WAEC SSCE - Physics - 2007
Question 1 Report
Calculate the effective resistance between P and Q in the diagram above
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Question 2 Report
A block of mass 4.0kg causes a spiral spring to extend by 0.16m from its unstretched position. The block is removed and another body of mass 0.50kg is hung from the same spiral spring. If the spring is then stretched and released, what is the angular frequency of the subsequent motion? [g = 10 ms -2]
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Question 3 Report
Which of the following statements about solid friction are correct? It i. is a force ii. occurs between the surfaces of two bodies in contact iii. depends on the area in contact
Question 4 Report
Which of the following phenomena is not a direct consequence of rectilinear propagation of light?
Answer Details
The phenomenon that is not a direct consequence of rectilinear propagation of light is diffraction of light. Rectilinear propagation of light means that light travels in a straight line in a uniform medium. Lunar and solar eclipses, images of objects in a pinhole camera, and shadows of opaque objects can all be explained by the principle of rectilinear propagation of light. However, diffraction of light is a bending of light waves around obstacles or through narrow slits, and it cannot be explained by the principle of rectilinear propagation of light. Diffraction is a result of the wave nature of light and occurs when light waves encounter obstacles that are comparable in size to the wavelength of light.
Question 5 Report
An object of mass 2 kg moves with a uniform speed of 10ms-1 for 5s along an straight path. Determine the magnitude of its acceleration
Answer Details
Since the object is moving with a constant velocity, its acceleration is zero. This is because acceleration is the rate of change of velocity over time, and if the velocity is constant, there is no change and hence, no acceleration. Therefore, the answer is (a) 0 ms-2.
Question 6 Report
A sounding tuning fork is brought near the open end of a pipe containing air and the loudness of the sound is observed to increase. This observation is due to
Answer Details
The observed increase in the loudness of the sound when a tuning fork is brought near the open end of a pipe containing air is due to resonance. Resonance occurs when an object is forced to vibrate at its natural frequency by an external force. In this case, the tuning fork produces sound waves at a specific frequency, which matches the natural frequency of the air column in the pipe. As a result, the air column in the pipe begins to vibrate with greater amplitude, which in turn increases the loudness of the sound. This phenomenon is similar to how pushing a child on a swing at the right time will cause them to swing higher and higher. In both cases, the external force is applied at just the right frequency to match the natural frequency of the object, resulting in increased amplitude. Therefore, the correct option is "resonance."
Question 7 Report
A cube made of metal of linear expansivity \(\alpha\) is heated through a temperature \(\theta\). If the initial volume of the cube is Vo, the correct expression for the increase in volume of the cube is
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Question 8 Report
Five 80-W and three 100W are run for 8 hours. If the cost of energy is N5.00 per unit, calculate the cost of running the lamp. [1 unit = 1 kWh]
Answer Details
The total energy consumed by the lamps can be calculated by first finding the total power rating of the lamps, and then multiplying it by the time they were used. Total power = (5 x 80W) + (3 x 100W) = 400W + 300W = 700W Total energy = Power x Time = 700W x 8 hours = 5600 Wh = 5.6 kWh Since 1 unit = 1 kWh, the energy consumed is 5.6 units. The cost of running the lamps can then be calculated by multiplying the energy consumed by the cost per unit of energy. Cost = Energy consumed x Cost per unit = 5.6 units x N5.00 per unit = N28.00 Therefore, the cost of running the lamps is N28.00. Answer is the correct answer.
Question 9 Report
A mass of mass m, experiences a viscous drag, F and an upthrust, U as he descends towards the ground at a steady velocity, Using a parachute. If the acceleration of free fall is g, which of the following expressions is correct?
Answer Details
When an object is descending at a steady velocity, the forces acting on it must be balanced. Therefore, the net force on the object must be zero. In this case, the forces acting on the mass are the viscous drag force, F, the upthrust force, U, and the force due to gravity, mg. Using Newton's second law, we know that the net force on an object is equal to its mass times its acceleration. Since the object is descending at a steady velocity, its acceleration is zero, so the net force on the object must also be zero. Therefore, we can write the following equation: U + F - mg = 0 This equation shows that the upthrust force and the viscous drag force must balance the force due to gravity in order for the object to descend at a steady velocity with a parachute. So, the correct expression is: U + F - mg = 0.
Question 10 Report
The distance between two points P and Q, along a wave is 0.05m. If the wave length of the wave is 0.10m, determine the phase angle between P and Q in radians
Answer Details
The phase angle between two points on a wave is given by the fraction of the wavelength between the points, measured in radians. In this case, the distance between points P and Q is half the wavelength (0.05m is half of 0.10m). Therefore, the phase angle between P and Q is half of the complete wave cycle, which is equal to π radians. So, the answer is: \(\pi\)
Question 11 Report
Which of the following surfaces will radiate heat energy best?
Answer Details
A black surface will radiate heat energy best. This is because black surfaces are good absorbers of all wavelengths of light, including infrared radiation, which is the type of radiation associated with heat energy. As a result, they also emit more radiation at these wavelengths, making them better at radiating heat energy compared to surfaces that reflect or transmit more light, such as white, yellow, or red surfaces.
Question 12 Report
The number of protons in an element increased by one after a radioactive decay. The element must have decay by emitting
Answer Details
When the number of protons in an element increases by one, it means that a proton has been added to the nucleus of the atom. This can only happen through the emission of a beta particle, which is essentially an electron that is emitted from the nucleus. During beta decay, a neutron in the nucleus of the atom is converted into a proton and an electron. The proton remains in the nucleus, increasing the atomic number by one, while the electron is emitted from the nucleus as a beta particle. Therefore, if the number of protons in an element increases by one, it must have undergone beta decay by emitting a beta particle. Alpha decay involves the emission of an alpha particle, which is made up of two protons and two neutrons. Gamma decay, on the other hand, involves the emission of a gamma ray, which is a high-energy photon. Neutron emission occurs when an unstable nucleus has too many neutrons, but this does not change the atomic number of the element.
Question 13 Report
A bulb marked 240V, 40W is used for 30 minutes. Calculate the heat generated
Answer Details
The formula for calculating electrical energy is given by: E = P x t Where E is the energy in joules, P is the power in watts, and t is the time in seconds. Firstly, we need to convert the time of 30 minutes to seconds. 30 minutes = 30 x 60 seconds = 1800 seconds The power of the bulb is given as 40W, and the voltage is given as 240V. We can use these values to calculate the current (I) flowing through the bulb: Power (P) = Voltage (V) x Current (I) Current (I) = Power (P) / Voltage (V) Current (I) = 40 / 240 = 0.1667 A (to 4 decimal places) Now we can use the formula for energy to calculate the heat generated: E = P x t E = 40 x 1800 E = 72000 J Therefore, the answer is 72000J. So, the correct option is 4) 72000J.
Question 14 Report
A swinging pendulum between the rest position and its maximum displacement possesses
Answer Details
A swinging pendulum between the rest position and its maximum displacement possesses both kinetic and potential energy. As the pendulum moves away from its rest position, it gains potential energy due to the increased height above the rest position. As the pendulum swings back towards the rest position, it loses potential energy and gains kinetic energy due to its increased speed. At the rest position, the pendulum has its maximum kinetic energy and minimum potential energy, and at the maximum displacement, it has its maximum potential energy and minimum kinetic energy. Thus, the pendulum continuously converts potential energy to kinetic energy and back again, and possesses both types of energy at any point during its motion.
Question 15 Report
a body moving at a constant speed accelerates when it is in
Question 16 Report
Which of the following statements about sound wave is not correct? Sound waves can be
Answer Details
The statement that sound waves can be polarized is not correct. Polarization refers to the orientation of the electric field component of a transverse wave. However, sound waves are longitudinal waves, which means that they oscillate in the same direction as the wave propagation. Therefore, sound waves do not have an electric field component and cannot be polarized. On the other hand, sound waves can be reflected, refracted, and diffracted. Reflection occurs when a sound wave bounces off a surface, such as an echo in a room. Refraction occurs when a sound wave bends as it passes through a medium with a different density, such as when sound waves pass through air with different temperatures. Diffraction occurs when a sound wave bends around obstacles in its path, such as when sound waves from a speaker spread out in all directions.
Question 17 Report
Which of the following physical quantities is correctly paired with it's corresponding S.I unit?
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Question 18 Report
Which of the following electromagnetic wave can be detected by its heating effect?
Answer Details
Infrared radiation can be detected by its heating effect. Infrared radiation is a type of electromagnetic radiation that has longer wavelengths than visible light. When infrared radiation falls on an object, the energy is absorbed by the object, causing the object's temperature to increase. This effect is commonly used in devices such as infrared heaters and thermal imaging cameras, which use infrared radiation to detect the heat signature of objects. Therefore, option D (infrared radiation) is the correct answer to the question.
Question 19 Report
The correct sequence of energy transformation in the operation of an x-ray tube is i. X-rays ii. Kinetic energy of thermo electrons iii. Potential energy of the target atoms iv. Heat energy of the emitted electrons
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Question 20 Report
A dry plastic comb used in combing hair was found to attract pieces of paper and dust. The most probable explanation for this phenomenon is that the comb has been given
Answer Details
The most probable explanation for the phenomenon of a dry plastic comb attracting pieces of paper and dust is that the comb has been electrically charged by friction. When the comb is rubbed against hair, electrons are transferred from the hair to the comb, leaving the comb with a negative charge and the hair with a positive charge. The negative charge on the comb can then attract positively charged pieces of paper and dust. This process is called charging by friction or triboelectric charging. Therefore, "electric charges by friction" is the correct answer.
Question 21 Report
A car of mass 800kg moves from rest on a horizontal track and travels 60m in 20s with uniform acceleration. Assuming there were no frictional forces, calculate the accelerating force
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Question 22 Report
The diagram above illustrates three forces acting on an object at point O. If the object is in equilibrium, determine the magnitude of the force P.
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Question 23 Report
According to the kinetic theory of gasses, which of the following property of a gas can be affected by the collision of the gas molecule with the walls of the container?
Answer Details
According to the kinetic theory of gases, the pressure of a gas can be affected by the collision of gas molecules with the walls of the container. As the gas molecules move randomly, they collide with the walls of the container and exert a force on it. This force per unit area is defined as the pressure of the gas. The more frequent and intense the collisions, the higher the pressure. Therefore, the pressure of a gas can be increased or decreased by changing the number of collisions with the walls of the container. The other options listed (temperature, energy, viscosity) are also properties of gases, but they are not directly affected by collisions with the walls of the container.
Question 24 Report
A body of mass 200g and specific heat capacity 0.4Jg-1k-1 cools from 37oC to 31oC. Calculate the quantity of the heat released by the body.
Answer Details
The quantity of heat released by the body can be calculated using the formula: Q = mcΔT where Q is the quantity of heat, m is the mass of the body, c is the specific heat capacity, and ΔT is the change in temperature. Substituting the given values: m = 200g = 0.2kg c = 0.4Jg^-1k^-1 ΔT = (31 - 37)°C = -6°C Q = (0.2kg) x (0.4Jg^-1k^-1) x (-6°C) = -0.48J The negative sign indicates that the body released heat to the surroundings, which is the expected outcome since the body cooled down. Thus, the correct answer is 0.48J. However, none of the given options match this value exactly. The closest option is 480J, but this is off by a factor of 1000.
Question 25 Report
An object is placed 10cm in front of a plane mirror. If it is moved 8cm away father away from the mirror, determine the distance of the final image from the mirror
Answer Details
When an object is placed in front of a plane mirror, its image is formed behind the mirror at the same distance as the object is from the mirror. In this case, the object is placed at a distance of 10 cm from the mirror. Therefore, the image will also be formed at a distance of 10 cm behind the mirror. When the object is moved 8 cm farther away from the mirror, its new distance from the mirror will be 18 cm (10 cm + 8 cm). Since the image distance from the mirror is always equal to the object distance from the mirror, the distance of the final image from the mirror will also be 18 cm. Therefore, the correct answer is: 18cm
Question 26 Report
The ice and steam point on a mercury-in-glass thermometer are 10cm and 30cm respectively. Calculate the temperature in decree Celsius when the mercury meniscus is the 14cm mark
Answer Details
We can use the principle of the thermometer, which states that the volume of the liquid (in this case, mercury) in the bulb and stem of the thermometer changes with temperature, and this change in volume is linearly related to the temperature. Therefore, we can set up a proportion: change in volume of mercury / total length of thermometer stem = change in temperature / temperature interval between ice and steam points Let x be the length of the mercury column when the temperature is at the desired point. Then the change in volume of mercury is 14 - 10 = 4 cm. The total length of the stem is 30 - 10 = 20 cm. The temperature interval between ice and steam points is 100oC. Substituting these values into the proportion, we get: 4 cm / 20 cm = change in temperature / 100oC Solving for the change in temperature, we get: change in temperature = (4 cm / 20 cm) * 100oC = 20oC To get the actual temperature at the 14 cm mark, we add the change in temperature to the ice point temperature (which is 0oC): temperature at 14 cm mark = 0oC + 20oC = 20oC Therefore, the answer is 20oC.
Question 27 Report
An external force of magnitude 100N acts on a particle of mass 0.15kg for 0.03s. Calculate the change in the speed of the particle
Answer Details
The change in the speed of the particle can be calculated using the formula: Δv = FΔt/m Where Δv is the change in velocity (speed), F is the external force applied, Δt is the time for which the force is applied, and m is the mass of the particle. Substituting the given values, we get: Δv = (100N) x (0.03s) / 0.15kg Δv = 20ms-1 Therefore, the change in the speed of the particle is 20ms-1. This means that the speed of the particle increased by 20 meters per second during the time when the external force was applied.
Question 28 Report
Which of the following factors can affect the speed of sound in air? i. The temperature of the surrounding ii. The direction of the wind iii. The pitch of the sound.
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Question 29 Report
A rectangular glass prism of thickness d and absolute refractive index n is placed on a point object which is viewed vertically downward from above the prism. Which of the following expressions correctly defines the apparent upward displacement of the object?
Question 30 Report
There is always an uncertainty involved in any attempt to measure the position and momentum of an electron simultaneously. This statement is known as the
Answer Details
The statement "There is always an uncertainty involved in any attempt to measure the position and momentum of an electron simultaneously" is known as the Heisenberg uncertainty principle. It states that it is impossible to simultaneously determine the precise position and momentum of a subatomic particle such as an electron. The more precisely we know the position of an electron, the less precisely we can know its momentum, and vice versa. This principle is a fundamental concept in quantum mechanics and has important implications for our understanding of the behavior of particles at the subatomic level.
Question 31 Report
Which of the following statements distinguishes thermal conduction for convection? i. Conduction requires a material medium while convection does not ii. In convection, there is actual motion of hot material, while in conduction, molecules molecules vibrates faster about their mean positions iii. Conduction takes place in solids while convection takes place in fluids.
Answer Details
Statement ii and iii distinguish thermal conduction from convection. Thermal conduction refers to the transfer of heat energy through a medium without any actual motion of the medium itself. In conduction, heat is transferred by the transfer of kinetic energy from one molecule to another through collisions. As a result, in conduction, molecules vibrate faster about their mean positions. On the other hand, convection is the transfer of heat energy by the actual movement of a fluid. In convection, hotter and less dense fluids rise while cooler and denser fluids sink, creating a circular motion that transfers heat energy from one place to another. Statement i is incorrect because both conduction and convection require a medium through which heat can be transferred. Statement iii is also incorrect because both thermal conduction and convection can occur in both solids and fluids. Therefore, the correct answer is ii and iii only.
Question 32 Report
In which of the following piece of apparatus is the radius of curvature equal to twice the focal length?
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Question 33 Report
Which of the following observations is not an effect of surface tension?
Answer Details
The observation that is not an effect of surface tension is "water flowing out more easily than engine oil from a container." Surface tension is the cohesive force that exists between molecules at the surface of a liquid, which causes the surface to behave like an elastic film. This property of liquids has several effects, such as the formation of droplets, the spherical shape of mercury droplets, and the ability of small insects to walk on the surface of water. However, the flow rate of a liquid out of a container is not directly related to surface tension. It is primarily determined by the viscosity of the liquid, which is a measure of its resistance to flow. Engine oil is more viscous than water, so it flows more slowly out of a container. This difference in viscosity is why engine oil is thicker and more sluggish than water. In summary, the rate of flow of a liquid out of a container is determined by its viscosity and not surface tension, which only affects the surface properties of the liquid.
Question 34 Report
The team rectilinear acceleration means the rate of increase of
Answer Details
The term "rectilinear acceleration" refers to the rate of change or increase in velocity along a straight-line path in a unit time. It is also known as linear acceleration or simply acceleration. Therefore, the correct option is: velocity along a straight-line path in a unit time.
Question 35 Report
A body accelerates uniformly from rest at 2ms-2. Calculate the magnitude of its velocity after travelling 9m
Answer Details
To solve this problem, we can use the kinematic equation: v2 = u2 + 2as Where: v = final velocity (what we want to find) u = initial velocity (in this case, 0 since the body starts from rest) a = acceleration (given as 2 ms-2) s = distance travelled (given as 9m) Substituting the given values, we get: v2 = 02 + 2 x 2 ms-2 x 9m v2 = 36 ms-1 v = √(36) ms-1 Therefore, the magnitude of the velocity of the body after travelling 9m with a uniform acceleration of 2ms-2 is 6.0 ms-1.
Question 36 Report
In a nuclear reactor, chain reactions result from the release of
Answer Details
In a nuclear reactor, chain reactions result from the release of neutrons. A nuclear chain reaction occurs when a neutron collides with a nucleus of a fissile material, such as uranium-235 or plutonium-239, which then splits into two smaller nuclei, releasing energy in the form of heat and more neutrons. These newly released neutrons can then collide with other fissile nuclei, causing them to split and release more neutrons, which continue the chain reaction. This chain reaction is carefully controlled in a nuclear reactor to produce a steady supply of heat that is used to generate electricity.
Question 37 Report
Which of the following statements is correct about resonance? Resonance occurs when a body
Question 38 Report
Which of the following operations does not represent an action of a force field?
Answer Details
The operation that does not represent an action of a force field is the "pushing of a wheel-barrow on a level ground". A force field is a region where a force acts on an object, and the object experiences a force because of the interaction with the field. In the case of the pushing of a wheel-barrow, the force is not a result of a field but is applied by the person pushing the wheelbarrow. Therefore, this operation does not represent an action of a force field.
Question 39 Report
Which of the following components is used for storing electric charges?
Answer Details
A capacitor is the component used for storing electric charges. A capacitor consists of two conductors separated by an insulator, called a dielectric. When a potential difference is applied across the conductors, electrons accumulate on one conductor and are depleted from the other, resulting in an electric field between the conductors. The insulator prevents the electrons from flowing between the conductors, so the charge is stored on the conductors, effectively storing electric charges. The ability of a capacitor to store charge is measured in farads, and capacitors are commonly used in electronic circuits for a variety of applications, including filtering, timing, and power storage.
Question 40 Report
A relative density bottle of volume 50cm3 is completely filled with a liquid at 30oC. It is then heated to 80oC such that 0.75cm3 of the liquid is expelled. Calculate the apparent cubic expansivity of the liquid.
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Question 41 Report
A body moving with an initial velocity U accelerates until it attains a velocity of V within a time t. The distance, s covered by the body is given by the expressions
Answer Details
The equation that relates initial velocity, final velocity, acceleration, time and distance is given by s = ut + 0.5at^2 + \(\frac{1}{2}\) at^2. We are given that the body is moving with an initial velocity U and attains a final velocity V within time t. Hence, the acceleration of the body can be calculated using the equation: a = (V - U) / t. Now, substituting the value of acceleration in the above equation, we get: s = Ut + \(\frac{1}{2}\)(V - U)t = \(\frac{V+U}{2}\)t Therefore, the expression for the distance, s covered by the body is s = (\(\frac{V+U}{2}\))t. Hence, option (C) is the correct answer.
Question 42 Report
Mechanical energy can be either
Answer Details
Mechanical energy is the energy possessed by an object due to its motion or position. It can be either potential energy or kinetic energy. Potential energy is the energy an object possesses due to its position, shape, or state. Kinetic energy, on the other hand, is the energy an object possesses due to its motion. Therefore, the correct answer is "potential or kinetic."
Question 43 Report
Which of the following wave is electromagnetic?
Answer Details
Electromagnetic waves are waves that are capable of propagating through a vacuum or through different media. These waves are made up of changing electric and magnetic fields that are perpendicular to each other and to the direction of the wave's propagation. Sound waves, water waves, and tidal waves are all mechanical waves, meaning they require a medium to propagate through. X-rays, on the other hand, are electromagnetic waves, which means they do not require a medium to propagate through and can travel through a vacuum. Therefore, the correct answer is x-rays.
Question 44 Report
The pressure of a fixed mass of an ideal gas at 27°C is 3 pa. The gas is heated at a constant volume until its pressure is 5 pa. Determine the new temperature of the gas.
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Question 45 Report
If a gas is excited by high voltage to produce a discharge and the light is examined in a spectrometer
Answer Details
When a gas is excited by high voltage to produce a discharge and the light is examined in a spectrometer, an emission spectrum is observed. This is because the excited gas atoms emit energy in the form of light of specific wavelengths as they return to their ground state. These emitted wavelengths are characteristic of the element or compound being examined and appear as bright lines in the spectrum. Therefore, the emission spectrum can be used to identify the elements present in the gas being examined.
Question 46 Report
When an object is placed at the centre of curvature of a concave mirror, its image is formed at
Answer Details
When an object is placed at the centre of curvature of a concave mirror, its image is formed at the centre of curvature. The centre of curvature is a point on the principal axis of the mirror, located at a distance equal to the radius of curvature from the mirror's vertex. When light rays from the object fall on the concave mirror, they reflect and converge at the centre of curvature. This convergence of light rays forms a real, inverted image of the object at the centre of curvature. Since the image is formed by the actual convergence of light rays, it is a real image. In summary, when an object is placed at the centre of curvature of a concave mirror, its image is formed at the same location, which is the centre of curvature of the mirror.
Question 47 Report
A plane inclined at 30o to the horizontal has an efficiency of 50%. Calculate the force parallel to the plane required to push a load of 120N uniformly upn the plane
Answer Details
When an object is on an inclined plane, its weight is resolved into two components: one perpendicular to the plane, and the other parallel to the plane. The perpendicular component is counteracted by the normal force of the plane, while the parallel component is responsible for the object's motion along the plane. Since the plane has an efficiency of 50%, it means that only half of the force parallel to the plane is used to move the object up the plane. The other half is lost due to friction and other factors. Therefore, to calculate the required force parallel to the plane, we need to divide the weight of the load by the efficiency: force = (load weight / efficiency) * sin(theta) where theta is the angle of inclination of the plane. Substituting the given values, we get: force = (120 / 0.5) * sin(30) force = 240N * 0.5 force = 120N Therefore, the force required to push the load uniformly up the plane is 120N, which corresponds to.
Question 48 Report
A liquid of volume 2.00 m3 and density 1.00 x 103 is mixed with 3.00 m3 of another liquid of density 8.00 x 103kg m-3. Calculate the density of the mixture [Assuming there is no chemical reaction]
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Question 49 Report
(a) Define the boiling point of a liquid.
(b) With the aid of a sketch diagram, describe an experiment to determine the boiling point of a small quantity of a Iiquid
(c) A piece of copper of mass 300 g at a temperature of 950°C is quickly transferred into a vessel of negligible thermal capacity containing 250 g of water at 25°C. If the final steady temperature of the mixture is 100°C, calculate the mass of water that will boil away. Especific heat capacity of copper = 4.0 x 10\(^2\) J kg\(^{-1}\)K\(^{-1}\) specific heat capacity of water = 4.2 x 10\(^{3}\) J kg\(^{-1}\)K\(^{-1}\) specific latent heat of vaporization of steam = 2.26 x 10\(^6\)J kg\(^{-1}\)
(a) Boiling point of a liquid
The boiling point of a liquid is the constant temperature at which the liquid changes into vapour throughout its bulk, that is, the temperature at which the saturated vapour pressure of the liquid becomes equal to the external (atmospheric) pressure.
(b) Experiment to determine the boiling point of a small quantity of a liquid
Place the small quantity of the liquid in a clean ignition (test) tube and lower it into a beaker of water so that the liquid is well below the water surface. Support a thermometer with its bulb immersed in the liquid, and place a stirrer in the surrounding water. Stand the beaker on a wire gauze and tripod above a Bunsen burner.
Heat the water bath gently and stir the water continuously so that the heat reaches the small tube evenly. Watch the thermometer: its reading rises steadily and then becomes constant while the liquid is boiling vigorously. Record this steady temperature. The constant temperature at which the thermometer reading remains fixed while the liquid boils is the boiling point of the liquid.
(c) Mass of water that boils away
Data: mass of copper \(m_{Cu}=300\,\text{g}=0.300\,\text{kg}\), specific heat capacity of copper \(c_{Cu}=4.0\times10^{2}\,\text{J kg}^{-1}\text{K}^{-1}\), mass of water \(m_w=250\,\text{g}=0.250\,\text{kg}\), specific heat capacity of water \(c_w=4.2\times10^{3}\,\text{J kg}^{-1}\text{K}^{-1}\), specific latent heat of vaporization \(L=2.26\times10^{6}\,\text{J kg}^{-1}\).
Heat lost by the copper as it cools from \(950^{\circ}\text{C}\) to the final temperature \(100^{\circ}\text{C}\):
\[ Q_{Cu}=m_{Cu}\,c_{Cu}\,\Delta\theta = 0.300\times(4.0\times10^{2})\times(950-100)=0.300\times400\times850=102000\,\text{J} \]Heat needed to raise the \(250\,\text{g}\) of water from \(25^{\circ}\text{C}\) to \(100^{\circ}\text{C}\):
\[ Q_{1}=m_{w}\,c_{w}\,\Delta\theta = 0.250\times(4.2\times10^{3})\times(100-25)=0.250\times4200\times75=78750\,\text{J} \]By conservation of energy, the heat remaining after warming the water is used to boil some of it into steam:
\[ Q_{2}=Q_{Cu}-Q_{1}=102000-78750=23250\,\text{J} \]If \(m_v\) is the mass of water boiled away, then \(Q_{2}=m_v L\):
\[ m_v=\frac{Q_{2}}{L}=\frac{23250}{2.26\times10^{6}}=1.03\times10^{-2}\,\text{kg}\approx 10.3\,\text{g} \]The mass of water that boils away is about \(10.3\,\text{g}\) (\(1.03\times10^{-2}\,\text{kg}\)).
Answer Details
(a) Boiling point of a liquid
The boiling point of a liquid is the constant temperature at which the liquid changes into vapour throughout its bulk, that is, the temperature at which the saturated vapour pressure of the liquid becomes equal to the external (atmospheric) pressure.
(b) Experiment to determine the boiling point of a small quantity of a liquid
Place the small quantity of the liquid in a clean ignition (test) tube and lower it into a beaker of water so that the liquid is well below the water surface. Support a thermometer with its bulb immersed in the liquid, and place a stirrer in the surrounding water. Stand the beaker on a wire gauze and tripod above a Bunsen burner.
Heat the water bath gently and stir the water continuously so that the heat reaches the small tube evenly. Watch the thermometer: its reading rises steadily and then becomes constant while the liquid is boiling vigorously. Record this steady temperature. The constant temperature at which the thermometer reading remains fixed while the liquid boils is the boiling point of the liquid.
(c) Mass of water that boils away
Data: mass of copper \(m_{Cu}=300\,\text{g}=0.300\,\text{kg}\), specific heat capacity of copper \(c_{Cu}=4.0\times10^{2}\,\text{J kg}^{-1}\text{K}^{-1}\), mass of water \(m_w=250\,\text{g}=0.250\,\text{kg}\), specific heat capacity of water \(c_w=4.2\times10^{3}\,\text{J kg}^{-1}\text{K}^{-1}\), specific latent heat of vaporization \(L=2.26\times10^{6}\,\text{J kg}^{-1}\).
Heat lost by the copper as it cools from \(950^{\circ}\text{C}\) to the final temperature \(100^{\circ}\text{C}\):
\[ Q_{Cu}=m_{Cu}\,c_{Cu}\,\Delta\theta = 0.300\times(4.0\times10^{2})\times(950-100)=0.300\times400\times850=102000\,\text{J} \]Heat needed to raise the \(250\,\text{g}\) of water from \(25^{\circ}\text{C}\) to \(100^{\circ}\text{C}\):
\[ Q_{1}=m_{w}\,c_{w}\,\Delta\theta = 0.250\times(4.2\times10^{3})\times(100-25)=0.250\times4200\times75=78750\,\text{J} \]By conservation of energy, the heat remaining after warming the water is used to boil some of it into steam:
\[ Q_{2}=Q_{Cu}-Q_{1}=102000-78750=23250\,\text{J} \]If \(m_v\) is the mass of water boiled away, then \(Q_{2}=m_v L\):
\[ m_v=\frac{Q_{2}}{L}=\frac{23250}{2.26\times10^{6}}=1.03\times10^{-2}\,\text{kg}\approx 10.3\,\text{g} \]The mass of water that boils away is about \(10.3\,\text{g}\) (\(1.03\times10^{-2}\,\text{kg}\)).
Question 50 Report
Define (a) tensile Stress; (b) tensile strain; (c) yield point.
(a) Tensile stress: the stretching force acting per unit cross-sectional area of a material, \(\text{stress}=\dfrac{F}{A}\). Unit: \(\text{Nm}^{-2}\) (pascal).
(b) Tensile strain: the extension produced per unit original length of the material, \(\text{strain}=\dfrac{e}{L}\). It has no unit.
(c) Yield point: the point on the stress-strain (or load-extension) graph beyond which the material no longer returns to its original length when the load is removed; it begins to undergo permanent (plastic) deformation.
Answer Details
(a) Tensile stress: the stretching force acting per unit cross-sectional area of a material, \(\text{stress}=\dfrac{F}{A}\). Unit: \(\text{Nm}^{-2}\) (pascal).
(b) Tensile strain: the extension produced per unit original length of the material, \(\text{strain}=\dfrac{e}{L}\). It has no unit.
(c) Yield point: the point on the stress-strain (or load-extension) graph beyond which the material no longer returns to its original length when the load is removed; it begins to undergo permanent (plastic) deformation.
Question 51 Report
Define
(a) electrolysis;
(b) electrolyte;
(c) electrode.
(a) Electrolysis: the chemical decomposition of an electrolyte (in molten or aqueous solution) brought about by passing an electric current through it.
(b) Electrolyte: a compound which, when molten or dissolved in water, conducts electricity and is chemically decomposed by it (because it contains free-moving ions).
(c) Electrode: a conductor (rod or plate) through which the electric current enters or leaves the electrolyte. The one connected to the positive terminal is the anode; the one connected to the negative terminal is the cathode.
Answer Details
(a) Electrolysis: the chemical decomposition of an electrolyte (in molten or aqueous solution) brought about by passing an electric current through it.
(b) Electrolyte: a compound which, when molten or dissolved in water, conducts electricity and is chemically decomposed by it (because it contains free-moving ions).
(c) Electrode: a conductor (rod or plate) through which the electric current enters or leaves the electrolyte. The one connected to the positive terminal is the anode; the one connected to the negative terminal is the cathode.
Question 52 Report
A particle is projected at an angle of 30° to the horizontal with a speed of 250 ms\(^{-1}\). Calculate the;
(a) total time of flight of the particle;
(b) speed of the particle at its maximum height. [ g = 10 ms\(^{-2}\)]
Given \(u=250\,\text{ms}^{-1}\), \(\theta=30^{\circ}\), \(g=10\,\text{ms}^{-2}\).
(a) Total time of flight.
\[ T=\frac{2u\sin\theta}{g}=\frac{2\times250\times\sin30^{\circ}}{10}=\frac{2\times250\times0.5}{10}=\frac{250}{10}=25\,\text{s} \](b) Speed at maximum height. At the highest point the vertical component of velocity is zero, so the speed there is just the (unchanged) horizontal component:
\[ v=u\cos\theta=250\times\cos30^{\circ}=250\times0.866=216.5\,\text{ms}^{-1} \]Answer Details
Given \(u=250\,\text{ms}^{-1}\), \(\theta=30^{\circ}\), \(g=10\,\text{ms}^{-2}\).
(a) Total time of flight.
\[ T=\frac{2u\sin\theta}{g}=\frac{2\times250\times\sin30^{\circ}}{10}=\frac{2\times250\times0.5}{10}=\frac{250}{10}=25\,\text{s} \](b) Speed at maximum height. At the highest point the vertical component of velocity is zero, so the speed there is just the (unchanged) horizontal component:
\[ v=u\cos\theta=250\times\cos30^{\circ}=250\times0.866=216.5\,\text{ms}^{-1} \]Question 53 Report
(a) State the conditions of equilibrium for a number of coplanar parallel forces.
(b) A metre rule is found to balance horizontally at the 48 cm mark. When a body of mass 60 g is suspended at the 6 cm mark, the balance point is found to be at the 30 cm mark. Calculate the;
(i) mass of the metre rule;
(ii) distance of the balance point from the zero end, if the body were moved to the 13 cm mark.
(c) a man pulls up a box of mass 70 kg using an inclined plane of effective length 5 m unto a platform 2.5 m high at a uniform speed. If the frictional force between the box and the plane is 1000 N;
(i) draw a diagram to illustrate all the forces acting on the box while in motion;
(ii) calculate the I. minimum effort applied in pulling up the box; II. velocity ratio of the plane, if it is inclined at 30° to the horizontal; Ill. force ratio of the plane.
(a) Conditions of equilibrium for coplanar parallel forces
A number of coplanar parallel forces are in equilibrium when:
- The algebraic sum of the forces is zero, i.e. the sum of the forces acting in one direction equals the sum of the forces acting in the opposite direction (the resultant force is zero).
- The algebraic sum of the moments of the forces about any point is zero, i.e. the sum of the clockwise moments about the point equals the sum of the anticlockwise moments about that same point (the principle of moments).
(b) Balancing a metre rule
The rule balances by itself at the \(48\,\text{cm}\) mark, so the whole weight of the rule acts at the \(48\,\text{cm}\) mark.
(i) Mass of the metre rule
With the \(60\,\text{g}\) body hung at the \(6\,\text{cm}\) mark, the new balance point (fulcrum) is at the \(30\,\text{cm}\) mark. The body sits on one side of the fulcrum and the weight of the rule acts on the other side. Taking moments about the \(30\,\text{cm}\) fulcrum:
\[ 60\times(30-6)=m\times(48-30) \]\[ 60\times24=m\times18 \]\[ m=\frac{60\times24}{18}=\frac{1440}{18}=80\,\text{g} \]The mass of the metre rule is 80 g.
(ii) New balance point with the body at the 13 cm mark
Let the new balance point be at the \(x\,\text{cm}\) mark. The body (\(60\,\text{g}\)) now acts at \(13\,\text{cm}\) and the rule's weight (\(80\,\text{g}\)) still acts at \(48\,\text{cm}\). Taking moments about the fulcrum at \(x\):
\[ 60\times(x-13)=80\times(48-x) \]\[ 60x-780=3840-80x \]\[ 140x=4620 \]\[ x=\frac{4620}{140}=33\,\text{cm} \]The new balance point is 33 cm from the zero end.
(c) Pulling a box up an inclined plane
Box mass \(=70\,\text{kg}\), so its weight \(W=mg=70\times10=700\,\text{N}\); length of plane \(L=5\,\text{m}\); height of platform \(h=2.5\,\text{m}\); frictional force \(F=1000\,\text{N}\); angle of incline \(\theta=30^{\circ}\).
(i) Diagram of the forces acting on the box while in motion
The four forces acting on the box are: its weight \(W\) acting vertically downwards; the normal (reaction) force \(N\) acting perpendicular to the surface of the plane; the effort \(E\) applied up along the plane; and the frictional force \(F\) acting down along the plane, opposing the upward motion.
(ii) Calculations
I. Minimum effort applied in pulling up the box
Since the box moves up at uniform (constant) speed, the effort must balance the component of the weight along the plane together with the friction acting down the plane:
\[ E=W\sin\theta+F=700\sin30^{\circ}+1000 \]\[ E=700\times0.5+1000=350+1000=1350\,\text{N} \]The minimum effort is 1350 N.
II. Velocity ratio of the plane
\[ V.R.=\frac{1}{\sin\theta}=\frac{1}{\sin30^{\circ}}=\frac{1}{0.5}=2 \](This agrees with \(\dfrac{\text{length}}{\text{height}}=\dfrac{5}{2.5}=2\).) The velocity ratio is 2.
III. Force ratio (mechanical advantage) of the plane
\[ \text{Force ratio}=M.A.=\frac{\text{load}}{\text{effort}}=\frac{700}{1350}=0.52 \]The force ratio of the plane is 0.52.
Answer Details
(a) Conditions of equilibrium for coplanar parallel forces
A number of coplanar parallel forces are in equilibrium when:
- The algebraic sum of the forces is zero, i.e. the sum of the forces acting in one direction equals the sum of the forces acting in the opposite direction (the resultant force is zero).
- The algebraic sum of the moments of the forces about any point is zero, i.e. the sum of the clockwise moments about the point equals the sum of the anticlockwise moments about that same point (the principle of moments).
(b) Balancing a metre rule
The rule balances by itself at the \(48\,\text{cm}\) mark, so the whole weight of the rule acts at the \(48\,\text{cm}\) mark.
(i) Mass of the metre rule
With the \(60\,\text{g}\) body hung at the \(6\,\text{cm}\) mark, the new balance point (fulcrum) is at the \(30\,\text{cm}\) mark. The body sits on one side of the fulcrum and the weight of the rule acts on the other side. Taking moments about the \(30\,\text{cm}\) fulcrum:
\[ 60\times(30-6)=m\times(48-30) \]\[ 60\times24=m\times18 \]\[ m=\frac{60\times24}{18}=\frac{1440}{18}=80\,\text{g} \]The mass of the metre rule is 80 g.
(ii) New balance point with the body at the 13 cm mark
Let the new balance point be at the \(x\,\text{cm}\) mark. The body (\(60\,\text{g}\)) now acts at \(13\,\text{cm}\) and the rule's weight (\(80\,\text{g}\)) still acts at \(48\,\text{cm}\). Taking moments about the fulcrum at \(x\):
\[ 60\times(x-13)=80\times(48-x) \]\[ 60x-780=3840-80x \]\[ 140x=4620 \]\[ x=\frac{4620}{140}=33\,\text{cm} \]The new balance point is 33 cm from the zero end.
(c) Pulling a box up an inclined plane
Box mass \(=70\,\text{kg}\), so its weight \(W=mg=70\times10=700\,\text{N}\); length of plane \(L=5\,\text{m}\); height of platform \(h=2.5\,\text{m}\); frictional force \(F=1000\,\text{N}\); angle of incline \(\theta=30^{\circ}\).
(i) Diagram of the forces acting on the box while in motion
The four forces acting on the box are: its weight \(W\) acting vertically downwards; the normal (reaction) force \(N\) acting perpendicular to the surface of the plane; the effort \(E\) applied up along the plane; and the frictional force \(F\) acting down along the plane, opposing the upward motion.
(ii) Calculations
I. Minimum effort applied in pulling up the box
Since the box moves up at uniform (constant) speed, the effort must balance the component of the weight along the plane together with the friction acting down the plane:
\[ E=W\sin\theta+F=700\sin30^{\circ}+1000 \]\[ E=700\times0.5+1000=350+1000=1350\,\text{N} \]The minimum effort is 1350 N.
II. Velocity ratio of the plane
\[ V.R.=\frac{1}{\sin\theta}=\frac{1}{\sin30^{\circ}}=\frac{1}{0.5}=2 \](This agrees with \(\dfrac{\text{length}}{\text{height}}=\dfrac{5}{2.5}=2\).) The velocity ratio is 2.
III. Force ratio (mechanical advantage) of the plane
\[ \text{Force ratio}=M.A.=\frac{\text{load}}{\text{effort}}=\frac{700}{1350}=0.52 \]The force ratio of the plane is 0.52.
Question 54 Report
(a)(i) With the aid of a labelled diagram describe the mode of operation of a modern X-ray tube.
(ii)State the energy transformations that take place during the operation of the X-ray tube.
(b) Define, as applied to X-rays, the following terms:
(i) hardness;
(ii) intensity.
(c) State (i) four uses of X-rays;
(ii) one hazard of over-exposure to X-rays in a radiological laboratory.
(a)(i) Mode of operation of a modern (Coolidge) X-ray tube
The tube is a highly evacuated glass envelope carrying a tungsten filament (cathode) at one end and a target-anode at the other, as shown in the labelled diagram below.
A low-voltage supply heats the tungsten filament, which by thermionic emission releases electrons. A concave focusing cup gathers these electrons into a narrow beam and directs them toward the target. A very high potential difference (the E.H.T., tens of kilovolts) is applied between the cathode and the anode, with the anode positive; this accelerates the electrons to a very high speed across the evacuated tube.
The fast electron beam strikes a small tungsten target embedded in a massive copper anode. On being suddenly stopped, the electrons give up their kinetic energy: only about 1 % is emitted as X-rays, which leave the tube through the window, while the greater part appears as heat. The copper anode and its cooling fins (aided in practice by circulating oil or water) conduct this heat away and prevent the target from melting.
The hardness (penetrating power) of the X-rays is controlled by the accelerating voltage across the tube, while the intensity (quantity of X-rays) is controlled by the filament heating current, which fixes the number of electrons produced.
(a)(ii) Energy transformations
\[\text{Electrical energy}\;\rightarrow\;\text{kinetic energy of the accelerated electrons}\;\rightarrow\;\text{X-ray (electromagnetic) energy}\;+\;\text{heat (thermal) energy at the target.}\]
(b) Definitions (as applied to X-rays)
(i) Hardness: the penetrating power of the X-rays. Hard X-rays have short wavelength and high frequency and penetrate deeply; the hardness increases with the accelerating potential difference across the tube.
(ii) Intensity: the energy radiated per unit area per unit time, i.e. the number of X-ray photons arriving per second on unit area. It depends on the number of electrons striking the target, and hence on the filament heating current.
(c)(i) Four uses of X-rays
- Medical diagnosis (radiography) to reveal fractured bones and the condition of internal organs.
- Radiotherapy, to destroy cancerous cells and tumours.
- Detection of hidden flaws, cracks and air bubbles in metals and welded joints (industrial radiography).
- Security screening of luggage at airports, and the study of crystal structure by X-ray crystallography.
(c)(ii) One hazard of over-exposure
Over-exposure to X-rays destroys living body cells, causing skin burns and cancer (for example leukaemia), as well as tissue damage, cataracts, sterility and genetic mutation.
Answer Details
(a)(i) Mode of operation of a modern (Coolidge) X-ray tube
The tube is a highly evacuated glass envelope carrying a tungsten filament (cathode) at one end and a target-anode at the other, as shown in the labelled diagram below.
A low-voltage supply heats the tungsten filament, which by thermionic emission releases electrons. A concave focusing cup gathers these electrons into a narrow beam and directs them toward the target. A very high potential difference (the E.H.T., tens of kilovolts) is applied between the cathode and the anode, with the anode positive; this accelerates the electrons to a very high speed across the evacuated tube.
The fast electron beam strikes a small tungsten target embedded in a massive copper anode. On being suddenly stopped, the electrons give up their kinetic energy: only about 1 % is emitted as X-rays, which leave the tube through the window, while the greater part appears as heat. The copper anode and its cooling fins (aided in practice by circulating oil or water) conduct this heat away and prevent the target from melting.
The hardness (penetrating power) of the X-rays is controlled by the accelerating voltage across the tube, while the intensity (quantity of X-rays) is controlled by the filament heating current, which fixes the number of electrons produced.
(a)(ii) Energy transformations
\[\text{Electrical energy}\;\rightarrow\;\text{kinetic energy of the accelerated electrons}\;\rightarrow\;\text{X-ray (electromagnetic) energy}\;+\;\text{heat (thermal) energy at the target.}\]
(b) Definitions (as applied to X-rays)
(i) Hardness: the penetrating power of the X-rays. Hard X-rays have short wavelength and high frequency and penetrate deeply; the hardness increases with the accelerating potential difference across the tube.
(ii) Intensity: the energy radiated per unit area per unit time, i.e. the number of X-ray photons arriving per second on unit area. It depends on the number of electrons striking the target, and hence on the filament heating current.
(c)(i) Four uses of X-rays
- Medical diagnosis (radiography) to reveal fractured bones and the condition of internal organs.
- Radiotherapy, to destroy cancerous cells and tumours.
- Detection of hidden flaws, cracks and air bubbles in metals and welded joints (industrial radiography).
- Security screening of luggage at airports, and the study of crystal structure by X-ray crystallography.
(c)(ii) One hazard of over-exposure
Over-exposure to X-rays destroys living body cells, causing skin burns and cancer (for example leukaemia), as well as tissue damage, cataracts, sterility and genetic mutation.
Question 55 Report
(a) Explain thermiopic emission
(b) State two applications of electrical conduction through gases.
(a) Thermionic emission: the emission (giving off) of electrons from the surface of a metal when it is heated to a high temperature. The heat gives the free surface electrons enough kinetic energy to overcome the work function (the surface attractive forces) and escape from the metal.
(b) Two applications of electrical conduction through gases:
- Fluorescent tubes and neon/discharge advertising signs.
- Sodium and mercury-vapour street lamps (the cathode-ray/television tube and lightning are also acceptable examples).
Answer Details
(a) Thermionic emission: the emission (giving off) of electrons from the surface of a metal when it is heated to a high temperature. The heat gives the free surface electrons enough kinetic energy to overcome the work function (the surface attractive forces) and escape from the metal.
(b) Two applications of electrical conduction through gases:
- Fluorescent tubes and neon/discharge advertising signs.
- Sodium and mercury-vapour street lamps (the cathode-ray/television tube and lightning are also acceptable examples).
Question 56 Report
(a)(i) What is a wave motion?
(ii) State two differences between a radio wave and a sound wave.
(b)(i) Given that you are provided with a tuning fork, a burette and other necessary apparatus, describe with the aid of a diagram, an experiment to determine the frequency of a note emitted by a source of sound. [assume the velocity of sound in air is known]
(ii)State two precautions necessary to obtain accurate result in the experiment described in (b)(i) above
(c) A pipe closed at one end is 100 cm long. If the air in the pipe is set into vibration and a fundamental note is produced, calculate the frequency of the note. [ velocity of sound in air = 340 ms\(^{-1}\)]
(a)(i) Wave motion. A wave motion is a disturbance that travels through a medium (or through space) transferring energy from one point to another without any net transfer of the particles of the medium.
(a)(ii) Two differences between a radio wave and a sound wave.
| Radio wave | Sound wave |
|---|---|
| It is an electromagnetic (transverse) wave. | It is a mechanical (longitudinal) wave. |
| Can travel through a vacuum (needs no material medium). | Requires a material medium for propagation. |
| Travels at the speed of light, \(3\times10^{8}\,\text{ms}^{-1}\). | Travels much more slowly, about \(340\,\text{ms}^{-1}\) in air. |
(b)(i) Experiment to determine the frequency of a note (resonance method using a burette).
The burette is clamped vertically and filled with water, so that the water and the tube walls enclose a short column of air at the open top. The vibrating tuning fork is held horizontally, with its prongs just above (not touching) the open mouth of the burette, as shown below.
The tap of the burette is opened so that water runs out slowly, gradually lengthening the air column. As the length increases, a position is reached where the sound suddenly becomes loudest; this is the first resonance. The length of the air column from the water surface to the open mouth is measured and recorded as \(L_1\).
Water is run out further, keeping the fork sounding over the mouth, until the sound is again loudest; this is the second resonance. This longer air column is measured and recorded as \(L_2\). The whole procedure is repeated a few times and the mean values of \(L_1\) and \(L_2\) taken.
Between the first and second resonances the air column has increased by exactly half a wavelength, so end-correction is eliminated:
\[ L_2-L_1=\frac{\lambda}{2}\;\Rightarrow\;\lambda=2\left(L_2-L_1\right) \]The frequency of the note is then obtained from \(f=\dfrac{v}{\lambda}\), giving
\[ f=\frac{v}{2\left(L_2-L_1\right)} \]where \(v\) is the known velocity of sound in air.
(b)(ii) Two precautions.
- The tuning fork should be struck gently on a soft rubber pad (not on a hard surface) and held with its prongs just above the mouth of the burette without touching it.
- The resonance lengths should be read at the position of loudest sound with the eye level with the water meniscus to avoid parallax error; the readings should be repeated and averaged.
(c) Fundamental frequency of a pipe closed at one end.
For a pipe closed at one end, the fundamental note has an air column equal to a quarter of a wavelength: \(L=\dfrac{\lambda}{4}\), so \(\lambda=4L\). With \(L=100\,\text{cm}=1.0\,\text{m}\),
\[ \lambda=4L=4\times1.0=4.0\,\text{m} \] \[ f=\frac{v}{\lambda}=\frac{340}{4.0}=85\,\text{Hz} \]The frequency of the fundamental note is 85 Hz.
Answer Details
(a)(i) Wave motion. A wave motion is a disturbance that travels through a medium (or through space) transferring energy from one point to another without any net transfer of the particles of the medium.
(a)(ii) Two differences between a radio wave and a sound wave.
| Radio wave | Sound wave |
|---|---|
| It is an electromagnetic (transverse) wave. | It is a mechanical (longitudinal) wave. |
| Can travel through a vacuum (needs no material medium). | Requires a material medium for propagation. |
| Travels at the speed of light, \(3\times10^{8}\,\text{ms}^{-1}\). | Travels much more slowly, about \(340\,\text{ms}^{-1}\) in air. |
(b)(i) Experiment to determine the frequency of a note (resonance method using a burette).
The burette is clamped vertically and filled with water, so that the water and the tube walls enclose a short column of air at the open top. The vibrating tuning fork is held horizontally, with its prongs just above (not touching) the open mouth of the burette, as shown below.
The tap of the burette is opened so that water runs out slowly, gradually lengthening the air column. As the length increases, a position is reached where the sound suddenly becomes loudest; this is the first resonance. The length of the air column from the water surface to the open mouth is measured and recorded as \(L_1\).
Water is run out further, keeping the fork sounding over the mouth, until the sound is again loudest; this is the second resonance. This longer air column is measured and recorded as \(L_2\). The whole procedure is repeated a few times and the mean values of \(L_1\) and \(L_2\) taken.
Between the first and second resonances the air column has increased by exactly half a wavelength, so end-correction is eliminated:
\[ L_2-L_1=\frac{\lambda}{2}\;\Rightarrow\;\lambda=2\left(L_2-L_1\right) \]The frequency of the note is then obtained from \(f=\dfrac{v}{\lambda}\), giving
\[ f=\frac{v}{2\left(L_2-L_1\right)} \]where \(v\) is the known velocity of sound in air.
(b)(ii) Two precautions.
- The tuning fork should be struck gently on a soft rubber pad (not on a hard surface) and held with its prongs just above the mouth of the burette without touching it.
- The resonance lengths should be read at the position of loudest sound with the eye level with the water meniscus to avoid parallax error; the readings should be repeated and averaged.
(c) Fundamental frequency of a pipe closed at one end.
For a pipe closed at one end, the fundamental note has an air column equal to a quarter of a wavelength: \(L=\dfrac{\lambda}{4}\), so \(\lambda=4L\). With \(L=100\,\text{cm}=1.0\,\text{m}\),
\[ \lambda=4L=4\times1.0=4.0\,\text{m} \] \[ f=\frac{v}{\lambda}=\frac{340}{4.0}=85\,\text{Hz} \]The frequency of the fundamental note is 85 Hz.
Question 57 Report
Copper of thickness d is plated on the cathode of a copper voltameter. If the total surface area of the cathode is 60 cm\(^2\) and a steady current of 5.0 A is maintained in the voltameter for 1 hour. calculate the value of d. [density of copper = 8.9 x 10\(^3\)kg m\(^{-3}\) electro chemical equivalent of copper = 3.3 x 10\(^{-7}\) kg C\(^{-1}\)
Mass of copper deposited (Faraday's first law): \(m=zIt\).
\[ m=(3.3\times10^{-7})(5.0)(3600)=5.94\times10^{-3}\,\text{kg} \]Volume of this copper: \(V=\dfrac{m}{\rho}\).
\[ V=\frac{5.94\times10^{-3}}{8.9\times10^{3}}=6.67\times10^{-7}\,\text{m}^3 \]The copper is spread over the cathode area \(A=60\,\text{cm}^2=60\times10^{-4}\,\text{m}^2=6.0\times10^{-3}\,\text{m}^2\), and \(V=A\times d\), so
\[ d=\frac{V}{A}=\frac{6.67\times10^{-7}}{6.0\times10^{-3}}=1.11\times10^{-4}\,\text{m} \]So the plating thickness \(d\approx1.1\times10^{-4}\,\text{m}\) (about \(0.11\,\text{mm}\)).
Answer Details
Mass of copper deposited (Faraday's first law): \(m=zIt\).
\[ m=(3.3\times10^{-7})(5.0)(3600)=5.94\times10^{-3}\,\text{kg} \]Volume of this copper: \(V=\dfrac{m}{\rho}\).
\[ V=\frac{5.94\times10^{-3}}{8.9\times10^{3}}=6.67\times10^{-7}\,\text{m}^3 \]The copper is spread over the cathode area \(A=60\,\text{cm}^2=60\times10^{-4}\,\text{m}^2=6.0\times10^{-3}\,\text{m}^2\), and \(V=A\times d\), so
\[ d=\frac{V}{A}=\frac{6.67\times10^{-7}}{6.0\times10^{-3}}=1.11\times10^{-4}\,\text{m} \]So the plating thickness \(d\approx1.1\times10^{-4}\,\text{m}\) (about \(0.11\,\text{mm}\)).
Question 58 Report
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 plane (linearly) polarized light the vibrations of the electric field take place in only one plane perpendicular to the direction of travel, whereas in ordinary (unpolarized) light the vibrations occur in all planes perpendicular to the direction of travel.
(b) Two uses of polaroids:
- In sunglasses and camera filters to cut down glare from reflected light.
- In liquid-crystal displays (LCDs); also in polarimeters to measure sugar concentration, in stress analysis of materials, and in 3-D glasses.
Answer Details
(a) Difference between plane polarized light and ordinary light: in plane (linearly) polarized light the vibrations of the electric field take place in only one plane perpendicular to the direction of travel, whereas in ordinary (unpolarized) light the vibrations occur in all planes perpendicular to the direction of travel.
(b) Two uses of polaroids:
- In sunglasses and camera filters to cut down glare from reflected light.
- In liquid-crystal displays (LCDs); also in polarimeters to measure sugar concentration, in stress analysis of materials, and in 3-D glasses.
Question 59 Report
A wire of length 5.0 m and diameter 2.0 mm extends by 0.25 mm when a force of 50 N was used to stretch it from its end. Calculate the;
(a) stress on the wire;
(b) strain in, the wire. [\(\pi = 3.142\)]
Length \(L=5.0\,\text{m}\), diameter \(=2.0\,\text{mm}\Rightarrow r=1.0\,\text{mm}=1.0\times10^{-3}\,\text{m}\), extension \(e=0.25\,\text{mm}=0.25\times10^{-3}\,\text{m}\), force \(F=50\,\text{N}\).
(a) Stress \(=\dfrac{F}{A}\), where \(A=\pi r^2\).
\[ A=3.142\times(1.0\times10^{-3})^2=3.142\times10^{-6}\,\text{m}^2 \]\[ \text{Stress}=\frac{50}{3.142\times10^{-6}}=1.59\times10^{7}\,\text{Nm}^{-2} \](b) Strain \(=\dfrac{\text{extension}}{\text{original length}}\).
\[ \text{Strain}=\frac{0.25\times10^{-3}}{5.0}=5.0\times10^{-5} \](Strain has no unit as it is a ratio of two lengths.)
Answer Details
Length \(L=5.0\,\text{m}\), diameter \(=2.0\,\text{mm}\Rightarrow r=1.0\,\text{mm}=1.0\times10^{-3}\,\text{m}\), extension \(e=0.25\,\text{mm}=0.25\times10^{-3}\,\text{m}\), force \(F=50\,\text{N}\).
(a) Stress \(=\dfrac{F}{A}\), where \(A=\pi r^2\).
\[ A=3.142\times(1.0\times10^{-3})^2=3.142\times10^{-6}\,\text{m}^2 \]\[ \text{Stress}=\frac{50}{3.142\times10^{-6}}=1.59\times10^{7}\,\text{Nm}^{-2} \](b) Strain \(=\dfrac{\text{extension}}{\text{original length}}\).
\[ \text{Strain}=\frac{0.25\times10^{-3}}{5.0}=5.0\times10^{-5} \](Strain has no unit as it is a ratio of two lengths.)
Question 60 Report
State two effects to show (a) the existence of matter waves;
(b) that radiation behaves like particles.
Answer Details
None
Question 61 Report
(a) With the aid of a simple diagram, explain how a step down transformer works.
(b)(i) State three ways by which energy is lost in a transformer
(ii) Mention how each of the losses in (b)(i) above can be minimized
(c) A 95% efficient transformer is used to operate a lamp rated 60W, 220 V from a 4400 V a.c supply. Calculate the;
(i) ratio of the number of turns in the primary coil to the number of turns in the secondary coil of the transformer
(ii) current taken from the main circuit.
(a) How a step-down transformer works
A transformer consists of two coils, a primary (input) coil and a secondary (output) coil, wound on a common laminated soft-iron core. In a step-down transformer the primary coil has more turns than the secondary coil (\(N_p > N_s\)).
When an alternating voltage is applied to the primary coil, the alternating current flowing in it sets up a continuously changing magnetic flux in the soft-iron core. The core carries this changing flux round to the secondary coil, so the secondary is linked by the same changing flux. By electromagnetic induction, the changing flux induces an alternating e.m.f. in the secondary coil. Because the secondary has fewer turns than the primary, the e.m.f. induced in it is smaller than the applied primary voltage, i.e. the voltage is stepped down (while the output current is correspondingly stepped up).
The simple diagram below shows the arrangement:
(b)(i) Three ways energy is lost in a transformer
- Eddy-current loss in the iron core.
- Hysteresis loss in the iron core.
- Copper (resistance/heating) loss in the windings.
(b)(ii) How each loss is minimized
| Energy loss | How it is minimized |
|---|---|
| Eddy-current loss in the core | Use a laminated core (thin iron sheets insulated from one another) to break up the induced eddy-current paths. |
| Hysteresis loss in the core | Make the core of soft iron, which is easily magnetized and demagnetized, so little energy is used up in each magnetization cycle. |
| Copper (heating) loss in the windings | Use thick, low-resistance copper wire for the coils so that \(I^2R\) heating is reduced. |
(c) Calculations
Efficiency \(=95\%=0.95\); lamp (secondary side) rated \(P_{out}=60\,\text{W},\; V_s=220\,\text{V}\); supply \(V_p=4400\,\text{V}\).
(i) Ratio of primary turns to secondary turns.
For a transformer the turns ratio equals the voltage ratio:
\[ \frac{N_p}{N_s}=\frac{V_p}{V_s}=\frac{4400}{220}=20 \]Therefore \(\;N_p:N_s = \mathbf{20:1}.\)
(ii) Current taken from the main circuit.
Using efficiency \(=\dfrac{\text{output power}}{\text{input power}}=\dfrac{V_s I_s}{V_p I_p}\):
\[ 0.95=\frac{P_{out}}{V_p\,I_p}=\frac{60}{4400\times I_p} \]\[ I_p=\frac{60}{0.95\times 4400}=\frac{60}{4180}=0.0144\,\text{A} \]The current taken from the mains is \(I_p \approx \mathbf{0.014\,A}\;(1.44\times10^{-2}\,\text{A},\ \text{about }14.4\,\text{mA}).\)
Answer Details
(a) How a step-down transformer works
A transformer consists of two coils, a primary (input) coil and a secondary (output) coil, wound on a common laminated soft-iron core. In a step-down transformer the primary coil has more turns than the secondary coil (\(N_p > N_s\)).
When an alternating voltage is applied to the primary coil, the alternating current flowing in it sets up a continuously changing magnetic flux in the soft-iron core. The core carries this changing flux round to the secondary coil, so the secondary is linked by the same changing flux. By electromagnetic induction, the changing flux induces an alternating e.m.f. in the secondary coil. Because the secondary has fewer turns than the primary, the e.m.f. induced in it is smaller than the applied primary voltage, i.e. the voltage is stepped down (while the output current is correspondingly stepped up).
The simple diagram below shows the arrangement:
(b)(i) Three ways energy is lost in a transformer
- Eddy-current loss in the iron core.
- Hysteresis loss in the iron core.
- Copper (resistance/heating) loss in the windings.
(b)(ii) How each loss is minimized
| Energy loss | How it is minimized |
|---|---|
| Eddy-current loss in the core | Use a laminated core (thin iron sheets insulated from one another) to break up the induced eddy-current paths. |
| Hysteresis loss in the core | Make the core of soft iron, which is easily magnetized and demagnetized, so little energy is used up in each magnetization cycle. |
| Copper (heating) loss in the windings | Use thick, low-resistance copper wire for the coils so that \(I^2R\) heating is reduced. |
(c) Calculations
Efficiency \(=95\%=0.95\); lamp (secondary side) rated \(P_{out}=60\,\text{W},\; V_s=220\,\text{V}\); supply \(V_p=4400\,\text{V}\).
(i) Ratio of primary turns to secondary turns.
For a transformer the turns ratio equals the voltage ratio:
\[ \frac{N_p}{N_s}=\frac{V_p}{V_s}=\frac{4400}{220}=20 \]Therefore \(\;N_p:N_s = \mathbf{20:1}.\)
(ii) Current taken from the main circuit.
Using efficiency \(=\dfrac{\text{output power}}{\text{input power}}=\dfrac{V_s I_s}{V_p I_p}\):
\[ 0.95=\frac{P_{out}}{V_p\,I_p}=\frac{60}{4400\times I_p} \]\[ I_p=\frac{60}{0.95\times 4400}=\frac{60}{4180}=0.0144\,\text{A} \]The current taken from the mains is \(I_p \approx \mathbf{0.014\,A}\;(1.44\times10^{-2}\,\text{A},\ \text{about }14.4\,\text{mA}).\)
Question 62 Report
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 25 m from the foot of the tower. Calculate the height of the tower. g = 10 ms\(^{-2}\)]
Horizontal projection: initial horizontal speed \(u=4\,\text{ms}^{-1}\), horizontal range \(x=25\,\text{m}\), \(g=10\,\text{ms}^{-2}\).
Horizontal motion (constant velocity) gives the time of flight:
\[ x=ut\Rightarrow t=\frac{x}{u}=\frac{25}{4}=6.25\,\text{s} \]Vertical motion (falling from rest, initial vertical velocity zero) gives the height of the tower:
\[ h=\tfrac{1}{2}gt^2=\tfrac{1}{2}\times10\times(6.25)^2=5\times39.06=195.3\,\text{m} \]So the tower is about \(195\,\text{m}\) high.
Answer Details
Horizontal projection: initial horizontal speed \(u=4\,\text{ms}^{-1}\), horizontal range \(x=25\,\text{m}\), \(g=10\,\text{ms}^{-2}\).
Horizontal motion (constant velocity) gives the time of flight:
\[ x=ut\Rightarrow t=\frac{x}{u}=\frac{25}{4}=6.25\,\text{s} \]Vertical motion (falling from rest, initial vertical velocity zero) gives the height of the tower:
\[ h=\tfrac{1}{2}gt^2=\tfrac{1}{2}\times10\times(6.25)^2=5\times39.06=195.3\,\text{m} \]So the tower is about \(195\,\text{m}\) high.
Question 63 Report
(a) List two factors that can affect the rate of diffusion
(b) State two examples to illustrate the effects of surface tension.
(a) Two factors that affect the rate of diffusion:
- Temperature (higher temperature gives faster diffusion).
- The mass/size of the diffusing molecules (lighter molecules diffuse faster). The concentration gradient is also acceptable.
(b) Two examples illustrating the effects of surface tension:
- Small liquid drops (e.g. water or mercury) take up a spherical shape.
- A dry steel needle or a small insect can rest on the surface of water without sinking. (Capillary rise of liquid in a narrow tube is also acceptable.)
Answer Details
(a) Two factors that affect the rate of diffusion:
- Temperature (higher temperature gives faster diffusion).
- The mass/size of the diffusing molecules (lighter molecules diffuse faster). The concentration gradient is also acceptable.
(b) Two examples illustrating the effects of surface tension:
- Small liquid drops (e.g. water or mercury) take up a spherical shape.
- A dry steel needle or a small insect can rest on the surface of water without sinking. (Capillary rise of liquid in a narrow tube is also acceptable.)
