WAEC SSCE - Physics - 2003
Tambaya 1 Rahoto
Which of the following condition is necessary for the occurrence of total internal reflection of light?
Bayanin Amsa
Total internal reflection occurs when light travels from a denser medium to a less dense medium, such as from water to air or from glass to air, and the angle of incidence is greater than the critical angle. The critical angle is the angle of incidence that produces an angle of refraction of 90 degrees. Therefore, the correct option is: "the angle of incidence must be greater than the critical angle".
Tambaya 2 Rahoto
A piece of metal of relative density 5.0 weighs 60N in air. Calculate its weight when fully immersed in water
Bayanin Amsa
The weight of an object is equal to the force of gravity acting on it. When an object is submerged in a fluid, it experiences a buoyant force, which is equal to the weight of the fluid displaced by the object. To calculate the weight of the metal when fully immersed in water, we need to first determine the volume of water displaced by the metal. We can do this by using the principle of Archimedes, which states that the buoyant force on an object is equal to the weight of the fluid displaced by the object. The weight of the metal in air is 60N, which is equal to its mass times the acceleration due to gravity. We can calculate the volume of the metal using its density and mass. Density is defined as mass per unit volume. Since the relative density of the metal is 5.0, its density is 5 times that of water, which has a density of 1000 kg/m^3. Density of metal = 5 x Density of water = 5 x 1000 kg/m^3 = 5000 kg/m^3 Mass of metal = Weight of metal / Acceleration due to gravity = 60N / 9.81 m/s^2 = 6.11 kg Volume of metal = Mass of metal / Density of metal = 6.11 kg / 5000 kg/m^3 = 0.001222 m^3 Since the metal is fully submerged in water, the volume of water displaced is equal to the volume of the metal. Weight of water displaced = Density of water x Volume of water displaced x Acceleration due to gravity = 1000 kg/m^3 x 0.001222 m^3 x 9.81 m/s^2 = 12N Therefore, the weight of the metal when fully immersed in water is 60N - 12N = 48N. Answer: 48N
Tambaya 3 Rahoto
Which of the following features of an electrostatic line of force is not correct?
Bayanin Amsa
The following feature of an electrostatic line of force is not correct: "it can cross another line of force in a region of intense electric field". Electrostatic lines of force are used to represent the direction and strength of an electric field. They are drawn such that at any point on a line of force, the tangent to the line gives the direction of the electric field at that point. Importantly, two lines of force cannot cross each other. This is because the direction of the electric field at any point must be uniquely defined, and if two lines crossed, there would be two different directions at the point of intersection. Therefore, the statement "it can cross another line of force in a region of intense electric field" is not correct.
Tambaya 4 Rahoto
The period of an oscillatory motion is defined as the
Bayanin Amsa
The period of an oscillatory motion is defined as the time taken to complete one oscillation. This means that if an object oscillates, the period is the time it takes to complete one cycle of its motion. For example, if a pendulum swings back and forth, the period is the time it takes to go from one side to the other and back again. The period is an important quantity in oscillatory motion because it can be used to determine the frequency of the motion, which is the number of oscillations per unit time.
Tambaya 5 Rahoto
The image which cannot be formed on a screen is said to be
Bayanin Amsa
The image which cannot be formed on a screen is said to be a virtual image. A virtual image is an image that appears to be behind the mirror or lens and cannot be projected onto a screen. It is formed by the apparent intersection of the light rays that are reflected or refracted by a mirror or lens. The rays of light do not actually converge to a point but only appear to do so. Virtual images are always erect (upright) and are usually smaller than the object. Examples of objects that produce virtual images include mirrors, lenses, and some optical instruments such as microscopes and telescopes.
Tambaya 6 Rahoto
A moving object is said to have a uniform acceleration if its
Bayanin Amsa
A moving object is said to have a uniform acceleration if its velocity increases by equal amount in equal time intervals. Uniform acceleration refers to a constant rate of change in velocity over time. This means that for every unit of time that passes, the velocity of the object changes by the same amount. For example, if an object has a uniform acceleration of 5 meters per second squared, its velocity will increase by 5 meters per second every second. This is also known as a constant acceleration, as opposed to a variable acceleration where the rate of change in velocity is not constant.
Tambaya 7 Rahoto
An object is placed on the principal axis and at the centre of curvature of a concave mirror, the image of the image formed by the mirror is
Bayanin Amsa
When an object is placed at the center of curvature of a concave mirror, its image will be formed at the center of curvature itself. In this case, the image formed by the mirror will be real and inverted. Therefore, the correct answer is real and inverted.
Tambaya 8 Rahoto
The mass on a loaded spiral spring oscillates vertically between two extreme positions P and R equidistant from the equilibrium position Q. Which of the following statements about the system is not correct
Bayanin Amsa
Tambaya 9 Rahoto
When the pressure of a fixed mass of gas is doubled at constant temperature, the volume of the gas is
Bayanin Amsa
According to Boyle's law, when the temperature of a fixed mass of gas is constant, the pressure and volume of the gas are inversely proportional. Therefore, if the pressure of a fixed mass of gas is doubled at constant temperature, the volume of the gas must be halved to maintain a constant temperature. This is because the gas molecules are forced to occupy a smaller volume due to the increased pressure. Therefore, the correct option is: - halved
Tambaya 11 Rahoto
The maximum power dissipated by a 100\(\Omega\) resistor in a circuit is 4W. Calculate the voltage across the resistor
Bayanin Amsa
The maximum power dissipated by a resistor in a circuit occurs when the resistor is operating at its maximum voltage. The power dissipated by a resistor can be calculated as: P = V^2 / R where P is the power in watts, V is the voltage across the resistor in volts, and R is the resistance in ohms. In this case, the power is given as 4W and the resistance is 100Ω, so we can rearrange the formula to solve for V: V^2 = P * R V = sqrt(P * R) Substituting the values given, we get: V = sqrt(4W * 100Ω) = sqrt(400V^2) = 20V Therefore, the voltage across the resistor is 20V. Answer: 20V
Tambaya 12 Rahoto
Which of the following statements correctly defines a simple machine? a device
Bayanin Amsa
The statement that correctly defines a simple machine is: "a device with which work can be done easily". A simple machine is a mechanical device that can change the magnitude or direction of a force, making it easier to perform work. Simple machines include the lever, pulley, wheel and axle, inclined plane, wedge, and screw. By using a simple machine, the force needed to perform a task can be reduced, but the distance through which that force must be applied is increased. This trade-off allows us to apply smaller forces over longer distances to accomplish tasks that would otherwise require much greater forces over shorter distances.
Tambaya 13 Rahoto
Two identical cells each of e.m.f 2v and internal resistance 1.0\(\Omega\) are connected in parallel. The combination is connected to an external load of 1.5\(\Omega\). Calculate the current in the circuit
Bayanin Amsa
To calculate the current in the circuit, we can use the following steps: 1. Find the equivalent internal resistance of the two cells in parallel: When two resistors are connected in parallel, their equivalent resistance is given by: 1/Req = 1/R1 + 1/R2 + ... Here, R1 = R2 = 1.0\(\Omega\) So, 1/Req = 1/1.0\(\Omega\) + 1/1.0\(\Omega\) => 1/Req = 2/1.0\(\Omega\) => Req = 0.5\(\Omega\) 2. Find the total e.m.f of the two cells in parallel: When two e.m.f sources are connected in parallel, their total e.m.f is the same as the e.m.f of each cell, i.e., 2V in this case. 3. Calculate the total current in the circuit: Using Ohm's law, the total current in the circuit is given by: I = E / (Rext + Req) Here, E is the total e.m.f (2V), Rext is the external resistance (1.5\(\Omega\)), and Req is the equivalent internal resistance of the two cells (0.5\(\Omega\)). => I = 2V / (1.5\(\Omega\) + 0.5\(\Omega\)) => I = 2V / 2\(\Omega\) => I = 1.0A Therefore, the current in the circuit is 1.0A. The correct answer is.
Tambaya 14 Rahoto
The bar magnet illustrated in the diagram above picks up an unmagnetised steel nail X. In turn, X picks up a similar nail Y and so on. The nail are said to be magnetised by
Bayanin Amsa
The nail X is picked up by the bar magnet because of the magnetic field produced by the magnet. When nail X comes into contact with nail Y, it induces a temporary magnetism in nail Y due to the magnetic field of nail X. This temporary magnetism in nail Y allows it to attract another similar nail Z, and the process continues. The process by which nail X magnetizes nail Y is called "induction." Induction is the process by which a magnetic field induces a magnetic field in a nearby object without physical contact. In this case, the magnetic field of nail X induces a temporary magnetic field in nail Y, making it temporarily magnetized. Therefore, the nails are magnetized by induction. The correct answer is.
Tambaya 15 Rahoto
A converging lens has a focal length of 5cm. Determine its power
Bayanin Amsa
Power is defined as the reciprocal of the focal length of a lens, expressed in meters and denoted by the unit dioptre (D). To convert from centimeters to meters, we divide by 100. Thus, the power of the lens is: P = 1/f = 1/0.05m = 20 D Therefore, the answer is +20.0.
Tambaya 16 Rahoto
A radioactive element has a decay constant of 0.077s-1, calculate its half life
Bayanin Amsa
The decay constant (λ) of a radioactive element is defined as the probability of decay of an individual nucleus per unit time. It can be mathematically shown that the number of nuclei (N) of a radioactive element remaining after time (t) is given by: N = N0 e-λt where N0 is the initial number of nuclei. The half-life of a radioactive element is the time taken for half of the radioactive nuclei to decay. Therefore, when N = 1/2 N0, we can rearrange the above equation to obtain the half-life (t1/2): t1/2 = ln2/λ Substituting the given value of the decay constant, we get: t1/2 = ln2/0.077 s-1 ≈ 9.0 s Therefore, the half-life of the radioactive element is approximately 9.0 s. Answer: 9.0s
Tambaya 17 Rahoto
A sonometer wire of linear density 0.08kg m-1 subjected to a tension of 800N is plucked. Calculate the speed of a pulse which moves from one end of the wire to the other
Bayanin Amsa
The speed of a pulse that moves through a wire is given by the formula: v = √(T/μ) where T is the tension in the wire and μ is the linear density of the wire. Substituting the given values into the formula, we get: v = √(800 N / 0.08 kg/m) v = √10000 m2/s2 v = 100 m/s Therefore, the speed of the pulse that moves from one end of the wire to the other is 100.0 m/s. Answer: 100.0 ms-1
Tambaya 18 Rahoto
A platinum resistance thermometer has a resistance 4\(\Omega\) at 0oC and 10\(\Omega\) at 100oC. Assuming the resistance changes uniformly with temperature. Calculate the resistance of the thermometer when the temperature is 45oC.
Bayanin Amsa
The resistance of a platinum resistance thermometer changes uniformly with temperature. Let's use the formula for the resistance-temperature relationship of a platinum resistance thermometer: R = R₀(1 + αT) Where R is the resistance at a given temperature T, R₀ is the resistance at 0°C (4Ω in this case), and α is the temperature coefficient of resistance of platinum. We can find α by using the resistance values at two different temperatures, say 0°C and 100°C: α = (R₁ - R₀) / (R₀ × ΔT) Where ΔT is the temperature difference between the two resistance values (100°C - 0°C = 100°C), and R₁ is the resistance at the higher temperature (10Ω in this case). α = (10Ω - 4Ω) / (4Ω × 100°C) = 0.015Ω/°C Now we can use the formula to find the resistance at 45°C: R = R₀(1 + αT) = 4Ω(1 + 0.015Ω/°C × 45°C) ≈ 6.7Ω Therefore, the answer is 6.7Ω.
Tambaya 19 Rahoto
Which of the following statements is not true of the isotopes of an element? They
Bayanin Amsa
The statement that is not true of the isotopes of an element is "have the same mass number". Isotopes are atoms of the same element that have the same atomic number (same number of protons in the nucleus), but a different mass number (different number of neutrons in the nucleus). Therefore, isotopes have the same chemical properties, since chemical properties are determined by the number and arrangement of electrons in the atom, which is determined by the atomic number. However, because isotopes have a different number of neutrons, they may have slightly different physical properties, such as different atomic masses, densities, and radioactive properties.
Tambaya 20 Rahoto
The volume of a given mass of an ideal gas at 327k and 9.52 x 104 Pa is 40cm3. calculate the volume of the gas at 273k and 1.034 x 105Pa.
Bayanin Amsa
Tambaya 21 Rahoto
Which of the following instruments gives a pure note when sounded?
Bayanin Amsa
A pure note is a sound wave of a single frequency or pitch without any harmonics. The instrument that can produce a pure note when sounded is a tuning fork. A tuning fork is a small metal object with two tines that vibrate when struck against a surface. The vibration of the tines creates a pure tone with a specific frequency, which is determined by the size, shape, and material of the tuning fork. This pure tone makes a tuning fork useful in tuning musical instruments or testing the frequency response of audio equipment.
Tambaya 23 Rahoto
A lamp is rated 240V 60W, calculate the resistance of its filament
Bayanin Amsa
To calculate the resistance of the filament in a lamp rated 240V and 60W, we can use Ohm's Law and the formula for electrical power. Ohm's Law states that resistance (R) is equal to voltage (V) divided by current (I), or R = V/I. The formula for electrical power (P) is given by P = IV, where I is the current in amperes. First, we can use the formula for electrical power to find the current in the circuit. We know that the lamp is rated 60W and 240V, so we can write: P = IV 60W = 240V x I I = 60W / 240V I = 0.25A Next, we can use Ohm's Law to find the resistance of the filament: R = V/I R = 240V / 0.25A R = 960Ω Therefore, the resistance of the filament in the lamp is 960Ω. The correct answer is (iii) 960Ω.
Tambaya 24 Rahoto
Nuclear fission is preferred to nuclear fusion in the generation of energy because?
Bayanin Amsa
Tambaya 25 Rahoto
The heat capacity of a calorimeter is the amount of energy required to
Bayanin Amsa
The heat capacity of a calorimeter is the amount of energy required to change the temperature of the calorimeter itself. Specifically, it is the amount of energy required to raise the temperature of the calorimeter by 1 Kelvin (or 1 degree Celsius). A calorimeter is a device used to measure the heat absorbed or released during a chemical or physical change. To do this, the calorimeter must be able to measure the change in temperature of the substances inside it. However, the calorimeter itself can also absorb or release heat during the process. The heat capacity of the calorimeter takes this into account by telling us how much energy is needed to change its own temperature. So, the correct answer is (c) change the temperature of the calorimeter.
Tambaya 26 Rahoto
The direction of induced current in a straight wire placed in magnetic field is determined by using?
Bayanin Amsa
The direction of the induced current in a straight wire placed in a magnetic field is determined by Lenz's law. Lenz's law states that the direction of the induced current in a conductor is such that it opposes the change in the magnetic flux that produced it. When a wire is placed in a magnetic field and the magnetic field is changing, an electromotive force (emf) is induced in the wire. This emf causes a current to flow in the wire, which creates its own magnetic field. According to Lenz's law, the direction of this induced current is such that it opposes the change in the magnetic flux that produced it. To determine the direction of the induced current in a straight wire placed in a magnetic field, we use Lenz's law along with the right-hand rule. The right-hand rule states that if we hold our right hand with our thumb pointing in the direction of the magnetic field and our fingers curled in the direction of the motion of the conductor, then the direction of the induced current can be determined by the direction in which the fingers of the right hand point. In summary, Lenz's law is used to determine the direction of the induced current in a straight wire placed in a magnetic field. The right-hand rule is used in conjunction with Lenz's law to determine the direction of the induced current by using the direction of the magnetic field and the direction of motion of the conductor.
Tambaya 27 Rahoto
A ball f mass 5.0kg hits a smooth vertical wall normally with a speed of 2ms-1. Determine the magnitude of the resulting impulse
Tambaya 28 Rahoto
The minimum energy required to remove an electron from an atom is known as
Bayanin Amsa
The minimum energy required to remove an electron from an atom is known as ionization energy. It is the energy needed to overcome the attractive force between the negatively charged electrons and positively charged nucleus of an atom. Once an electron is removed from the atom, the atom becomes a positively charged ion. The ionization energy varies depending on the atomic number of the element and the number of electrons in the outermost shell. The ionization energy is an important property of atoms and is used to determine their reactivity and chemical behavior.
Tambaya 29 Rahoto
The diagram above shows the resultant electric field pattern due to two electric point charges P and S. Which of the following statements is correct?
Bayanin Amsa
Tambaya 30 Rahoto
One difference between a reactance and a resistance in an alternating current circuit is
Bayanin Amsa
In an alternating current circuit, resistance and reactance are two important properties that determine the flow of current. Resistance is a property that resists the flow of current in a circuit and dissipates energy in the form of heat. On the other hand, reactance is a property that opposes the flow of current due to the presence of capacitance or inductance in the circuit. One main difference between resistance and reactance is that energy is dissipated in a resistance, while energy is not dissipated in a reactance. In a resistance, the energy of the current is converted into heat, and some of the electrical energy is lost in the form of heat. In contrast, a reactance stores energy in the form of an electric or magnetic field and releases it back into the circuit when the polarity of the current changes. Therefore, a reactance does not dissipate energy but stores and releases it. Another difference is that the current in a reactance can be lower than in a resistance, depending on the frequency of the alternating current and the type of reactance (capacitive or inductive). In a capacitive reactance, the current leads the voltage, while in an inductive reactance, the current lags the voltage. This means that the reactance can cause the current to be out of phase with the voltage, resulting in a lower net current flowing through the circuit. In summary, resistance dissipates energy in the form of heat while reactance stores and releases energy. The current in a reactance can be lower than in a resistance depending on the frequency of the alternating current and the type of reactance.
Tambaya 31 Rahoto
A converging lens produces an image four times as large as an object placed 25cm from the lens. Calculate its focal length
Bayanin Amsa
Tambaya 33 Rahoto
A positively charged glass rod is placed by near the cap of a positively charge electroscope. The divergence of the leaf is observed to
Bayanin Amsa
When a positively charged glass rod is placed near the cap of a positively charged electroscope, it will repel the positive charges in the electroscope, causing the charges to move away from the cap and towards the leaves. As a result, the leaves will experience a repulsive force and will diverge, indicating that the electroscope is charged. If the glass rod is brought closer to the cap, the repulsive force will become stronger, causing the leaves to diverge further. Therefore, the divergence of the leaves will increase. So the correct answer is "increase".
Tambaya 35 Rahoto
The tendency of a body to remain at least when a force is supplied to it is called
Bayanin Amsa
The tendency of a body to remain at rest or in a state of uniform motion in a straight line unless acted upon by an external force is called inertia. In simpler terms, it means that a stationary object will stay at rest, and a moving object will continue to move in a straight line unless a force is applied to it. Therefore, the correct option is "inertia." - Impulse is the change in momentum resulting from a force applied to an object for a specific amount of time. - Momentum is the product of an object's mass and velocity, and it describes how difficult it is to stop the object. - Friction is a force that opposes motion between two surfaces in contact.
Tambaya 36 Rahoto
a metal rod of length 100cm is heated through 100oC, calculate the change in length of the rod [linear expansivity of the material of the rod is 3 x 10-6K-1]
Bayanin Amsa
The change in length of a material can be determined using the formula: ΔL = LαΔT where ΔL is the change in length, L is the original length of the material, α is the coefficient of linear expansion of the material, and ΔT is the change in temperature. In this case, the length of the rod is 100cm, the coefficient of linear expansion is 3 x 10^-6 K^-1, and the change in temperature is 100oC. Substituting these values into the formula, we have: ΔL = (100cm)(3 x 10^-6 K^-1)(100oC) ΔL = 0.3cm or 3mm Therefore, the change in length of the rod is 3mm. The answer is (B) 3mm.
Tambaya 37 Rahoto
A wave of weight length 0.30m travels 900m in 3.0s, calculate its frequency
Bayanin Amsa
The formula for the wave speed is given by: wave speed = frequency x wavelength We know that the wavelength of the wave is 0.30m, and it travels a distance of 900m in 3.0s. Therefore, the wave speed can be calculated as: wave speed = distance/time = 900m/3.0s = 300m/s We can then rearrange the formula to solve for the frequency: frequency = wave speed / wavelength Substituting the given values: frequency = 300m/s / 0.30m = 1000 Hz Therefore, the frequency of the wave is 1000 Hz. Answer: 3) 1000Hz
Tambaya 38 Rahoto
A current of 10A passes through a conductor for 10s, calculate the charge flowing through the conductor
Bayanin Amsa
The charge flowing through a conductor is given by the product of current and time. The equation for charge is Q = I x t. In this case, the current is 10A and the time is 10s, therefore: Q = 10A x 10s = 100C Hence, the charge flowing through the conductor is 100C. Therefore, the correct option is (a) 100.0C.
Tambaya 39 Rahoto
Which of the following reactions represents nuclear fusion?
Bayanin Amsa
Nuclear fusion is a process where two or more atomic nuclei combine to form a heavier nucleus. In the given options, the reaction that represents nuclear fusion is: \( ^2_1H + ^2_1H \to ^3_2He + ^1_0n\) In this reaction, two hydrogen nuclei (also called protons) combine to form a helium nucleus (also called an alpha particle) and a neutron. This is the process that occurs in the sun and other stars, where the high temperature and pressure allow nuclear fusion to take place, releasing large amounts of energy.
Tambaya 40 Rahoto
Which of the following statements about the human eye and the camera is correct?
Tambaya 41 Rahoto
A force acts on a body for 0.5s changing its momentum from 16.0kg ms-1 to 21.0kgms-1, calculate the magnitude of the force
Bayanin Amsa
The magnitude of the force F can be calculated using the formula: F = Δp/Δt where Δp is the change in momentum and Δt is the time taken for the change. In this case, Δp = final momentum - initial momentum = 21.0 kg ms^-1 - 16.0 kg ms^-1 = 5.0 kg ms^-1 Δt = 0.5 s Substituting these values into the formula: F = Δp/Δt = 5.0 kg ms^-1 / 0.5 s = 10.0 N Therefore, the magnitude of the force is 10.0 N. Answer: 10.0N
Tambaya 42 Rahoto
Which of the following statement about latent heat of vaporization is correct? It
Bayanin Amsa
Tambaya 43 Rahoto
A tuber of cassava can be processed into powder. This explains the face that
Bayanin Amsa
Tambaya 44 Rahoto
Which of the following statements is true of ultra violet radiations? It
Bayanin Amsa
Ultra violet radiation is an electromagnetic wave. Electromagnetic waves include radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. These waves have different frequencies and wavelengths, with ultraviolet radiation having a higher frequency and shorter wavelength than visible light. Ultra violet radiation is capable of causing sunburns because it has enough energy to break the bonds of molecules in our skin, causing damage to our cells. This can lead to skin cancer and other health problems. Therefore, the statement that is true about ultraviolet radiation is that it is an electromagnetic wave. The correct answer is.
Tambaya 45 Rahoto
Which of the following factors does not affect the the electric resistance of a wire?
Bayanin Amsa
The factor that does not affect the electric resistance of a wire is "mass". Mass is not a factor that determines the resistance of a wire. The resistance of a wire is dependent on its length, cross-sectional area, temperature, and the type of material it is made of. The length of a wire is directly proportional to its resistance; the longer the wire, the higher the resistance. The cross-sectional area of a wire is inversely proportional to its resistance; the larger the area, the lower the resistance. The temperature of a wire affects its resistance because resistance increases with temperature for most materials. Finally, different materials have different resistance properties, and that affects the resistance of the wire.
Tambaya 46 Rahoto
The horizontal floor of a water reservoir appears to be 1.0m deep when viewed vertically from above. If the refractive index of water is 1.35, calculate the real depth of the reservoir
Bayanin Amsa
Tambaya 47 Rahoto
(a) List two types of waves, apart from light, that can be plane polarized.
(b) State two applications of plane polarized light.
(a) Waves (other than light) that can be plane polarized
Only transverse waves can be polarized. Examples apart from light are:
- Radio waves (and microwaves) - transverse electromagnetic waves.
- Transverse mechanical waves on a stretched rope or string (X-rays are also acceptable).
(b) Two applications of plane polarized light
- In polaroid sunglasses and camera filters to cut down glare from reflected light.
- In liquid-crystal displays (LCDs); also used in stress analysis of transparent models (photoelasticity) and in the polarimeter to measure sugar concentration.
Bayanin Amsa
(a) Waves (other than light) that can be plane polarized
Only transverse waves can be polarized. Examples apart from light are:
- Radio waves (and microwaves) - transverse electromagnetic waves.
- Transverse mechanical waves on a stretched rope or string (X-rays are also acceptable).
(b) Two applications of plane polarized light
- In polaroid sunglasses and camera filters to cut down glare from reflected light.
- In liquid-crystal displays (LCDs); also used in stress analysis of transparent models (photoelasticity) and in the polarimeter to measure sugar concentration.
Tambaya 48 Rahoto
a) (i) Illustrate, using a ray diagram, how an image can be formed by a convex mirror.
(ii) State one advantage rid one disadvantage of using a convex mirror as a iving mirror.
(iii) Explain the action of a compound microscope.
(b) Illustrate using labelled diagrams only, sonometer wire of length I, vibrating at its fundamental (ii) first overtone (iii) second overtone
(c) A tuning fork vibrating at a frequency of 512 Hz is held over the top of a jar filled with water and fitted with a tap at the buttom. If the jar is 60 cm tall and the speed of sound is 350 ms\(^{-1}\), determine the possible resonance position(s).
(a)(i) Formation of an image by a convex mirror
The reflected rays diverge, but their backward extensions meet behind the mirror. Hence the image is virtual, erect and diminished, and is formed between the pole, P, and the principal focus, F.
(a)(ii) Convex mirror as a driving mirror
- Advantage: It provides a wide field of view and gives an erect image of vehicles behind.
- Disadvantage: The image is diminished, so objects appear farther away than they actually are and distance may be misjudged.
(a)(iii) Action of a compound microscope
A compound microscope consists of two converging lenses of short focal lengths: the objective lens and the eyepiece lens. The object is placed just beyond the focal point of the objective lens. The objective forms a real, inverted and magnified intermediate image within the focal length of the eyepiece. The eyepiece acts as a simple magnifying glass and produces a final image which is virtual, highly magnified and inverted relative to the original object.
(b) Modes of vibration of a sonometer wire
(c) Resonance positions
The air column above the water is closed at the water surface and open at the top. Thus, resonance occurs at air-column lengths:
\[L=\frac{\lambda}{4},\quad \frac{3\lambda}{4},\quad \frac{5\lambda}{4},\ldots\]
\[\lambda=\frac{v}{f}=\frac{350}{512}=0.6836\ \text{m}\approx0.684\ \text{m}\]
First resonance position:
\[L_1=\frac{\lambda}{4}=\frac{0.6836}{4}=0.1709\ \text{m}\approx17.1\ \text{cm}\]
Second resonance position:
\[L_2=\frac{3\lambda}{4}=\frac{3(0.6836)}{4}=0.5127\ \text{m}\approx51.3\ \text{cm}\]
Third resonance position:
\[L_3=\frac{5\lambda}{4}=\frac{5(0.6836)}{4}=0.8545\ \text{m}\approx85.5\ \text{cm}\]
Since the jar is only \(60\ \text{cm}\) high, the third resonance is not possible. Therefore, the possible resonance air-column lengths are 17.1 cm and 51.3 cm, measured downwards from the open top of the jar.
Bayanin Amsa
(a)(i) Formation of an image by a convex mirror
The reflected rays diverge, but their backward extensions meet behind the mirror. Hence the image is virtual, erect and diminished, and is formed between the pole, P, and the principal focus, F.
(a)(ii) Convex mirror as a driving mirror
- Advantage: It provides a wide field of view and gives an erect image of vehicles behind.
- Disadvantage: The image is diminished, so objects appear farther away than they actually are and distance may be misjudged.
(a)(iii) Action of a compound microscope
A compound microscope consists of two converging lenses of short focal lengths: the objective lens and the eyepiece lens. The object is placed just beyond the focal point of the objective lens. The objective forms a real, inverted and magnified intermediate image within the focal length of the eyepiece. The eyepiece acts as a simple magnifying glass and produces a final image which is virtual, highly magnified and inverted relative to the original object.
(b) Modes of vibration of a sonometer wire
(c) Resonance positions
The air column above the water is closed at the water surface and open at the top. Thus, resonance occurs at air-column lengths:
\[L=\frac{\lambda}{4},\quad \frac{3\lambda}{4},\quad \frac{5\lambda}{4},\ldots\]
\[\lambda=\frac{v}{f}=\frac{350}{512}=0.6836\ \text{m}\approx0.684\ \text{m}\]
First resonance position:
\[L_1=\frac{\lambda}{4}=\frac{0.6836}{4}=0.1709\ \text{m}\approx17.1\ \text{cm}\]
Second resonance position:
\[L_2=\frac{3\lambda}{4}=\frac{3(0.6836)}{4}=0.5127\ \text{m}\approx51.3\ \text{cm}\]
Third resonance position:
\[L_3=\frac{5\lambda}{4}=\frac{5(0.6836)}{4}=0.8545\ \text{m}\approx85.5\ \text{cm}\]
Since the jar is only \(60\ \text{cm}\) high, the third resonance is not possible. Therefore, the possible resonance air-column lengths are 17.1 cm and 51.3 cm, measured downwards from the open top of the jar.
Tambaya 49 Rahoto
(a) Define (i) Linear momentum; (ii) Impulse
(b) State the principle of conservation of linear momentum.
(c) A tractor of mass 5.0 x 10\(^{3}\)kg is used a tow a car of mass 2.5 x 103 kg. The tractor moved with a speed of 3.0 ms\(^{-1}\) just before the towing rope becomes taut. Calculate the:
(i) Speed of the tractor immediately the rope becomes taut
(ii) Loss in kinetic energy of the system just after the car has started Moving;
(iii) Impulse in the rope when it jerks the car into motion.
Bayanin Amsa
None
Tambaya 50 Rahoto
(a)(i) What is meant by neutral point in a magnetic field?
(ii) Draw and label a diagram to show the pattern and direction of the magnetic field produced around a straight current-carrying wire.
(b) When is an ammeter said to be (i) Sensitive (ii) accurate?
(c)(i) Explain, using a labelled diagram, how a delicate magnetic material could be protected, from the Earth's magnetic field.
(ii) A charge of 1.6 x 10\(^{-19}\) C enters a magnetic field of flux density 2.0 T with a velocity of 2.5 x 10\(^7\) ms\(^{-1}\) at an angle of 30° with the field. Calculate the magnitude of the force exerted on the charge by the field.
(d) State the laws of electro-magnetic induction.
(a)(i) Neutral point
A neutral point is a point at which the resultant magnetic flux density is zero. At this point, the magnetic fields acting there are equal in magnitude and opposite in direction.
(a)(ii) Magnetic field around a straight current-carrying wire
The field consists of concentric circles centred on the wire. For a current out of the page, the direction of the field is anticlockwise, as shown.
(b) Ammeter
(i) Sensitive: An ammeter is sensitive when it produces an appreciable or large deflection for a very small current.
(ii) Accurate: An ammeter is accurate when the current it measures is very close to the true value of the current flowing in the circuit.
(c)(i) Protection from the Earth's magnetic field
The delicate magnetic material is enclosed in a thick soft-iron or other high-permeability magnetic shield. The shield provides an easier path for the Earth's magnetic field lines, so that very little magnetic field exists in the cavity containing the material.
(c)(ii) Force on the charge
The magnetic force is given by:
\[F=qvB\sin\theta\]
\[F=(1.6\times10^{-19})(2.5\times10^{7})(2.0)\sin30^\circ\]
\[F=(1.6\times10^{-19})(2.5\times10^{7})(2.0)(0.5)\]
\[\boxed{F=4.0\times10^{-12}\ \text{N}}\]
(d) Laws of electromagnetic induction
- Whenever the magnetic flux linkage through a circuit changes, an e.m.f. is induced in the circuit. The magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linkage.
- The induced current flows in such a direction as to oppose the change in magnetic flux that produces it.
Bayanin Amsa
(a)(i) Neutral point
A neutral point is a point at which the resultant magnetic flux density is zero. At this point, the magnetic fields acting there are equal in magnitude and opposite in direction.
(a)(ii) Magnetic field around a straight current-carrying wire
The field consists of concentric circles centred on the wire. For a current out of the page, the direction of the field is anticlockwise, as shown.
(b) Ammeter
(i) Sensitive: An ammeter is sensitive when it produces an appreciable or large deflection for a very small current.
(ii) Accurate: An ammeter is accurate when the current it measures is very close to the true value of the current flowing in the circuit.
(c)(i) Protection from the Earth's magnetic field
The delicate magnetic material is enclosed in a thick soft-iron or other high-permeability magnetic shield. The shield provides an easier path for the Earth's magnetic field lines, so that very little magnetic field exists in the cavity containing the material.
(c)(ii) Force on the charge
The magnetic force is given by:
\[F=qvB\sin\theta\]
\[F=(1.6\times10^{-19})(2.5\times10^{7})(2.0)\sin30^\circ\]
\[F=(1.6\times10^{-19})(2.5\times10^{7})(2.0)(0.5)\]
\[\boxed{F=4.0\times10^{-12}\ \text{N}}\]
(d) Laws of electromagnetic induction
- Whenever the magnetic flux linkage through a circuit changes, an e.m.f. is induced in the circuit. The magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linkage.
- The induced current flows in such a direction as to oppose the change in magnetic flux that produces it.
Tambaya 51 Rahoto
(a) Explain the term electrolyte and give two examples.
(b) State the relationship between the mass of a substance liberated during electrolysis and the charge passed.
(a) Electrolyte
An electrolyte is a substance (usually a molten ionic compound or an aqueous solution of an acid, base or salt) that conducts electric current by the movement of its free ions and is chemically decomposed in the process.
Two examples:
- Dilute tetraoxosulphate(VI) acid (dilute H2SO4).
- Copper(II) tetraoxosulphate(VI) solution (CuSO4 solution); acidified water or sodium chloride solution are also acceptable.
(b) Mass liberated and charge passed
By Faraday's first law of electrolysis, the mass m of a substance liberated at an electrode is directly proportional to the quantity of electric charge Q that passes through the electrolyte: \[ m \propto Q \Rightarrow m = z Q = z I t \]
where z is the electrochemical equivalent of the substance, I the current and t the time.
Bayanin Amsa
(a) Electrolyte
An electrolyte is a substance (usually a molten ionic compound or an aqueous solution of an acid, base or salt) that conducts electric current by the movement of its free ions and is chemically decomposed in the process.
Two examples:
- Dilute tetraoxosulphate(VI) acid (dilute H2SO4).
- Copper(II) tetraoxosulphate(VI) solution (CuSO4 solution); acidified water or sodium chloride solution are also acceptable.
(b) Mass liberated and charge passed
By Faraday's first law of electrolysis, the mass m of a substance liberated at an electrode is directly proportional to the quantity of electric charge Q that passes through the electrolyte: \[ m \propto Q \Rightarrow m = z Q = z I t \]
where z is the electrochemical equivalent of the substance, I the current and t the time.
Tambaya 52 Rahoto
(a) Using the kinetic theory of matter, explain why;
(i) Evaporation causes cooling
(ii) Boiling water changes to steam without any change in temperature, although heat is being supplied to the water.
(b) (i) State Boyle's law.
(ii) With the aid of a labelled diagram, describe an experiment to illustrate the relationship between the volume and pressure of a given mass of gas at constant temperature. (iii) State two precautions necessary to obtain accurate results.
(a) Kinetic theory of matter
(i) Evaporation causes cooling: Molecules in a liquid are in continual random motion and possess different kinetic energies. The molecules with the greatest kinetic energy near the surface escape from the liquid during evaporation. Their escape reduces the average kinetic energy of the molecules left in the liquid. Since temperature is proportional to the average kinetic energy of the molecules, the temperature of the liquid falls; hence evaporation causes cooling.
(ii) Water boils without a rise in temperature: At its boiling point, the heat supplied to water is used as latent heat of vaporization. It is used to overcome the attractive intermolecular forces and to separate the water molecules to form steam. It does not increase the average kinetic energy of the molecules. Therefore, the temperature remains constant until all the water has changed to steam.
(b)(i) Boyle's law
For a fixed mass of gas at constant temperature, the volume of the gas is inversely proportional to its pressure.
Thus,
\[V\propto\frac{1}{P}\qquad\text{or}\qquad PV=\text{constant}.\]
(b)(ii) Experiment
The apparatus is arranged as shown below. A quantity of dry air is trapped in the graduated burette B by closing the tap. The burette is kept vertical. The mercury reservoir R is raised slowly to compress the trapped air. For each setting, read the volume, \(V\), of trapped air in the burette and the difference, \(h\), between the mercury levels. The barometer reading, \(H\), is also noted.
When the mercury level in the reservoir is higher than that in the burette, the pressure of the trapped air is:
\[P=H+h.\]
For example, if \(H=760\ \text{mmHg}\), the following readings may be obtained:
| Mercury head, \(h\) (mm) | Pressure, \(P=760+h\) (mmHg) | Volume, \(V\) (cm3) | \(1/V\) (cm−3) | \(PV\) (mmHg cm3) |
|---|---|---|---|---|
| 0 | 760 | 30.00 | 0.03333 | 22 800 |
| 95 | 855 | 26.67 | 0.03750 | 22 800 |
| 190 | 950 | 24.00 | 0.04167 | 22 800 |
| 380 | 1140 | 20.00 | 0.05000 | 22 800 |
| 760 | 1520 | 15.00 | 0.06667 | 22 800 |
A plot of \(P\) against \(1/V\) is a straight line through the origin:
The gradient of the graph is \(22\,800\ \text{mmHg cm}^3\). Hence \(P=(22\,800)(1/V)\), so \(PV=22\,800\ \text{mmHg cm}^3\), a constant. This verifies Boyle's law.
(b)(iii) Precautions
- Use dry air and ensure that the apparatus is airtight, so that the mass of trapped gas remains constant.
- Raise or lower the reservoir slowly and wait for the gas to return to room temperature before taking each reading.
- Take all scale readings at eye level to avoid parallax error.
Bayanin Amsa
(a) Kinetic theory of matter
(i) Evaporation causes cooling: Molecules in a liquid are in continual random motion and possess different kinetic energies. The molecules with the greatest kinetic energy near the surface escape from the liquid during evaporation. Their escape reduces the average kinetic energy of the molecules left in the liquid. Since temperature is proportional to the average kinetic energy of the molecules, the temperature of the liquid falls; hence evaporation causes cooling.
(ii) Water boils without a rise in temperature: At its boiling point, the heat supplied to water is used as latent heat of vaporization. It is used to overcome the attractive intermolecular forces and to separate the water molecules to form steam. It does not increase the average kinetic energy of the molecules. Therefore, the temperature remains constant until all the water has changed to steam.
(b)(i) Boyle's law
For a fixed mass of gas at constant temperature, the volume of the gas is inversely proportional to its pressure.
Thus,
\[V\propto\frac{1}{P}\qquad\text{or}\qquad PV=\text{constant}.\]
(b)(ii) Experiment
The apparatus is arranged as shown below. A quantity of dry air is trapped in the graduated burette B by closing the tap. The burette is kept vertical. The mercury reservoir R is raised slowly to compress the trapped air. For each setting, read the volume, \(V\), of trapped air in the burette and the difference, \(h\), between the mercury levels. The barometer reading, \(H\), is also noted.
When the mercury level in the reservoir is higher than that in the burette, the pressure of the trapped air is:
\[P=H+h.\]
For example, if \(H=760\ \text{mmHg}\), the following readings may be obtained:
| Mercury head, \(h\) (mm) | Pressure, \(P=760+h\) (mmHg) | Volume, \(V\) (cm3) | \(1/V\) (cm−3) | \(PV\) (mmHg cm3) |
|---|---|---|---|---|
| 0 | 760 | 30.00 | 0.03333 | 22 800 |
| 95 | 855 | 26.67 | 0.03750 | 22 800 |
| 190 | 950 | 24.00 | 0.04167 | 22 800 |
| 380 | 1140 | 20.00 | 0.05000 | 22 800 |
| 760 | 1520 | 15.00 | 0.06667 | 22 800 |
A plot of \(P\) against \(1/V\) is a straight line through the origin:
The gradient of the graph is \(22\,800\ \text{mmHg cm}^3\). Hence \(P=(22\,800)(1/V)\), so \(PV=22\,800\ \text{mmHg cm}^3\), a constant. This verifies Boyle's law.
(b)(iii) Precautions
- Use dry air and ensure that the apparatus is airtight, so that the mass of trapped gas remains constant.
- Raise or lower the reservoir slowly and wait for the gas to return to room temperature before taking each reading.
- Take all scale readings at eye level to avoid parallax error.
Tambaya 53 Rahoto
(a)(i) Explain photoelectric emission.
(ii) State four applications of photoelectric emission.
(b) Draw and label a diagram showing the structure of a simple type of a photocell and explain its mode of operation.
(c) In a photocell, no electrons are emitted until the threshold frequency of light is reached.
(i) Explain what happens to the energy of the light before emission of electrons begin.
(ii) State one factor that may affect the number of emitted electrons.
(a)(i) Photoelectric emission
Photoelectric emission is the emission of electrons from the surface of a metal when electromagnetic radiation of sufficiently high frequency falls on it. The minimum frequency capable of producing emission is called the threshold frequency.
(a)(ii) Applications of photoelectric emission
- Burglar alarm systems.
- Photographic exposure meters.
- Automatic doors and automatic street-light switches.
- Reproduction of sound from the sound track of films.
(b) Structure and operation of a simple photoelectric cell
The photocell consists of an evacuated glass bulb containing a curved photosensitive cathode and a central anode. The cathode is coated with a photosensitive material such as caesium. The anode is connected to the positive terminal of a battery through a microammeter.
When light of frequency at least equal to the threshold frequency falls on the cathode, photoelectrons are emitted. The positively charged anode attracts and collects these electrons. Their movement through the external circuit constitutes a photoelectric current, which is indicated by the microammeter. For radiation above the threshold frequency, increasing the intensity of the light increases the photoelectric current.
(c)(i) Before photoelectric emission begins, the light energy is absorbed by electrons in the photosensitive surface. If the frequency is below the threshold frequency, the energy supplied to each electron is insufficient to overcome the work function or potential barrier of the metal. The absorbed energy is therefore dissipated, mainly as heat, and no electrons are emitted.
(c)(ii) One factor affecting the number of electrons emitted is the intensity of the incident light. Greater intensity supplies more photons per second and hence emits more electrons, provided that the frequency is above the threshold frequency.
Bayanin Amsa
(a)(i) Photoelectric emission
Photoelectric emission is the emission of electrons from the surface of a metal when electromagnetic radiation of sufficiently high frequency falls on it. The minimum frequency capable of producing emission is called the threshold frequency.
(a)(ii) Applications of photoelectric emission
- Burglar alarm systems.
- Photographic exposure meters.
- Automatic doors and automatic street-light switches.
- Reproduction of sound from the sound track of films.
(b) Structure and operation of a simple photoelectric cell
The photocell consists of an evacuated glass bulb containing a curved photosensitive cathode and a central anode. The cathode is coated with a photosensitive material such as caesium. The anode is connected to the positive terminal of a battery through a microammeter.
When light of frequency at least equal to the threshold frequency falls on the cathode, photoelectrons are emitted. The positively charged anode attracts and collects these electrons. Their movement through the external circuit constitutes a photoelectric current, which is indicated by the microammeter. For radiation above the threshold frequency, increasing the intensity of the light increases the photoelectric current.
(c)(i) Before photoelectric emission begins, the light energy is absorbed by electrons in the photosensitive surface. If the frequency is below the threshold frequency, the energy supplied to each electron is insufficient to overcome the work function or potential barrier of the metal. The absorbed energy is therefore dissipated, mainly as heat, and no electrons are emitted.
(c)(ii) One factor affecting the number of electrons emitted is the intensity of the incident light. Greater intensity supplies more photons per second and hence emits more electrons, provided that the frequency is above the threshold frequency.
Tambaya 54 Rahoto
(a) What is the principle upon which the lighting in fluorescent tubes operate?
(b) State two factors which determine the colour of projection, g the acceleration of free fall due to gravity light from a fluorescent tube
(a) Principle of the fluorescent tube
It works on the principle of fluorescence. An electric discharge is passed through mercury vapour in the tube; the excited mercury atoms emit mostly ultraviolet radiation. This ultraviolet light strikes the phosphor (fluorescent) powder coating the inside of the tube, which absorbs it and re-emits the energy as visible light.
(b) Two factors that determine the colour of the emitted light
- The nature (chemical composition) of the phosphor coating on the inside of the tube.
- The type of gas or vapour enclosed in the tube. (The energy/wavelength of the exciting radiation is also acceptable.)
Bayanin Amsa
(a) Principle of the fluorescent tube
It works on the principle of fluorescence. An electric discharge is passed through mercury vapour in the tube; the excited mercury atoms emit mostly ultraviolet radiation. This ultraviolet light strikes the phosphor (fluorescent) powder coating the inside of the tube, which absorbs it and re-emits the energy as visible light.
(b) Two factors that determine the colour of the emitted light
- The nature (chemical composition) of the phosphor coating on the inside of the tube.
- The type of gas or vapour enclosed in the tube. (The energy/wavelength of the exciting radiation is also acceptable.)
Tambaya 55 Rahoto
Define;
(i) Elasticity
(ii) Young's modulus
(iii) Force constant
(i) Elasticity
Elasticity is the property of a material by which it returns to its original size and shape after the deforming force applied to it has been removed.
(ii) Young's modulus
Young's modulus (the modulus of elasticity) is the ratio of the tensile stress to the tensile strain of a material, provided the elastic limit is not exceeded: \[ E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{e/L} = \frac{FL}{Ae} \] Its S.I. unit is the pascal (N m-2).
(iii) Force constant
The force constant (stiffness) of an elastic material is the force required to produce unit extension in it: \[ k = \frac{F}{e} \] Its S.I. unit is newton per metre (N m-1).
Bayanin Amsa
(i) Elasticity
Elasticity is the property of a material by which it returns to its original size and shape after the deforming force applied to it has been removed.
(ii) Young's modulus
Young's modulus (the modulus of elasticity) is the ratio of the tensile stress to the tensile strain of a material, provided the elastic limit is not exceeded: \[ E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{e/L} = \frac{FL}{Ae} \] Its S.I. unit is the pascal (N m-2).
(iii) Force constant
The force constant (stiffness) of an elastic material is the force required to produce unit extension in it: \[ k = \frac{F}{e} \] Its S.I. unit is newton per metre (N m-1).
Tambaya 56 Rahoto
A force of 40 N applied at the end of a wire of length 4m and diameter 2.00 mm produces an extension of 0.24 mm. Calculate the;
(a) stress on the wire;
(b) strain in the wire (\(\pi = 3.142\))
Tambaya 57 Rahoto
Explain the rise of water in a glass capillary tube using the kinetic theory.
Rise of water in a glass capillary tube (kinetic theory)
The molecules of water and the molecules of glass are in constant motion and exert intermolecular forces. Two forces are involved: the adhesive force between water molecules and glass molecules, and the cohesive force between water molecules themselves.
For water and glass the adhesive force is stronger than the cohesive force. The water molecules are therefore attracted more strongly to the glass wall than to one another, so they climb up the sides of the tube and the water wets the glass, giving a concave meniscus.
This upward pull produces a surface-tension force acting all round the line of contact between the water and the glass. This force lifts the column of water, and the liquid rises up the narrow tube until the weight of the raised column exactly balances the upward surface-tension force. The narrower the tube, the smaller the weight to be supported, so the higher the water rises.
(For a liquid such as mercury, where cohesion exceeds adhesion, the reverse happens and the liquid is depressed below the outside level.)
Bayanin Amsa
Rise of water in a glass capillary tube (kinetic theory)
The molecules of water and the molecules of glass are in constant motion and exert intermolecular forces. Two forces are involved: the adhesive force between water molecules and glass molecules, and the cohesive force between water molecules themselves.
For water and glass the adhesive force is stronger than the cohesive force. The water molecules are therefore attracted more strongly to the glass wall than to one another, so they climb up the sides of the tube and the water wets the glass, giving a concave meniscus.
This upward pull produces a surface-tension force acting all round the line of contact between the water and the glass. This force lifts the column of water, and the liquid rises up the narrow tube until the weight of the raised column exactly balances the upward surface-tension force. The narrower the tube, the smaller the weight to be supported, so the higher the water rises.
(For a liquid such as mercury, where cohesion exceeds adhesion, the reverse happens and the liquid is depressed below the outside level.)
Tambaya 58 Rahoto
(a) State two objects each used in sports and warfare which may be considered as projectiles.
(b) the horizontal range R, of a projectile is given by the expression;
R = \(\frac{u^2 sin 2\theta}{g}\) where \(\theta\) is the angle of projection, g the acceleration of free fall due to gravity and u the initial velocity. At what value of \(\theta\) will R be maximium?
(a) Objects that behave as projectiles
- In sports: a thrown javelin, a kicked football (also a shot-put, discus, or golf ball).
- In warfare: a bullet fired from a gun, a launched bomb or missile (also a cannon ball or arrow).
(b) Angle for maximum range
The range is \[ R = \frac{u^2 \sin 2\theta}{g} \] For a fixed initial speed u and fixed g, R is greatest when \(\sin 2\theta\) is greatest. The maximum value of a sine function is 1, so \[ \sin 2\theta = 1 \Rightarrow 2\theta = 90^{\circ} \Rightarrow \theta = 45^{\circ} \]
The range is maximum when the angle of projection is 45°.
Bayanin Amsa
(a) Objects that behave as projectiles
- In sports: a thrown javelin, a kicked football (also a shot-put, discus, or golf ball).
- In warfare: a bullet fired from a gun, a launched bomb or missile (also a cannon ball or arrow).
(b) Angle for maximum range
The range is \[ R = \frac{u^2 \sin 2\theta}{g} \] For a fixed initial speed u and fixed g, R is greatest when \(\sin 2\theta\) is greatest. The maximum value of a sine function is 1, so \[ \sin 2\theta = 1 \Rightarrow 2\theta = 90^{\circ} \Rightarrow \theta = 45^{\circ} \]
The range is maximum when the angle of projection is 45°.
Tambaya 59 Rahoto
(a) What is meant by the wave-particles duality of matter?
(b) mention one physical phenomenon, in each case, that can be explained in terms of the wave nature and particle nature of light.
Tambaya 60 Rahoto
(a) Explain diffusion
(b) State one factor that can affect the rate of diffusion.
(a) Diffusion
Diffusion is the net movement of the molecules of a substance from a region of higher concentration to a region of lower concentration until they are evenly spread out. It occurs because molecules are in continuous random motion; the greater number in the crowded region means more of them, on average, drift towards the less crowded region, so the substance gradually spreads throughout the space available.
(b) One factor affecting the rate of diffusion
- Temperature (higher temperature gives faster molecular motion and quicker diffusion). Other acceptable factors: the concentration gradient, or the mass/density of the diffusing molecules.
Bayanin Amsa
(a) Diffusion
Diffusion is the net movement of the molecules of a substance from a region of higher concentration to a region of lower concentration until they are evenly spread out. It occurs because molecules are in continuous random motion; the greater number in the crowded region means more of them, on average, drift towards the less crowded region, so the substance gradually spreads throughout the space available.
(b) One factor affecting the rate of diffusion
- Temperature (higher temperature gives faster molecular motion and quicker diffusion). Other acceptable factors: the concentration gradient, or the mass/density of the diffusing molecules.
Tambaya 61 Rahoto
(a) What is surface tension?
(b) State two methods by which the surface tension of a liquid can be reduced.
(a) Surface tension
Surface tension is the property of the free surface of a liquid by which it behaves like a stretched elastic membrane, tending to contract to the smallest possible area. It is measured as the force acting at right angles per unit length along a line drawn in the surface: \[ \gamma = \frac{F}{L} \] with S.I. unit newton per metre (N m-1). It arises from the net inward cohesive pull on molecules at the surface.
(b) Two methods of reducing surface tension
- Adding a detergent, soap or other surface-active impurity to the liquid.
- Raising the temperature of the liquid.
Bayanin Amsa
(a) Surface tension
Surface tension is the property of the free surface of a liquid by which it behaves like a stretched elastic membrane, tending to contract to the smallest possible area. It is measured as the force acting at right angles per unit length along a line drawn in the surface: \[ \gamma = \frac{F}{L} \] with S.I. unit newton per metre (N m-1). It arises from the net inward cohesive pull on molecules at the surface.
(b) Two methods of reducing surface tension
- Adding a detergent, soap or other surface-active impurity to the liquid.
- Raising the temperature of the liquid.
