) Work done by the force in causing the shrink
Reading the contraction from the metre rule. The pointer attached to the load marks the position of the lower end of the spring against the fixed rule:
- In air (fig. a) the pointer reads \(20\ \text{cm}\) (the spring is fully stretched by the whole weight \(W_a\)).
- In water (fig. b) the upthrust lifts the load, so the pointer rises to \(18\ \text{cm}\).
The contraction (shrink) of the spring is the difference of the two readings:
\[x = 20\ \text{cm}-18\ \text{cm}=2\ \text{cm}=2\times10^{-2}\ \text{m}.\]
Work done. For a spring obeying Hooke's law, the work done in changing its length by \(x\) against a force constant \(k\) is
\[W=\tfrac{1}{2}kx^{2}.\]
With \(k=2.0\times10^{11}\ \text{N m}^{-1}\) and \(x=2\times10^{-2}\ \text{m}\):
\[W=\tfrac{1}{2}\times(2.0\times10^{11})\times(2\times10^{-2})^{2}\]
\[W=\tfrac{1}{2}\times(2.0\times10^{11})\times(4\times10^{-4})\]
\[W=\mathbf{4.0\times10^{7}\ \text{J}}.\]
The key step is to read \(x\) as the difference of the two rule readings (\(20\ \text{cm}\) in air and \(18\ \text{cm}\) in water), giving \(x = 2\ \text{cm}\), and only then substitute into \(W=\tfrac12 kx^{2}\).
Question 53 Rapport
State: (a) the difference between plane polarized light and ordinary light;
(b) two uses of polaroids.
(a) Difference between plane polarized light and ordinary light: In ordinary (unpolarized) light the vibrations of the electric field take place in all directions (all planes) perpendicular to the direction of propagation. In plane polarized light the vibrations are restricted to a single plane containing the direction of propagation.
(b) Two uses of polaroids:
- In polaroid sunglasses, to reduce glare from light reflected off surfaces such as water and roads.
- In the control of light intensity and in photography (polarizing filters), and in liquid-crystal displays (LCDs) and stress analysis of transparent materials.
Détails de la réponse
(a) Difference between plane polarized light and ordinary light: In ordinary (unpolarized) light the vibrations of the electric field take place in all directions (all planes) perpendicular to the direction of propagation. In plane polarized light the vibrations are restricted to a single plane containing the direction of propagation.
(b) Two uses of polaroids:
- In polaroid sunglasses, to reduce glare from light reflected off surfaces such as water and roads.
- In the control of light intensity and in photography (polarizing filters), and in liquid-crystal displays (LCDs) and stress analysis of transparent materials.
Question 54 Rapport
(a)(i) What is an eclipse?
(ii) List the three types of eclipse.
(b) A student in a lecture theatre can read from the board clearly but requires a pair of spectacles to read from a book.
(i) What eye defect has this student?
(ii) What type of lens is needed to correct the eye defect?
(iii) The focal length of the lens used to correct this defect is 10cm. Calculate the power of the lens.
(c) A car B moves towards a stationary car A. If B produces an ultrasonic sound at a point and it takes 5.6 x 10\(^{-3}\)s for a beep to be heard in B, calculate the distance between the two cars at that instant. (Speed. sound in air = 340 ms\(^-\))
(d) The image of an object is located 9 cm behind a convex mirror. If magnification produced is 0.6, calculate the focal length of the mirror.
(a)(i) Eclipse: An eclipse is the total or partial disappearance of a heavenly body (the sun or the moon) from view, caused when the shadow of one body falls on another as the sun, earth and moon come to lie nearly in a straight line.
(a)(ii) Three types of eclipse:
- Eclipse of the sun (solar eclipse)
- Eclipse of the moon (lunar eclipse)
- Annular eclipse
(b)(i) Eye defect: The student sees distant objects (the board) clearly but cannot focus on near objects (the book), so the defect is long sight (hypermetropia).
(b)(ii) Correcting lens: A convex (converging) lens.
(b)(iii) Power of the lens: \(f = 10\,\text{cm} = 0.10\,\text{m}\).
\[ P = \frac{1}{f} = \frac{1}{0.10} = +10\,\text{D (dioptres)}. \]
(c) Distance between the cars: The ultrasonic beep travels from B to A and is reflected back to B, so it covers a total path of \(2d\) in \(t = 5.6\times10^{-3}\,\text{s}\).
\[ 2d = vt = 340\times5.6\times10^{-3} = 1.904\,\text{m}, \]
\[ d = \frac{1.904}{2} = 0.952\,\text{m}. \]
The two cars are 0.952 m apart at that instant.
(d) Focal length of the convex mirror: The image is virtual, formed \(9\,\text{cm}\) behind the mirror, so \(v = -9\,\text{cm}\). The magnification \(m = \dfrac{-v}{u} = 0.6\) gives \(u = \dfrac{9}{0.6} = 15\,\text{cm}\).
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-9} + \frac{1}{15} = -0.1111 + 0.0667 = -0.0444, \]
\[ f = -22.5\,\text{cm}. \]
The focal length of the convex mirror is 22.5 cm, the negative sign confirming that it is a diverging (convex) mirror.
Détails de la réponse
(a)(i) Eclipse: An eclipse is the total or partial disappearance of a heavenly body (the sun or the moon) from view, caused when the shadow of one body falls on another as the sun, earth and moon come to lie nearly in a straight line.
(a)(ii) Three types of eclipse:
- Eclipse of the sun (solar eclipse)
- Eclipse of the moon (lunar eclipse)
- Annular eclipse
(b)(i) Eye defect: The student sees distant objects (the board) clearly but cannot focus on near objects (the book), so the defect is long sight (hypermetropia).
(b)(ii) Correcting lens: A convex (converging) lens.
(b)(iii) Power of the lens: \(f = 10\,\text{cm} = 0.10\,\text{m}\).
\[ P = \frac{1}{f} = \frac{1}{0.10} = +10\,\text{D (dioptres)}. \]
(c) Distance between the cars: The ultrasonic beep travels from B to A and is reflected back to B, so it covers a total path of \(2d\) in \(t = 5.6\times10^{-3}\,\text{s}\).
\[ 2d = vt = 340\times5.6\times10^{-3} = 1.904\,\text{m}, \]
\[ d = \frac{1.904}{2} = 0.952\,\text{m}. \]
The two cars are 0.952 m apart at that instant.
(d) Focal length of the convex mirror: The image is virtual, formed \(9\,\text{cm}\) behind the mirror, so \(v = -9\,\text{cm}\). The magnification \(m = \dfrac{-v}{u} = 0.6\) gives \(u = \dfrac{9}{0.6} = 15\,\text{cm}\).
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-9} + \frac{1}{15} = -0.1111 + 0.0667 = -0.0444, \]
\[ f = -22.5\,\text{cm}. \]
The focal length of the convex mirror is 22.5 cm, the negative sign confirming that it is a diverging (convex) mirror.
Question 55 Rapport
(a) Explain the terms:
(i) transmutation as it relates to radioactivity; (ii) stopping potential.
(b) \(^{23}_{11}A + ^2_1B\) —> \(^p_qC\) + proton
\(^p_qC\) —> \(^r_sE\) + beta
A nucleus C, formed artificially from A and B radioactive and quickly decays to another nucli E as indicated in the nuclear equations abc Datermine the values of p, q, r and s.
(c) A certain metal of work function 1.6 eV is irradiated with ultra-violet light of wavelength 3.6 x 10\(^{-7}\) Calculate the maximum
(i) kinetic energy of ejected electron in joules;
(ii) speed of an emitted electron. (1eV = 1.6 x 10\(^{-18}\) J; C = 3.0 x 10\(^{8}\) ms\(^{-1}\); m, = 9.1 x 10\(^{-31}\) kg; h = 6.6 x 10\(^{-34}\) Js)
(d) If source of the ultra-violet light in (c) above is mo away from the surface of the metal, state the of on the maximum speed of the ejected electron
(a)(i) Transmutation: Transmutation is the change of one element into another as a result of a nuclear reaction, in which the number of protons in the nucleus (the atomic number) changes, for example during radioactive decay or artificial nuclear bombardment.
(a)(ii) Stopping potential: The stopping potential is the minimum negative (retarding) potential that must be applied to the collecting electrode of a photocell to just stop the most energetic photoelectrons from reaching it, so that the photocurrent becomes zero. \(eV_s = KE_{max}\).
(b) Nuclear equations:
\[ {}^{23}_{11}A + {}^{2}_{1}B \rightarrow {}^{p}_{q}C + {}^{1}_{1}\text{(proton)}. \]
Conserving nucleon number: \(23 + 2 = p + 1 \Rightarrow p = 24\). Conserving charge: \(11 + 1 = q + 1 \Rightarrow q = 11\).
\[ {}^{24}_{11}C \rightarrow {}^{r}_{s}E + {}^{0}_{-1}\beta. \]
Conserving nucleon number: \(24 = r + 0 \Rightarrow r = 24\). Conserving charge: \(11 = s + (-1) \Rightarrow s = 12\).
Therefore \(p = 24,\; q = 11,\; r = 24,\; s = 12\).
(c) Energy of the incident photon:
\[ E = \frac{hc}{\lambda} = \frac{(6.6\times10^{-34})(3.0\times10^{8})}{3.6\times10^{-7}} = 5.5\times10^{-19}\,\text{J}. \]
Work function \(W_0 = 1.6\,\text{eV} = 1.6\times(1.6\times10^{-19}) = 2.56\times10^{-19}\,\text{J}\).
(i) Maximum kinetic energy:
\[ KE_{max} = E - W_0 = 5.5\times10^{-19} - 2.56\times10^{-19} = 2.94\times10^{-19}\,\text{J}. \]
(ii) Speed of the emitted electron: from \(KE = \tfrac{1}{2}mv^{2}\),
\[ v = \sqrt{\frac{2\,KE}{m}} = \sqrt{\frac{2\times2.94\times10^{-19}}{9.1\times10^{-31}}} = \sqrt{6.46\times10^{11}} \approx 8.04\times10^{5}\,\text{m s}^{-1}. \]
(d) Moving the source of ultra-violet light farther from the metal only reduces the intensity (number of photons per second arriving), so fewer electrons are emitted per second. It does not change the frequency/energy of each photon, so the maximum kinetic energy and hence the maximum speed of the ejected electron remain unchanged.
Détails de la réponse
(a)(i) Transmutation: Transmutation is the change of one element into another as a result of a nuclear reaction, in which the number of protons in the nucleus (the atomic number) changes, for example during radioactive decay or artificial nuclear bombardment.
(a)(ii) Stopping potential: The stopping potential is the minimum negative (retarding) potential that must be applied to the collecting electrode of a photocell to just stop the most energetic photoelectrons from reaching it, so that the photocurrent becomes zero. \(eV_s = KE_{max}\).
(b) Nuclear equations:
\[ {}^{23}_{11}A + {}^{2}_{1}B \rightarrow {}^{p}_{q}C + {}^{1}_{1}\text{(proton)}. \]
Conserving nucleon number: \(23 + 2 = p + 1 \Rightarrow p = 24\). Conserving charge: \(11 + 1 = q + 1 \Rightarrow q = 11\).
\[ {}^{24}_{11}C \rightarrow {}^{r}_{s}E + {}^{0}_{-1}\beta. \]
Conserving nucleon number: \(24 = r + 0 \Rightarrow r = 24\). Conserving charge: \(11 = s + (-1) \Rightarrow s = 12\).
Therefore \(p = 24,\; q = 11,\; r = 24,\; s = 12\).
(c) Energy of the incident photon:
\[ E = \frac{hc}{\lambda} = \frac{(6.6\times10^{-34})(3.0\times10^{8})}{3.6\times10^{-7}} = 5.5\times10^{-19}\,\text{J}. \]
Work function \(W_0 = 1.6\,\text{eV} = 1.6\times(1.6\times10^{-19}) = 2.56\times10^{-19}\,\text{J}\).
(i) Maximum kinetic energy:
\[ KE_{max} = E - W_0 = 5.5\times10^{-19} - 2.56\times10^{-19} = 2.94\times10^{-19}\,\text{J}. \]
(ii) Speed of the emitted electron: from \(KE = \tfrac{1}{2}mv^{2}\),
\[ v = \sqrt{\frac{2\,KE}{m}} = \sqrt{\frac{2\times2.94\times10^{-19}}{9.1\times10^{-31}}} = \sqrt{6.46\times10^{11}} \approx 8.04\times10^{5}\,\text{m s}^{-1}. \]
(d) Moving the source of ultra-violet light farther from the metal only reduces the intensity (number of photons per second arriving), so fewer electrons are emitted per second. It does not change the frequency/energy of each photon, so the maximum kinetic energy and hence the maximum speed of the ejected electron remain unchanged.
Question 56 Rapport
A stone projected horizontally from the top of a tower with a speed of 4 ms\(^{-1}\) lands on the level ground at a horizontal distance of 25 m from the foot of the tower. Calculate the height of the tower. = 10 ms\(^{-2}\)]
For a projectile launched horizontally, the horizontal and vertical motions are independent.
Step 1 - find the time of flight from the horizontal motion. Horizontally the speed is constant at \(u = 4\,\text{m s}^{-1}\):
\[ R = u\,t \;\Rightarrow\; 25 = 4t \;\Rightarrow\; t = 6.25\,\text{s}. \]
Step 2 - find the height from the vertical motion. The stone starts with zero vertical velocity and falls freely:
\[ h = \tfrac{1}{2} g t^{2} = \tfrac{1}{2}\times10\times(6.25)^{2}. \]
\[ h = 5\times39.06 \approx 195.3\,\text{m}. \]
The height of the tower is approximately 195.3 m.
Détails de la réponse
For a projectile launched horizontally, the horizontal and vertical motions are independent.
Step 1 - find the time of flight from the horizontal motion. Horizontally the speed is constant at \(u = 4\,\text{m s}^{-1}\):
\[ R = u\,t \;\Rightarrow\; 25 = 4t \;\Rightarrow\; t = 6.25\,\text{s}. \]
Step 2 - find the height from the vertical motion. The stone starts with zero vertical velocity and falls freely:
\[ h = \tfrac{1}{2} g t^{2} = \tfrac{1}{2}\times10\times(6.25)^{2}. \]
\[ h = 5\times39.06 \approx 195.3\,\text{m}. \]
The height of the tower is approximately 195.3 m.
Question 57 Rapport
The uncertainty in determining the duration during which an electron remains in a particular energy level before returning to the ground state is 2.0 x 10\(^{-9}\)s. Calculate the uncertainty in determining its energy at that level. (\(\frac{h}{2 \pi}\) = h = 1.054 x 10\(^{-34}\)Js)
By Heisenberg's uncertainty principle for energy and time,
\[ \Delta E \,\Delta t \geq \frac{h}{2\pi} = \hbar. \]
Taking the limiting (minimum) case, the uncertainty in energy is
\[ \Delta E = \frac{\hbar}{\Delta t} = \frac{1.054\times10^{-34}}{2.0\times10^{-9}}. \]
\[ \Delta E = 5.27\times10^{-26}\,\text{J}. \]
The uncertainty in determining the energy of that level is about \(5.27\times10^{-26}\,\text{J}\).
Détails de la réponse
By Heisenberg's uncertainty principle for energy and time,
\[ \Delta E \,\Delta t \geq \frac{h}{2\pi} = \hbar. \]
Taking the limiting (minimum) case, the uncertainty in energy is
\[ \Delta E = \frac{\hbar}{\Delta t} = \frac{1.054\times10^{-34}}{2.0\times10^{-9}}. \]
\[ \Delta E = 5.27\times10^{-26}\,\text{J}. \]
The uncertainty in determining the energy of that level is about \(5.27\times10^{-26}\,\text{J}\).
Question 58 Rapport
(a) What is a vector quantity?
(b) Three vectors \(3\ \text{ms}^{-1}\) N \(45^\circ\) W, \(12\ \text{ms}^{-1}\) W and \(5\ \text{ms}^{-1}\) S act at a point.
(i) Sketch a vector diagram to illustrate the given information.
(ii) Calculate the resultant of the vectors.
(c) In a laboratory experiment to determine the force constant of a spiral spring, the mass or, the spring was varied and the corresponding extensions were measured and recorded as shown in the table below.
| Mass M/g | Weight W/N | Extension e/cm |
| 50 | 6.5 | |
| 100 | 11.0 | |
| 150 | 15.0 | |
| 200 | 20.0 | |
| 250 | 25.0 |
(i) Copy and complete the table. (Take \(g = 10\ \text{ms}^{-2}\))
(ii) Plot a graph with weight, W, on the vertical axis and extension, e, on the horizontal axis.
(iii) Using the graph, determine the force constant of the spring.
(iv) Determine the natural length of the spring if its length was 38.0 cm when loaded with 250 g mass.
(a) A vector quantity is a physical quantity which has both magnitude and direction. Examples include velocity, force and displacement.
(b)(i) Vector diagram
(b)(ii) Resultant of the vectors
Take east as the positive x-direction and north as the positive y-direction.
| Vector | x-component / m s-1 | y-component / m s-1 |
|---|---|---|
| 3 m s-1, N45°W | \(-3\sin45^\circ=-2.12\) | \(+3\cos45^\circ=+2.12\) |
| 12 m s-1, W | \(-12.00\) | \(0\) |
| 5 m s-1, S | \(0\) | \(-5.00\) |
| Resultant components | \(-14.12\) | \(-2.88\) |
\[R=\sqrt{(-14.12)^2+(-2.88)^2}=14.4\ \text{m s}^{-1}\]
\[\theta=\tan^{-1}\left(\frac{2.88}{14.12}\right)=11.5^\circ\]
Hence, the resultant is \(14.4\ \text{m s}^{-1}\), \(11.5^\circ\) south of west, or equivalently S78.5°W.
(c)(i) Completed table
Since \(W=Mg\), with \(g=10\ \text{m s}^{-2}\):
| Mass, \(M\) / g | Weight, \(W\) / N | Extension, \(e\) / cm |
|---|---|---|
| 50 | 0.5 | 6.5 |
| 100 | 1.0 | 11.0 |
| 150 | 1.5 | 15.0 |
| 200 | 2.0 | 20.0 |
| 250 | 2.5 | 25.0 |
(c)(ii) Graph of weight against extension
(c)(iii) The force constant, \(k\), is the gradient of the graph:
Using two widely separated points on the line of best fit, \((10.0\ \text{cm},0.903\ \text{N})\) and \((25.0\ \text{cm},2.530\ \text{N})\),
\[k=\frac{\Delta W}{\Delta e}=\frac{2.530-0.903}{25.0-10.0}=0.1085\ \text{N cm}^{-1}\]
\[k=0.1085\times100=10.85\ \text{N m}^{-1}\approx \boxed{10.8\ \text{N m}^{-1}}\]
(c)(iv) At a mass of 250 g, the extension is 25.0 cm. Therefore,
\[\text{natural length}=38.0-25.0=\boxed{13.0\ \text{cm}}\]
Détails de la réponse
(a) A vector quantity is a physical quantity which has both magnitude and direction. Examples include velocity, force and displacement.
(b)(i) Vector diagram
(b)(ii) Resultant of the vectors
Take east as the positive x-direction and north as the positive y-direction.
| Vector | x-component / m s-1 | y-component / m s-1 |
|---|---|---|
| 3 m s-1, N45°W | \(-3\sin45^\circ=-2.12\) | \(+3\cos45^\circ=+2.12\) |
| 12 m s-1, W | \(-12.00\) | \(0\) |
| 5 m s-1, S | \(0\) | \(-5.00\) |
| Resultant components | \(-14.12\) | \(-2.88\) |
\[R=\sqrt{(-14.12)^2+(-2.88)^2}=14.4\ \text{m s}^{-1}\]
\[\theta=\tan^{-1}\left(\frac{2.88}{14.12}\right)=11.5^\circ\]
Hence, the resultant is \(14.4\ \text{m s}^{-1}\), \(11.5^\circ\) south of west, or equivalently S78.5°W.
(c)(i) Completed table
Since \(W=Mg\), with \(g=10\ \text{m s}^{-2}\):
| Mass, \(M\) / g | Weight, \(W\) / N | Extension, \(e\) / cm |
|---|---|---|
| 50 | 0.5 | 6.5 |
| 100 | 1.0 | 11.0 |
| 150 | 1.5 | 15.0 |
| 200 | 2.0 | 20.0 |
| 250 | 2.5 | 25.0 |
(c)(ii) Graph of weight against extension
(c)(iii) The force constant, \(k\), is the gradient of the graph:
Using two widely separated points on the line of best fit, \((10.0\ \text{cm},0.903\ \text{N})\) and \((25.0\ \text{cm},2.530\ \text{N})\),
\[k=\frac{\Delta W}{\Delta e}=\frac{2.530-0.903}{25.0-10.0}=0.1085\ \text{N cm}^{-1}\]
\[k=0.1085\times100=10.85\ \text{N m}^{-1}\approx \boxed{10.8\ \text{N m}^{-1}}\]
(c)(iv) At a mass of 250 g, the extension is 25.0 cm. Therefore,
\[\text{natural length}=38.0-25.0=\boxed{13.0\ \text{cm}}\]
Question 59 Rapport
(a) Explain mutual induction
(b) State four use of electromagnets
The diagram above illustrates two coils X an arranged so that their axes are collinear. X is connected to an a.c. supply and has an amme in series with it while Y is connected to a lamp Explain the following observations. The (i) lamp is lit when the a.c. supply is switched on;
(ii) brightness of the light from the lamp increases when distance between X. and Y is decreased;
(iii) filament of the lamp glows brighter whet bundle of insulatttriron wires is placed along ccmmon axis of the coils.
(d) Two cells, one have an emf of 2.0 V and an internal resistance of 0.4 and the other having an emf of 2.0 V and an internal resistance of 0.1 \(\Omega\), are connected in parallel. The combination is then connected in series with a resistor.
(i) Draw a circuit diagram of the arrangement.
(ii) Calculate the current through 5\(\Omega\) resistor.
(a) Mutual induction is the production of an e.m.f. in one coil or circuit when the current in a nearby coil or circuit is changing. The changing current produces a changing magnetic flux which links the second coil.
(b) Uses of electromagnets
- Electric bells.
- Electromagnetic relays and switches.
- Electromagnetic cranes for lifting iron and steel.
- Magnetic separation of iron scraps from non-magnetic materials.
(c)
- The alternating current in coil X produces a continuously changing magnetic flux. Some of this flux links coil Y and induces an alternating e.m.f. in it. A current therefore flows through the lamp and it lights.
- When the coils are brought closer, more of the magnetic flux from X links coil Y. The induced e.m.f. and current in Y increase, so the lamp becomes brighter.
- The bundle of insulated soft-iron wires has high magnetic permeability and provides a path that concentrates the magnetic flux through both coils. The flux linkage and induced e.m.f. in Y increase, producing a larger current and a brighter lamp. The wires are insulated to reduce eddy currents.
(d)(i) Circuit diagram
(d)(ii) Current through the \(5\,\Omega\) resistor
Since the two cells have equal e.m.f.s and are connected in parallel, the resultant e.m.f. is
\[E = 2.0\text{ V}\]
The equivalent internal resistance is
\[ \frac{1}{r}=\frac{1}{0.4}+\frac{1}{0.1}=2.5+10=12.5 \]
\[ r=\frac{1}{12.5}=0.08\,\Omega \]
Total resistance of the circuit:
\[ R_{\text{total}}=5.0+0.08=5.08\,\Omega \]
Hence,
\[ I=\frac{E}{R_{\text{total}}}=\frac{2.0}{5.08}=0.394\text{ A} \]
Current through the \(5\,\Omega\) resistor \(=0.39\text{ A}\) (approximately).
Détails de la réponse
(a) Mutual induction is the production of an e.m.f. in one coil or circuit when the current in a nearby coil or circuit is changing. The changing current produces a changing magnetic flux which links the second coil.
(b) Uses of electromagnets
- Electric bells.
- Electromagnetic relays and switches.
- Electromagnetic cranes for lifting iron and steel.
- Magnetic separation of iron scraps from non-magnetic materials.
(c)
- The alternating current in coil X produces a continuously changing magnetic flux. Some of this flux links coil Y and induces an alternating e.m.f. in it. A current therefore flows through the lamp and it lights.
- When the coils are brought closer, more of the magnetic flux from X links coil Y. The induced e.m.f. and current in Y increase, so the lamp becomes brighter.
- The bundle of insulated soft-iron wires has high magnetic permeability and provides a path that concentrates the magnetic flux through both coils. The flux linkage and induced e.m.f. in Y increase, producing a larger current and a brighter lamp. The wires are insulated to reduce eddy currents.
(d)(i) Circuit diagram
(d)(ii) Current through the \(5\,\Omega\) resistor
Since the two cells have equal e.m.f.s and are connected in parallel, the resultant e.m.f. is
\[E = 2.0\text{ V}\]
The equivalent internal resistance is
\[ \frac{1}{r}=\frac{1}{0.4}+\frac{1}{0.1}=2.5+10=12.5 \]
\[ r=\frac{1}{12.5}=0.08\,\Omega \]
Total resistance of the circuit:
\[ R_{\text{total}}=5.0+0.08=5.08\,\Omega \]
Hence,
\[ I=\frac{E}{R_{\text{total}}}=\frac{2.0}{5.08}=0.394\text{ A} \]
Current through the \(5\,\Omega\) resistor \(=0.39\text{ A}\) (approximately).
Question 60 Rapport
(a). State: (a) two applications of electrolysis in an industry
(b) one application of electrolysis in a school laboratory.
(a) Two applications of electrolysis in industry:
- Electroplating - coating objects with a thin layer of metal (for example chromium, silver or gold plating) to protect them from corrosion or to improve appearance.
- Extraction and purification (refining) of metals - such as the extraction of aluminium from molten alumina and the purification of copper by electrolytic refining. (Electrolytic production of chemicals such as caustic soda and chlorine is also acceptable.)
(b) One application of electrolysis in a school laboratory: Electrolysis of acidified water to prepare and collect hydrogen and oxygen gases (for example in a Hofmann voltameter). (Electroplating of a small object or determining the chemical equivalent of a metal are also acceptable.)
Détails de la réponse
(a) Two applications of electrolysis in industry:
- Electroplating - coating objects with a thin layer of metal (for example chromium, silver or gold plating) to protect them from corrosion or to improve appearance.
- Extraction and purification (refining) of metals - such as the extraction of aluminium from molten alumina and the purification of copper by electrolytic refining. (Electrolytic production of chemicals such as caustic soda and chlorine is also acceptable.)
(b) One application of electrolysis in a school laboratory: Electrolysis of acidified water to prepare and collect hydrogen and oxygen gases (for example in a Hofmann voltameter). (Electroplating of a small object or determining the chemical equivalent of a metal are also acceptable.)
Question 61 Rapport
Explain why water in a narrow glass tube has a concave meniscus while mercury, in the same tube, has a convex meniscus.
The shape of a meniscus depends on the relative strengths of two forces: cohesion (the attraction between molecules of the same liquid) and adhesion (the attraction between the liquid molecules and the molecules of the glass).
Water (concave meniscus): For water in a glass tube, the adhesive force between water and glass is greater than the cohesive force between the water molecules. The water is therefore pulled up the walls of the tube, wetting the glass, and its surface curves upward at the edges to give a concave meniscus.
Mercury (convex meniscus): For mercury, the cohesive force between the mercury molecules is greater than the adhesive force between mercury and glass. The mercury therefore does not wet the glass; it is pulled together and depressed at the walls, so its surface bulges upward in the middle to give a convex meniscus.
Détails de la réponse
The shape of a meniscus depends on the relative strengths of two forces: cohesion (the attraction between molecules of the same liquid) and adhesion (the attraction between the liquid molecules and the molecules of the glass).
Water (concave meniscus): For water in a glass tube, the adhesive force between water and glass is greater than the cohesive force between the water molecules. The water is therefore pulled up the walls of the tube, wetting the glass, and its surface curves upward at the edges to give a concave meniscus.
Mercury (convex meniscus): For mercury, the cohesive force between the mercury molecules is greater than the adhesive force between mercury and glass. The mercury therefore does not wet the glass; it is pulled together and depressed at the walls, so its surface bulges upward in the middle to give a convex meniscus.
Question 62 Rapport
State three methods of polarizing an unpolarized light
Three methods of polarizing an unpolarized beam of light are:
- Polarization by reflection - when ordinary light is reflected from a transparent surface (for example water or glass) at the polarizing (Brewster) angle, the reflected light becomes plane polarized.
- Polarization by selective absorption (using a polaroid/dichroic material) - a polaroid sheet transmits vibrations in only one plane and absorbs those in all other planes, producing plane polarized light.
- Polarization by double refraction (using a doubly-refracting crystal such as calcite/Nicol prism) - the crystal splits the light into two plane polarized rays vibrating in mutually perpendicular planes.
(Polarization by scattering, as occurs with sunlight in the atmosphere, is also acceptable.)
Détails de la réponse
Three methods of polarizing an unpolarized beam of light are:
- Polarization by reflection - when ordinary light is reflected from a transparent surface (for example water or glass) at the polarizing (Brewster) angle, the reflected light becomes plane polarized.
- Polarization by selective absorption (using a polaroid/dichroic material) - a polaroid sheet transmits vibrations in only one plane and absorbs those in all other planes, producing plane polarized light.
- Polarization by double refraction (using a doubly-refracting crystal such as calcite/Nicol prism) - the crystal splits the light into two plane polarized rays vibrating in mutually perpendicular planes.
(Polarization by scattering, as occurs with sunlight in the atmosphere, is also acceptable.)
Question 63 Rapport
(a) Explain diffusion.
(b) Give one reason why the rate of diffusion is higher in gases than in liquids at the same temperature.
(a) Diffusion: Diffusion is the net movement of the molecules of a substance from a region where they are more concentrated to a region where they are less concentrated, arising from the continuous random motion of the molecules, until they become uniformly distributed.
(b) Why diffusion is faster in gases than in liquids (at the same temperature): In a gas the molecules are much farther apart and the intermolecular forces are very weak, so the molecules move freely at high speeds over large distances between collisions. In a liquid the molecules are close together with strong intermolecular forces, so their movement is slower and more restricted. Hence, at the same temperature, molecules of a gas diffuse much faster than those of a liquid.
Détails de la réponse
(a) Diffusion: Diffusion is the net movement of the molecules of a substance from a region where they are more concentrated to a region where they are less concentrated, arising from the continuous random motion of the molecules, until they become uniformly distributed.
(b) Why diffusion is faster in gases than in liquids (at the same temperature): In a gas the molecules are much farther apart and the intermolecular forces are very weak, so the molecules move freely at high speeds over large distances between collisions. In a liquid the molecules are close together with strong intermolecular forces, so their movement is slower and more restricted. Hence, at the same temperature, molecules of a gas diffuse much faster than those of a liquid.
Question 64 Rapport
An electron of charge 1.60 x 10\(^{-19}\) C is accelerated under a potential difference of 1.0 x 10\(^5\) V. Calculate the energy of the electron in joules.
The energy gained by a charge \(Q\) accelerated through a potential difference \(V\) is the work done on it:
\[ E = QV. \]
Substituting \(Q = 1.60\times10^{-19}\,\text{C}\) and \(V = 1.0\times10^{5}\,\text{V}\):
\[ E = (1.60\times10^{-19})\times(1.0\times10^{5}) = 1.60\times10^{-14}\,\text{J}. \]
The energy of the electron is \(1.60\times10^{-14}\,\text{J}\).
Détails de la réponse
The energy gained by a charge \(Q\) accelerated through a potential difference \(V\) is the work done on it:
\[ E = QV. \]
Substituting \(Q = 1.60\times10^{-19}\,\text{C}\) and \(V = 1.0\times10^{5}\,\text{V}\):
\[ E = (1.60\times10^{-19})\times(1.0\times10^{5}) = 1.60\times10^{-14}\,\text{J}. \]
The energy of the electron is \(1.60\times10^{-14}\,\text{J}\).
