WAEC SSCE - Physics - 2001
Question 1 Report
Humidity is used to describe the amount of
Answer Details
Humidity is used to describe the amount of water vapour present in the atmosphere. It is the measure of the water vapour content in the air, which can affect how comfortable we feel in a particular environment. Humidity can be measured in various ways, including absolute humidity, relative humidity, and specific humidity. Knowing the humidity level is important for a range of applications, from weather forecasting to indoor comfort control.
Question 2 Report
The magnitude of the gravitational force between two particles 0.10m apart is 10N. If the distance between them is increased to 0.20m, calculate the magnitude of the new force
Answer Details
The gravitational force between two particles is given by the formula F = Gm1m2/r2, where G is the gravitational constant, m1 and m2 are the masses of the particles, and r is the distance between them. In this case, the distance between the particles is initially 0.10m and the force is 10N. If the distance is increased to 0.20m, we can use the formula to calculate the new force. Let's assume that the masses of the particles remain constant. The distance between them is doubled, so the new distance is 2r = 0.20m. Plugging this into the formula and solving for the new force, we get: F' = Gm1m2/(2r)2 = Gm1m2/4r2 Since the masses of the particles and the gravitational constant are constants, the only thing that has changed is the distance between the particles. Therefore, the new force is proportional to 1/r2. Since r has doubled, the new force will be (1/2)2 = 1/4 of the original force. So, the new force is: F' = (1/4)F = (1/4)10N = 2.5N Therefore, the magnitude of the new force is 2.5N. Option (C) is the correct answer.
Question 3 Report
The diagram above shows the wire of a sonometer box set into resonance by a vibrating tuning fork of frequency f, when the wire is stretched by a tension T, the resonating length of the wire is I while the wavelength of the sound produced is \(\lambda\). Which of the following statements is not correct? The wave produced on the wire is
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Question 5 Report
The material used for constructing the core of an electromagnet is
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The material used for constructing the core of an electromagnet is iron. An electromagnet is made by wrapping a coil of wire around a ferromagnetic core. When electric current flows through the wire, it creates a magnetic field around the wire. The ferromagnetic core helps to increase the strength of the magnetic field created by the current flowing through the wire. Iron is a good material for the core of an electromagnet because it is a ferromagnetic material that is easily magnetized and demagnetized. It also has a high magnetic permeability, which means it can conduct magnetic flux effectively.
Question 6 Report
Which of the following physical qualities affects the saturated vapour pressure of a liquid ?
Answer Details
The physical quality that affects the saturated vapour pressure of a liquid is temperature. Saturated vapour pressure is the pressure exerted by the vapour present in a system at equilibrium with its condensed phase, usually a liquid. As temperature increases, the average kinetic energy of the liquid molecules increases, causing more molecules to escape from the liquid phase and enter the gas phase. This leads to an increase in the number of gas molecules, which in turn results in an increase in pressure. As a result, the saturated vapour pressure of a liquid increases with increasing temperature, and decreases with decreasing temperature. Therefore, temperature is the correct answer. The other options - volume, mass, and density - do not directly affect the saturated vapour pressure of a liquid.
Question 7 Report
A piece of cloth appears green in sunlight. When held in red light, it will appear
Answer Details
When an object appears a certain color, it is because it reflects that particular color of light and absorbs all other colors. Therefore, if a piece of cloth appears green in sunlight, it means that it reflects green light and absorbs all other colors. When the same piece of cloth is held in red light, it will appear black. This is because the red light being shone on the cloth contains only red light waves, and the cloth will absorb all the red light, as it does not reflect red light. Therefore, it appears black, as there is no reflected color for our eyes to see.
Question 8 Report
The Earth's magnetic equator passes through Jos in Nigeria. At Jos. the
Answer Details
The Earth's magnetic field has a north and south pole, and it is tilted with respect to the Earth's rotational axis. This means that at different points on the Earth's surface, the magnetic field has different characteristics. The magnetic equator is an imaginary line that circles the Earth, and it is the point where the magnetic field is horizontal. At Jos in Nigeria, the magnetic equator passes through, and this means that the magnetic field is horizontal at this point. Therefore, the angle of dip (the angle between the magnetic field lines and the Earth's surface) is zero.
Question 9 Report
A nitrogen nucleus bombarded with an alpha particle produces an oxygen nucleus and a proton. The nuclear reaction for this process is \(^4_2H + ^{14}_2N \to ^{17}_{8}O + ^1_1H + Q\). Which of the following statements about the reaction is not correct?
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Question 10 Report
The amplitude of the motion of a body performing simple harmonic motion decreases with time because
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Question 11 Report
In the circuit diagram above. Calculate the energy stored in the inductor at resonance
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Question 12 Report
The diagram above shows the wire of a sonometer box set into resonance by a vibrating tuning fork of frequency f, when the wire is stretched by a tension T, the resonating length of the wire is I while the wavelength of the sound produced is \(\lambda\). Which of the following statements about the frequency of vibration of the wire is not correct?
Answer Details
Question 14 Report
Surface waves travelling in deep water at 15ms-1 are incident at a shallow water boundary. If the angles of incidence and refraction are 45o and 30o respectively. Calculate the speed of the waves in shallow water
Answer Details
When surface waves travelling in deep water meet a shallow water boundary, they experience a change in speed and direction. This is known as refraction. The speed of the waves is dependent on the medium they are travelling through, so when they enter shallow water, their speed will change. We can use Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the velocities in the two media. In this case, we have: \begin{align*} \frac{\sin i}{\sin r} &= \frac{v_i}{v_r} \\ \frac{\sin 45}{\sin 30} &= \frac{15}{v_r} \\ \frac{\sqrt{2}}{0.5} &= \frac{15}{v_r} \\ v_r &= \frac{15}{\frac{\sqrt{2}}{0.5}} \\ v_r &= 10.6\text{ms}^{-1} \end{align*} Therefore, the speed of the waves in shallow water is approximately 10.6ms-1. The correct option is (c) 10.6ms-1.
Question 15 Report
The volume of an ideal gas at a pressure of 77cmHg and temperature 60oC is 24cm3. If the temperature and pressure are increased to 98oC
and 81cmhg respectively, calculate the new volume of the gas
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Question 16 Report
Electrical resistance is a property of an electrical energy to be converted into
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Electrical resistance is a property of a material to oppose the flow of electrical current through it. When electrical current flows through a conductor, some of the electrical energy is converted into other forms of energy. One of the forms of energy that can be generated is heat energy, which is the result of collisions between the electrons and the atoms in the material. This is known as Joule heating, and the amount of heat generated is proportional to the resistance of the material and the square of the current flowing through it. Therefore, when electrical energy flows through a material with resistance, some of the energy is converted into heat energy due to the resistance of the material.
Question 18 Report
A screw jack with a Tommy bar of length 12cm is used to raise a car through a vertical height of 25cm by turning the Tommy bar through 50 revolutions. Calculate the approximate velocity ratio of the jack [\(\pi = 3.142\)]
Answer Details
A screw jack is a simple machine used to lift heavy loads with little effort. The velocity ratio of a screw jack is defined as the ratio of the distance moved by the effort (Tommy bar) to the distance moved by the load (car). In this case, the car is raised through a vertical height of 25cm by turning the Tommy bar through 50 revolutions. The distance moved by the effort is the circumference of the circle traced out by the end of the Tommy bar, which is given by: circumference = 2πr = 2 x 3.142 x 0.12m = 0.754m (approx.) Since the Tommy bar is turned through 50 revolutions, the total distance moved by the effort is: distance moved by effort = 0.754m x 50 = 37.7m (approx.) Therefore, the velocity ratio is: velocity ratio = distance moved by effort / distance moved by load = 37.7m / 0.25m = 150.8 (approx.) So the approximate velocity ratio of the jack is 151, which is closest to.
Question 19 Report
Which of the following physical processes cannot be explained by the molecular theory of matter?
Answer Details
The physical processes of evaporation, thermal conduction, radiation of heat, and convectional current of fluids can all be explained by the molecular theory of matter. The theory explains that all matter is made up of tiny particles (atoms or molecules) which are constantly in motion. The motion of these particles can cause heat transfer through conduction, convection, and radiation. Evaporation is a process where some of the particles at the surface of a liquid have enough energy to escape and become a gas. Therefore, all the options listed can be explained by the molecular theory of matter.
Question 20 Report
Which of the following statements about elastic collision is correct?
Answer Details
In an elastic collision, both kinetic energy and momentum are conserved. This means that the total kinetic energy before the collision is equal to the total kinetic energy after the collision, and the total momentum before the collision is equal to the total momentum after the collision. There is no loss of kinetic energy or momentum in an elastic collision. Therefore, the statement "both kinetic energy and momentum are conserved" is correct. The other statements are incorrect as they imply a loss of either kinetic energy or momentum in an elastic collision, which is not the case.
Question 21 Report
An object weight 10N immersed in a liquid displaces a quantity of the liquid. If the liquid displaced weighs 6N, determine the upthrust of the object
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Question 22 Report
Convex spherical mirrors are preferred to plane mirrors as driving mirrors because
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Convex spherical mirrors are preferred to plane mirrors as driving mirrors because they provide a wider field of view. Convex mirrors are curved outward and the reflected light rays diverge, resulting in a smaller image that appears farther away. This makes it easier for drivers to see objects behind them and in their blind spots. The wider field of view provided by convex mirrors is particularly useful for large vehicles such as buses and trucks, which have larger blind spots than smaller cars.
Question 23 Report
If the frequency of the tuning fork at resonance is 256Hz, calculate the frequency of vibration of the wire when the tension in he wire is increased to 196N
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Question 24 Report
A simple machine with an efficiency of 75% lifts a load of 5000N when a force of 500N is applied to it. Calculate the velocity ratio of the machine
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Question 25 Report
A block weighing 15N rests on a flat surface and a horizontal force of 3N is exerted on it. Determine the friction force on the block
Answer Details
The force exerted on the block is 3N, and the block is at rest, meaning that the force of friction is equal in magnitude and opposite in direction to the applied force. Therefore, the friction force on the block is 3N. Answer: 3.0N
Question 26 Report
Production of x-ray tube begins with
Question 27 Report
An electric bulb is rated 60W, 220V. Calculate the resistance of its filament when it is operating normally
Answer Details
The power of an electrical appliance can be calculated by the formula P = V\(^2\)/R, where P is the power in watts, V is the voltage in volts, and R is the resistance in ohms. In this case, the bulb is rated 60W and 220V. So, we can rearrange the formula to find the resistance: R = V\(^2\)/P. Substituting the values gives us: R = 220\(^2\)/60. Evaluating this expression gives us: R = 806.7\(\Omega\). Therefore, the resistance of the filament when the bulb is operating normally is 806.7\(\Omega\). Note: It is important to remember that the actual resistance of the filament may vary due to factors such as temperature, age, and manufacturing tolerances.
Question 28 Report
As the plates of a charged variable capacitor are moved closer together, the potential difference between them
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Question 29 Report
Which of the following statements about magnetic lines of force is not correct?
Answer Details
The correct option is "magnetic lines of force are closely packed together at neutral point". This statement is incorrect because neutral points are regions where the magnetic field is zero, and hence, there are no magnetic lines of force in these regions. Magnetic lines of force are an imaginary concept used to represent the direction and intensity of magnetic fields. They are assumed to be continuous and form closed loops, and are said to originate from the north pole and terminate at the south pole of a magnet. They also represent the path along which a free north pole would move in the presence of a magnetic field. Magnetic lines of force do not cross one another, and the presence of magnetic lines of force in a region indicates the presence of a magnetic field.
Question 30 Report
Which of the following equations represents the distance x, travelled by a body moving on a straight road with a constant speed? [The other symbol have their usual meanings]
Answer Details
The equation that represents the distance x, travelled by a body moving on a straight road with a constant speed is "x = Ut". In this equation, "U" represents the initial velocity of the body and "t" represents the time taken to cover the distance "x". Since the body is moving with a constant speed, its velocity remains the same throughout the journey and can be represented by "U". Therefore, the distance covered by the body can be calculated by multiplying the velocity "U" with the time taken "t", which is given by the equation "x = Ut". The other given equations represent the distance covered by a body in different scenarios like constant acceleration, initial velocity with constant acceleration, etc.
Question 31 Report
Calculate the reactance of the inductor in the circuit diagram shown above [\(\pi = \frac{22}{7}\)]
Question 32 Report
The apparent weight of a body fully 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-2]
Answer Details
The apparent weight of a body in a fluid is equal to the weight of the fluid displaced by the body. Using this principle, we can calculate the volume of the body as follows: Let V be the volume of the body Density of water = 1000kgm-3 Weight of the body in air = 96N Weight of the body in water = 32N We know that: Weight of the body in air - Weight of the body in water = Upthrust = Weight of the water displaced Therefore, 96N - 32N = Weight of the water displaced 64N = Weight of the water displaced Weight of water displaced = Volume of body x Density of water x g 64N = V x 1000kgm-3 x 10ms-2 V = 64N / (1000kgm-3 x 10ms-2) V = 6.4 x 10-3m3 Therefore, the volume of the body is 6.4 x 10-3m3. Answer is correct.
Question 33 Report
Three identical lamps each power 100W are connected in parallel across a potential difference of 250V. Calculate the current in the circuit
Answer Details
When lamps are connected in parallel, each lamp has the same voltage across it. Therefore, the potential difference across each of the three lamps is 250V. The power of each lamp is given as 100W. We know that power (P) is equal to voltage (V) multiplied by current (I), i.e., P = VI. Therefore, the current through each lamp can be calculated as I = P/V = 100/250 = 0.4 A. Since the lamps are connected in parallel, the total current in the circuit is the sum of the currents through each lamp. Hence, the total current in the circuit is 3 times the current through one lamp, which is 3 x 0.4 A = 1.2 A. Therefore, the correct option is 1.2A.
Question 34 Report
Electrons passing through crystals are diffracted because they
Answer Details
Electrons, which are usually considered as particles, can also exhibit wave-like properties, a concept known as wave-particle duality. When electrons are passed through crystals, their wave-like nature allows them to diffract, meaning they bend and scatter as they encounter the atoms in the crystal. This is similar to the way light diffracts when it passes through a narrow opening or encounters an obstacle. Therefore, the correct answer is that electrons passing through crystals are diffracted because they possess wave properties.
Question 35 Report
The amplitude of a particle executing simple harmonic motion in 5cm while its angular frequency is 10 rad s-1. calculate the magnitude of the maximum acceleration of the particle
Answer Details
The maximum acceleration of a particle in simple harmonic motion is given by the product of the square of the angular frequency and the displacement of the particle from the equilibrium position. That is, amax = -ω2A Where, A is the amplitude of the particle, and ω is the angular frequency. Substituting the given values, we get: amax = - (10 rad/s)2 x 0.05 m amax = - 5 m/s2 or 5 ms-2 Since the maximum acceleration is a scalar quantity, we take its magnitude, which is simply 5 ms-2. Therefore, the correct option is 5.00ms-2.
Question 36 Report
Dispersion of white light by a glass prism occurs because
Question 37 Report
a beam PQ pivoted at P carries a load of 80N as shown above. Calculate the effort E, required to keep it horizontal. [neglect the weight of the beam]
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Question 38 Report
Which of the following statements about pressure is not correct? pressure
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Question 39 Report
A spring balance,which is suspended from the roof of a lift, carries a mass of 1kg at its free end. If the lift accelerates upwards at 2.5ms-2, determine the reading on the spring balance. [g = 10ms-2]
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Question 40 Report
An incident ray is reflected normally by a plane mirror onto a screen where it forms a bright spot. The mirror and screen are parallel and 1m apart. If the mirror is rotated through 5o, calculate the displacement of the spot
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Question 41 Report
The time rate of change of momentum is
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The time rate of change of momentum is called "force". Force is a vector quantity that causes an object to accelerate or change its velocity. When a force acts on an object for a certain amount of time, it produces a change in the object's momentum, which is called impulse. Therefore, force can also be defined as the rate of change of momentum, that is, the amount of impulse produced per unit of time.
Question 42 Report
An object is placed 20cm from a lens. If an image is formed on a screen 260cm away from the lens, calculate the magnification of the image
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Question 43 Report
The amount of heat given out or absorbed when a substance changes its state at a constant temperature is known as
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The amount of heat given out or absorbed when a substance changes its state at a constant temperature is known as latent heat. This means that even though the temperature remains constant during the change of state, heat is still being absorbed or released. For example, when ice is melted into water or water is evaporated into steam, there is no change in temperature, but heat is still absorbed by the substance during the change of state. The latent heat is specific to the substance and the change of state, and can be calculated by multiplying the mass of the substance by the specific latent heat of the substance for the particular change of state.
Question 44 Report
A given mass of an ideal gas has a pressure of 500Nm-2 at -13oC. If its volume remains constant. Calculate its pressure at 247oC.
Answer Details
The pressure, volume, and temperature of an ideal gas are related by the ideal gas law, which is given by: P V = n R T where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the absolute temperature. If the volume of the gas remains constant, then the ideal gas law becomes: P = n R T / V Since the mass of the gas is constant, the number of moles n is also constant. Therefore, we can write: P/T = constant This means that the pressure of the gas is directly proportional to its temperature, when the volume is constant. To solve the problem, we can use the above formula to find the constant, and then use it to find the pressure at the new temperature: P1 / T1 = P2 / T2 where P1 is the initial pressure, T1 is the initial temperature, P2 is the final pressure, and T2 is the final temperature. Converting the temperatures to kelvin scale: T1 = 273 - 13 = 260 K T2 = 273 + 247 = 520 K Substituting the values: 500 / 260 = P2 / 520 Solving for P2: P2 = (500 / 260) x 520 = 1000 Nm-2 Therefore, the pressure of the gas at the new temperature is 1000 Nm-2. The correct option is: 1000Nm-2.
Question 45 Report
A resistance thermometer has a resistance of 20\(\Omega\) at 0oC and 85\(\Omega\) at 100oC. If its resistance is 52\(\Omega\)in a medium, calculate the corresponding temperature
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Question 46 Report
(a) Explain how a gas can be made to conduct electricity.
(b) Name the electric charge carriers in gases.
(a) How a gas can be made to conduct electricity
At ordinary pressure a gas is an insulator. It can be made to conduct by ionising it, that is by producing charged particles within it. This is done by applying a very high potential difference across the gas while it is at low pressure (as in a discharge tube), or by heating it strongly, or by exposing it to X-rays, ultraviolet or radioactive radiation. The high field/energy strips electrons from some of the gas atoms, producing free electrons and positive ions; under the applied field these charges drift to the electrodes and so a current flows.
(b) Electric charge carriers in gases
- Positive ions.
- Electrons (together with negative ions).
Answer Details
(a) How a gas can be made to conduct electricity
At ordinary pressure a gas is an insulator. It can be made to conduct by ionising it, that is by producing charged particles within it. This is done by applying a very high potential difference across the gas while it is at low pressure (as in a discharge tube), or by heating it strongly, or by exposing it to X-rays, ultraviolet or radioactive radiation. The high field/energy strips electrons from some of the gas atoms, producing free electrons and positive ions; under the applied field these charges drift to the electrodes and so a current flows.
(b) Electric charge carriers in gases
- Positive ions.
- Electrons (together with negative ions).
Question 47 Report
(a)(i) What is an echo? (ii) State two useful applications of echoes.
(iii) Why are the walls, floors and ceilings of a recording studio heavily padded?
(b)(i) Explain timbre and overtones.
(ii) What is resonance?
(c) As a ship approaches a cliff, its siren is sounded and the echo is heard in the ship after 12 seconds. 2.1 minutes later the siren was sounded again and the echo was heard 8 seconds later. If the speed of sound in air is 340 ms\(^{-1}\), calculate the velocity at which the ship was approaching the cliff.
(a)(i) An echo is a reflected sound wave heard distinctly after the original sound, produced when sound is reflected from a hard, distant surface back to the listener.
(a)(ii) Two useful applications:
- Depth sounding (echo sounding/SONAR) to measure the depth of the sea or locate shoals of fish and submarines.
- Detecting flaws in metals and, in medicine, ultrasonic scanning of the body.
(a)(iii) The walls, floors and ceilings of a recording studio are heavily padded with soft, porous material to absorb sound and prevent reflection (echoes and reverberation), so that only clear, direct sound is recorded.
(b)(i) Timbre (quality) is the property of a musical note that lets the ear distinguish two notes of the same pitch and loudness from different sources; it depends on the number and relative strength of the overtones present. Overtones are the higher frequencies (above the fundamental) that accompany a note.
(b)(ii) Resonance is the condition in which a body is set into vibration with large amplitude by a periodic force whose frequency equals the natural frequency of the body.
(c) Velocity of the ship
Distance of the cliff at the first sounding \(= \tfrac{1}{2}(340)(12) = 2040\,\text{m}\).
Distance at the second sounding \(= \tfrac{1}{2}(340)(8) = 1360\,\text{m}\).
Distance travelled by the ship between the two soundings:
\[ 2040 - 1360 = 680\,\text{m} \]Time between soundings \(= 2.1\,\text{min} = 126\,\text{s}\). Hence:
\[ v = \frac{680}{126} \approx 5.4\,\text{ms}^{-1} \]The ship approaches the cliff at about \(5.4\,\text{ms}^{-1}\).
Answer Details
(a)(i) An echo is a reflected sound wave heard distinctly after the original sound, produced when sound is reflected from a hard, distant surface back to the listener.
(a)(ii) Two useful applications:
- Depth sounding (echo sounding/SONAR) to measure the depth of the sea or locate shoals of fish and submarines.
- Detecting flaws in metals and, in medicine, ultrasonic scanning of the body.
(a)(iii) The walls, floors and ceilings of a recording studio are heavily padded with soft, porous material to absorb sound and prevent reflection (echoes and reverberation), so that only clear, direct sound is recorded.
(b)(i) Timbre (quality) is the property of a musical note that lets the ear distinguish two notes of the same pitch and loudness from different sources; it depends on the number and relative strength of the overtones present. Overtones are the higher frequencies (above the fundamental) that accompany a note.
(b)(ii) Resonance is the condition in which a body is set into vibration with large amplitude by a periodic force whose frequency equals the natural frequency of the body.
(c) Velocity of the ship
Distance of the cliff at the first sounding \(= \tfrac{1}{2}(340)(12) = 2040\,\text{m}\).
Distance at the second sounding \(= \tfrac{1}{2}(340)(8) = 1360\,\text{m}\).
Distance travelled by the ship between the two soundings:
\[ 2040 - 1360 = 680\,\text{m} \]Time between soundings \(= 2.1\,\text{min} = 126\,\text{s}\). Hence:
\[ v = \frac{680}{126} \approx 5.4\,\text{ms}^{-1} \]The ship approaches the cliff at about \(5.4\,\text{ms}^{-1}\).
Question 48 Report
(a) Define specific heat capacity.
(b)(i) With the aid of a labelled diagram, describe an experiment to determine the specific heat capacity of copper using a copper ball.
(ii) State two precautions necessary to obtain accurate results.
(c) A piece of copper block of mass 24 g at 230°C is placed in a copper calorimeter of mass 60 g containing 54 g of water at 31°C. Assuming heat losses are negligible, calculate the final steady temperature of the mixture. [specific heat capacity of water = 4200 J kg\(^{-1}\) K\(^{-1}\)] [specific heat capacity of copper = 400 J kg\(^{-1}\) K\(^{-1}\)]
(a) Specific heat capacity is the quantity of heat required to raise the temperature of unit mass (1 kg) of a substance by one kelvin (1 K). Its unit is \(\text{J kg}^{-1}\,\text{K}^{-1}\).
(b)(i) Experiment to determine the specific heat capacity of copper (method of mixtures)
Procedure:
- Weigh the empty copper calorimeter together with its stirrer; record the mass \(m_c\).
- Half-fill the calorimeter with water, weigh again, and obtain the mass of water \(m_w\) by subtraction. Note the initial temperature of the water and calorimeter, \(\theta_1\).
- Weigh the copper ball; record its mass \(m_b\). Suspend it by a thread inside a beaker of water and heat until the water boils steadily. Record the steady temperature of the ball, \(\theta\) (about \(100^\circ\text{C}\)).
- Shake off the adhering water and quickly transfer the hot ball into the calorimeter. Stir gently and record the highest steady temperature of the mixture, \(\theta_2\).
Since heat losses are negligible, heat lost by the copper ball = heat gained by the water + heat gained by the calorimeter and stirrer:
\[ m_b c_b(\theta-\theta_2) = m_w c_w(\theta_2-\theta_1) + m_c c_b(\theta_2-\theta_1) \]Making \(c_b\), the specific heat capacity of copper, the subject:
\[ c_b = \frac{m_w c_w(\theta_2-\theta_1)}{m_b(\theta-\theta_2) - m_c(\theta_2-\theta_1)} \](b)(ii) Two precautions
- Transfer the hot copper ball to the calorimeter quickly and shake off any adhering water first, to minimise heat loss to the surroundings.
- Lag (insulate) the calorimeter and stir the water gently before reading the highest steady temperature.
(c) Final steady temperature of the mixture
Data: copper block \(m_b=24\text{ g}=0.024\text{ kg}\) at \(230^\circ\text{C}\); calorimeter \(m_c=60\text{ g}=0.060\text{ kg}\); water \(m_w=54\text{ g}=0.054\text{ kg}\) at \(31^\circ\text{C}\); \(c_w=4200\ \text{J kg}^{-1}\text{K}^{-1}\); \(c_b=400\ \text{J kg}^{-1}\text{K}^{-1}\). Let the final steady temperature be \(\theta\).
Heat lost by copper block = heat gained by water + heat gained by copper calorimeter:
\[ (0.024)(400)(230-\theta) = (0.054)(4200)(\theta-31) + (0.060)(400)(\theta-31) \] \[ 9.6(230-\theta) = (226.8 + 24)(\theta-31) = 250.8(\theta-31) \] \[ 2208 - 9.6\theta = 250.8\theta - 7774.8 \] \[ 2208 + 7774.8 = 250.8\theta + 9.6\theta \] \[ 9982.8 = 260.4\,\theta \] \[ \theta = \frac{9982.8}{260.4} = 38.34^\circ\text{C} \]The final steady temperature of the mixture is \(\theta \approx 38.3^\circ\text{C}\).
Answer Details
(a) Specific heat capacity is the quantity of heat required to raise the temperature of unit mass (1 kg) of a substance by one kelvin (1 K). Its unit is \(\text{J kg}^{-1}\,\text{K}^{-1}\).
(b)(i) Experiment to determine the specific heat capacity of copper (method of mixtures)
Procedure:
- Weigh the empty copper calorimeter together with its stirrer; record the mass \(m_c\).
- Half-fill the calorimeter with water, weigh again, and obtain the mass of water \(m_w\) by subtraction. Note the initial temperature of the water and calorimeter, \(\theta_1\).
- Weigh the copper ball; record its mass \(m_b\). Suspend it by a thread inside a beaker of water and heat until the water boils steadily. Record the steady temperature of the ball, \(\theta\) (about \(100^\circ\text{C}\)).
- Shake off the adhering water and quickly transfer the hot ball into the calorimeter. Stir gently and record the highest steady temperature of the mixture, \(\theta_2\).
Since heat losses are negligible, heat lost by the copper ball = heat gained by the water + heat gained by the calorimeter and stirrer:
\[ m_b c_b(\theta-\theta_2) = m_w c_w(\theta_2-\theta_1) + m_c c_b(\theta_2-\theta_1) \]Making \(c_b\), the specific heat capacity of copper, the subject:
\[ c_b = \frac{m_w c_w(\theta_2-\theta_1)}{m_b(\theta-\theta_2) - m_c(\theta_2-\theta_1)} \](b)(ii) Two precautions
- Transfer the hot copper ball to the calorimeter quickly and shake off any adhering water first, to minimise heat loss to the surroundings.
- Lag (insulate) the calorimeter and stir the water gently before reading the highest steady temperature.
(c) Final steady temperature of the mixture
Data: copper block \(m_b=24\text{ g}=0.024\text{ kg}\) at \(230^\circ\text{C}\); calorimeter \(m_c=60\text{ g}=0.060\text{ kg}\); water \(m_w=54\text{ g}=0.054\text{ kg}\) at \(31^\circ\text{C}\); \(c_w=4200\ \text{J kg}^{-1}\text{K}^{-1}\); \(c_b=400\ \text{J kg}^{-1}\text{K}^{-1}\). Let the final steady temperature be \(\theta\).
Heat lost by copper block = heat gained by water + heat gained by copper calorimeter:
\[ (0.024)(400)(230-\theta) = (0.054)(4200)(\theta-31) + (0.060)(400)(\theta-31) \] \[ 9.6(230-\theta) = (226.8 + 24)(\theta-31) = 250.8(\theta-31) \] \[ 2208 - 9.6\theta = 250.8\theta - 7774.8 \] \[ 2208 + 7774.8 = 250.8\theta + 9.6\theta \] \[ 9982.8 = 260.4\,\theta \] \[ \theta = \frac{9982.8}{260.4} = 38.34^\circ\text{C} \]The final steady temperature of the mixture is \(\theta \approx 38.3^\circ\text{C}\).
Question 49 Report
(a) In his first attempt, a long jumper took off from the springboard with a speed of 8 ms\(^{-1}\) at 30° to the horizontal. He makes a second attempt with the same speed at 45° to the horizontal. Given that the expression for the horizontal range of a projectile is \(\frac{v^2 sin \theta}{g}\) where all the symbols have their usual meanings, show that he gains a distance of 0.8576 m in his second attempt.
(b)(i) State Hooke's law of elasticity.
(ii) Describe an experiment to verify Hooke's law.
(iii) State two precautions you would take if you were to perform this experiment in the laboratory.
(c) A spiral spring of natural length 20.00 cm has a scale pan hanging freely in its lower end. When an object of mass 40 g is placed in the pan, its length becomes 21.80 cm. When another object of mass 60g Is placed in the pan, the length becomes 22.05cm. Calculate the mass of the scale pan. [g = 10 ms\(^{-2}\)]
(a) Gain in the long jump
The horizontal range of a projectile is \(R = \dfrac{v^2\sin 2\theta}{g}\). With \(v = 8\,\text{ms}^{-1}\) and \(g = 10\,\text{ms}^{-2}\), \(\dfrac{v^2}{g} = \dfrac{64}{10} = 6.4\,\text{m}\).
First attempt (\(\theta = 30^\circ\)): \(R_1 = 6.4\sin 60^\circ = 6.4(0.8660) = 5.5424\,\text{m}\).
Second attempt (\(\theta = 45^\circ\)): \(R_2 = 6.4\sin 90^\circ = 6.4(1) = 6.4000\,\text{m}\).
\[ R_2 - R_1 = 6.4000 - 5.5424 = 0.8576\,\text{m} \quad\text{(shown)} \](b)(i) Hooke's law: Provided the elastic limit is not exceeded, the extension of an elastic material is directly proportional to the applied force (load).
(b)(ii) Experiment to verify Hooke's law: A spiral spring is clamped vertically with a pointer against a metre rule. The unstretched pointer reading is noted. Known masses are added one at a time; for each load the new pointer reading is taken and the extension found by subtraction. A graph of load against extension is plotted; a straight line through the origin verifies that extension is proportional to load, confirming Hooke's law.
(b)(iii) Two precautions:
- Do not exceed the elastic limit of the spring (avoid overloading so it returns to its original length).
- Take readings with the eye level to the pointer to avoid parallax error; allow the spring to settle before reading.
(c) Mass of the scale pan
Extension is proportional to the total load (pan mass \(m\) plus object). Natural length \(= 20.00\,\text{cm}\).
With 40 g: length \(21.80\,\text{cm}\), extension \(e_1 = 1.80\,\text{cm}\), load \(=(m+40)\).
With 60 g: length \(22.05\,\text{cm}\), extension \(e_2 = 2.05\,\text{cm}\), load \(=(m+60)\).
\[ \frac{m+40}{1.80} = \frac{m+60}{2.05} \] \[ 2.05(m+40) = 1.80(m+60) \] \[ 2.05m + 82 = 1.80m + 108 \;\Rightarrow\; 0.25m = 26 \;\Rightarrow\; m = 104\,\text{g} \]The mass of the scale pan is \(104\,\text{g}\ (0.104\,\text{kg})\).
Answer Details
(a) Gain in the long jump
The horizontal range of a projectile is \(R = \dfrac{v^2\sin 2\theta}{g}\). With \(v = 8\,\text{ms}^{-1}\) and \(g = 10\,\text{ms}^{-2}\), \(\dfrac{v^2}{g} = \dfrac{64}{10} = 6.4\,\text{m}\).
First attempt (\(\theta = 30^\circ\)): \(R_1 = 6.4\sin 60^\circ = 6.4(0.8660) = 5.5424\,\text{m}\).
Second attempt (\(\theta = 45^\circ\)): \(R_2 = 6.4\sin 90^\circ = 6.4(1) = 6.4000\,\text{m}\).
\[ R_2 - R_1 = 6.4000 - 5.5424 = 0.8576\,\text{m} \quad\text{(shown)} \](b)(i) Hooke's law: Provided the elastic limit is not exceeded, the extension of an elastic material is directly proportional to the applied force (load).
(b)(ii) Experiment to verify Hooke's law: A spiral spring is clamped vertically with a pointer against a metre rule. The unstretched pointer reading is noted. Known masses are added one at a time; for each load the new pointer reading is taken and the extension found by subtraction. A graph of load against extension is plotted; a straight line through the origin verifies that extension is proportional to load, confirming Hooke's law.
(b)(iii) Two precautions:
- Do not exceed the elastic limit of the spring (avoid overloading so it returns to its original length).
- Take readings with the eye level to the pointer to avoid parallax error; allow the spring to settle before reading.
(c) Mass of the scale pan
Extension is proportional to the total load (pan mass \(m\) plus object). Natural length \(= 20.00\,\text{cm}\).
With 40 g: length \(21.80\,\text{cm}\), extension \(e_1 = 1.80\,\text{cm}\), load \(=(m+40)\).
With 60 g: length \(22.05\,\text{cm}\), extension \(e_2 = 2.05\,\text{cm}\), load \(=(m+60)\).
\[ \frac{m+40}{1.80} = \frac{m+60}{2.05} \] \[ 2.05(m+40) = 1.80(m+60) \] \[ 2.05m + 82 = 1.80m + 108 \;\Rightarrow\; 0.25m = 26 \;\Rightarrow\; m = 104\,\text{g} \]The mass of the scale pan is \(104\,\text{g}\ (0.104\,\text{kg})\).
Question 50 Report
A ball thrown vertically upward reaches a maximum height of 50 m above the level of projection. Calculate the;
(i) time taken to reach the maximum height,
(ii) speed of the throw. [g = 10 ms\(^{-2}\)]
Given: maximum height \(h = 50\ \text{m}\), \(g = 10\ \text{m s}^{-2}\), and velocity at the top \(= 0\).
(ii) Speed of throw (find first): using \(v^{2} = u^{2} - 2gh\) with \(v = 0\):
\[ 0 = u^{2} - 2(10)(50) \Rightarrow u^{2} = 1000 \] \[ u = \sqrt{1000} = 31.6\ \text{m s}^{-1} \](i) Time to reach maximum height: using \(v = u - gt\) with \(v = 0\):
\[ t = \frac{u}{g} = \frac{31.6}{10} = 3.16\ \text{s} \]The ball takes about 3.16 s to reach the top, having been thrown at about 31.6 m s\(^{-1}\).
Answer Details
Given: maximum height \(h = 50\ \text{m}\), \(g = 10\ \text{m s}^{-2}\), and velocity at the top \(= 0\).
(ii) Speed of throw (find first): using \(v^{2} = u^{2} - 2gh\) with \(v = 0\):
\[ 0 = u^{2} - 2(10)(50) \Rightarrow u^{2} = 1000 \] \[ u = \sqrt{1000} = 31.6\ \text{m s}^{-1} \](i) Time to reach maximum height: using \(v = u - gt\) with \(v = 0\):
\[ t = \frac{u}{g} = \frac{31.6}{10} = 3.16\ \text{s} \]The ball takes about 3.16 s to reach the top, having been thrown at about 31.6 m s\(^{-1}\).
Question 51 Report
(a) What is meant by 'a beam of polarised light?
(b) With the aid of well labelled diagrams, illustrate the action in of a polaroid spectacle on a beam of sunlight.
(a) A beam of polarised light
Light is a transverse wave. In an ordinary (unpolarised) beam the electric-field vibrations occur in all planes perpendicular to the direction in which the light travels. A beam of polarised light is a beam in which these vibrations have been restricted to one single plane perpendicular to the direction of travel. In other words, the vibration of the wave takes place in only one plane (the plane of polarisation), so the light is said to be plane-polarised.
(b) Action of a polaroid spectacle on a beam of sunlight
Sunlight reaching the eye is unpolarised: its vibrations lie in every plane at right angles to the beam. A polaroid lens contains long parallel molecular chains that create a single transmission (polarising) axis. When sunlight strikes the polaroid:
- only the vibration parallel to the transmission axis is allowed through, and
- the vibrations in every other plane are absorbed.
The light that emerges is therefore plane-polarised and much reduced in intensity. This is why polaroid spectacles cut down the dazzling glare of sunlight (especially reflected glare, which is already partly polarised), so the wearer sees more comfortably. The action is illustrated below.
In the diagram, the incoming sunlight vibrates in all directions (shown by the double-headed arrows pointing every way). After passing through the polaroid, whose transmission axis is vertical, only the vertical component of vibration survives, so the beam reaching the eye is plane-polarised in the vertical plane and dimmer than the original sunlight.
Answer Details
(a) A beam of polarised light
Light is a transverse wave. In an ordinary (unpolarised) beam the electric-field vibrations occur in all planes perpendicular to the direction in which the light travels. A beam of polarised light is a beam in which these vibrations have been restricted to one single plane perpendicular to the direction of travel. In other words, the vibration of the wave takes place in only one plane (the plane of polarisation), so the light is said to be plane-polarised.
(b) Action of a polaroid spectacle on a beam of sunlight
Sunlight reaching the eye is unpolarised: its vibrations lie in every plane at right angles to the beam. A polaroid lens contains long parallel molecular chains that create a single transmission (polarising) axis. When sunlight strikes the polaroid:
- only the vibration parallel to the transmission axis is allowed through, and
- the vibrations in every other plane are absorbed.
The light that emerges is therefore plane-polarised and much reduced in intensity. This is why polaroid spectacles cut down the dazzling glare of sunlight (especially reflected glare, which is already partly polarised), so the wearer sees more comfortably. The action is illustrated below.
In the diagram, the incoming sunlight vibrates in all directions (shown by the double-headed arrows pointing every way). After passing through the polaroid, whose transmission axis is vertical, only the vertical component of vibration survives, so the beam reaching the eye is plane-polarised in the vertical plane and dimmer than the original sunlight.
Question 52 Report
(a) State two properties of cathode rays.
(b) Explain how the intensity and energy of cathode rays may be increased.
(a) Two properties of cathode rays
- They travel in straight lines from the cathode and cast sharp shadows of objects placed in their path.
- They are streams of negatively charged particles (electrons), so they are deflected by both electric and magnetic fields (towards the positive plate).
Other acceptable properties: they carry energy and momentum, they produce heat and fluorescence on striking a screen, and they can produce X-rays on hitting a metal target.
(b) Increasing the intensity and energy of cathode rays
Intensity (number of electrons per second) is increased by raising the filament (heater) current. A hotter cathode emits more electrons by thermionic emission, so the beam current and hence the brightness increase.
Energy (speed of each electron) is increased by raising the accelerating potential difference between the cathode and the anode. Each electron gains kinetic energy \(eV\) as it is accelerated, so a larger anode voltage gives faster, more energetic electrons.
Answer Details
(a) Two properties of cathode rays
- They travel in straight lines from the cathode and cast sharp shadows of objects placed in their path.
- They are streams of negatively charged particles (electrons), so they are deflected by both electric and magnetic fields (towards the positive plate).
Other acceptable properties: they carry energy and momentum, they produce heat and fluorescence on striking a screen, and they can produce X-rays on hitting a metal target.
(b) Increasing the intensity and energy of cathode rays
Intensity (number of electrons per second) is increased by raising the filament (heater) current. A hotter cathode emits more electrons by thermionic emission, so the beam current and hence the brightness increase.
Energy (speed of each electron) is increased by raising the accelerating potential difference between the cathode and the anode. Each electron gains kinetic energy \(eV\) as it is accelerated, so a larger anode voltage gives faster, more energetic electrons.
Question 53 Report
(a) When nitrogen (atomic mass = 14, atomic number = 7) is bombarded with neutrons, the collisions result in disintegrations in which alpha particles are produced. Represent this transmutation in a symbolic equation.
(b)(i) How does a radioactive atom differ from a stable one?
(ii) Explain 'half life'.
(iii) A sample of radioactive material has a haft life of 35 days. Calculate the fraction of the original quantity that will remain after 105 days.
(c) Light of wavelength 5.00 x 10\(^{-7}\)m is incident on a material of work function 1.90 eV. Calculate
(i) photon energy.
(ii) kinetic energy of the most energetic photo electron.
(iii) stopping potential [Plancks constant h =6.6 x 10\(^{-34}\)Js] [c= 3.0 x10\(^8\)ms\(^{-2}\), leV= 1.6 x 10\(^{19}\)J]
(a) Transmutation equation
Nitrogen-14 bombarded by a neutron emits an alpha particle:
\[ {}^{14}_{7}\text{N} + {}^{1}_{0}\text{n} \;\rightarrow\; {}^{4}_{2}\text{He} + {}^{11}_{5}\text{B} \](Mass numbers: \(14+1 = 4+11\); atomic numbers: \(7+0 = 2+5\). The residual nucleus is boron-11.)
(b)(i) A radioactive atom has an unstable nucleus that spontaneously disintegrates, emitting radiation (alpha, beta or gamma), whereas a stable atom has a nucleus that does not decay.
(b)(ii) Half life is the time taken for half the atoms (or nuclei) in a given sample of a radioactive material to decay.
(b)(iii) \(105\,\text{days} = 3\times 35\,\text{days} = 3\) half lives. Fraction remaining:
\[ \left(\tfrac{1}{2}\right)^3 = \frac{1}{8} \](c) Photoelectric calculations (\(\lambda = 5.00\times10^{-7}\,\text{m}\), \(W_0 = 1.90\,\text{eV}\)):
(i) Photon energy:
\[ E = \frac{hc}{\lambda} = \frac{(6.6\times10^{-34})(3.0\times10^{8})}{5.0\times10^{-7}} = 3.96\times10^{-19}\,\text{J} \ (\approx 2.48\,\text{eV}) \](ii) Maximum kinetic energy: work function \(W_0 = 1.90\times1.6\times10^{-19} = 3.04\times10^{-19}\,\text{J}\).
\[ E_k = E - W_0 = 3.96\times10^{-19} - 3.04\times10^{-19} = 9.2\times10^{-20}\,\text{J} \ (0.575\,\text{eV}) \](iii) Stopping potential:
\[ V_s = \frac{E_k}{e} = \frac{9.2\times10^{-20}}{1.6\times10^{-19}} = 0.575\,\text{V} \]Answer Details
(a) Transmutation equation
Nitrogen-14 bombarded by a neutron emits an alpha particle:
\[ {}^{14}_{7}\text{N} + {}^{1}_{0}\text{n} \;\rightarrow\; {}^{4}_{2}\text{He} + {}^{11}_{5}\text{B} \](Mass numbers: \(14+1 = 4+11\); atomic numbers: \(7+0 = 2+5\). The residual nucleus is boron-11.)
(b)(i) A radioactive atom has an unstable nucleus that spontaneously disintegrates, emitting radiation (alpha, beta or gamma), whereas a stable atom has a nucleus that does not decay.
(b)(ii) Half life is the time taken for half the atoms (or nuclei) in a given sample of a radioactive material to decay.
(b)(iii) \(105\,\text{days} = 3\times 35\,\text{days} = 3\) half lives. Fraction remaining:
\[ \left(\tfrac{1}{2}\right)^3 = \frac{1}{8} \](c) Photoelectric calculations (\(\lambda = 5.00\times10^{-7}\,\text{m}\), \(W_0 = 1.90\,\text{eV}\)):
(i) Photon energy:
\[ E = \frac{hc}{\lambda} = \frac{(6.6\times10^{-34})(3.0\times10^{8})}{5.0\times10^{-7}} = 3.96\times10^{-19}\,\text{J} \ (\approx 2.48\,\text{eV}) \](ii) Maximum kinetic energy: work function \(W_0 = 1.90\times1.6\times10^{-19} = 3.04\times10^{-19}\,\text{J}\).
\[ E_k = E - W_0 = 3.96\times10^{-19} - 3.04\times10^{-19} = 9.2\times10^{-20}\,\text{J} \ (0.575\,\text{eV}) \](iii) Stopping potential:
\[ V_s = \frac{E_k}{e} = \frac{9.2\times10^{-20}}{1.6\times10^{-19}} = 0.575\,\text{V} \]Question 54 Report
The value of the e.m.f of a voltaic cell, which has dilute tetraoxosulphate (VI) acid as its electrolyte and copper and zinc as its electrodes becomes less with use. Explain this observation and state how it can be corrected.
The fall in e.m.f. with use is caused by two defects of the simple voltaic cell:
- Polarisation. As the cell works, hydrogen gas is liberated and clings to the copper (positive) plate, forming an insulating layer. This layer increases the internal resistance and sets up a back e.m.f., so the useful e.m.f. falls.
- Local action. Impurities (such as carbon and iron) in the commercial zinc plate form tiny local cells with the zinc, so the zinc is eaten away even when no current is being drawn, weakening the cell.
Corrections:
- Polarisation is removed by adding a depolariser (an oxidising agent such as manganese(IV) oxide or potassium dichromate) which oxidises the hydrogen to water.
- Local action is prevented by amalgamating the zinc plate, i.e. coating it with mercury so that the surface presents a clean layer of pure zinc only.
Answer Details
The fall in e.m.f. with use is caused by two defects of the simple voltaic cell:
- Polarisation. As the cell works, hydrogen gas is liberated and clings to the copper (positive) plate, forming an insulating layer. This layer increases the internal resistance and sets up a back e.m.f., so the useful e.m.f. falls.
- Local action. Impurities (such as carbon and iron) in the commercial zinc plate form tiny local cells with the zinc, so the zinc is eaten away even when no current is being drawn, weakening the cell.
Corrections:
- Polarisation is removed by adding a depolariser (an oxidising agent such as manganese(IV) oxide or potassium dichromate) which oxidises the hydrogen to water.
- Local action is prevented by amalgamating the zinc plate, i.e. coating it with mercury so that the surface presents a clean layer of pure zinc only.
Question 55 Report
(a) The mass and wavelength of a moving electron are 9.0 x 10\(^{-31}\) kg and 1.0 x 10\(^{-10}\)m respectively. Calculate the kinetic energy of the electron. [ h = 6.6 x 10\(^{-34}\) Js]
Kinetic energy of the electron from its de Broglie wavelength
The de Broglie relation links wavelength to momentum:
\[ \lambda = \frac{h}{mv} \quad\Rightarrow\quad v = \frac{h}{m\lambda} \]Substituting \(h = 6.6\times10^{-34}\,\text{Js}\), \(m = 9.0\times10^{-31}\,\text{kg}\), \(\lambda = 1.0\times10^{-10}\,\text{m}\):
\[ v = \frac{6.6\times10^{-34}}{(9.0\times10^{-31})(1.0\times10^{-10})} = \frac{6.6\times10^{-34}}{9.0\times10^{-41}} = 7.33\times10^{6}\,\text{ms}^{-1} \]Then the kinetic energy is:
\[ E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(9.0\times10^{-31})(7.33\times10^{6})^2 \] \[ E_k = \tfrac{1}{2}(9.0\times10^{-31})(5.38\times10^{13}) \approx 2.42\times10^{-17}\,\text{J} \]Equivalently, using \(E_k = \dfrac{h^2}{2m\lambda^2}\) gives the same result, \(E_k \approx 2.4\times10^{-17}\,\text{J}\).
Answer Details
Kinetic energy of the electron from its de Broglie wavelength
The de Broglie relation links wavelength to momentum:
\[ \lambda = \frac{h}{mv} \quad\Rightarrow\quad v = \frac{h}{m\lambda} \]Substituting \(h = 6.6\times10^{-34}\,\text{Js}\), \(m = 9.0\times10^{-31}\,\text{kg}\), \(\lambda = 1.0\times10^{-10}\,\text{m}\):
\[ v = \frac{6.6\times10^{-34}}{(9.0\times10^{-31})(1.0\times10^{-10})} = \frac{6.6\times10^{-34}}{9.0\times10^{-41}} = 7.33\times10^{6}\,\text{ms}^{-1} \]Then the kinetic energy is:
\[ E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(9.0\times10^{-31})(7.33\times10^{6})^2 \] \[ E_k = \tfrac{1}{2}(9.0\times10^{-31})(5.38\times10^{13}) \approx 2.42\times10^{-17}\,\text{J} \]Equivalently, using \(E_k = \dfrac{h^2}{2m\lambda^2}\) gives the same result, \(E_k \approx 2.4\times10^{-17}\,\text{J}\).
Question 56 Report
A lead shot is projected from tha ground level with a velocity u at an angle \(\theta\) to the horizontal. Given the time, t for the lead shot to reach its maximum height as; t = \(\frac{u \sin \theta}{g}\) where "g" is the acceleration of free fall due to gravity, show that the greatest height reached by the body is h\(_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g}\)
For the vertical motion of the projectile, the initial vertical velocity is \(u\sin\theta\) and the acceleration due to gravity is \(-g\).
Using
\[ h=(u\sin\theta)t-\frac{1}{2}gt^{2} \]At maximum height,
\[ t=\frac{u\sin\theta}{g} \]Therefore,
\[ h_{\text{max}}=(u\sin\theta)\left(\frac{u\sin\theta}{g}\right) -\frac{1}{2}g\left(\frac{u\sin\theta}{g}\right)^2 \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{g} -\frac{1}{2}g\left(\frac{u^2\sin^2\theta}{g^2}\right) \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{g} -\frac{u^2\sin^2\theta}{2g} \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{2g} \]Hence, the greatest height reached is
\[ \boxed{h_{\text{max}}=\frac{u^2\sin^2\theta}{2g}} \]Answer Details
For the vertical motion of the projectile, the initial vertical velocity is \(u\sin\theta\) and the acceleration due to gravity is \(-g\).
Using
\[ h=(u\sin\theta)t-\frac{1}{2}gt^{2} \]At maximum height,
\[ t=\frac{u\sin\theta}{g} \]Therefore,
\[ h_{\text{max}}=(u\sin\theta)\left(\frac{u\sin\theta}{g}\right) -\frac{1}{2}g\left(\frac{u\sin\theta}{g}\right)^2 \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{g} -\frac{1}{2}g\left(\frac{u^2\sin^2\theta}{g^2}\right) \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{g} -\frac{u^2\sin^2\theta}{2g} \] \[ h_{\text{max}}=\frac{u^2\sin^2\theta}{2g} \]Hence, the greatest height reached is
\[ \boxed{h_{\text{max}}=\frac{u^2\sin^2\theta}{2g}} \]Question 57 Report
(a) What is electrolysis?
(b) A current of 2A is passed through a copper voltameter for 5 minutes. If the electrochemical equivalent of copper is 3.27 x 10\(^{-7}\) kg. Determine the mass of the corper deposited.
(a) What is electrolysis?
Electrolysis is the chemical decomposition of an electrolyte (a molten ionic compound or an aqueous solution of one) caused by the passage of a direct electric current through it. The current makes the positive ions (cations) move to the cathode and the negative ions (anions) move to the anode, where they are discharged and new substances are set free.
(b) Mass of copper deposited
This uses Faraday's first law of electrolysis, which states that the mass \(m\) of a substance liberated at an electrode is proportional to the quantity of charge \(Q\) passed, so that \(m = zQ = zIt\), where \(z\) is the electrochemical equivalent.
Given: current \(I = 2\ \text{A}\); time \(t = 5\ \text{min} = 5 \times 60 = 300\ \text{s}\); electrochemical equivalent \(z = 3.27 \times 10^{-7}\ \text{kg C}^{-1}\).
Working:
\[ m = zIt = (3.27 \times 10^{-7}) \times 2 \times 300 \] \[ m = (3.27 \times 10^{-7}) \times 600 \] \[ m = 1.962 \times 10^{-4}\ \text{kg} \]The mass of copper deposited is \(1.962 \times 10^{-4}\ \text{kg}\), which is about \(0.196\ \text{g}\).
Key point: the time must be converted from minutes to seconds before substituting, because the ampere is one coulomb per second; forgetting to change 5 min into 300 s is the most common error and would give an answer 60 times too small.
Answer Details
(a) What is electrolysis?
Electrolysis is the chemical decomposition of an electrolyte (a molten ionic compound or an aqueous solution of one) caused by the passage of a direct electric current through it. The current makes the positive ions (cations) move to the cathode and the negative ions (anions) move to the anode, where they are discharged and new substances are set free.
(b) Mass of copper deposited
This uses Faraday's first law of electrolysis, which states that the mass \(m\) of a substance liberated at an electrode is proportional to the quantity of charge \(Q\) passed, so that \(m = zQ = zIt\), where \(z\) is the electrochemical equivalent.
Given: current \(I = 2\ \text{A}\); time \(t = 5\ \text{min} = 5 \times 60 = 300\ \text{s}\); electrochemical equivalent \(z = 3.27 \times 10^{-7}\ \text{kg C}^{-1}\).
Working:
\[ m = zIt = (3.27 \times 10^{-7}) \times 2 \times 300 \] \[ m = (3.27 \times 10^{-7}) \times 600 \] \[ m = 1.962 \times 10^{-4}\ \text{kg} \]The mass of copper deposited is \(1.962 \times 10^{-4}\ \text{kg}\), which is about \(0.196\ \text{g}\).
Key point: the time must be converted from minutes to seconds before substituting, because the ampere is one coulomb per second; forgetting to change 5 min into 300 s is the most common error and would give an answer 60 times too small.
Question 58 Report
(a) Distinguish between stress and strain as used in elasticity.
(b) When a force of 40 N is applied to the free end of an elastic cord, an extension of 5cm is produced in the cord. Calculate the work done on the cord.
(a) Stress and strain
Stress is the force acting per unit cross-sectional area of the material:
\[ \text{stress} = \frac{\text{force}}{\text{area}} \quad (\text{unit: N m}^{-2}\ \text{or Pa}) \]Strain is the extension produced per unit original length of the material:
\[ \text{strain} = \frac{\text{extension}}{\text{original length}} \quad (\text{no unit, it is a ratio}) \]Thus stress measures the intensity of the applied load while strain measures the relative deformation it produces.
(b) Work done on the cord
For an elastic cord obeying Hooke's law, the work done equals the elastic potential energy stored, given by the area under the force-extension graph:
\[ W = \tfrac{1}{2}Fe \]With \(F = 40\,\text{N}\) and \(e = 5\,\text{cm} = 0.05\,\text{m}\):
\[ W = \tfrac{1}{2}(40)(0.05) = 1.0\,\text{J} \]Answer Details
(a) Stress and strain
Stress is the force acting per unit cross-sectional area of the material:
\[ \text{stress} = \frac{\text{force}}{\text{area}} \quad (\text{unit: N m}^{-2}\ \text{or Pa}) \]Strain is the extension produced per unit original length of the material:
\[ \text{strain} = \frac{\text{extension}}{\text{original length}} \quad (\text{no unit, it is a ratio}) \]Thus stress measures the intensity of the applied load while strain measures the relative deformation it produces.
(b) Work done on the cord
For an elastic cord obeying Hooke's law, the work done equals the elastic potential energy stored, given by the area under the force-extension graph:
\[ W = \tfrac{1}{2}Fe \]With \(F = 40\,\text{N}\) and \(e = 5\,\text{cm} = 0.05\,\text{m}\):
\[ W = \tfrac{1}{2}(40)(0.05) = 1.0\,\text{J} \]Question 59 Report
The above reprt the graph electron energy against the frequency of the radiation incident on a metal surface. lnterprete the;
(i) slope of the graph;
(ii) intercept, OC;
(iii) intercept, OK.
The graph is a straight line of the maximum kinetic energy \(E\) of the emitted photoelectrons (vertical axis, in joules) against the frequency \(f\) of the incident radiation (horizontal axis, in hertz). The line crosses the frequency axis at the point \(K\) and, when produced backwards, cuts the energy axis below the origin at the point \(C\). The graph is a plot of Einstein's photoelectric equation,
\[ E = hf - hf_0 \]
where \(h\) is Planck's constant and \(f_0\) is the threshold frequency of the metal. Comparing this with the equation of a straight line \(y = mx + c\): \(m = h\) is the gradient and \(c = -hf_0\) is the intercept on the energy axis.
(i) The slope of the graph
Comparing \(E = hf - hf_0\) with \(y = mx + c\), the gradient is
\[ \text{slope} = \frac{\Delta E}{\Delta f} = h \]
So the slope represents Planck's constant, \(h\) \(\left(\approx 6.6\times10^{-34}\,\text{J s}\right)\). It has the same value for every metal.
(ii) The intercept OC
OC is the intercept on the energy axis, obtained when \(f = 0\). From the equation, \(E = -hf_0\), so this intercept is negative and equal in magnitude to the work function \(\varphi\) of the metal:
\[ OC = -hf_0 = -\varphi \]
Thus \(|OC|\) is the minimum energy needed to release an electron from the surface of the metal, i.e. the work function.
(iii) The intercept OK
OK is the intercept on the frequency axis, obtained when the photoelectron energy \(E = 0\). Setting \(0 = hf - hf_0\) gives \(f = f_0\). Therefore OK represents the threshold (cut-off) frequency \(f_0\), the minimum frequency of the incident radiation below which no electrons are emitted from the metal.
Answer Details
The graph is a straight line of the maximum kinetic energy \(E\) of the emitted photoelectrons (vertical axis, in joules) against the frequency \(f\) of the incident radiation (horizontal axis, in hertz). The line crosses the frequency axis at the point \(K\) and, when produced backwards, cuts the energy axis below the origin at the point \(C\). The graph is a plot of Einstein's photoelectric equation,
\[ E = hf - hf_0 \]
where \(h\) is Planck's constant and \(f_0\) is the threshold frequency of the metal. Comparing this with the equation of a straight line \(y = mx + c\): \(m = h\) is the gradient and \(c = -hf_0\) is the intercept on the energy axis.
(i) The slope of the graph
Comparing \(E = hf - hf_0\) with \(y = mx + c\), the gradient is
\[ \text{slope} = \frac{\Delta E}{\Delta f} = h \]
So the slope represents Planck's constant, \(h\) \(\left(\approx 6.6\times10^{-34}\,\text{J s}\right)\). It has the same value for every metal.
(ii) The intercept OC
OC is the intercept on the energy axis, obtained when \(f = 0\). From the equation, \(E = -hf_0\), so this intercept is negative and equal in magnitude to the work function \(\varphi\) of the metal:
\[ OC = -hf_0 = -\varphi \]
Thus \(|OC|\) is the minimum energy needed to release an electron from the surface of the metal, i.e. the work function.
(iii) The intercept OK
OK is the intercept on the frequency axis, obtained when the photoelectron energy \(E = 0\). Setting \(0 = hf - hf_0\) gives \(f = f_0\). Therefore OK represents the threshold (cut-off) frequency \(f_0\), the minimum frequency of the incident radiation below which no electrons are emitted from the metal.
Question 60 Report
(a) Explain the statement the capacitance of a capacitor is 5\(\mu\)F.
(b)(i) State the factors upon which the capacitance of a parallel plate capacitor depend.
(ii) State how the capacitance depends on each of these factors stated in (b)(i).
(c) A series arrangement of three capacitors of values 8uF, 12\(\mu\)F, and 24\(\mu\)F is connected in series with 90-V battery.
(i) Draw an open-circuit diagram for this arrangement.
(ii) Calculate the effective capacitance in the circuit.
(iii) On closed circuit, calculate the charge on each capacitor when fully charged.
(iv) Determine the p.d across the 8\(\mu\)F capacitor.
(a) To say that the capacitance of a capacitor is \(5\,\mu\text{F}\) means that the capacitor stores a charge of \(5\,\mu\text{C}\) \((5\times10^{-6}\,\text{C})\) on each plate for every \(1\,\text{V}\) of potential difference applied across its plates. In other words, the ratio of the charge stored to the potential difference across the plates is \(5\,\mu\text{C V}^{-1}\), since \(C=\dfrac{Q}{V}\).
(b)(i) & (ii) The capacitance of a parallel-plate capacitor is given by \(C=\dfrac{\varepsilon A}{d}\). The factors on which it depends and the form of dependence are:
| Factor | How the capacitance depends on it |
| Common (overlapping) area of the plates, \(A\) | Directly proportional: \(C\propto A\). Increasing the area increases the capacitance. |
| Distance between the plates, \(d\) | Inversely proportional: \(C\propto \dfrac{1}{d}\). Increasing the separation decreases the capacitance. |
| Permittivity (nature) of the dielectric, \(\varepsilon\) | Directly proportional: \(C\propto \varepsilon\). A dielectric of higher permittivity increases the capacitance. |
(c)(i) Open-circuit diagram (switch open) showing the three capacitors in series with the \(90\,\text{V}\) battery:
(c)(ii) Effective capacitance. For capacitors in series the reciprocals add:
\[\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{8}+\frac{1}{12}+\frac{1}{24}=\frac{3+2+1}{24}=\frac{6}{24}=\frac{1}{4}\]\[C=4\,\mu\text{F}\](c)(iii) Charge on each capacitor. In a series arrangement the same charge flows onto every capacitor, and it equals the charge on the equivalent capacitor:
\[Q=CV=4\times10^{-6}\times90=3.6\times10^{-4}\,\text{C}=360\,\mu\text{C}\]Therefore the charge on each of the \(8\,\mu\text{F}\), \(12\,\mu\text{F}\) and \(24\,\mu\text{F}\) capacitors is the same: \(Q=360\,\mu\text{C}\).
(c)(iv) P.d. across the \(8\,\mu\text{F}\) capacitor.
\[V_{8}=\frac{Q}{C_1}=\frac{360\,\mu\text{C}}{8\,\mu\text{F}}=45\,\text{V}\]The potential difference across the \(8\,\mu\text{F}\) capacitor is \(45\,\text{V}\). (As a check: \(V_{12}=360/12=30\,\text{V}\) and \(V_{24}=360/24=15\,\text{V}\); \(45+30+15=90\,\text{V}\), which equals the battery voltage.)
Answer Details
(a) To say that the capacitance of a capacitor is \(5\,\mu\text{F}\) means that the capacitor stores a charge of \(5\,\mu\text{C}\) \((5\times10^{-6}\,\text{C})\) on each plate for every \(1\,\text{V}\) of potential difference applied across its plates. In other words, the ratio of the charge stored to the potential difference across the plates is \(5\,\mu\text{C V}^{-1}\), since \(C=\dfrac{Q}{V}\).
(b)(i) & (ii) The capacitance of a parallel-plate capacitor is given by \(C=\dfrac{\varepsilon A}{d}\). The factors on which it depends and the form of dependence are:
| Factor | How the capacitance depends on it |
| Common (overlapping) area of the plates, \(A\) | Directly proportional: \(C\propto A\). Increasing the area increases the capacitance. |
| Distance between the plates, \(d\) | Inversely proportional: \(C\propto \dfrac{1}{d}\). Increasing the separation decreases the capacitance. |
| Permittivity (nature) of the dielectric, \(\varepsilon\) | Directly proportional: \(C\propto \varepsilon\). A dielectric of higher permittivity increases the capacitance. |
(c)(i) Open-circuit diagram (switch open) showing the three capacitors in series with the \(90\,\text{V}\) battery:
(c)(ii) Effective capacitance. For capacitors in series the reciprocals add:
\[\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{8}+\frac{1}{12}+\frac{1}{24}=\frac{3+2+1}{24}=\frac{6}{24}=\frac{1}{4}\]\[C=4\,\mu\text{F}\](c)(iii) Charge on each capacitor. In a series arrangement the same charge flows onto every capacitor, and it equals the charge on the equivalent capacitor:
\[Q=CV=4\times10^{-6}\times90=3.6\times10^{-4}\,\text{C}=360\,\mu\text{C}\]Therefore the charge on each of the \(8\,\mu\text{F}\), \(12\,\mu\text{F}\) and \(24\,\mu\text{F}\) capacitors is the same: \(Q=360\,\mu\text{C}\).
(c)(iv) P.d. across the \(8\,\mu\text{F}\) capacitor.
\[V_{8}=\frac{Q}{C_1}=\frac{360\,\mu\text{C}}{8\,\mu\text{F}}=45\,\text{V}\]The potential difference across the \(8\,\mu\text{F}\) capacitor is \(45\,\text{V}\). (As a check: \(V_{12}=360/12=30\,\text{V}\) and \(V_{24}=360/24=15\,\text{V}\); \(45+30+15=90\,\text{V}\), which equals the battery voltage.)
