WAEC SSCE - Physics - 2000
Frage 1 Bericht
A uniform cylindrical hydrometer of mass 20g and cross sectional area 0.54cm2 floats upright in a liquid. If 25cm of its length is submerged, calculate the relative density of the liquid. (Density of water = 1 gcm-3)
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When a hydrometer floats in a liquid, it displaces an amount of liquid equal to its own volume, and the weight of the liquid displaced acts as an upthrust to balance the weight of the hydrometer. If the hydrometer floats with 25 cm of its length submerged, then the volume of liquid displaced by the hydrometer is equal to the volume of the submerged part of the hydrometer, which is given by the formula V = Al, where A is the cross-sectional area of the hydrometer and l is the length submerged. So, the volume of liquid displaced is V = 0.54 cm2 x 25 cm = 13.5 cm3. The weight of the liquid displaced is equal to the weight of the hydrometer, which is given by the formula W = mg, where m is the mass of the hydrometer and g is the acceleration due to gravity. So, W = 0.02 kg x 9.81 m/s2 = 0.1962 N. The relative density of the liquid is given by the formula ρliquid/ρwater = Fupthrust/W, where Fupthrust is the upthrust on the hydrometer due to the liquid. The upthrust is equal to the weight of the liquid displaced, which we have already calculated to be 0.1962 N. So, ρliquid/ρwater = 0.1962 N / (0.54 cm2 x 25 cm x 9.81 m/s2 x 1 g/cm3) = 1.48 Therefore, the relative density of the liquid is 1.48. Answer (b) 1.48.
Frage 2 Bericht
A material of mass 1.0 x 10-2kg undergoes a fission process which decreases its mass by 0.02 percent. Calculate the amount of energy released in the process. (c = 3.0 x 108ms-1)
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The question involves calculating the amount of energy released during a fission process, given the initial mass and the percentage decrease in mass. The formula for calculating energy released during fission is E = Δmc^2, where Δm is the decrease in mass, c is the speed of light, and E is the energy released. In this case, the mass decreased by 0.02 percent, which means Δm = 0.02/100 * 1.0 x 10^-2 kg = 2 x 10^-6 kg. Plugging this into the formula along with the speed of light (c = 3.0 x 10^8 m/s), we get: E = (2 x 10^-6 kg) x (3.0 x 10^8 m/s)^2 E = 1.8 x 10^10 J Therefore, the amount of energy released in the fission process is 1.8 x 10^10 J, which is option D.
Frage 3 Bericht
A turning fork of frequency 600Hz is sounded over a closed resonance tube. If the first and second resonant positions are 0.130m and 0.413m respectively, calculate the speed of sound in air
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Frage 4 Bericht
When white light passes through a triangular glass prism, there is dispersion because of
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When white light passes through a triangular glass prism, it separates into its constituent colors because of the difference in speed of the components of light, a phenomenon known as dispersion. The amount of refraction depends on the wavelength of the light, so different colors are refracted differently as they pass through the prism, causing them to spread out and creating a rainbow-like spectrum of colors. This happens because the refractive index of the glass prism is different for different colors of light. The higher the refractive index of a material, the slower light travels through it. Thus, the shorter wavelength blue and violet light refract more than the longer wavelength red and orange light. This phenomenon is the basis for many optical instruments, including spectrometers and rainbows.
Frage 5 Bericht
A wooden block of mass 1.6kg rest on a rough horizontal surface, if the limiting friction force between the block and the surface is 8N. Calculate the coefficient of friction. (g = 10ms-2)
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The coefficient of friction (μ) is the ratio of the frictional force (F) to the normal force (N) between two surfaces in contact. In this case, the normal force is equal to the weight of the block, which is given by: N = mg where m is the mass of the block and g is the acceleration due to gravity (10ms-2). N = 1.6 kg x 10 ms-2 = 16 N The limiting friction force (F) is given as 8N. Therefore, we have: F = μN Substituting the values we have obtained, we can solve for μ as follows: μ = F/N = 8N/16N = 0.5 Therefore, the coefficient of friction is 0.5. Answer: 0.5
Frage 6 Bericht
The nucleon number and the proton number of a neutral atom of an element are 23 and 11 respectively. How many neutrons are present in the atom?
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The nucleon number of an atom is the sum of the number of protons and the number of neutrons in the nucleus of the atom. Therefore, the number of neutrons in the atom can be calculated by subtracting the number of protons (which is given as 11) from the nucleon number (which is given as 23). So, Number of neutrons = Nucleon number - Proton number = 23 - 11 = 12 Therefore, there are 12 neutrons present in the atom. The correct answer is (ii) 12.
Frage 7 Bericht
An immersion heater is rated 120W. How long does it take the heater to raise the temperature of 1.2kg of water by 15oC. (Assuming heat lost to the surrounding is negligible. Specific heat capacity of water = 4200 Jkg-1K-1C)
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We can use the formula: Q = mcΔT where Q is the amount of heat required to raise the temperature of the water, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature. First, let's calculate Q: Q = (1.2 kg) x (4200 Jkg^-1K^-1) x (15°C) = 75600 J This means we need to supply the water with 75600 J of heat in order to raise its temperature by 15°C. Now, let's use the formula: P = Q/t where P is the power of the immersion heater (in watts), t is the time taken (in seconds), and Q is the amount of heat supplied to the water. Converting 120W to J/s, we have: P = 120W = 120 J/s Plugging in the values for P and Q, we get: 120 J/s = 75600 J/t Solving for t, we get: t = 75600 J / 120 J/s = 630 s Converting to minutes, we get: t = 630 s / 60 s/min = 10.5 min Therefore, it would take 10.5 minutes for the immersion heater to raise the temperature of 1.2 kg of water by 15°C. Answer: 10.5 minutes
Frage 8 Bericht
An inelastic collision takes place between ball of known masses, just before the collision one of the ball is moving with a known velocity while the other is stationary. Which of the following physical quantities can be determined from the information given?
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Frage 9 Bericht
The resultant of two forces acting on an object is maximum when the angle between them is?
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Frage 10 Bericht
A simple microscope forms an image 10cm from an eye close to the lens. If the object is 6cm from the eye, calculate the focal length of the lens
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Frage 11 Bericht
In which of the following transitions is the largest quantum of energy liberated by an hydrogen atom when the electron changes energy levels? [n is the quantum number]
Frage 12 Bericht
The bob of a simple pendulum takes 0.25s to swing from its equilibrium position to one extreme end. Calculate its period
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Frage 13 Bericht
A periodic pulse travels a distance of 20.0m in 1.00s. If its frequency is 2.0 x 103Hz. Calculate the wavelength.
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The speed of a wave is given by the product of its wavelength and frequency. Therefore, we can use the formula: v = λf where v is the speed of the wave, λ is the wavelength, and f is the frequency. In this problem, we are given the frequency of the wave as 2.0 x 10^3 Hz and the distance it travels in 1.00 second as 20.0 m. We can calculate the speed of the wave as: v = distance/time = 20.0 m/1.00 s = 20.0 m/s Now, we can rearrange the formula to solve for the wavelength: λ = v/f = 20.0 m/s / 2.0 x 10^3 Hz = 1.0 x 10^-2 m Therefore, the wavelength of the wave is 1.0 x 10^-2 m. Answer: 1.0 x 10^-2 m
Frage 14 Bericht
A device used to prevent wearing away of the make and-brake contacts of an induction coil is called a\an
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The device used to prevent wearing away of the make and-break contacts of an induction coil is called a capacitor. A capacitor is an electrical component that stores energy in an electric field. It consists of two conductive plates separated by an insulating material called a dielectric. In the context of an induction coil, a capacitor is used to reduce the arcing that occurs at the make and-break contacts. This is because when the contacts are separated, the voltage across them rises rapidly, and without a capacitor, this can cause arcing and damage to the contacts. The capacitor absorbs this voltage, preventing arcing and extending the life of the contacts.
Frage 15 Bericht
A machine of efficiency 8% is used to raise a body of mass 75kg through a vertical height of 3m in 30s. Calculate the power input. (g = 10ms-2)
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Efficiency of a machine is defined as the ratio of output work to the input work. Mathematically, efficiency = (output work / input work) x 100%. In this question, the machine has an efficiency of 8%, which means that only 8% of the input work is converted into useful work. The rest is lost as heat or other forms of energy. Therefore, the output work of the machine is 8% of the input work. The machine is used to raise a body of mass 75kg through a vertical height of 3m in 30s. The work done by the machine is equal to the gravitational potential energy gained by the body. The gravitational potential energy gained by an object of mass m raised through a height h is given by the formula mgh, where g is the acceleration due to gravity. Therefore, the work done by the machine is W = mgh = 75 x 10 x 3 = 2250 J. We are asked to calculate the power input, which is the rate at which work is done by the machine. Power is given by the formula power = work done / time taken. Substituting the values, we get power = 2250 / 30 = 75 W. However, this is the output power of the machine. We need to calculate the input power, which is the power required to produce this output power, taking into account the efficiency of the machine. The input power is given by the formula input power = output power / efficiency. Substituting the values, we get input power = 75 / 0.08 = 937.5 W. Therefore, the power input is approximately 937.5 W, which is closest to option D: 93.8 W.
Frage 16 Bericht
The duality of matter implies that matter
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The duality of matter implies that matter has both wave and particle properties. In physics, duality refers to the idea that matter can exhibit characteristics of both waves and particles. This concept is best illustrated by the famous double-slit experiment, which showed that particles, such as electrons, can exhibit wave-like behavior, such as diffraction and interference. Conversely, waves, such as light, can exhibit particle-like behavior, such as the photoelectric effect. The duality of matter is an important concept in quantum mechanics and helps explain the behavior of matter on both a microscopic and macroscopic scale.
Frage 17 Bericht
In a nuclear reactor, electricity can be generated through the following processes. Arrange the processes in the correct order. i. The steam is used to drive turbines. ii. The heat energy released is removed by passing water through the reactor. iii. The turbines in turn generate electricity. iv. The water then passes through some form of heat exchanger to produce steam
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The correct order of processes for generating electricity in a nuclear reactor is: ii. The heat energy released is removed by passing water through the reactor. iv. The water then passes through some form of heat exchanger to produce steam. i. The steam is used to drive turbines. iii. The turbines in turn generate electricity. This is because in a nuclear reactor, the heat energy is produced by the nuclear fission reactions occurring within the fuel rods. This heat energy is then used to heat up water, which is circulated through the reactor core to remove the heat energy. This is the second process (ii) mentioned in the options. The hot water then passes through a heat exchanger, which is the fourth process (iv) mentioned in the options. The heat exchanger uses the hot water to produce steam, which drives the turbines. This is the first process (i) mentioned in the options. Finally, the turbines generate electricity by rotating a generator, which is the third process (iii) mentioned in the options. Therefore, the correct order of processes is ii, iv, i, and iii.
Frage 18 Bericht
In the diagram above, an incident ray AY makes an angle of 30o with the normal XY. If the mirror is rotated anticlockwise about Y through an angle of 20o, while AY is fixed, what angle will the reflected ray now make with the incident ray?
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Frage 19 Bericht
The volume and pressure of a given mass of gas at 27oC are 76cm3 and 80cm of mercury respectively. Calculate the volume at s.t.p
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To solve the problem, we need to use the combined gas law, which relates the pressure, volume, and temperature of a gas. The combined gas law states that: P1V1/T1 = P2V2/T2 where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively. We are given the initial pressure, volume, and temperature of the gas as P1 = 80 cmHg, V1 = 76 cm^3, and T1 = 27°C. We want to find the volume of the gas at STP, which is a temperature of 0°C and a pressure of 1 atm (or 760 mmHg). To use the combined gas law, we need to convert the initial temperature to kelvin (K), since temperature must be in kelvin in this equation. We do this by adding 273 to the initial temperature: T1 = 27°C + 273 = 300 K We also need to convert the initial pressure from cmHg to atm, using the conversion factor: 1 atm = 760 mmHg = 101.325 kPa = 76 cmHg So, the initial pressure in atm is: P1 = 80 cmHg × (1 atm/76 cmHg) = 1.05 atm Now we can plug in the values into the combined gas law, with the final pressure P2 = 1 atm and the final temperature T2 = 0°C + 273 = 273 K: P1V1/T1 = P2V2/T2 (1.05 atm)(76 cm^3)/(300 K) = (1 atm)(V2)/(273 K) Solving for V2 gives: V2 = (1.05 atm)(76 cm^3)(273 K)/(300 K) = 72.8 cm^3 Therefore, the volume of the gas at STP is 72.8 cm^3, which is.
Frage 20 Bericht
An iron rod is moved from the earth to the moon. Which of the following properties of the rod would remain unchanged? i. mass ii. weight iii. relative density
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Frage 21 Bericht
An inductor of inductance 10H carries a current of 0.2A. Calculate the energy stored in the inductor
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Frage 22 Bericht
Two bodies P and Q, are in thermal equilibrium, which of the following statements about the bodies is correct?
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When two objects are in thermal equilibrium, it means that they are at the same temperature. This means that the heat energy flows between them at equal rates, and as a result, the temperatures of the two objects will remain constant over time. Therefore, the correct statement about the bodies is that P and Q are at the same temperature. The other options may or may not be true, depending on the specific situation, but they are not necessarily related to the fact that the bodies are in thermal equilibrium.
Frage 23 Bericht
A bus travelling at 15ms-1 accelerates uniformly at 4ms-2, what is the distance covered in 10s?
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The initial velocity of the bus, u = 15m/s, the acceleration, a = 4m/s2, and the time, t = 10s. We can use the formula: distance = ut + (1/2)at2 Substituting the values given: distance = (15m/s)(10s) + (1/2)(4m/s2)(10s)2 distance = 150m + 200m distance = 350m Therefore, the distance covered in 10 seconds is 350m. Answer is the correct answer.
Frage 24 Bericht
A piece of metal of mass 50g is cooled from 80oC to 20oC. Calculate the amount of heat lost. (specific heat capacity of the material of metal = 450Jkg-1K-1)
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The amount of heat lost by an object is given by the formula: Q = mcΔT where Q is the amount of heat lost or gained, m is the mass of the object, c is the specific heat capacity of the material, and ΔT is the change in temperature. In this question, the mass of the metal is given as 50g, which is 0.05kg. The specific heat capacity of the material of the metal is given as 450Jkg-1K-1. The change in temperature is 80oC - 20oC = 60oC. Substituting the given values into the formula: Q = (0.05kg)(450Jkg-1K-1)(60K) Q = 1.35 x 103J Therefore, the amount of heat lost by the piece of metal is 1.35 x 103J. Option (d) is the correct answer.
Frage 25 Bericht
The diagram shows the variation of volume V of a glass with temperature at the point X is
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The correct answer is -273oC. The diagram shows the variation of volume V of a glass with temperature. We can see that as the temperature approaches -273oC, the volume of the glass decreases rapidly. This is because at -273oC, all molecular motion ceases and the glass has zero volume, which is also known as absolute zero. Therefore, point X on the graph represents -273oC.
Frage 26 Bericht
If a bar magnet is accidentally broken into three pieces as shown in the diagram above, the polarities of P and Q respectively are
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When a bar magnet is broken, each piece of the magnet will become an independent magnet with its own north (N) and south (S) pole. To determine the polarities of P and Q, we can use the fact that opposite poles of a magnet attract each other, while like poles repel each other. Looking at the diagram, we can see that P is attracted to the south pole of the larger piece of the magnet, indicating that P must have a north pole. Similarly, Q is attracted to the north pole of the larger piece of the magnet, indicating that Q must have a south pole. Therefore, the correct answer is (B) N and S.
Frage 28 Bericht
The half life of a radioactive substance is 14 days. If 48g of this substance is stored, after how many days will 1.5g of the original substance remain?
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The half-life of a radioactive substance is the time taken for the substance to reduce to half of its original quantity. In this case, the half-life of the substance is 14 days, which means that after 14 days, half of the substance would have decayed. Let's calculate how many half-lives are needed for 48g to reduce to 1.5g: 48g → 24g (1 half-life) 24g → 12g (2 half-lives) 12g → 6g (3 half-lives) 6g → 3g (4 half-lives) 3g → 1.5g (5 half-lives) Therefore, it will take 5 half-lives for 48g of the substance to decay to 1.5g. The total time taken for 5 half-lives is 5 x 14 = 70 days. Therefore, after 70 days, 1.5g of the original substance will remain. Hence, the answer is 70 days.
Frage 29 Bericht
The driver of a car moving with uniform speed of 40ms-1 observes a truck approaching from the opposite direction with a speed of 20ms-1. Calculate the speed of the car relative to that of the truck
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The relative speed of the car with respect to the truck is the difference between their speeds. Since the truck is moving in the opposite direction, its speed is taken as negative. Thus: Relative speed = speed of car - speed of truck = 40ms^-1 - (-20ms^-1) = 40ms^-1 + 20ms^-1 = 60ms^-1 Therefore, the speed of the car relative to that of the truck is 60ms-1.
Frage 30 Bericht
In the diagram given, the current passing through the 6\(\Omega\) resistor is 1.5A. Calculate the terminal p.d. of the battery in the diagram
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To solve this problem, we need to use Kirchhoff's voltage law (KVL) which states that the sum of the potential differences around any closed loop in a circuit is zero. Starting from the positive terminal of the battery and moving clockwise around the circuit, we encounter a potential drop across the 6Ω resistor and a potential rise across the battery. Using Ohm's law, we can calculate the potential drop across the 6Ω resistor as: V = IR = 1.5A × 6Ω = 9V This means that the potential difference across the battery is: V_battery = V_6Ω + V_rise = 9V + V_rise We don't know the potential rise yet, but we can use KVL to find it. Moving clockwise around the circuit, we encounter a potential rise across the battery, a potential drop across the 4Ω resistor, and another potential drop across the 2Ω resistor. Using Ohm's law and KVL, we can write: V_rise - I × R_4Ω - I × R_2Ω = 0 where I is the current passing through the 6Ω resistor, which is also the current passing through the entire circuit. Substituting the given values, we get: V_rise - 1.5A × 4Ω - 1.5A × 2Ω = 0 V_rise = 1.5A × 6Ω = 9V Therefore, the potential difference across the battery is: V_battery = V_6Ω + V_rise = 9V + 9V = 18V So the terminal potential difference of the battery in the circuit is 18V. Therefore, the correct option is 10.80V.
Frage 31 Bericht
In the diagram above, the current passing through the 6\(\Omega\) resistor is 1.5A. Calculate the current in the 30\(\Omega\) resistor
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Frage 32 Bericht
The cubic expansivity of mercury is 1.8 x 10-4K-1 and the linear expansivity of glass 8.0 x 10-6K-1, calculate the apparent expansivity of mercury in a glass container
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Frage 33 Bericht
the factor which enables the ear to distinguish between a note played on different instruments is the
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Frage 34 Bericht
The apparent weight of a body wholly immersed in water is 32N and its weight in air is 96N. Calculate the volume of the body. (Density of water = 1000kgm-3, g = 10ms-3)
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When a body is immersed in water, it experiences an upthrust force, which reduces its weight. According to Archimedes' principle, this upthrust force is equal to the weight of the water displaced by the body. Therefore, we can use the given data to find the volume of the body as follows: Weight of body in air = 96N Apparent weight of body in water = 32N Upthrust force = Weight of water displaced = Weight of body in air - Apparent weight of body in water Upthrust force = 96N - 32N = 64N Now, we can use the formula for upthrust force to find the volume of the body: Upthrust force = Volume of body x Density of water x g 64N = Volume of body x 1000kgm-3 x 10ms-2 Volume of body = 64N / (1000kgm-3 x 10ms-2) = 6.4 x 10-3m3 Therefore, the volume of the body is 6.4 x 10-3m3.
Frage 35 Bericht
loudness increases as a vibrating tuning fork is brought nearer the end of a pipe containing air column due to
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The correct answer is "resonance". When a vibrating tuning fork is brought near the end of a pipe containing an air column, the frequency of the sound produced by the tuning fork matches the natural frequency of the air column in the pipe. This causes the air column to vibrate with a large amplitude, producing a louder sound. This phenomenon is known as resonance. As the tuning fork is brought closer to the end of the pipe, the resonance effect becomes stronger, causing the sound to become louder. This effect is commonly used in musical instruments such as flutes and organs to produce a desired tone or pitch.
Frage 36 Bericht
calculate the inductance of an inductor whose reactance is one ohm at 50 Hz.
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Frage 37 Bericht
A ball bearing is gently released from rest and allowed to fall through a viscous fluid. Which of the following statements about the motion is correct?
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As the ball bearing falls through the viscous fluid, the fluid exerts a resistive force against the motion of the ball bearing. This resistive force opposes the motion of the ball bearing and increases with its speed until it becomes equal to the weight of the ball bearing. At this point, the ball bearing stops accelerating and reaches a constant speed called the terminal velocity. Therefore, the correct statement about the motion is: when terminal velocity is attained, the acceleration of the ball bearing becomes zero. Option (a) is incorrect because the acceleration of the ball bearing decreases as it approaches the terminal velocity. Option (c) is incorrect because the velocity of the ball bearing increases initially, but as it approaches terminal velocity, the increase in velocity slows down until it becomes constant. Option (d) is incorrect because there is a resultant force acting on the ball bearing before it attains terminal velocity, which is the difference between the weight of the ball bearing and the resistive force of the fluid.
Frage 38 Bericht
The temperature at which the saturated vapour pressure of a liquid is equal to the external atmospheric pressure is known as its
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The temperature at which the saturated vapour pressure of a liquid becomes equal to the external atmospheric pressure is called its boiling point. At this temperature, the liquid changes its phase from liquid to gas or vapor. The boiling point of a liquid is a characteristic physical property that depends on factors such as the nature of the liquid, external atmospheric pressure, and intermolecular forces between the molecules of the liquid.
Frage 39 Bericht
Borh's theory provides evidence for the
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Bohr's theory provides evidence for the existence of energy levels in the atom. According to Bohr's theory, electrons orbit around the nucleus in specific energy levels or shells, and each shell has a fixed energy level. When an electron jumps from a higher energy level to a lower energy level, it releases energy in the form of light. This explains the observed line spectra of elements, and supports the idea of quantized energy levels in the atom. Therefore, is correct.
Frage 40 Bericht
The period of oscillation pf a particle executing simple harmonic motion is 4\(\pi\) seconds. If the amplitude of oscillation is 3.0m. Calculate its period
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Frage 41 Bericht
A concave mirror can be used to produce a parallel beam of light if a lighted bulb is placed
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A concave mirror is a mirror that curves inward, like the inside of a spoon. The property of a concave mirror that allows it to produce a parallel beam of light is its ability to reflect incoming light rays in such a way that they all converge to a single point called the focal point. When an object is placed between the focus and the pole of a concave mirror, a virtual, upright, and magnified image is formed behind the mirror. However, if a lighted bulb is placed at the focus of a concave mirror, the incoming light rays are reflected parallel to the principal axis of the mirror. Therefore, the correct answer is "at its focus."
Frage 42 Bericht
A nuclide \(^{202}_{84} Y\) emits in succession an \(\alpha-particle\) and a \(\beta-particle\). The atomic number of the resulting nuclide is
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The nuclide \(^{202}_{84} Y\) emits an \(\alpha-particle\) which consists of two protons and two neutrons, leading to a decrease in the atomic number by 2 and mass number by 4. Therefore, the resulting nuclide is \(^{198}_{82} Pb\). This nuclide then emits a \(\beta-particle\) which is an electron emitted from the nucleus, leading to an increase in the atomic number by 1 and no change in the mass number. Therefore, the atomic number of the resulting nuclide is 83. So, the answer is (b) 83.
Frage 43 Bericht
The region around a magnet in which the magnetic influence is experienced is called
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The correct answer is "magnetic field." A magnetic field is the region around a magnet or a moving electric charge where the magnetic influence or force can be felt. It is a vector field, meaning it has both magnitude and direction, and is measured in units of tesla (T) or gauss (G). In the case of a magnet, the magnetic field is strongest at the poles and weaker as you move away from them. Magnetic fields are used in many everyday applications, such as in speakers, electric motors, and MRI machines.
Frage 44 Bericht
Absorption line spectra exhibited by atoms is a result of
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Absorption line spectra exhibited by atoms is a result of the transition of an electron from a higher to a lower energy level. When energy is absorbed by an atom, an electron moves from a lower energy level to a higher one. Conversely, when an electron moves from a higher energy level to a lower one, energy is emitted in the form of electromagnetic radiation. This radiation is observed as a spectrum of colored lines with specific wavelengths that correspond to the specific energy levels involved in the transition. The spectrum produced is unique for each atom, providing a way to identify the atoms present in a sample. Therefore, the correct option is (d) transition of an electron from a higher to a lower energy level.
Frage 45 Bericht
A proton of wavelength \(\lambda_o\) is emitted when an electron in an atom makes a transition from a level of energy 2E. If the electron transits from \(\frac{4}{3}E_k\) to E level, determine the wavelength of the photon that would be emitted
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The energy of a photon can be calculated using the formula E = hc/λ where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon. When an electron transitions from a higher energy level to a lower energy level, the energy difference is released as a photon. Therefore, the energy of the emitted photon is equal to the difference in energy between the two levels. In this case, the electron transitions from \(\frac{4}{3}E_k\) to E level, which means the energy of the emitted photon is given by: E = (4/3)Ek - E Now, using the formula E = hc/λ, we can find the wavelength of the photon: λ = hc/E Substituting the expression for E: λ = hc/[(4/3)Ek - E] We know that the wavelength of the proton emitted by the electron transition from 2E to E is given by λ0. Therefore, we can write: 2E = hc/λ0 Rearranging for λ0: λ0 = hc/2E Substituting this expression for 2E in the formula for the energy of the emitted photon, we get: λ = hc/[(4/3)(hc/2λ0) - (hc/2λ0)] Simplifying: λ = 3λ0/4 Therefore, the wavelength of the photon that would be emitted is 3/4 of the wavelength of the proton emitted when the electron transitions from 2E to E. Answer: \(\frac{3}{4}\)\(\lambda_o\)
Frage 46 Bericht
The mass of water vapour in a given volume of air is 0.05g at 20oC, while the mass of water vapour required to saturate it at the same temperature is 0.15g. Calculate the relative humidity of the air
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The relative humidity is the ratio of the actual amount of water vapour present in the air to the maximum amount of water vapour that can be present in the air at a given temperature expressed as a percentage. In this question, the actual amount of water vapour present in the air is given as 0.05g and the mass of water vapour required to saturate it at the same temperature is given as 0.15g. So, the relative humidity can be calculated as: Relative humidity = (actual amount of water vapour/maximum amount of water vapour) x 100% Maximum amount of water vapour is the mass of water vapour required to saturate the air, which is given as 0.15g. Therefore, Relative humidity = (0.05/0.15) x 100% = 33.33% Hence, the relative humidity of the air is 33.33%. Option (c) is the correct answer.
Frage 47 Bericht
The diagram above illustrates three forces T1, T2 and 100N in equilibrium. Determine the magnitude of T1
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Frage 48 Bericht
Using the force displacement diagram shown above. Calculate the work done
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The work done is equal to the area under the force-displacement graph. The graph shows a rectangle with base 5m and height 400N, which represents the work done to displace an object by 5m against a constant force of 400N. Therefore, the work done is: Work done = Base × Height = 5m × 400N = 2000J Hence, the answer is 2000J.
Frage 49 Bericht
A wave has an amplitude equal to 4.0m, angular speed \(\frac{1}{3} \pi\) rads-1 and phase angle \(\frac{2}{3} \pi\) rad. The displacement y of the wave particle is given as
Frage 50 Bericht
The capacitance of a parallel-plate capacitor is increased by making the area of the plates
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The capacitance of a parallel-plate capacitor is directly proportional to the area of the plates and inversely proportional to the distance between the plates. This means that by increasing the area of the plates, the capacitance of the capacitor will increase. This can be seen from the capacitance equation, which is given as C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric material between the plates, A is the area of the plates, and d is the distance between the plates. Thus, the larger the area of the plates, the larger the capacitance of the capacitor. Therefore, the correct option is: large and their separation small.
Frage 51 Bericht
An object is situated within the Earth's gravitational field. Which of the following factor does not affect the acceleration of free fall g?
Antwortdetails
The factor that does not affect the acceleration of free fall, g, is the mass of the object. According to Newton's law of gravitation, the acceleration of free fall is dependent on the mass of the Earth and the distance between the center of the Earth and the object. Therefore, the distance of the object from the center of the Earth, the latitude of the Earth on which the object is situated, and the rotation of the Earth will affect the acceleration of free fall, but the mass of the object will not. This means that objects with different masses will experience the same acceleration of free fall when dropped from the same height, ignoring air resistance.
Frage 52 Bericht
A uniform metre rule of mass 90g is provided at the 40cm mark. If the rule is in equilibrium with an unknown mass m placed at the 10cm mark and a 72g mass at the 70cm mark, determine m
Antwortdetails
Frage 53 Bericht
(a)(i) State the laws of refraction of light.
(ii) Describe an experiment to determine the refractive index, n, of the material of an equilaleral triangular glass prism using the minimum deviation method.
(b) A rectangular glass prism of thickness 12cm is placed on a mark on a piece of paper resting on a horizontal bench:
(i) Draw a ray diagram to show the apparent position of the mark in the glass prism.
(ii) If the refractive-index of the material of the prism is 1.5, calculate the apparent displacement of the mark.
(a)(i) Laws of refraction of light
- The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
- The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media (Snell's law): \( \dfrac{\sin i}{\sin r} = n \).
(a)(ii) Determining the refractive index by the minimum-deviation method
Place the equilateral triangular glass prism flat on a sheet of paper and trace its outline. Remove the prism, and on one refracting face mark a point; draw the normal at this point and rule an incident ray meeting the face at a chosen angle of incidence \(i\). Replace the prism exactly on its outline. Stick two pins \(P_1\) and \(P_2\) vertically on the incident ray. Looking through the second refracting face, stick two more pins \(P_3\) and \(P_4\) so that they appear in a straight line with the images of \(P_1\) and \(P_2\). Remove the prism, ring the pin holes, and join \(P_3P_4\) to the face to give the emergent ray. Produce the incident and emergent rays forward until they meet; the angle between them is the angle of deviation \(d\). Measure \(i\) and \(d\) with a protractor.
Repeat the whole procedure for at least five values of \(i\), then plot a graph of the deviation \(d\) (vertical axis) against the angle of incidence \(i\) (horizontal axis). The graph is a smooth U-shaped curve; the deviation falls to a lowest value and then rises again. The value of \(d\) at the lowest point of the curve is the minimum deviation \(D\), read here as \(D = 37.2^\circ\).
With the refracting (apex) angle of the equilateral prism \(A = 60^\circ\), the refractive index is obtained from
\[ n = \frac{\sin\!\left(\dfrac{A + D}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)} = \frac{\sin\!\left(\dfrac{60^\circ + 37.2^\circ}{2}\right)}{\sin 30^\circ} = \frac{\sin 48.6^\circ}{\sin 30^\circ} = \frac{0.750}{0.500} = 1.50. \]Precautions: keep pin traces neat and thin; fix all pins truly vertical; and avoid parallax when reading the protractor.
(b)(i) Ray diagram showing the apparent position of the mark
The mark \(O\) is on the paper directly beneath the glass. Rays leaving \(O\) strike the top surface and are refracted away from the normal as they pass from glass to air, so they diverge more steeply. To the eye above, these emergent rays appear to come from a point \(I\) which lies vertically above \(O\) but nearer the top surface. Thus the mark is seen raised: \(I\) is its apparent (virtual) position.
(b)(ii) Apparent displacement of the mark
For viewing normally through a parallel-sided block,
\[ n = \frac{\text{real depth}}{\text{apparent depth}} \quad\Rightarrow\quad \text{apparent depth} = \frac{\text{real depth}}{n} = \frac{12}{1.5} = 8\ \text{cm}. \]Therefore the apparent displacement (the amount by which the mark appears raised) is
\[ \text{apparent displacement} = \text{real depth} - \text{apparent depth} = 12 - 8 = 4\ \text{cm}. \]Antwortdetails
(a)(i) Laws of refraction of light
- The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
- The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media (Snell's law): \( \dfrac{\sin i}{\sin r} = n \).
(a)(ii) Determining the refractive index by the minimum-deviation method
Place the equilateral triangular glass prism flat on a sheet of paper and trace its outline. Remove the prism, and on one refracting face mark a point; draw the normal at this point and rule an incident ray meeting the face at a chosen angle of incidence \(i\). Replace the prism exactly on its outline. Stick two pins \(P_1\) and \(P_2\) vertically on the incident ray. Looking through the second refracting face, stick two more pins \(P_3\) and \(P_4\) so that they appear in a straight line with the images of \(P_1\) and \(P_2\). Remove the prism, ring the pin holes, and join \(P_3P_4\) to the face to give the emergent ray. Produce the incident and emergent rays forward until they meet; the angle between them is the angle of deviation \(d\). Measure \(i\) and \(d\) with a protractor.
Repeat the whole procedure for at least five values of \(i\), then plot a graph of the deviation \(d\) (vertical axis) against the angle of incidence \(i\) (horizontal axis). The graph is a smooth U-shaped curve; the deviation falls to a lowest value and then rises again. The value of \(d\) at the lowest point of the curve is the minimum deviation \(D\), read here as \(D = 37.2^\circ\).
With the refracting (apex) angle of the equilateral prism \(A = 60^\circ\), the refractive index is obtained from
\[ n = \frac{\sin\!\left(\dfrac{A + D}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)} = \frac{\sin\!\left(\dfrac{60^\circ + 37.2^\circ}{2}\right)}{\sin 30^\circ} = \frac{\sin 48.6^\circ}{\sin 30^\circ} = \frac{0.750}{0.500} = 1.50. \]Precautions: keep pin traces neat and thin; fix all pins truly vertical; and avoid parallax when reading the protractor.
(b)(i) Ray diagram showing the apparent position of the mark
The mark \(O\) is on the paper directly beneath the glass. Rays leaving \(O\) strike the top surface and are refracted away from the normal as they pass from glass to air, so they diverge more steeply. To the eye above, these emergent rays appear to come from a point \(I\) which lies vertically above \(O\) but nearer the top surface. Thus the mark is seen raised: \(I\) is its apparent (virtual) position.
(b)(ii) Apparent displacement of the mark
For viewing normally through a parallel-sided block,
\[ n = \frac{\text{real depth}}{\text{apparent depth}} \quad\Rightarrow\quad \text{apparent depth} = \frac{\text{real depth}}{n} = \frac{12}{1.5} = 8\ \text{cm}. \]Therefore the apparent displacement (the amount by which the mark appears raised) is
\[ \text{apparent displacement} = \text{real depth} - \text{apparent depth} = 12 - 8 = 4\ \text{cm}. \]Frage 54 Bericht
A body of mass 0.6kg is thrown vertically upward from the ground with a speed of 20ms\(^{-2}\). Calculate its;
(i) potential energy at the maximum height reached.
(ii) kinetic energy just before it hits the ground.
A body of mass \(m = 0.6\,\text{kg}\) is thrown vertically upward from the ground with initial speed \(u = 20\,\text{ms}^{-1}\).
(i) Potential energy at maximum height
By conservation of energy, at the highest point all the initial kinetic energy has become potential energy (the body is momentarily at rest):
\[ \text{P.E.}_{max} = \frac{1}{2}mu^{2} = \frac{1}{2}(0.6)(20)^{2} = \frac{1}{2}(0.6)(400) = 120\,\text{J} \](ii) Kinetic energy just before it hits the ground
Neglecting air resistance, the body returns to the ground with the same speed it left, so its kinetic energy just before impact equals its initial kinetic energy (equal also to the P.E. at the top):
\[ \text{K.E.} = \frac{1}{2}mu^{2} = 120\,\text{J} \]Antwortdetails
A body of mass \(m = 0.6\,\text{kg}\) is thrown vertically upward from the ground with initial speed \(u = 20\,\text{ms}^{-1}\).
(i) Potential energy at maximum height
By conservation of energy, at the highest point all the initial kinetic energy has become potential energy (the body is momentarily at rest):
\[ \text{P.E.}_{max} = \frac{1}{2}mu^{2} = \frac{1}{2}(0.6)(20)^{2} = \frac{1}{2}(0.6)(400) = 120\,\text{J} \](ii) Kinetic energy just before it hits the ground
Neglecting air resistance, the body returns to the ground with the same speed it left, so its kinetic energy just before impact equals its initial kinetic energy (equal also to the P.E. at the top):
\[ \text{K.E.} = \frac{1}{2}mu^{2} = 120\,\text{J} \]Frage 55 Bericht
(a)(i) Mention two modes of heat transfer other than convection.
(ii) Explain land and sea breezes.
(b) An iron rod of length 30cm is heated through 50 kelvin. Calculate its increase in length. (linear expansivity of iron = 1.2 x 10\(^{-5}\)K\(^{-1}\)
(c) An electric heater immersed in some water raises the temperature of the water from 40°C to 100°C in 6 minutes. After another 25 minutes, it is noticed that half the water has boiled away. Neglecting heat losses to the surrounding, calculate the specific latent heat of vaporisation of water.
Antwortdetails
None
Frage 56 Bericht
A force of 40 N is applied at the end of a wire 4m long and produces an extension of 0.24mm. If the diameter of the wire is 2.00mm, calculate the;
(i) stress on the wire; (ii) strain in the wire.
A force \(F = 40\,\text{N}\) stretches a wire of length \(L = 4\,\text{m}\), producing an extension \(e = 0.24\,\text{mm} = 0.24\times10^{-3}\,\text{m}\); the wire diameter is \(2.00\,\text{mm}\), so radius \(r = 1.00\times10^{-3}\,\text{m}\).
Cross-sectional area:
\[ A = \pi r^{2} = 3.142\times(1.00\times10^{-3})^{2} = 3.142\times10^{-6}\,\text{m}^{2} \](i) Stress on the wire:
\[ \text{stress} = \frac{F}{A} = \frac{40}{3.142\times10^{-6}} = 1.27\times10^{7}\,\text{Nm}^{-2} \](ii) Strain in the wire:
\[ \text{strain} = \frac{e}{L} = \frac{0.24\times10^{-3}}{4} = 6.0\times10^{-5} \](Strain is a ratio and has no unit.)
Antwortdetails
A force \(F = 40\,\text{N}\) stretches a wire of length \(L = 4\,\text{m}\), producing an extension \(e = 0.24\,\text{mm} = 0.24\times10^{-3}\,\text{m}\); the wire diameter is \(2.00\,\text{mm}\), so radius \(r = 1.00\times10^{-3}\,\text{m}\).
Cross-sectional area:
\[ A = \pi r^{2} = 3.142\times(1.00\times10^{-3})^{2} = 3.142\times10^{-6}\,\text{m}^{2} \](i) Stress on the wire:
\[ \text{stress} = \frac{F}{A} = \frac{40}{3.142\times10^{-6}} = 1.27\times10^{7}\,\text{Nm}^{-2} \](ii) Strain in the wire:
\[ \text{strain} = \frac{e}{L} = \frac{0.24\times10^{-3}}{4} = 6.0\times10^{-5} \](Strain is a ratio and has no unit.)
Frage 57 Bericht
During the electrolysis of copper (II) tetraoxosulphate (VI) solution, a steady current of 4.0 x 10\(^2\) A flowing for one hour liberated 0.48 g of copper. Calculate the mass of copper liberated by one coulomb of charge.
Frage 58 Bericht
Define (i) Elasticity; (ii) Young's modulus; (iii) Force constant.
(i) Elasticity
Elasticity is the ability of a material to regain its original shape and size after the deforming force applied to it has been removed.
(ii) Young's modulus
Young's modulus is the ratio of tensile stress to tensile strain for a material stretched within its elastic (proportional) limit:
\[ Y = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/L} = \frac{FL}{Ae} \]Its unit is \(\text{Nm}^{-2}\) (pascal).
(iii) Force constant
The force constant (elastic/spring constant) is the force required to produce unit extension in a material obeying Hooke's law:
\[ k = \frac{F}{e} \]Its unit is \(\text{Nm}^{-1}\).
Antwortdetails
(i) Elasticity
Elasticity is the ability of a material to regain its original shape and size after the deforming force applied to it has been removed.
(ii) Young's modulus
Young's modulus is the ratio of tensile stress to tensile strain for a material stretched within its elastic (proportional) limit:
\[ Y = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/L} = \frac{FL}{Ae} \]Its unit is \(\text{Nm}^{-2}\) (pascal).
(iii) Force constant
The force constant (elastic/spring constant) is the force required to produce unit extension in a material obeying Hooke's law:
\[ k = \frac{F}{e} \]Its unit is \(\text{Nm}^{-1}\).
Frage 59 Bericht
a) Define the term surface tension
(b) Calculate the force required to lift a needle 4cm long off the surface of water if the surface tension of water is 7.3 x 10\(^{-4}\) NM\(^{-1}\)
(a) Surface tension
Surface tension is the force per unit length acting along (tangent to) the surface of a liquid, at right angles to a line drawn in the surface. It arises from the net inward attraction on the surface molecules and makes the surface behave like a stretched elastic skin. Its unit is \(\text{Nm}^{-1}\).
(b) Force to lift the needle
A floating needle is held by the surface film on both sides (two surfaces), each of length \(L\). The force due to surface tension is:
\[ F = \gamma \times 2L \]Data: \(\gamma = 7.3\times10^{-4}\,\text{Nm}^{-1}\), \(L = 4\,\text{cm} = 0.04\,\text{m}\).
\[ F = 7.3\times10^{-4} \times 2 \times 0.04 = 7.3\times10^{-4}\times0.08 = 5.84\times10^{-5}\,\text{N} \]The force required is about \(5.84\times10^{-5}\,\text{N}\).
Antwortdetails
(a) Surface tension
Surface tension is the force per unit length acting along (tangent to) the surface of a liquid, at right angles to a line drawn in the surface. It arises from the net inward attraction on the surface molecules and makes the surface behave like a stretched elastic skin. Its unit is \(\text{Nm}^{-1}\).
(b) Force to lift the needle
A floating needle is held by the surface film on both sides (two surfaces), each of length \(L\). The force due to surface tension is:
\[ F = \gamma \times 2L \]Data: \(\gamma = 7.3\times10^{-4}\,\text{Nm}^{-1}\), \(L = 4\,\text{cm} = 0.04\,\text{m}\).
\[ F = 7.3\times10^{-4} \times 2 \times 0.04 = 7.3\times10^{-4}\times0.08 = 5.84\times10^{-5}\,\text{N} \]The force required is about \(5.84\times10^{-5}\,\text{N}\).
Frage 60 Bericht
(a) Differentiate between interference and polarisation as applied to waves.
(b) Mention two uses of polaroids
(a) Interference vs polarisation
Interference is the superposition of two waves of the same frequency travelling in the same region, so that they reinforce one another (constructive interference) or cancel one another (destructive interference) depending on their path difference. It occurs in all types of waves (both transverse and longitudinal).
Polarisation is the restriction of the vibrations of a wave to a single plane containing the direction of travel. It occurs only in transverse waves, because it depends on the vibrations being at right angles to the direction of propagation. The fact that light can be polarised is direct evidence that light is a transverse wave.
(b) Two uses of polaroids
- In polaroid sunglasses to cut down glare (reflected, horizontally polarised light).
- In photographic filters and in liquid-crystal displays (LCDs) to control the intensity and plane of transmitted light; also in stress analysis of transparent materials.
Antwortdetails
(a) Interference vs polarisation
Interference is the superposition of two waves of the same frequency travelling in the same region, so that they reinforce one another (constructive interference) or cancel one another (destructive interference) depending on their path difference. It occurs in all types of waves (both transverse and longitudinal).
Polarisation is the restriction of the vibrations of a wave to a single plane containing the direction of travel. It occurs only in transverse waves, because it depends on the vibrations being at right angles to the direction of propagation. The fact that light can be polarised is direct evidence that light is a transverse wave.
(b) Two uses of polaroids
- In polaroid sunglasses to cut down glare (reflected, horizontally polarised light).
- In photographic filters and in liquid-crystal displays (LCDs) to control the intensity and plane of transmitted light; also in stress analysis of transparent materials.
Frage 61 Bericht
(a) Define the following terms:
(i) Electric field intensity
(ii) Electric potential
(b) The diagram below illustrates two collinear electric charges of magnitudes + Q and -Q. The charges are equidistant from a point P at which a rest charge is placed.
Copy the diagram and use arrows to indicate, from the point P, the direction of the;
(i) electric force F\(_1\) due to + Q.
(ii) electric force F\(_2\) due to -Q.
(iii) electric field intensity E.
(c) What is meant by dielectric substance?
(ii) List the factors which determine the capacitance of a parallel plate capacitor and state the effect each of them has on the capacitance.The diagram above represents a section of a circuit. Calculate the effective capacitance in the section.
(iii)
The diagram above represents a section of a circuit. Calculate the effective capacitance in the section.
(a)(i) Electric field intensity at a point is the electric force experienced per unit positive charge placed at that point, \( E = \dfrac{F}{Q} \), measured in \( \text{N C}^{-1} \) (or \( \text{V m}^{-1} \)). It is a vector quantity.
(a)(ii) Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point against the electric field, \( V = \dfrac{W}{Q} \), measured in volts (\( \text{J C}^{-1} \)). It is a scalar quantity.
(b) The charges \(+Q\) and \(-Q\) are collinear, with the rest charge at \(P\) lying on the same line midway between them. Taking the rest charge as positive:
- (i) Force \(F_1\) due to \(+Q\): like charges repel, so the arrow points away from \(+Q\), that is from \(P\) directed towards \(-Q\).
- (ii) Force \(F_2\) due to \(-Q\): unlike charges attract, so the arrow also points from \(P\) directed towards \(-Q\).
- (iii) Resultant field intensity \(E\): both forces act in the same direction along the line, so \(E\) is directed from the \(+Q\) side towards the \(-Q\) side, with magnitude equal to the sum of the two contributions.
(c)(i) A dielectric substance is an insulating (non-conducting) material placed between the plates of a capacitor; it becomes polarised in the field and increases the capacitance.
(c)(ii) Factors determining the capacitance of a parallel-plate capacitor \(\left(C = \dfrac{K\,\varepsilon_0 A}{d}\right)\):
| Factor | Effect on capacitance |
|---|---|
| Common (overlap) area of the plates, \(A\) | Capacitance increases as the area increases (directly proportional). |
| Distance between the plates, \(d\) | Capacitance decreases as the separation increases (inversely proportional). |
| Nature of the dielectric (dielectric constant \(K\)) | Capacitance increases as the dielectric constant increases. |
(c)(iii) Effective capacitance of the section. \(C_2\) and \(C_3\) are in parallel:
\[ C_{2,3} = C_2 + C_3 = 20 + 20 = 40\ \mu\text{F}. \]
This combination is in series with \(C_1 = 40\ \mu\text{F}\):
\[ \frac{1}{C_T} = \frac{1}{C_{2,3}} + \frac{1}{C_1} \quad\Rightarrow\quad C_T = \frac{C_{2,3}\times C_1}{C_{2,3}+C_1} = \frac{40\times 40}{40+40} = 20\ \mu\text{F}. \]
The effective capacitance of the section is \(20\ \mu\text{F}\).
Antwortdetails
(a)(i) Electric field intensity at a point is the electric force experienced per unit positive charge placed at that point, \( E = \dfrac{F}{Q} \), measured in \( \text{N C}^{-1} \) (or \( \text{V m}^{-1} \)). It is a vector quantity.
(a)(ii) Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point against the electric field, \( V = \dfrac{W}{Q} \), measured in volts (\( \text{J C}^{-1} \)). It is a scalar quantity.
(b) The charges \(+Q\) and \(-Q\) are collinear, with the rest charge at \(P\) lying on the same line midway between them. Taking the rest charge as positive:
- (i) Force \(F_1\) due to \(+Q\): like charges repel, so the arrow points away from \(+Q\), that is from \(P\) directed towards \(-Q\).
- (ii) Force \(F_2\) due to \(-Q\): unlike charges attract, so the arrow also points from \(P\) directed towards \(-Q\).
- (iii) Resultant field intensity \(E\): both forces act in the same direction along the line, so \(E\) is directed from the \(+Q\) side towards the \(-Q\) side, with magnitude equal to the sum of the two contributions.
(c)(i) A dielectric substance is an insulating (non-conducting) material placed between the plates of a capacitor; it becomes polarised in the field and increases the capacitance.
(c)(ii) Factors determining the capacitance of a parallel-plate capacitor \(\left(C = \dfrac{K\,\varepsilon_0 A}{d}\right)\):
| Factor | Effect on capacitance |
|---|---|
| Common (overlap) area of the plates, \(A\) | Capacitance increases as the area increases (directly proportional). |
| Distance between the plates, \(d\) | Capacitance decreases as the separation increases (inversely proportional). |
| Nature of the dielectric (dielectric constant \(K\)) | Capacitance increases as the dielectric constant increases. |
(c)(iii) Effective capacitance of the section. \(C_2\) and \(C_3\) are in parallel:
\[ C_{2,3} = C_2 + C_3 = 20 + 20 = 40\ \mu\text{F}. \]
This combination is in series with \(C_1 = 40\ \mu\text{F}\):
\[ \frac{1}{C_T} = \frac{1}{C_{2,3}} + \frac{1}{C_1} \quad\Rightarrow\quad C_T = \frac{C_{2,3}\times C_1}{C_{2,3}+C_1} = \frac{40\times 40}{40+40} = 20\ \mu\text{F}. \]
The effective capacitance of the section is \(20\ \mu\text{F}\).
Frage 62 Bericht
What is a projectile? Give four examples of projectiles in everyday life
Projectile
A projectile is any object that is given an initial velocity and then moves freely under the action of gravity alone (after release, no driving force acts on it apart from its weight), following a curved (parabolic) path.
Four everyday examples of projectiles
- A ball kicked or thrown into the air (e.g. a football or cricket ball).
- A bullet fired from a gun.
- An arrow shot from a bow.
- A stone thrown horizontally or at an angle; water spurting from a hose; a long-jumper in flight.
Antwortdetails
Projectile
A projectile is any object that is given an initial velocity and then moves freely under the action of gravity alone (after release, no driving force acts on it apart from its weight), following a curved (parabolic) path.
Four everyday examples of projectiles
- A ball kicked or thrown into the air (e.g. a football or cricket ball).
- A bullet fired from a gun.
- An arrow shot from a bow.
- A stone thrown horizontally or at an angle; water spurting from a hose; a long-jumper in flight.
Frage 63 Bericht
(a) Define angle of contact
(b) Draw sketches to show angles of contact for a capillary tube dipped vertically in (i) water; (ii) mercury.
(a) Angle of contact
The angle of contact for a liquid in contact with a solid is the angle, measured through the liquid, between the solid surface (the wall of the container) and the tangent to the liquid surface (the meniscus) at the point where the liquid, the solid and the air meet.
Its value shows whether the liquid wets the solid or not:
- If the angle of contact is acute (\(\theta < 90^\circ\)) the liquid wets the surface, the adhesive forces between liquid and solid being stronger than the cohesive forces within the liquid.
- If the angle of contact is obtuse (\(\theta > 90^\circ\)) the liquid does not wet the surface, the cohesive forces being stronger than the adhesive forces.
(b) Sketches of the angles of contact
(i) Water in a clean glass capillary tube. Water wets glass because adhesion (glass–water) exceeds cohesion (water–water). The liquid therefore climbs the walls, giving a concave meniscus that curves upward at the edges, and the water rises in the tube. The tangent to the meniscus at the wall lies almost along the glass, so the angle of contact is acute, close to \(0^\circ\) for pure water on clean glass.
(ii) Mercury in a glass capillary tube. Mercury does not wet glass because cohesion (mercury–mercury) exceeds adhesion (glass–mercury). The liquid is pulled away from the walls, giving a convex (dome-shaped) meniscus, and the mercury is depressed below the outside level. The tangent to the meniscus at the wall turns steeply away from the glass, so the angle of contact is obtuse, about \(140^\circ\).
Antwortdetails
(a) Angle of contact
The angle of contact for a liquid in contact with a solid is the angle, measured through the liquid, between the solid surface (the wall of the container) and the tangent to the liquid surface (the meniscus) at the point where the liquid, the solid and the air meet.
Its value shows whether the liquid wets the solid or not:
- If the angle of contact is acute (\(\theta < 90^\circ\)) the liquid wets the surface, the adhesive forces between liquid and solid being stronger than the cohesive forces within the liquid.
- If the angle of contact is obtuse (\(\theta > 90^\circ\)) the liquid does not wet the surface, the cohesive forces being stronger than the adhesive forces.
(b) Sketches of the angles of contact
(i) Water in a clean glass capillary tube. Water wets glass because adhesion (glass–water) exceeds cohesion (water–water). The liquid therefore climbs the walls, giving a concave meniscus that curves upward at the edges, and the water rises in the tube. The tangent to the meniscus at the wall lies almost along the glass, so the angle of contact is acute, close to \(0^\circ\) for pure water on clean glass.
(ii) Mercury in a glass capillary tube. Mercury does not wet glass because cohesion (mercury–mercury) exceeds adhesion (glass–mercury). The liquid is pulled away from the walls, giving a convex (dome-shaped) meniscus, and the mercury is depressed below the outside level. The tangent to the meniscus at the wall turns steeply away from the glass, so the angle of contact is obtuse, about \(140^\circ\).
Frage 64 Bericht
State any three properties of matter which are common to all substances.
Properties common to all substances (matter)
- It has mass.
- It occupies space (has volume).
- It is made up of tiny particles (molecules/atoms) and has weight in a gravitational field.
Antwortdetails
Properties common to all substances (matter)
- It has mass.
- It occupies space (has volume).
- It is made up of tiny particles (molecules/atoms) and has weight in a gravitational field.
Frage 65 Bericht
An electron of mass 9. 1 x 10\(^{-31}\) kg moves with a velocity of 4.2 x 10 ms\(^{-1}\) between the cathode and anode of an x-ray tube. Calculate the wavelength. (Take Planck's constant h = 6.6 x 10\(^{-34}\) J S)
The wavelength associated with a moving electron is the de Broglie wavelength
\[ \lambda = \frac{h}{mv}, \]where \(h = 6.6\times10^{-34}\ \text{J s}\), \(m = 9.1\times10^{-31}\ \text{kg}\) and \(v = 4.2\times10^{7}\ \text{m s}^{-1}\) (the electron speed in an x-ray tube).
\[ mv = 9.1\times10^{-31} \times 4.2\times10^{7} = 3.82\times10^{-23}\ \text{kg m s}^{-1}. \] \[ \lambda = \frac{6.6\times10^{-34}}{3.82\times10^{-23}} = 1.73\times10^{-11}\ \text{m}. \]Wavelength \(\approx 1.7\times10^{-11}\ \text{m}\) (about \(0.017\ \text{nm}\)), which lies in the X-ray region.
Antwortdetails
The wavelength associated with a moving electron is the de Broglie wavelength
\[ \lambda = \frac{h}{mv}, \]where \(h = 6.6\times10^{-34}\ \text{J s}\), \(m = 9.1\times10^{-31}\ \text{kg}\) and \(v = 4.2\times10^{7}\ \text{m s}^{-1}\) (the electron speed in an x-ray tube).
\[ mv = 9.1\times10^{-31} \times 4.2\times10^{7} = 3.82\times10^{-23}\ \text{kg m s}^{-1}. \] \[ \lambda = \frac{6.6\times10^{-34}}{3.82\times10^{-23}} = 1.73\times10^{-11}\ \text{m}. \]Wavelength \(\approx 1.7\times10^{-11}\ \text{m}\) (about \(0.017\ \text{nm}\)), which lies in the X-ray region.
Frage 66 Bericht
(a)(i) Define the term linear momentum.
(ii) State the law of conservation of linear momentum.
(b) A ball P of mass 0.25 kg, loses one-third of its velocity when it makes a head on collision with an identical ball Q at rest. After the collision, Q moves off with a speed of 2ms\(^{-1}\) in the original direction of P. Calculate the initial velocity of R
(c)(i) State Newton's second law of motion.
(ii) Show that F = ma where F is the magnitude of the force acting on a body of mass m to give it an acceleration of magnitude a.
(iii) The engine of a vehicle moves it forward with a force of 9600 N against a resistive force of 2200 N. If the mass of the vehicle is 3400 kg, calculate the acceleration produced.
(a)(i) Linear momentum is the product of the mass of a body and its velocity, \(p = mv\). It is a vector quantity with SI unit \(\text{kg m s}^{-1}\).
(a)(ii) Law of conservation of linear momentum: In a closed system on which no external resultant force acts, the total linear momentum before an interaction (e.g. a collision) equals the total linear momentum after it.
(b) Let the initial velocity of P be \(u\). Both balls have equal mass \(m = 0.25\ \text{kg}\). P loses one-third of its velocity, so after collision P moves with \(v_P = u - \tfrac{1}{3}u = \tfrac{2}{3}u\). Q starts at rest and moves off at \(v_Q = 2\ \text{m s}^{-1}\).
Conservation of momentum (mass cancels):
\[ u = \tfrac{2}{3}u + 2 \;\Rightarrow\; \tfrac{1}{3}u = 2 \;\Rightarrow\; u = 6\ \text{m s}^{-1}. \]Initial velocity of P \(= 6\ \text{m s}^{-1}\).
(c)(i) Newton's second law: The rate of change of momentum of a body is directly proportional to the resultant force acting on it and takes place in the direction of that force.
(c)(ii) For a body of mass \(m\) whose velocity changes from \(u\) to \(v\) in time \(t\),
\[ F \propto \frac{mv - mu}{t} = \frac{m(v-u)}{t} = ma, \]since \(a = \dfrac{v-u}{t}\). Writing the proportionality with a constant \(k\), \(F = kma\); the newton is defined so that \(k = 1\), giving \(F = ma\).
(c)(iii) Net (resultant) force \(= 9600 - 2200 = 7400\ \text{N}\). Acceleration
\[ a = \frac{F}{m} = \frac{7400}{3400} = 2.18\ \text{m s}^{-2}. \]Antwortdetails
(a)(i) Linear momentum is the product of the mass of a body and its velocity, \(p = mv\). It is a vector quantity with SI unit \(\text{kg m s}^{-1}\).
(a)(ii) Law of conservation of linear momentum: In a closed system on which no external resultant force acts, the total linear momentum before an interaction (e.g. a collision) equals the total linear momentum after it.
(b) Let the initial velocity of P be \(u\). Both balls have equal mass \(m = 0.25\ \text{kg}\). P loses one-third of its velocity, so after collision P moves with \(v_P = u - \tfrac{1}{3}u = \tfrac{2}{3}u\). Q starts at rest and moves off at \(v_Q = 2\ \text{m s}^{-1}\).
Conservation of momentum (mass cancels):
\[ u = \tfrac{2}{3}u + 2 \;\Rightarrow\; \tfrac{1}{3}u = 2 \;\Rightarrow\; u = 6\ \text{m s}^{-1}. \]Initial velocity of P \(= 6\ \text{m s}^{-1}\).
(c)(i) Newton's second law: The rate of change of momentum of a body is directly proportional to the resultant force acting on it and takes place in the direction of that force.
(c)(ii) For a body of mass \(m\) whose velocity changes from \(u\) to \(v\) in time \(t\),
\[ F \propto \frac{mv - mu}{t} = \frac{m(v-u)}{t} = ma, \]since \(a = \dfrac{v-u}{t}\). Writing the proportionality with a constant \(k\), \(F = kma\); the newton is defined so that \(k = 1\), giving \(F = ma\).
(c)(iii) Net (resultant) force \(= 9600 - 2200 = 7400\ \text{N}\). Acceleration
\[ a = \frac{F}{m} = \frac{7400}{3400} = 2.18\ \text{m s}^{-2}. \]Frage 67 Bericht
(a)(1) State the energy transformations which take place during the operation of a modern x-ray tube.
(ii) Distinguish between hard and soft x-rays.
(iii) State three uses of x-rays.
(iv) Mention one hazard of over-exposure to x-rays in a radiological laboratory, and indicate any two safety precautions.
(b) A possible fusion reaction is \(^2_1 H + ^2_1 H \to ^3_1H + ^1_1H + Q\)
where Q is the energy released as a result of the reaction. If Q = 4.03 MeV, calculate the atomic mass of \(^3_1H\) in atomic mass units. (\(^2_1 H = 2.01410 U; ^1_1H = 1.00783 U; 1U = 931 MeV\))
(a)(i) Energy transformations in an x-ray tube: electrical energy at the cathode heats the filament (electrical → heat → light), the emitted electrons gain kinetic energy in the accelerating field (electrical → kinetic), and on striking the target this kinetic energy is converted mainly to heat with a small fraction to X-rays (kinetic → heat + X-ray/electromagnetic energy).
(a)(ii) Hard vs soft X-rays: Hard X-rays have very short wavelength, high frequency and high penetrating power (produced by high tube voltage); soft X-rays have longer wavelength, lower frequency and low penetrating power (produced by lower tube voltage).
(a)(iii) Three uses: medical diagnosis (imaging bones/fractures); radiotherapy (treatment of cancer); detection of flaws in metals and inspection of luggage/crystals (crystallography).
(a)(iv) Hazard: over-exposure damages living tissue and can cause cancer or radiation burns. Precautions: shield operators with lead screens/aprons, and limit exposure time / keep a safe distance.
(b) For \(^2_1H + {}^2_1H \to {}^3_1H + {}^1_1H + Q\), the mass defect equals the energy released:
\[ \Delta m = \frac{Q}{931} = \frac{4.03}{931} = 4.33\times10^{-3}\ \text{u}. \]Mass balance: \((2\times2.01410) - \big[m(^3_1H) + 1.00783\big] = \Delta m\).
\[ 4.02820 - 1.00783 - m(^3_1H) = 4.33\times10^{-3}, \] \[ m(^3_1H) = 3.02037 - 0.00433 = 3.01604\ \text{u}. \]Atomic mass of \(^3_1H \approx 3.016\ \text{u}\).
Antwortdetails
(a)(i) Energy transformations in an x-ray tube: electrical energy at the cathode heats the filament (electrical → heat → light), the emitted electrons gain kinetic energy in the accelerating field (electrical → kinetic), and on striking the target this kinetic energy is converted mainly to heat with a small fraction to X-rays (kinetic → heat + X-ray/electromagnetic energy).
(a)(ii) Hard vs soft X-rays: Hard X-rays have very short wavelength, high frequency and high penetrating power (produced by high tube voltage); soft X-rays have longer wavelength, lower frequency and low penetrating power (produced by lower tube voltage).
(a)(iii) Three uses: medical diagnosis (imaging bones/fractures); radiotherapy (treatment of cancer); detection of flaws in metals and inspection of luggage/crystals (crystallography).
(a)(iv) Hazard: over-exposure damages living tissue and can cause cancer or radiation burns. Precautions: shield operators with lead screens/aprons, and limit exposure time / keep a safe distance.
(b) For \(^2_1H + {}^2_1H \to {}^3_1H + {}^1_1H + Q\), the mass defect equals the energy released:
\[ \Delta m = \frac{Q}{931} = \frac{4.03}{931} = 4.33\times10^{-3}\ \text{u}. \]Mass balance: \((2\times2.01410) - \big[m(^3_1H) + 1.00783\big] = \Delta m\).
\[ 4.02820 - 1.00783 - m(^3_1H) = 4.33\times10^{-3}, \] \[ m(^3_1H) = 3.02037 - 0.00433 = 3.01604\ \text{u}. \]Atomic mass of \(^3_1H \approx 3.016\ \text{u}\).
