WAEC SSCE - Physics - 2004
Tambaya 1 Rahoto
The main function of the mouth piece of a telephone is that it converts sound energy to
Bayanin Amsa
The main function of the mouthpiece of a telephone is to convert sound energy to electrical energy. When a person speaks into the mouthpiece, the sound waves from their voice cause a diaphragm or microphone in the mouthpiece to vibrate. These vibrations are then converted into electrical signals that can be transmitted through the telephone wires to the receiver at the other end. The electrical signals can then be converted back into sound waves by the receiver, allowing the person on the other end to hear the speaker's voice. Therefore, option B - electrical energy - is the correct answer.
Tambaya 2 Rahoto
A body is said to be moving with uniform acceleration if it experiences equal
Bayanin Amsa
A body is said to be moving with uniform acceleration if it experiences an equal increase in velocity at equal time intervals. In other words, the rate at which the velocity of the body changes is constant over time. This means that if we plot the velocity of the body against time, we will get a straight line with a constant slope. The acceleration can be calculated by dividing the change in velocity by the change in time. It is important to note that uniform acceleration refers to a constant rate of change in velocity and not speed, since the velocity of a body is a vector quantity that takes into account both its speed and direction.
Tambaya 3 Rahoto
When compared with water, mercury is preferred as a thermometric liquid because
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Mercury is preferred as a thermometric liquid over water because it has a much lower specific heat capacity. This means that it requires less heat energy to change its temperature than water. This makes it easier to measure small temperature changes accurately. Additionally, mercury has a much higher density than water, which allows it to rise and fall more easily in narrow capillary tubes used in thermometers. The other options are not correct - mercury does not exhibit anomalous thermal expansion and does not have molecules with very strong adhesive forces.
Tambaya 4 Rahoto
Calculate the change in volume when 90g of ice is completely melted. [density of water = 1g cm-3, density of ice = 0.9cm-3]
Tambaya 5 Rahoto
Which of the following values on the absolute scale of temperature is the ice point?
Bayanin Amsa
The ice point is the temperature at which pure water melts or freezes. On the absolute scale of temperature, the ice point is equivalent to 273K. Therefore, the correct option is 273K.
Tambaya 6 Rahoto
A car fuse marked 3A operates optimally on a 12V battery, calculate the resistance of the fuse
Bayanin Amsa
The formula for calculating resistance is: R = V/I where R is resistance, V is voltage and I is current. Given the fuse is marked 3A, it means it can carry a maximum current of 3A. Also, the voltage of the battery is given as 12V. Using Ohm's law, we can calculate the resistance of the fuse as follows: R = V/I = 12V / 3A = 4.0Ω Therefore, the resistance of the fuse is 4.0Ω.
Tambaya 7 Rahoto
Which of the following concept is not an evidence of the particle nature of matter?
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Tambaya 8 Rahoto
On which of the following factor does the viscosity of a liquid depend? i. Nature of the liquid. ii. Temperature of the liquid. iii. Area of the liquid surface
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Tambaya 9 Rahoto
An elastic string of force constant 200N m-1 is stretched through 0.8m within its elastic limit. Calculate the energy stored in the string
Bayanin Amsa
The energy stored in an elastic string that is stretched is given by the equation: E = 0.5kx² where E is the energy stored, k is the force constant of the string, and x is the displacement from its original length. In this case, the force constant k is given as 200 N/m, and the displacement x is 0.8 m. Substituting these values in the above equation, we get: E = 0.5 × 200 N/m × (0.8 m)² E = 0.5 × 200 N/m × 0.64 m² E = 64 J Therefore, the energy stored in the elastic string is 64J. Hence, the correct option is (a) 64.0J.
Tambaya 10 Rahoto
Calculate the energy stored in a 20\(\mu F\) capacitor. If the p.d between the plates is 40V
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Tambaya 11 Rahoto
A charge of 1.6 x 10-10C is placed in a uniform electric field of intensity 2.0 x 105 NC-1. calculate the magnitude of the electric force exerted on the change
Bayanin Amsa
The magnitude of electric force can be calculated using the formula: Electric force (F) = charge (Q) x electric field intensity (E) Substituting the given values, we get: F = Q x E F = (1.6 x 10^-10 C) x (2.0 x 10^5 NC^-1) F = 3.2 x 10^-5 N Therefore, the magnitude of the electric force exerted on the charge is 3.2 x 10^-5 N. The correct option is (C) 3.2 x 10^-5 N.
Tambaya 12 Rahoto
A body of mass 8kg moving with a speed of 30ms-1 is brought to rest in 2.5s by a constant retarding force. Calculate the magnitude of the force
Bayanin Amsa
The formula for force is F = ma, where F is force, m is mass and a is acceleration. The body is initially moving with a speed of 30ms-1 and is brought to rest, which means the final velocity is 0ms-1. The acceleration, a, can be calculated using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken. Rearranging the formula to make a the subject gives a = (v - u)/t. Substituting the values gives a = (0 - 30)/2.5 = -12ms-2. The negative sign indicates that the acceleration is in the opposite direction to the initial velocity. Substituting the mass and acceleration into the formula F = ma gives F = 8 x (-12) = -96N. The negative sign indicates that the force is in the opposite direction to the initial velocity. Taking the magnitude of the force gives 96N. Therefore, the magnitude of the force is 96N. Note that the negative sign indicates the direction of the force, and is not part of the magnitude of the force.
Tambaya 13 Rahoto
A machine of velocity ratio 5 is used in lifting a load with an effort of 500N. If the machine is 80% efficient. Determine the magnitude of load
Bayanin Amsa
The velocity ratio of a machine is defined as the ratio of the distance moved by the effort to the distance moved by the load. In general, it is given by: Velocity ratio = (distance moved by effort) / (distance moved by load) In this problem, the velocity ratio is given as 5, which means that for every 5 units of distance moved by the effort, the load moves 1 unit of distance. Let us assume that the load is lifted through a height of h meters. Then, the effort is applied through a distance of 5h meters. The work done by the effort is given by the product of the effort and the distance moved by it. Similarly, the work done on the load is given by the product of the load and the distance moved by it. Since the machine is 80% efficient, the work done by the effort is equal to 80% of the work done on the load. Mathematically, Effort x Distance moved by effort = Load x Distance moved by load x Efficiency 500N x 5h = Load x h x 0.8 Simplifying the above equation, we get: Load = (500N x 5 x 0.8) / 1 Load = 2000N Therefore, the magnitude of the load is 2000N. Hence, the correct option is: 2000N.
Tambaya 14 Rahoto
An atom is normally electrically neutral because the
Bayanin Amsa
An atom is composed of protons, neutrons, and electrons. The protons are positively charged, the electrons are negatively charged, and the neutrons are neutral. In an atom that is normally electrically neutral, the number of positively charged protons is equal to the number of negatively charged electrons. This means that the positive charge of the protons is exactly balanced by the negative charge of the electrons, resulting in a net charge of zero. Therefore, the correct option is that the number of electrons is equal to the number of protons.
Tambaya 15 Rahoto
A diver is 5.2m below the surface of water of density 1000Kgm-3. If the atmospheric pressure is 1.02 x 105pa. Calculate the pressure on the diver [g = 10ms-2]
Bayanin Amsa
The pressure on the diver at a depth of 5.2m can be calculated using the formula: P = P0 + ρgh where P0 is the atmospheric pressure, ρ is the density of the water, g is the acceleration due to gravity, and h is the depth of the diver. Substituting the given values, we get: P = 1.02 x 105 + (1000 kg/m3) x (10 m/s2) x (5.2 m) P = 1.02 x 105 + 5.2 x 104 Pa P = 1.54 x 105 Pa Therefore, the pressure on the diver is 1.54 x 105 Pa. The correct option is: 1.54 x 105 pa.
Tambaya 16 Rahoto
The image of an optical pin placed at the principal focus of a concave mirror will be formed
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When an object is placed at the principal focus of a concave mirror, the reflected rays will be parallel to each other. This happens because the light rays coming from the object get refracted by the mirror in such a way that they all converge at the principal focus of the mirror. Therefore, the image of the object will be formed at infinity. So, the correct option is "at infinity".
Tambaya 17 Rahoto
A ball bearing is projected vertically upwards from the ground with a velocity of 15ms-1. Calculate the time taken by the ball to return to the ground [g = 10ms-1]
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The ball bearing was projected upwards with an initial velocity of 15ms-1. At the highest point, the velocity of the ball becomes zero before it starts falling back to the ground. Using the equation of motion for free-falling objects: h = ut + 1/2 gt^2 where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time. At the highest point, the height of the ball bearing is given by: h = (15)^2 / (2 × 10) = 11.25m (using v² = u² + 2gh) The time taken for the ball bearing to reach the highest point can be obtained by using the equation of motion: v = u + gt 0 = 15 - 10t t = 1.5s The total time taken for the ball bearing to reach the ground is twice the time taken to reach the highest point, which is: 2t = 2 × 1.5s = 3.0s Therefore, the time taken by the ball bearing to return to the ground is 3.0s. Hence, the answer is 3.0s.
Tambaya 18 Rahoto
Which of the following representations is correct for an atom X with 28 electrons and 30 neutrons?
Bayanin Amsa
The correct representation for an atom X with 28 electrons and 30 neutrons is \(^{58}_{28} X\). The atomic number of an element is the number of protons in its nucleus. Since the element X has 28 electrons, which are equal to the number of protons in its nucleus, the atomic number is 28. The mass number of an element is the total number of protons and neutrons in its nucleus. Since the element X has 30 neutrons in addition to its 28 protons, the mass number is 58. Therefore, the correct representation of the atom X is \(^{58}_{28} X\). Option (d) is the correct representation.
Tambaya 19 Rahoto
A steel needle floating on water, sinks when kerosine is added to the water. This is because the kerosine
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Tambaya 20 Rahoto
A man 1.8m tall stands 3m away from a pin hole camera. If the distance between the pin hole and the screen of the camera is 0.3m. Calculate the height of the image of the man produced by the camera
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The pinhole camera model assumes that light travels in straight lines and that the aperture is so small that the light rays pass through it in parallel lines. Based on this model, we can use the similar triangle relationship between the object and image to find the height of the image produced. We have a right-angled triangle with the object, pinhole and screen of the camera as the vertices. Let's call the height of the man "h" and the height of the image "h'". From the triangle, we have: h / 3m = h' / 0.3m Cross-multiplying, we get: h' = (h / 3) x 0.3 Substituting h = 1.8m, we get: h' = (1.8 / 3) x 0.3 = 0.18m Therefore, the height of the image of the man produced by the camera is 0.18m. The correct option is (a) 0.18m.
Tambaya 22 Rahoto
A girl stand 80m away from a tall cliff and blows a whistle. If the speed of sound in air is 330ms-1, how long would it take for her to hear the echo of the sound?
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Tambaya 23 Rahoto
When a metal ball is heated through 30oC,its volume becomes 1.0018cm3. If the linear expansivity of the material of the ball is 2.0 x 10-5 K-1, calculate its original volume
Bayanin Amsa
When a material is heated, it expands due to increase in temperature. The amount of expansion depends on the linear expansivity of the material which is the fractional increase in length per degree rise in temperature. Let V0 be the original volume of the ball and ΔT = 30°C be the temperature rise. Then, the new volume V after heating is given by: V = V0(1 + γΔT) where γ is the linear expansivity of the material. Substituting the given values, we have: V = 1.0018 cm3 = V0(1 + 2.0 x 10-5 x 30) Simplifying the expression, we get: V0 = V / (1 + γΔT) = 1.0018 cm3 / (1 + 2.0 x 10-5 x 30) Calculating this expression gives us: V0 = 1.0000 cm3 Therefore, the original volume of the metal ball is 1.0000 cm3. Hence, the answer is 1.0000 cm3.
Tambaya 24 Rahoto
A wave travelling from water to glass suffers a change in its speed at the common boundary. Which of the following properties explains this observation?
Bayanin Amsa
The property that explains the observation of a wave changing its speed when travelling from water to glass at the common boundary is refraction. Refraction occurs when a wave crosses the boundary between two media with different densities, causing a change in its speed and direction. In this case, the speed of the wave changes because the density of the medium changes, as glass is denser than water. This change in speed causes the wave to bend or change direction. Therefore, refraction is the property that explains the observation of a wave changing its speed at the common boundary between water and glass.
Tambaya 25 Rahoto
During a thunderstorm, lightning and thunder occurred simultaneously in the atmosphere. An observer on the earth surface's sees the lightning flash shortly before hearing the thunder. This is because the
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The reason why an observer on the earth's surface sees the lightning flash before hearing the thunder during a thunderstorm is that the speed of light is much faster than the speed of sound. When lightning occurs, it generates a large amount of energy that produces light, heat, and sound. Light travels at a very high speed of approximately 299,792,458 meters per second, whereas sound travels at a much lower speed of approximately 343 meters per second. Therefore, the observer sees the lightning flash almost instantaneously as it travels to the observer's eyes, while the sound waves from the thunder travel much slower and take some time to reach the observer's ears. This delay is known as the thunder's audible time delay, and the amount of time it takes for the sound waves to reach the observer is dependent on the distance between the observer and the lightning strike. Hence, the option that best explains why the observer sees the lightning flash before hearing the thunder is that the speed of sound is less than that of light.
Tambaya 26 Rahoto
The bob of a single pendulum has a mass of 0.02kg. Determine the weight of the bob [g = 10ms-2]
Bayanin Amsa
The weight of an object is given by the product of its mass and the acceleration due to gravity, that is, weight = mass x acceleration due to gravity Given that the mass of the bob of the pendulum is 0.02kg and acceleration due to gravity is 10ms-2, weight = 0.02kg x 10ms-2 weight = 0.2N Therefore, the weight of the bob of the single pendulum is 0.2N. So the correct option is: - 0.20N
Tambaya 27 Rahoto
Which of the following reasons explains why the walls and ceilings of a standard concert hall are usually covered with perforated pads? To
Bayanin Amsa
Perforated pads are used in concert halls to reduce the effect of reverberation of sound waves. Reverberation happens when sound waves reflect off surfaces multiple times, causing an echo-like effect that can distort the original sound. By covering walls and ceilings with perforated pads, the sound waves are absorbed by the pads, reducing their reflection and therefore reducing the amount of reverberation in the hall. This allows for clearer and more accurate sound, making it easier for performers and audience members to hear the music as intended. Therefore, the correct option is: "reduce the effect of reverberation of sound waves".
Tambaya 28 Rahoto
Find the cost of running a 60W lamp for 24 hours. If 1kW hr cost 5 naira
Bayanin Amsa
To find the cost of running a 60W lamp for 24 hours, we need to calculate the energy consumed by the lamp in kilowatt-hours (kWh) and then multiply by the cost of one kWh. First, let's convert the power of the lamp from watts to kilowatts by dividing it by 1000: 60 W ÷ 1000 = 0.06 kW Next, we can calculate the energy consumed by the lamp in kWh by multiplying the power by the time: 0.06 kW × 24 h = 1.44 kWh Finally, we can find the cost of running the lamp by multiplying the energy consumed by the cost per kWh: 1.44 kWh × 5 naira/kWh = 7.2 naira Therefore, the answer is option (C) 7.2 naira.
Tambaya 29 Rahoto
The diagram above represent displacement D, versus time, t graph of a progressive wave. Deduce the frequency of the wave
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Tambaya 30 Rahoto
Which of the following statements about the mass and the weight of a body is not correct? its
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The statement that "weigh at the equator and at the poles is the same" is not correct. Mass is a scalar quantity that measures the amount of matter in an object, and it remains the same regardless of the object's location. Weight, on the other hand, is a force that is exerted on an object due to the gravitational pull of the planet it is on. Therefore, weight depends on both the mass of the object and the strength of the gravitational field it is in. The gravitational pull on the surface of the Earth is not the same everywhere, as the Earth is not a perfect sphere and it rotates. The gravitational force at the poles is slightly stronger than at the equator because the poles are closer to the center of the Earth, where gravity is stronger. As a result, the weight of an object will be slightly different at the poles and the equator, even if its mass remains the same.
Tambaya 31 Rahoto
The speed of an object in rectilinear motion can be determine from the
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The speed of an object in rectilinear motion can be determined from the slope of a distance-time graph. A distance-time graph plots the distance covered by an object on the y-axis against the time taken on the x-axis. The slope of this graph represents the rate of change of distance with respect to time, which is the speed of the object. The steeper the slope, the faster the object is moving, and the shallower the slope, the slower the object is moving. Therefore, the slope of a distance-time graph can be used to determine the speed of an object in rectilinear motion.
Tambaya 32 Rahoto
A radioactive substance has a half-life of 3 days. If a mass of 1.55g of this substance is left after decaying for 15 days, determine the original value of the mass
Bayanin Amsa
The half-life of a radioactive substance is the time taken for half of the initial mass of the substance to decay. Let's denote the initial mass of the substance by m. After 3 days (one half-life), the mass of the substance remaining is m/2. After 6 days (two half-lives), the mass remaining is (m/2)/2 = m/4. Similarly, after 9 days (three half-lives), the mass remaining is m/8, and after 12 days (four half-lives), the mass remaining is m/16. After 15 days (five half-lives), we are told that there is 1.55g of the substance remaining. Therefore, we can set up the following equation: 1.55g = m/32 Solving for m gives: m = 49.6g Therefore, the original value of the mass of the radioactive substance was 49.6g.
Tambaya 33 Rahoto
On a cold day, a concrete floor feels colder to the bear foot then a mat on the same floor because the
Bayanin Amsa
A concrete floor feels colder to bare feet than a mat on the same floor because the concrete floor is a better conductor of heat than the mat. Heat flows from a warmer object to a colder object until they reach thermal equilibrium. When we stand on the floor, the heat from our body transfers to the surrounding area by conduction. Since concrete is a better conductor of heat than the mat, it takes more heat away from our feet, making the concrete floor feel colder. The mat, on the other hand, insulates our feet and does not conduct heat away from our feet as efficiently as the concrete floor. Therefore, it feels warmer than the concrete floor.
Tambaya 34 Rahoto
Which of the following statements is not correct?
Bayanin Amsa
The statement that is not correct is: "magnetic fields are scalar quantities." This statement is incorrect because magnetic fields are vector quantities, meaning they have both magnitude and direction. The direction of a magnetic field is given by the direction of the force that would be exerted on a moving charged particle in that field. Magnetic fields can also be visualized using magnetic field lines, which show the direction and strength of the field at different points in space.
Tambaya 35 Rahoto
In an experiment using a converging lens to produce real images on a screen, the linear magnification m, is plotted against the image distance v, as illustrated in the diagram below. The distance OP represents the
Tambaya 36 Rahoto
An object weighs 60.0N in air, 48.2N in a certain liquid X, and 44.9N in water, Calculate the relative density of X
Bayanin Amsa
Weight of object in air (W_air): 60.0 N
Weight of object in liquid X (W_X): 48.2 N
Weight of object in water (W_water): 44.9 N
The loss of weight in a fluid is equal to the buoyant force, which is equal to the weight of the displaced fluid.
Buoyant force in water: 𝐹buoyant, water=𝑊air−𝑊water=60.0 N−44.9 N=15.1 NFbuoyant, water=Wair−Wwater=60.0N−44.9N=15.1N
Since the buoyant force is equal to the weight of the water displaced and the density of water (ρ_water) is 1 g/cm³ (or 1000 kg/m³), the volume of the object (V) can be calculated using:
𝐹buoyant, water=𝜌water⋅𝑉⋅𝑔Fbuoyant, water=ρwater⋅V⋅g 15.1 N=1000 kg/m3⋅𝑉⋅9.8 m/s215.1N=1000kg/m3⋅V⋅9.8m/s2 𝑉=15.1 N1000 kg/m3⋅9.8 m/s2V=1000kg/m3⋅9.8m/s215.1N 𝑉≈1.54×10−3 m3V≈1.54×10−3m3
Buoyant force in liquid X:
𝐹buoyant, X=𝑊air−𝑊X=60.0 N−48.2 N=11.8 NFbuoyant, X=Wair−WX=60.0N−48.2N=11.8N
Using the same volume V (since the object's volume doesn't change), the density of liquid X (ρ_X) can be calculated:
𝐹buoyant, X=𝜌X⋅𝑉⋅𝑔Fbuoyant, X=ρX⋅V⋅g 11.8 N=𝜌X⋅1.54×10−3 m3⋅9.8 m/s211.8N=ρX⋅1.54×10−3m3⋅9.8m/s2 𝜌X=11.8 N1.54×10−3 m3⋅9.8 m/s2ρX=1.54×10−3m3⋅9.8m/s211.8N 𝜌X≈786.1 kg/m3ρX≈786.1kg/m3
Relative density of liquid X: The relative density (specific gravity) is the ratio of the density of liquid X to the density of water: Relative density=𝜌X𝜌water=786.1 kg/m31000 kg/m3≈0.786Relative density=ρwaterρX=1000kg/m3786.1kg/m3≈0.786
Therefore, the relative density of liquid X is approximately 0.786.
Tambaya 37 Rahoto
A loaded spring performs simple harmonic motion with an amplitude of 5cm. If the maximum acceleration of the load is 20cm s-1, calculate the angular frequency of the motion
Bayanin Amsa
The acceleration of an object in simple harmonic motion can be represented as a = -ω2x, where a is acceleration, x is displacement, and ω is the angular frequency. Given that the maximum acceleration of the load is 20cm s-1, we can substitute a = 20cm s-1, and x = 5cm (the amplitude) into the equation above to solve for ω. 20cm s-1 = -ω2(5cm) ω2 = -20cm s-1/5cm = -4 s-2 ω = ±2 s-1 Since the problem asks for the magnitude of ω, we take the positive value: ω = 2 rad s-1 Therefore, the answer is option (A) 2 rad s-1.
Tambaya 38 Rahoto
A faulty barometer reads 72.6cm Hg when the atmospheric pressure is 75.0cm Hg. Calculate the atmospheric pressure when the barometer reads 72.0cn Hg
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Tambaya 39 Rahoto
The quantity of electric charge stored in a capacitor is measured in
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The quantity of electric charge stored in a capacitor is measured in coulombs. A capacitor is an electronic component that is used to store electric charge. When a capacitor is charged, it stores electrical energy in the form of electric charge. The amount of electric charge stored in a capacitor is directly proportional to the voltage applied across the capacitor and the capacitance of the capacitor. The unit of electric charge is coulomb, which is defined as the charge transported by a constant current of one ampere in one second. Therefore, the quantity of electric charge stored in a capacitor is measured in coulombs.
Tambaya 40 Rahoto
The temperature of 900cm3 of an ideal gas at a pressure of 114cm Hg is 27oC. Calculate its volume at 76cm Hg and 0oC
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Tambaya 42 Rahoto
In which of the following devices is the principle of photoelectric effect not applicable?
Bayanin Amsa
The principle of photoelectric effect refers to the phenomenon where electrons are emitted from a material when light of certain frequencies shines on it. This means that in devices that use the photoelectric effect, the emission of electrons is triggered by light. Therefore, the device that does not use the photoelectric effect is the periscope. A periscope is an optical device that uses mirrors to reflect light and allows a viewer to see objects that are not in their direct line of sight. It does not use any photoelectric material or any process related to the emission of electrons by light.
Tambaya 43 Rahoto
The basic difference between a transverse wave and a longitudinal wave travelling in the same direction in a medium is in the
Bayanin Amsa
The basic difference between a transverse wave and a longitudinal wave is in the direction of vibration of the particles of the medium. In a transverse wave, the particles of the medium vibrate perpendicular to the direction of propagation of the wave. In contrast, in a longitudinal wave, the particles of the medium vibrate parallel to the direction of propagation of the wave. Therefore, the direction of vibration of the particles of the medium is the main difference between a transverse wave and a longitudinal wave traveling in the same direction in a medium. The amplitude, wavelength, and period of vibration of the particles of the medium can be the same in both types of waves, depending on the specific characteristics of the wave.
Tambaya 44 Rahoto
A magnified and virtual image of a near object is produced by
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None of the given options is correct. A magnified and virtual image of a near object is produced by a simple microscope. A simple microscope consists of a single converging lens that is used to magnify small objects, such as biological specimens or small components in electronic devices. When an object is placed close to the lens, a magnified and virtual image is formed on the same side of the lens as the object. The image is virtual because it cannot be projected onto a screen, and it is magnified because it appears larger than the object itself. Prism binoculars use two lenses and a prism to produce a magnified, upright image of a distant object. Astronomical telescopes are also used to view distant objects, such as stars and galaxies. A periscope is a device that uses mirrors to reflect light and allow the viewer to see objects around corners or over obstacles.
Tambaya 45 Rahoto
The sound from a vibrating tuning sound is when the tip of its stem is placed on a wooden table than when in air because
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Tambaya 46 Rahoto
A force acting on a body causes a change in the momentum of the body from 12kg ms-1 to 16kg ms-1 in 0.2s. calculate the magnitude of the force
Bayanin Amsa
The change in momentum of a body is given by the formula: Δp = mΔv where Δp is the change in momentum, m is the mass of the body, and Δv is the change in velocity. In this case, the change in momentum is: Δp = 16 kg ms^-1 - 12 kg ms^-1 = 4 kg ms^-1 The time taken for the change in momentum is: Δt = 0.2 s Therefore, the magnitude of the force is given by: F = Δp/Δt = (4 kg ms^-1)/(0.2 s) = 20 N So, the correct option is: 20.0N.
Tambaya 47 Rahoto
A body is dropped from a certain height above the ground level, falls with uniform
Bayanin Amsa
When a body is dropped from a height above the ground level, it begins to fall towards the ground due to the force of gravity acting on it. The acceleration due to gravity is a constant value, which means that the speed of the body increases uniformly as it falls towards the ground. Therefore, the body falls with a uniform speed, meaning that the speed at which it is falling remains constant throughout its journey towards the ground. The body does not have a constant velocity because velocity is a vector quantity that takes into account both the speed and direction of the body. In this case, the body is only moving in one direction, towards the ground, so its velocity is constantly changing as it gains speed. The acceleration of the body is due to gravity and is constant throughout its fall. This means that the rate at which the body gains speed is also constant, resulting in a uniform acceleration. Retardation, on the other hand, refers to a decrease in speed or a negative acceleration. Since the body is only experiencing a force in the downward direction, there is no force acting against its motion to cause a decrease in speed. Therefore, there is no retardation acting on the body during its fall.
Tambaya 48 Rahoto
Which of the following statements about X-rays is not correct? They
Bayanin Amsa
X-rays are not produced from the nucleus of an atom, so the statement "are produced from the nucleus of an atom" is not correct. X-rays are electromagnetic waves that have a high frequency and short wavelength. They can penetrate solid objects, making them useful for medical imaging, studying crystal lattice, and industrial applications. However, X-rays can also be harmful to living cells, as they have enough energy to ionize atoms and molecules, which can damage or destroy cells. X-rays are produced when high-energy electrons collide with a target material, such as tungsten, in an X-ray machine.
Tambaya 49 Rahoto
Which of following functions is performed by manganese IV oxide in the Lenclanche cell?
Bayanin Amsa
Manganese IV oxide in the Leclanché cell performs the function of depolarizing the cell. The Leclanché cell is a type of dry cell commonly used in household batteries. When a cell is discharging, chemical reactions occur inside the cell, which can lead to the accumulation of hydrogen gas on the negative electrode, hindering the flow of electricity. Manganese IV oxide acts as a depolarizer, meaning that it reacts with the hydrogen gas to form water, thereby preventing the accumulation of hydrogen gas and allowing the cell to continue functioning efficiently.
Tambaya 50 Rahoto
On a basis of the kinetic theory, an increase in the temperature of a fixed volume of an ideal gas causes
Bayanin Amsa
On the basis of the kinetic theory, an increase in the temperature of a fixed volume of an ideal gas causes an increase in the average speed of the gas molecules. This is because the kinetic theory states that the temperature of a gas is directly proportional to the average kinetic energy of its molecules. As the temperature increases, the molecules of the gas gain more kinetic energy and move faster, resulting in an increase in their average speed. This increase in the average speed of the gas molecules also results in an increase in the pressure exerted by the gas, as they collide more frequently with the walls of the container.
Tambaya 51 Rahoto
Given that the gravitational constant is G. Newton's universal law of gravitation states that the force of attraction between two masses M1 and M2 separation by a distance r is
Bayanin Amsa
Newton's universal law of gravitation states that the force of attraction between two masses M1 and M2 separated by a distance r is given by the formula: F = G * M1 * M2 / r^2, where G is the gravitational constant. This means that the force of attraction between the two masses is directly proportional to the product of their masses (M1 and M2), and inversely proportional to the square of the distance between them (r^2). In other words, the greater the masses of the objects and the closer they are, the stronger the force of attraction between them. is the correct answer as it is the formula that represents the force of attraction between two masses based on Newton's universal law of gravitation.
Tambaya 52 Rahoto
A Spiral spring of natural length 20.00cm has a scale hanging freely in its lower end. When an object ass 40 g is placed in the pan. its length becomes cm. When the object is replaced with another of 60g, the length becomes 22.05cm. Calculate the mass scale pan. [g = 10 ms\(^{-2}\)]
Using Hooke’s law: extension of the spring is proportional to the total mass attached.
For the 40 g mass, the spring length is \(21.80\text{ cm}\), hence extension:
\[ e_1 = 21.80 - 20.00 = 1.80\text{ cm} \] \[ (m + 40)g = 1.80k \qquad \text{... (i)} \]For the 60 g mass, the extension is:
\[ e_2 = 22.05 - 20.00 = 2.05\text{ cm} \] \[ (m + 60)g = 2.05k \qquad \text{... (ii)} \]Dividing equation (i) by equation (ii):
\[ \frac{m+40}{m+60}=\frac{1.80}{2.05} \] \[ 1.80(m+60)=2.05(m+40) \] \[ 1.80m+108=2.05m+82 \] \[ 26=0.25m \] \[ m=104\text{ g} \]Mass of the scale pan \(= 104\text{ g}\).
Bayanin Amsa
Using Hooke’s law: extension of the spring is proportional to the total mass attached.
For the 40 g mass, the spring length is \(21.80\text{ cm}\), hence extension:
\[ e_1 = 21.80 - 20.00 = 1.80\text{ cm} \] \[ (m + 40)g = 1.80k \qquad \text{... (i)} \]For the 60 g mass, the extension is:
\[ e_2 = 22.05 - 20.00 = 2.05\text{ cm} \] \[ (m + 60)g = 2.05k \qquad \text{... (ii)} \]Dividing equation (i) by equation (ii):
\[ \frac{m+40}{m+60}=\frac{1.80}{2.05} \] \[ 1.80(m+60)=2.05(m+40) \] \[ 1.80m+108=2.05m+82 \] \[ 26=0.25m \] \[ m=104\text{ g} \]Mass of the scale pan \(= 104\text{ g}\).
Tambaya 53 Rahoto
(a) For a water voltameter, identify the;
(i) electrolyte;
(ii) electrodes;
(iii) substances deposited on the electrodes
(b)State the ratio of the volume of the substances deposited in (a) (iii) above
(a)(i) Electrolyte: acidulated water (water containing a few drops of tetraoxosulphate(VI) acid, i.e. dilute sulphuric acid).
(ii) Electrodes: platinum electrodes (an anode and a cathode).
(iii) Substances deposited (liberated): hydrogen gas is liberated at the cathode and oxygen gas is liberated at the anode.
(b) Ratio of volumes: volume of hydrogen : volume of oxygen \(= 2 : 1\) (twice as much hydrogen as oxygen).
Bayanin Amsa
(a)(i) Electrolyte: acidulated water (water containing a few drops of tetraoxosulphate(VI) acid, i.e. dilute sulphuric acid).
(ii) Electrodes: platinum electrodes (an anode and a cathode).
(iii) Substances deposited (liberated): hydrogen gas is liberated at the cathode and oxygen gas is liberated at the anode.
(b) Ratio of volumes: volume of hydrogen : volume of oxygen \(= 2 : 1\) (twice as much hydrogen as oxygen).
Tambaya 54 Rahoto
State three methods of polarising an unpolarised light
Three methods of producing (plane) polarised light from unpolarised light:
- By reflection: reflecting light from a plane surface (e.g. glass or water) at the polarising angle (Brewster's angle) gives a completely plane-polarised reflected beam.
- By selective absorption: passing light through a Polaroid (dichroic material) which transmits vibrations in one plane and absorbs those in the perpendicular plane.
- By double refraction: passing light through a doubly refracting crystal such as calcite (Nicol prism), which separates the light into two plane-polarised beams.
(Polarisation by scattering, e.g. of sunlight by air molecules, is also accepted.)
Bayanin Amsa
Three methods of producing (plane) polarised light from unpolarised light:
- By reflection: reflecting light from a plane surface (e.g. glass or water) at the polarising angle (Brewster's angle) gives a completely plane-polarised reflected beam.
- By selective absorption: passing light through a Polaroid (dichroic material) which transmits vibrations in one plane and absorbs those in the perpendicular plane.
- By double refraction: passing light through a doubly refracting crystal such as calcite (Nicol prism), which separates the light into two plane-polarised beams.
(Polarisation by scattering, e.g. of sunlight by air molecules, is also accepted.)
Tambaya 55 Rahoto
Define upper fixed point and lower fixed point as used in thermometry
(b) The electrical resistances of the element in a platinum restistance thermometer at 100\(^o\)C, 0\(^o\) and room temperature are 75.000, 63.000 and 64.992 \(\Omega\) respectively. Use these data to determine the room temperature.
(c) (i) State Boyle's law
(ii) A uniform capillary tube, closed at one end contained dry air trapped by a thread of mecury 8.5 x 10\(^{-2}\)m long. When the tube was held horinzontally, the length of the air column was 5.0 x 10\(^{-2}\)m, when it was held vertically with the closed end downwards, the length was 4.5 x 10\(^{-2}\)m, Determine the value of the atmospheric pressure. [g = 10ms\(^{-2}\), density of mecury = 1.36 x 10\(^4\) kg m\(^{-3}\)]
(a) Upper fixed point: the temperature of pure boiling water (steam) at standard atmospheric pressure, taken as \(100^{\circ}\text{C}\). Lower fixed point: the temperature of pure melting ice at standard atmospheric pressure, taken as \(0^{\circ}\text{C}\).
(b) For a platinum resistance thermometer, temperature on the resistance scale is:
\[ \theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100 \]With \(R_0 = 63.000\ \Omega\), \(R_{100} = 75.000\ \Omega\), \(R_\theta = 64.992\ \Omega\):
\[ \theta = \frac{64.992 - 63.000}{75.000 - 63.000} \times 100 = \frac{1.992}{12.000} \times 100 \] \[ \theta = 16.6^{\circ}\text{C} \]The room temperature is \(16.6^{\circ}\text{C}\).
(c)(i) Boyle's law: the volume of a fixed mass of gas is inversely proportional to its pressure, provided the temperature is kept constant; i.e. \(PV = \text{constant}\).
(c)(ii) Let the atmospheric pressure be \(H\) (expressed as a length of mercury). The mercury thread is \(8.5\ \text{cm}\) long.
Horizontal: the mercury adds nothing, so the air pressure \(= H\) and the length is \(5.0\ \text{cm}\).
Vertical, closed end downwards: the mercury column presses down on the trapped air, so its pressure \(= H + 8.5\), and the length is \(4.5\ \text{cm}\).
Applying Boyle's law (constant cross-section, so length replaces volume):
\[ H \times 5.0 = (H + 8.5) \times 4.5 \] \[ 5.0H = 4.5H + 38.25 \Rightarrow 0.5H = 38.25 \Rightarrow H = 76.5\ \text{cm of mercury} \]Converting to pascals with \(\rho = 1.36 \times 10^{4}\ \text{kg m}^{-3}\), \(g = 10\ \text{ms}^{-2}\), \(H = 0.765\ \text{m}\):
\[ P = \rho g H = 1.36 \times 10^{4} \times 10 \times 0.765 = 1.04 \times 10^{5}\ \text{Pa} \]The atmospheric pressure is \(76.5\ \text{cmHg} \approx 1.04 \times 10^{5}\ \text{Pa}\).
Bayanin Amsa
(a) Upper fixed point: the temperature of pure boiling water (steam) at standard atmospheric pressure, taken as \(100^{\circ}\text{C}\). Lower fixed point: the temperature of pure melting ice at standard atmospheric pressure, taken as \(0^{\circ}\text{C}\).
(b) For a platinum resistance thermometer, temperature on the resistance scale is:
\[ \theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100 \]With \(R_0 = 63.000\ \Omega\), \(R_{100} = 75.000\ \Omega\), \(R_\theta = 64.992\ \Omega\):
\[ \theta = \frac{64.992 - 63.000}{75.000 - 63.000} \times 100 = \frac{1.992}{12.000} \times 100 \] \[ \theta = 16.6^{\circ}\text{C} \]The room temperature is \(16.6^{\circ}\text{C}\).
(c)(i) Boyle's law: the volume of a fixed mass of gas is inversely proportional to its pressure, provided the temperature is kept constant; i.e. \(PV = \text{constant}\).
(c)(ii) Let the atmospheric pressure be \(H\) (expressed as a length of mercury). The mercury thread is \(8.5\ \text{cm}\) long.
Horizontal: the mercury adds nothing, so the air pressure \(= H\) and the length is \(5.0\ \text{cm}\).
Vertical, closed end downwards: the mercury column presses down on the trapped air, so its pressure \(= H + 8.5\), and the length is \(4.5\ \text{cm}\).
Applying Boyle's law (constant cross-section, so length replaces volume):
\[ H \times 5.0 = (H + 8.5) \times 4.5 \] \[ 5.0H = 4.5H + 38.25 \Rightarrow 0.5H = 38.25 \Rightarrow H = 76.5\ \text{cm of mercury} \]Converting to pascals with \(\rho = 1.36 \times 10^{4}\ \text{kg m}^{-3}\), \(g = 10\ \text{ms}^{-2}\), \(H = 0.765\ \text{m}\):
\[ P = \rho g H = 1.36 \times 10^{4} \times 10 \times 0.765 = 1.04 \times 10^{5}\ \text{Pa} \]The atmospheric pressure is \(76.5\ \text{cmHg} \approx 1.04 \times 10^{5}\ \text{Pa}\).
Tambaya 56 Rahoto
In his first attempt, a long jumper took off from the spring board 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 \(\frac{u^2 sin 2\theta}{g}\) where all the symbols have their usual meanings, show that he gains a distance of 0.8576 m in his second attempt. [g = 10ms\(^{-2}\)]
Using \(R = \dfrac{u^2 \sin 2\theta}{g}\) with \(u = 8\ \text{ms}^{-1}\) and \(g = 10\ \text{ms}^{-2}\), so \(u^2 = 64\).
First attempt (\(\theta = 30^{\circ}\), \(2\theta = 60^{\circ}\), \(\sin 60^{\circ} = 0.8660\)):
\[ R_1 = \frac{64 \times 0.8660}{10} = \frac{55.425}{10} = 5.5425\ \text{m} \]Second attempt (\(\theta = 45^{\circ}\), \(2\theta = 90^{\circ}\), \(\sin 90^{\circ} = 1\)):
\[ R_2 = \frac{64 \times 1}{10} = 6.4\ \text{m} \]Gain in distance:
\[ R_2 - R_1 = 6.4 - 5.5425 = 0.8575 \approx 0.8576\ \text{m} \]Hence he gains about \(0.8576\ \text{m}\) in the second attempt, as required.
Bayanin Amsa
Using \(R = \dfrac{u^2 \sin 2\theta}{g}\) with \(u = 8\ \text{ms}^{-1}\) and \(g = 10\ \text{ms}^{-2}\), so \(u^2 = 64\).
First attempt (\(\theta = 30^{\circ}\), \(2\theta = 60^{\circ}\), \(\sin 60^{\circ} = 0.8660\)):
\[ R_1 = \frac{64 \times 0.8660}{10} = \frac{55.425}{10} = 5.5425\ \text{m} \]Second attempt (\(\theta = 45^{\circ}\), \(2\theta = 90^{\circ}\), \(\sin 90^{\circ} = 1\)):
\[ R_2 = \frac{64 \times 1}{10} = 6.4\ \text{m} \]Gain in distance:
\[ R_2 - R_1 = 6.4 - 5.5425 = 0.8575 \approx 0.8576\ \text{m} \]Hence he gains about \(0.8576\ \text{m}\) in the second attempt, as required.
Tambaya 57 Rahoto
(a) (i) Explain electromotive force
(ii) list two sources of electromotive force other than a chemical cell
(b) A chemical cell of electromotive force, E, and internal resistance, r, is connected in series with an ammeter, a plug key a plug key and an external load of resistance R. A volumeter is connected across the cell. Draw a circuit diagram to illustrate the arrangement,
(c) for the arrangement in (b) above, with the key opened and closed, the voltmeter readings are V\(_o\) and V respectively.
(i) Explain the physical meanings of V\(_o\) and V
(ii) Find an expression for the (I) current passing through the circuit (II) maximum power dissipated in the cell and external load respectively; (III) efficiency of the cell
(a)(i) Electromotive force (e.m.f.): the total energy supplied by a source in driving one coulomb of charge completely round the whole circuit (both the external load and the internal resistance). Equivalently, it is the terminal potential difference of the source when no current is being drawn from it (open circuit). It is measured in volts (V).
(a)(ii) Two sources of e.m.f. other than a chemical cell:
- a dynamo / electrical generator (electromagnetic induction);
- a thermocouple (thermoelectric effect).
(A solar cell / photocell is also acceptable.)
(b) Circuit diagram. The cell (e.m.f. \(E\), internal resistance \(r\)) is joined in series with a plug key, an ammeter and the external load of resistance \(R\), while the voltmeter is connected across the terminals of the cell:
(c)(i) Physical meaning of \(V_o\) and \(V\):
- With the key open, no current flows, so there is no voltage drop across the internal resistance \(r\). The voltmeter reading \(V_o\) is therefore the full e.m.f. of the cell, \(V_o = E\).
- With the key closed, a current \(I\) flows and a voltage \(Ir\) is lost inside the cell. The voltmeter reading \(V\) is then the terminal potential difference of the cell, \(V = E - Ir\), which is less than \(V_o\).
(c)(ii)(I) Current in the circuit. Applying the circuit equation \(E = I(R+r)\):
\[ I = \frac{E}{R + r} \](c)(ii)(II) Maximum power dissipated. The power delivered to the external load is
\[ P_R = I^2 R = \frac{E^2 R}{(R + r)^2}. \]Differentiating \(P_R\) with respect to \(R\) and setting \(\dfrac{dP_R}{dR}=0\) shows that this is greatest when the load is matched to the cell, i.e. when \(R = r\) (the maximum-power-transfer condition). Substituting \(R = r\):
\[ P_{R(\max)} = \frac{E^2 r}{(r + r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}. \]At this same condition the power dissipated inside the cell is \(P_r = I^2 r = \dfrac{E^2 r}{(2r)^2} = \dfrac{E^2}{4r}\) as well, so the total maximum power drawn from the cell is
\[ P_{\text{total}} = P_{R(\max)} + P_r = \frac{E^2}{4r} + \frac{E^2}{4r} = \frac{E^2}{2r}. \](c)(ii)(III) Efficiency of the cell. The efficiency is the ratio of the useful power delivered to the external load to the total power produced by the cell:
\[ \eta = \frac{\text{power in } R}{\text{total power}} = \frac{I^2 R}{I^2 (R + r)} = \frac{R}{R + r} \times 100\% = \frac{V}{V_o} \times 100\%. \]In particular, at the maximum-power-transfer condition \(R = r\), the efficiency is only \(\dfrac{r}{r+r}\times100\% = 50\%\): half of the energy is wasted inside the cell when the greatest power is transferred to the load.
Bayanin Amsa
(a)(i) Electromotive force (e.m.f.): the total energy supplied by a source in driving one coulomb of charge completely round the whole circuit (both the external load and the internal resistance). Equivalently, it is the terminal potential difference of the source when no current is being drawn from it (open circuit). It is measured in volts (V).
(a)(ii) Two sources of e.m.f. other than a chemical cell:
- a dynamo / electrical generator (electromagnetic induction);
- a thermocouple (thermoelectric effect).
(A solar cell / photocell is also acceptable.)
(b) Circuit diagram. The cell (e.m.f. \(E\), internal resistance \(r\)) is joined in series with a plug key, an ammeter and the external load of resistance \(R\), while the voltmeter is connected across the terminals of the cell:
(c)(i) Physical meaning of \(V_o\) and \(V\):
- With the key open, no current flows, so there is no voltage drop across the internal resistance \(r\). The voltmeter reading \(V_o\) is therefore the full e.m.f. of the cell, \(V_o = E\).
- With the key closed, a current \(I\) flows and a voltage \(Ir\) is lost inside the cell. The voltmeter reading \(V\) is then the terminal potential difference of the cell, \(V = E - Ir\), which is less than \(V_o\).
(c)(ii)(I) Current in the circuit. Applying the circuit equation \(E = I(R+r)\):
\[ I = \frac{E}{R + r} \](c)(ii)(II) Maximum power dissipated. The power delivered to the external load is
\[ P_R = I^2 R = \frac{E^2 R}{(R + r)^2}. \]Differentiating \(P_R\) with respect to \(R\) and setting \(\dfrac{dP_R}{dR}=0\) shows that this is greatest when the load is matched to the cell, i.e. when \(R = r\) (the maximum-power-transfer condition). Substituting \(R = r\):
\[ P_{R(\max)} = \frac{E^2 r}{(r + r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}. \]At this same condition the power dissipated inside the cell is \(P_r = I^2 r = \dfrac{E^2 r}{(2r)^2} = \dfrac{E^2}{4r}\) as well, so the total maximum power drawn from the cell is
\[ P_{\text{total}} = P_{R(\max)} + P_r = \frac{E^2}{4r} + \frac{E^2}{4r} = \frac{E^2}{2r}. \](c)(ii)(III) Efficiency of the cell. The efficiency is the ratio of the useful power delivered to the external load to the total power produced by the cell:
\[ \eta = \frac{\text{power in } R}{\text{total power}} = \frac{I^2 R}{I^2 (R + r)} = \frac{R}{R + r} \times 100\% = \frac{V}{V_o} \times 100\%. \]In particular, at the maximum-power-transfer condition \(R = r\), the efficiency is only \(\dfrac{r}{r+r}\times100\% = 50\%\): half of the energy is wasted inside the cell when the greatest power is transferred to the load.
Tambaya 58 Rahoto
(a) Define Young's modulus
(b) When a force of 50 N is applied to the free end of an elastic cord, an extension of 4 cm is produced in the cord. Calculate the work done on the cord.
(a) Young's modulus: the ratio of the tensile stress to the tensile strain of a material, within the limit of proportionality (elastic limit). That is:
\[ E = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/l} = \frac{Fl}{Ae} \]where \(F\) is the applied force, \(A\) the cross-sectional area, \(e\) the extension and \(l\) the original length.
(b) Force \(F = 50\ \text{N}\), extension \(e = 4\ \text{cm} = 0.04\ \text{m}\). For an elastic (Hooke's-law) cord the force builds up uniformly from 0 to \(F\), so the work done (energy stored) is:
\[ W = \frac{1}{2} F e = \frac{1}{2} \times 50 \times 0.04 \] \[ W = 1.0\ \text{J} \]The work done on the cord is \(1.0\ \text{J}\).
Bayanin Amsa
(a) Young's modulus: the ratio of the tensile stress to the tensile strain of a material, within the limit of proportionality (elastic limit). That is:
\[ E = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/l} = \frac{Fl}{Ae} \]where \(F\) is the applied force, \(A\) the cross-sectional area, \(e\) the extension and \(l\) the original length.
(b) Force \(F = 50\ \text{N}\), extension \(e = 4\ \text{cm} = 0.04\ \text{m}\). For an elastic (Hooke's-law) cord the force builds up uniformly from 0 to \(F\), so the work done (energy stored) is:
\[ W = \frac{1}{2} F e = \frac{1}{2} \times 50 \times 0.04 \] \[ W = 1.0\ \text{J} \]The work done on the cord is \(1.0\ \text{J}\).
Tambaya 59 Rahoto
(a) Define gravitational field intensity
(b) In an experiment to determine the acceleration of free-fall due to gravity, g, using a simple pendulum of length I, six different values of I were used to obtain six corresponding values of period T. If a graph of I along the vertical axis is plotted against T\(^2\) on the horizontal axis;
(i) make a sketch to show the nature of the graph,
(ii) write down the equation that relates T, I and g hence obtain an expression for the slope of the graph
(iii) given that the slope of the graph is 0.25, determine the value for g [Take \(\pi\) = 3.142]
(c) A stone, thrown horizontally from the top of a vertical wall with a velocity of 15 ms\(^{-1}\), hits the horizontal ground at a point 45m from the base of the wall. Calculate the
(i) times of light of the stone
(ii) height of the wall [g = 10ms\(^{-2}\)]
(a) Gravitational field intensity at a point is the gravitational force experienced per unit mass placed at that point; \(g = \dfrac{F}{m}\). It is a vector quantity directed towards the centre of the attracting body and is measured in \(\text{N kg}^{-1}\).
(b)(i) Since \(l = \dfrac{g}{4\pi^2}\,T^2\), the graph of \(l\) (vertical axis) against \(T^2\) (horizontal axis) is a straight line passing through the origin with a positive, constant slope:
(b)(ii) For a simple pendulum of length \(l\) oscillating through a small angle, the period is
\[ T = 2\pi \sqrt{\frac{l}{g}} \]Squaring both sides,
\[ T^2 = \frac{4\pi^2}{g}\, l \quad\Rightarrow\quad l = \frac{g}{4\pi^2}\, T^2 \]Comparing with \(l = (\text{slope})\,T^2\), the slope of the \(l\)-against-\(T^2\) graph is
\[ \text{slope} = \frac{g}{4\pi^2} \](b)(iii) Given slope \(= 0.25\):
\[ g = 4\pi^2 \times \text{slope} = 4 \times (3.142)^2 \times 0.25 \] \[ g = 4 \times 9.872 \times 0.25 = 9.87\ \text{ms}^{-2} \](c) Horizontal projection: initial horizontal velocity \(u = 15\ \text{ms}^{-1}\), horizontal range \(R = 45\ \text{m}\), \(g = 10\ \text{ms}^{-2}\).
(c)(i) The horizontal motion is at constant velocity, so the time of flight is
\[ R = u\,t \quad\Rightarrow\quad t = \frac{R}{u} = \frac{45}{15} = 3\ \text{s} \](c)(ii) Vertically the stone starts from rest and falls freely, so the height of the wall is
\[ h = \frac{1}{2}\,g\,t^2 = \frac{1}{2} \times 10 \times 3^2 = \frac{1}{2} \times 10 \times 9 = 45\ \text{m} \]Therefore the time of flight is \(3\ \text{s}\) and the wall is \(45\ \text{m}\) high.
Bayanin Amsa
(a) Gravitational field intensity at a point is the gravitational force experienced per unit mass placed at that point; \(g = \dfrac{F}{m}\). It is a vector quantity directed towards the centre of the attracting body and is measured in \(\text{N kg}^{-1}\).
(b)(i) Since \(l = \dfrac{g}{4\pi^2}\,T^2\), the graph of \(l\) (vertical axis) against \(T^2\) (horizontal axis) is a straight line passing through the origin with a positive, constant slope:
(b)(ii) For a simple pendulum of length \(l\) oscillating through a small angle, the period is
\[ T = 2\pi \sqrt{\frac{l}{g}} \]Squaring both sides,
\[ T^2 = \frac{4\pi^2}{g}\, l \quad\Rightarrow\quad l = \frac{g}{4\pi^2}\, T^2 \]Comparing with \(l = (\text{slope})\,T^2\), the slope of the \(l\)-against-\(T^2\) graph is
\[ \text{slope} = \frac{g}{4\pi^2} \](b)(iii) Given slope \(= 0.25\):
\[ g = 4\pi^2 \times \text{slope} = 4 \times (3.142)^2 \times 0.25 \] \[ g = 4 \times 9.872 \times 0.25 = 9.87\ \text{ms}^{-2} \](c) Horizontal projection: initial horizontal velocity \(u = 15\ \text{ms}^{-1}\), horizontal range \(R = 45\ \text{m}\), \(g = 10\ \text{ms}^{-2}\).
(c)(i) The horizontal motion is at constant velocity, so the time of flight is
\[ R = u\,t \quad\Rightarrow\quad t = \frac{R}{u} = \frac{45}{15} = 3\ \text{s} \](c)(ii) Vertically the stone starts from rest and falls freely, so the height of the wall is
\[ h = \frac{1}{2}\,g\,t^2 = \frac{1}{2} \times 10 \times 3^2 = \frac{1}{2} \times 10 \times 9 = 45\ \text{m} \]Therefore the time of flight is \(3\ \text{s}\) and the wall is \(45\ \text{m}\) high.
Tambaya 60 Rahoto
Give three observations in support of de Broglie's assumption that moving particles behave like waves.
Tambaya 61 Rahoto
(a) Explain;
(i) wave motion
(ii) stationary wave
(b)(i) List four physical properties of a wave
(ii) Define amplitude and use it to distinguish between the node and antinode of a stationary wave
(iii) List the factors on which the frequency of vibration in a stretched string depends
(c) The equation, y = 5 sin (3x - 4t), where y is in millimeters, x is in meres and t is in seconds represents a wave motion . Determine the;
(i) frequency
(ii) period
(iii) speed of the wave
(a)(i) Wave motion: a disturbance that travels through a medium (or space) transferring energy from one point to another without any bulk transfer of the particles of the medium; the particles merely vibrate about their fixed mean positions.
(ii) Stationary (standing) wave: a wave formed when two progressive waves of the same frequency and amplitude travelling in opposite directions superpose; it has fixed points of zero amplitude (nodes) and points of maximum amplitude (antinodes), and the wave profile does not advance.
(b)(i) Four physical properties of a wave: amplitude, wavelength, frequency (or period), and speed (velocity).
(ii) Amplitude is the maximum displacement of a particle from its equilibrium (mean) position. In a stationary wave, a node is a point of zero amplitude (the particles do not move), while an antinode is a point of maximum amplitude.
(iii) The frequency of vibration of a stretched string depends on its length, the tension in the string, and its mass per unit length (linear density).
(c) Comparing \(y = 5\sin(3x - 4t)\) with \(y = A\sin(kx - \omega t)\): \(A = 5\ \text{mm}\), \(k = 3\ \text{m}^{-1}\), \(\omega = 4\ \text{rad s}^{-1}\).
(i) Frequency:
\[ f = \frac{\omega}{2\pi} = \frac{4}{2\pi} = 0.637\ \text{Hz} \](ii) Period:
\[ T = \frac{1}{f} = \frac{2\pi}{\omega} = \frac{2\pi}{4} = 1.57\ \text{s} \](iii) Speed:
\[ v = \frac{\omega}{k} = \frac{4}{3} = 1.33\ \text{ms}^{-1} \](Equivalently \(v = f\lambda\), with \(\lambda = \tfrac{2\pi}{k} = 2.09\ \text{m}\).)
Bayanin Amsa
(a)(i) Wave motion: a disturbance that travels through a medium (or space) transferring energy from one point to another without any bulk transfer of the particles of the medium; the particles merely vibrate about their fixed mean positions.
(ii) Stationary (standing) wave: a wave formed when two progressive waves of the same frequency and amplitude travelling in opposite directions superpose; it has fixed points of zero amplitude (nodes) and points of maximum amplitude (antinodes), and the wave profile does not advance.
(b)(i) Four physical properties of a wave: amplitude, wavelength, frequency (or period), and speed (velocity).
(ii) Amplitude is the maximum displacement of a particle from its equilibrium (mean) position. In a stationary wave, a node is a point of zero amplitude (the particles do not move), while an antinode is a point of maximum amplitude.
(iii) The frequency of vibration of a stretched string depends on its length, the tension in the string, and its mass per unit length (linear density).
(c) Comparing \(y = 5\sin(3x - 4t)\) with \(y = A\sin(kx - \omega t)\): \(A = 5\ \text{mm}\), \(k = 3\ \text{m}^{-1}\), \(\omega = 4\ \text{rad s}^{-1}\).
(i) Frequency:
\[ f = \frac{\omega}{2\pi} = \frac{4}{2\pi} = 0.637\ \text{Hz} \](ii) Period:
\[ T = \frac{1}{f} = \frac{2\pi}{\omega} = \frac{2\pi}{4} = 1.57\ \text{s} \](iii) Speed:
\[ v = \frac{\omega}{k} = \frac{4}{3} = 1.33\ \text{ms}^{-1} \](Equivalently \(v = f\lambda\), with \(\lambda = \tfrac{2\pi}{k} = 2.09\ \text{m}\).)
Tambaya 62 Rahoto
(a) Define diffusion
(b) State Graham's law of diffusion
(a) Diffusion: the spontaneous movement (spreading) of the molecules of a substance from a region of higher concentration to a region of lower concentration until they are evenly distributed, arising from the random motion of the molecules.
(b) Graham's law of diffusion: at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or of its molar mass):
\[ R \propto \frac{1}{\sqrt{\rho}} \qquad \text{i.e.} \qquad \frac{R_1}{R_2} = \sqrt{\frac{\rho_2}{\rho_1}} \]where \(R\) is the rate of diffusion and \(\rho\) is the density of the gas.
Bayanin Amsa
(a) Diffusion: the spontaneous movement (spreading) of the molecules of a substance from a region of higher concentration to a region of lower concentration until they are evenly distributed, arising from the random motion of the molecules.
(b) Graham's law of diffusion: at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or of its molar mass):
\[ R \propto \frac{1}{\sqrt{\rho}} \qquad \text{i.e.} \qquad \frac{R_1}{R_2} = \sqrt{\frac{\rho_2}{\rho_1}} \]where \(R\) is the rate of diffusion and \(\rho\) is the density of the gas.
Tambaya 63 Rahoto
(a) State two;
(i) properties of x-rays
(ii) reasons to show that x-rays are waves
(iii) uses of x-rays other than those in medicine;
(iv) hazards of x-rays.
(b) The potential difference between the cathode and target of an x-ray tube is 5.00 x 10\(^4\)V and the current in the tube is 2.00 x 10\(^{-2}\)A. Given that only one percent of the total energy supplied is emitted as x-radiation, determine the ospheric pressure.
(i) maximum frequency of the emitted radiation
(ii) rate at which heat is removed from the target in order to keep it at steady temperature. [Planck's constant, h = 6.63 x 10\(^{-34}\) Js, electronic charge e = 1.60 x 10\(^{-19}\) C]
Bayanin Amsa
None
Tambaya 64 Rahoto
The diagram above illustrates the path ABC, in a vertical x z plane, of a bullet shot into the air at an angle above the horizontal. Copy the diagram, and, using arrows, indicate the relative magnitudes and directions of the vertical and horizontal components of the velocities of the bullet at the point A, B and C.
Bayanin Amsa
None
Tambaya 65 Rahoto
A parallel beam of unpolarised light is incident on a plane glass surface at an angle of 58\(^{o}\) to the normal. If the reflected beam is completely polarised, calculate the refractive index of the glass.
When the reflected beam is completely polarised, the angle of incidence is the polarising (Brewster) angle \(\theta_p = 58^{\circ}\), and by Brewster's law:
\[ n = \tan \theta_p = \tan 58^{\circ} \] \[ n = 1.60 \]The refractive index of the glass is about \(1.60\).
Bayanin Amsa
When the reflected beam is completely polarised, the angle of incidence is the polarising (Brewster) angle \(\theta_p = 58^{\circ}\), and by Brewster's law:
\[ n = \tan \theta_p = \tan 58^{\circ} \] \[ n = 1.60 \]The refractive index of the glass is about \(1.60\).
Tambaya 66 Rahoto
(a) List 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 and cast sharp shadows of objects placed in their path.
- They are negatively charged, so they are deflected by both electric and magnetic fields.
- They possess kinetic energy (produce heat) and cause certain substances to fluoresce; they travel from cathode to anode.
(b) Increasing the intensity and energy of cathode rays:
- Intensity (the number of electrons emitted per second) is increased by raising the temperature of the cathode, i.e. increasing the heating (filament) current, so that more electrons are emitted by thermionic emission.
- Energy (the speed/kinetic energy of the electrons) is increased by raising the accelerating potential difference between the cathode and the anode, since each electron gains kinetic energy \( \tfrac{1}{2}mv^2 = eV\).
Bayanin Amsa
(a) Two properties of cathode rays:
- They travel in straight lines and cast sharp shadows of objects placed in their path.
- They are negatively charged, so they are deflected by both electric and magnetic fields.
- They possess kinetic energy (produce heat) and cause certain substances to fluoresce; they travel from cathode to anode.
(b) Increasing the intensity and energy of cathode rays:
- Intensity (the number of electrons emitted per second) is increased by raising the temperature of the cathode, i.e. increasing the heating (filament) current, so that more electrons are emitted by thermionic emission.
- Energy (the speed/kinetic energy of the electrons) is increased by raising the accelerating potential difference between the cathode and the anode, since each electron gains kinetic energy \( \tfrac{1}{2}mv^2 = eV\).
