WAEC SSCE - Physics - 2012
Tambaya 1 Rahoto
A metal with a work function of 3.0ev is irradiated with ultraviolet light of energy 4.0 ev. The kinetic energy of the ejected electron is
Bayanin Amsa
The work function of a metal is the minimum amount of energy required to remove an electron from the surface of the metal. When the metal is irradiated with light of energy greater than the work function, electrons are emitted from the metal surface. The excess energy of the light is used to give kinetic energy to the ejected electron. In this case, the metal has a work function of 3.0eV and is irradiated with ultraviolet light of energy 4.0eV. Since the energy of the light is greater than the work function, electrons will be ejected from the metal surface. The kinetic energy of the ejected electron can be found by subtracting the work function from the energy of the incident light: Kinetic energy of ejected electron = Energy of incident light - Work function = 4.0eV - 3.0eV = 1.0eV Therefore, the kinetic energy of the ejected electron is 1.0eV.
Tambaya 2 Rahoto
In a R-C circuit
Bayanin Amsa
In an R-C circuit, a resistor R and a capacitor C are connected in series to an AC voltage source. When an AC voltage is applied to a capacitor in a circuit, it takes some time to fully charge, as the capacitor acts like an open circuit to DC, but passes AC through it. As a result, the current flowing through the circuit will lag the voltage by some angle. This angle is determined by the ratio of the resistance and capacitance in the circuit. Therefore, in a R-C circuit, Vrms leads Irms by 90o.
Tambaya 3 Rahoto
An object of height 2.5cm is places 20cm from a convex mirror of focal length 10cm. Calculate the height of its image.
Bayanin Amsa
The given problem is related to optics and involves the use of the mirror formula. According to the mirror formula, 1/f = 1/v + 1/u where, f = focal length of the mirror u = object distance v = image distance The magnification produced by the mirror is given by the ratio of the height of the image to the height of the object. magnification = height of image / height of object From the problem statement, height of object (h) = 2.5 cm distance of object from the mirror (u) = 20 cm focal length of the mirror (f) = 10 cm Using the mirror formula, we can find the image distance (v) as: 1/10 = 1/v + 1/20 => 1/v = 1/10 - 1/20 => v = 20 cm The image is formed at a distance of 20 cm from the mirror. Now, using the magnification formula, we can find the height of the image as: magnification = height of image / height of object => height of image = magnification * height of object The magnification produced by a convex mirror is always less than 1. Therefore, the image formed will be smaller than the object. For a convex mirror, magnification = -v/u Substituting the given values, we get: magnification = -20/20 = -1 Therefore, the height of the image is given by: height of image = magnification * height of object => height of image = -1 * 2.5 cm => height of image = -2.5 cm Since the height of the image cannot be negative, we take its absolute value: height of image = |-2.5 cm| = 2.5 cm Hence, the height of the image is 2.5 cm. Therefore, the correct option is (a) 2.5 cm.
Tambaya 4 Rahoto
An atom in an excited state is one whose
Bayanin Amsa
An atom in an excited state is one whose electrons have moved to higher energy levels. When an atom absorbs energy, its electrons move to higher energy levels, or orbitals, which are farther from the nucleus. When an electron moves from a lower energy level to a higher energy level, the atom is said to be in an excited state. Eventually, the electrons return to their original energy level by emitting energy in the form of electromagnetic radiation.
Tambaya 5 Rahoto
A block of wood of density 0.6 gcm\(^{-3}\), weighing 3.06N in air, floats freely in a liquid of density 0.9 gcm\(^{-3}\). Calculate volume of the portion immersed (g = 10ms\(^{-2}\))
Tambaya 6 Rahoto
Which of the following sets of coloured light is/are secondary colours i. Red, Blue and green ii. Blue, cyan and magneta iii. Green, Magenta and yellow, IV Yellow, Cyan and Magenta
Bayanin Amsa
The secondary colors of light are the colors that result from mixing two primary colors of light. The three primary colors of light are red, green, and blue. Option i consists of two primary colors of light (red and blue) and a secondary color of light (green). Therefore, it is not a set of secondary colors of light. Option ii consists of one primary color of light (blue) and two secondary colors of light (cyan and magenta). Therefore, it is a set of secondary colors of light. Option iii consists of two secondary colors of light (magenta and yellow) and a primary color of light (green). Therefore, it is not a set of secondary colors of light. Option iv consists of three secondary colors of light (cyan, magenta, and yellow). Therefore, it is a set of secondary colors of light. Thus, the correct answer is option ii and iv only.
Tambaya 7 Rahoto
Which of the following arrangements of radiations is in the order of decreasing penetrating power?
Bayanin Amsa
The penetrating power of radiation is a measure of its ability to penetrate through matter. The more penetrating the radiation is, the farther it can travel through matter. In general, the penetrating power of radiation is inversely proportional to its ionizing power. - Alpha particles (\(\alpha\)): They are positively charged and have a large mass. They have a low penetrating power and are stopped by a sheet of paper or a few centimeters of air. - Beta particles (\(\beta\)): They are negatively charged and have a smaller mass compared to alpha particles. They have a higher penetrating power than alpha particles and can penetrate through a few millimeters of aluminum or several meters of air. - Gamma rays (\(\gamma\)): They are neutral and have no mass. They have the highest penetrating power and can penetrate through several centimeters of lead or many meters of concrete. Therefore, the correct order of decreasing penetrating power is \(\gamma, \beta, \alpha\). lists the radiations in this order.
Tambaya 8 Rahoto
The time of flight for a projectile motion is given by the expression
Bayanin Amsa
The time of flight is the total time taken by a projectile to complete its motion before it returns to the same level from which it was projected. The given expression for time of flight involves the initial velocity of the projectile (U), the angle of projection (theta) and the acceleration due to gravity (g). Therefore, the expression for the time of flight of a projectile is: \(\frac{2U sin \theta}{g}\) Option (C) matches the given expression and is the correct answer.
Tambaya 9 Rahoto
A telescope is said to be in normal adjustment when the
Bayanin Amsa
A telescope is said to be in normal adjustment when the objective focal point coincides with that of the eyepiece. This means that the light rays coming in from the object being observed converge to a point at the same location where the eyepiece is positioned, allowing the viewer to see a magnified image. If the objective focal point is too far away from the eyepiece or if the focal lengths of the objective and eyepiece are not matched, the viewer will not be able to see a clear image. Therefore, adjusting the telescope to achieve normal adjustment is crucial for obtaining a clear, magnified image of the object being observed.
Tambaya 10 Rahoto
Two similar cells are used to light two similar lamps as illustrated in the diagrams above. In which of the circuit diagrams are the lamps brightest?
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Tambaya 11 Rahoto
A body is pulled through a distance of 500m by a force of 20N. If the power developed is 0.4kW, calculate the time for which the force acts
Tambaya 12 Rahoto
Which of the following devices is used to determine the relative density of an acid?
Bayanin Amsa
The device used to determine the relative density of an acid is a hydrometer. A hydrometer is an instrument used to measure the density or specific gravity of a liquid. It works based on Archimedes' principle, which states that the buoyant force on an object is equal to the weight of the fluid displaced by the object. The hydrometer consists of a glass tube with a bulb at the bottom, and a scale to measure the density. When the hydrometer is placed in the acid, it floats and sinks to a certain level depending on the density of the acid. The density is then read off the scale at the level where the surface of the liquid intersects the scale on the hydrometer. Therefore, a hydrometer is used to determine the relative density of an acid.
Tambaya 13 Rahoto
A girl stands on a scale in a lift. If the reading on the scale is less than her weight, then the lift is moving
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Tambaya 14 Rahoto
the core of a transformer is usually laminated in order to
Bayanin Amsa
The core of a transformer is laminated in order to reduce eddy currents. Eddy currents are created by a magnetic field that is changing over time, inducing a current in the core. This current can cause energy loss and heating of the core, leading to inefficiencies in the transformer. Laminating the core creates many thin layers of material, each separated by a thin layer of insulation. This structure reduces the size of the current loops that can form and reduces the resistance of the current flow in the material. This reduction in eddy current also reduces energy loss, leading to greater efficiency of the transformer.
Tambaya 15 Rahoto
The S.I unit of heat is
Bayanin Amsa
The SI unit of heat is the joule (J). Heat is a form of energy that is transferred from one body to another as a result of a difference in temperature. The amount of heat transferred is measured in joules, which is the amount of energy required to move an object with a force of one newton for a distance of one meter in the direction of the force. Therefore, the correct option is (a) joule. Kelvins (K) are used to measure temperature, watts (W) measure power and amperes (A) measure electric current.
Tambaya 16 Rahoto
The movement of fluid up or down a narrow tube is
Bayanin Amsa
The movement of fluid up or down a narrow tube is capillarity. Capillary action, also known as capillarity, is the ability of a liquid to flow in narrow spaces without the assistance of, or even in opposition to, external forces like gravity. This occurs because of the combination of two opposing forces: adhesion, the attraction between the liquid molecules and the inner surface of the tube, and cohesion, the attraction between the liquid molecules themselves. When the adhesion force is greater than the cohesion force, the liquid is pulled up the tube, and when the cohesion force is greater, the liquid is pushed down the tube. Capillarity is responsible for many natural phenomena, such as the ability of plants to transport water from their roots to their leaves, as well as some man-made ones, such as the operation of wicks in candles and the ink in fountain pens.
Tambaya 18 Rahoto
A resultant force of magnitude 15N acts on a body of mass 250g. Calculate the magnitude of the acceleration
Bayanin Amsa
The magnitude of the acceleration is given by the formula: acceleration = resultant force / mass In this question, the resultant force is given as 15N and the mass is 250g which is equivalent to 0.25kg. Therefore, acceleration = 15N / 0.25kg = 60.00ms-2 Hence, the correct option is 60.00ms-2.
Tambaya 19 Rahoto
An object is placed at a point X between the focal point F and the optical centre C of a diverging lens. If F is the focal point on the other side of the lens, the image of the object is formed between
Bayanin Amsa
For a diverging lens, the image formed is always virtual, upright and smaller than the object. When the object is placed between the focal point and the optical center of the lens, the image formed will always be virtual, upright and will appear on the same side of the object. Also, the image will be formed by the diverging lens itself, so it will be between the lens and the focal point F on the same side as the object. Therefore, the correct option is (B) X and C.
Tambaya 20 Rahoto
Which of the following substances is the most volatile at room temperature?
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Tambaya 21 Rahoto
A bird flies at 10ms\(^{-1}\) for 3s, 15ms\(^{-1}\) for 3s and 20 ms\(^{-1}\) for 4s. Calculate the bird's average speed
Bayanin Amsa
The average speed is defined as the total distance traveled divided by the total time taken. Let's first calculate the total distance traveled by the bird. Distance traveled in the first 3 seconds = speed × time = 10 ms\(^{-1}\) × 3 s = 30 m Distance traveled in the next 3 seconds = speed × time = 15 ms\(^{-1}\) × 3 s = 45 m Distance traveled in the next 4 seconds = speed × time = 20 ms\(^{-1}\) × 4 s = 80 m Total distance traveled = 30 m + 45 m + 80 m = 155 m Now let's calculate the total time taken: Total time taken = 3 s + 3 s + 4 s = 10 s Therefore, the average speed of the bird is: Average speed = Total distance traveled / Total time taken = 155 m / 10 s = 15.5 ms\(^{-1}\) So the correct option is: - 15.5 ms\(^{-1}\)
Tambaya 22 Rahoto
An object is placed 10cm from a converging lens of foal length 15cm. Calculate the magnification of the image formed
Tambaya 23 Rahoto
A simple pendulum makes 50 oscillations in one minute. Determine its period of oscillation
Bayanin Amsa
The period of a simple pendulum is the time it takes for one complete oscillation. In this case, we are given that the pendulum makes 50 oscillations in one minute. To find the period, we can use the formula: period = time / number of oscillations We know that the time for one minute is 60 seconds. So, we have: period = 60 seconds / 50 oscillations period = 1.2 seconds Therefore, the period of oscillation is 1.20 seconds, which corresponds to.
Tambaya 25 Rahoto
Which of the following properties of waves is exclusive to transverse waves?
Bayanin Amsa
Polarization is the property of waves that is exclusive to transverse waves. Polarization refers to the orientation of the oscillations of the wave, which can be either perpendicular or parallel to the direction of wave propagation. In a transverse wave, the oscillations are always perpendicular to the direction of propagation. This means that transverse waves can be polarized, while longitudinal waves cannot. Reflection, interference, and diffraction can occur in both transverse and longitudinal waves.
Tambaya 26 Rahoto
Which of the following properties is an advantage of lead-acid accumulator over an alkaline accumulator?
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Tambaya 27 Rahoto
An object is dropped from the top of a tower. If it takes 4s for it to reach the ground, calculate the height of the tower.(g = 10 ms-2, ignore air resistance)
Bayanin Amsa
When an object is dropped from rest, it falls under the influence of gravity. The distance it falls in time t is given by the formula: s = (1/2)gt² where g is the acceleration due to gravity. In this question, we are given t = 4s and g = 10 ms⁻². Substituting these values into the formula above, we get: s = (1/2) x 10 ms⁻² x (4 s)² = 80 m So, the height of the tower is 80 meters. Therefore, the correct answer is 80m.
Tambaya 28 Rahoto
A metal sheet of area 100cm\(^2\) was heated through 70°C. Calculate its new area if the linear expansivity of the metal is 0.000017K\(^{-1}\).
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Tambaya 29 Rahoto
Two capacitors, each of capacitance 2\(\mu\)F are connected in parallel. If the p.d across them is 120V, calculate the charge on each capacitor
Bayanin Amsa
The total capacitance of two capacitors connected in parallel is given by the sum of individual capacitances, which is 2μF + 2μF = 4μF. The charge on a capacitor is given by the product of its capacitance and the potential difference across it, which is Q = CV. Using the formula Q = CV and substituting the values, we get: Q = 4μF x 120V = 480μC Since there are two capacitors connected in parallel, the total charge is divided equally between them. Therefore, the charge on each capacitor is: Q/2 = 480μC/2 = 240μC So the answer is 2.4 x 10-4C.
Tambaya 30 Rahoto
A ray of light passes from air to water to glass to air. Given that the refractive index for light passing from air to water is \(\frac{4}{3}\) and air to glass is \(\frac{3}{2}\), calculate the refractive index of glass relative to water
Bayanin Amsa
Tambaya 31 Rahoto
The derived unit of pressure can be expressed as
Bayanin Amsa
Pressure is defined as force per unit area. Mathematically, pressure (P) can be expressed as P = F/A, where F is force and A is the area. The unit of force is newton (N) and the unit of area is square meter (m2). Therefore, the unit of pressure is N/m2. This can be further simplified as kg m/s2 / m2 which gives kg m-1s-2. Therefore, option (B) is the correct answer.
Tambaya 32 Rahoto
The sound heard by a person after the reflection of the sound generated by him is called
Bayanin Amsa
The sound heard by a person after the reflection of the sound generated by him is called an "echo." An echo occurs when sound waves bounce off a hard surface and return to the listener's ear after a noticeable delay. The delay between the original sound and the reflected sound is called the "echo time." The quality of the echo is determined by the distance between the sound source and the reflecting surface, as well as the shape and size of the reflecting surface.
Tambaya 34 Rahoto
The tendency for a stationary body to continue to remain at rest when a force is applied to it is known as
Bayanin Amsa
The tendency for a stationary body to continue to remain at rest when a force is applied to it is known as inertia. Inertia is a fundamental property of matter that resists changes in motion, whether it is a stationary object or an object in motion. It is often described as the resistance of an object to changes in its state of motion, and it is directly related to an object's mass. The greater the mass of an object, the greater its inertia and the more force it will take to move it from a stationary position. This property of inertia is the reason why a heavy object requires more force to move than a light one.
Tambaya 35 Rahoto
An inductor is connected to a 24V, 50Hz mains supply. If the current through the inductor is 1.5A, calculate the inductance of the inductor (\(\pi = \frac{22}{7}\))
Bayanin Amsa
The formula for the inductive reactance is XL = 2πfL, where XL is the inductive reactance in ohms, f is the frequency in Hertz and L is the inductance in Henrys. We are given that the mains supply has a frequency of 50Hz, and that the current through the inductor is 1.5A. We also know that the voltage across the inductor is V = I XL, where V is the voltage in volts. The voltage across the inductor is given as 24V, so we can use Ohm's law to find the inductive reactance: XL = V / I = 24V / 1.5A = 16Ω We can now use the formula XL = 2πfL to find the inductance: L = XL / 2πf = 16Ω / (2π x 50Hz) = 0.051H ≈ 5.1 x 10-2H Therefore, the inductance of the inductor is approximately 5.1 x 10-2H. The correct option is (d).
Tambaya 36 Rahoto
A piece of copper of mass 20g at a temperature of 110°C was dropped into a mixture of ice and water at 0°C. If the final steady temperature of the mixture is 0°C . Calculate the amount of ice that melted [Specific heat capacity of copper = 0.4 Jg\(^{-1}\)K\(^{-1}\), specific latent heat of fusion of ice = 330Jg\(^{-1}\)]
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Tambaya 37 Rahoto
The diagram above illustrates trapped air in a syringe placed in water that is gradually heated. It is observed that the piston rises as the temperature of the water rises. Which of the following statements explains this observation?
Bayanin Amsa
As the temperature of the water in the syringe increases, the water molecules transfer some of their thermal energy to the trapped air molecules. This causes the air molecules to move faster, collide more frequently, and exert more pressure on the walls of the syringe. The increase in pressure of the trapped air forces the piston to move upwards, against the external atmospheric pressure. Therefore, the correct statement explaining the observation is "molecules of trapped air expands more".
Tambaya 38 Rahoto
Which of the following factors decreases the rate of evaporation of a liquid?
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Tambaya 39 Rahoto
When two waves are superimposed on each other, the following occurrences are possible, except
Bayanin Amsa
Tambaya 40 Rahoto
The magnetic relay is a device used for
Bayanin Amsa
The magnetic relay is a device used for controlling another circuit carrying a larger current. It consists of an electromagnet, a spring-loaded armature, and a set of contacts. When a current passes through the coil of the electromagnet, it generates a magnetic field which attracts the armature and causes it to move, closing or opening the contacts. This can be used to switch on or off a circuit carrying a larger current, or to switch between different circuits. The magnetic relay is commonly used in applications such as industrial automation, power systems, and telecommunications.
Tambaya 41 Rahoto
A 12\(\Omega\) resistor dissipates 8kJ of heat in 20s. Calculate the current through the resistor
Tambaya 42 Rahoto
The thermopile is a device for detecting
Bayanin Amsa
A thermopile is a device that is used to detect radiant energy. When radiant energy is absorbed by a thermopile, it generates a small voltage, which can be measured by a voltmeter. This is because the thermopile is made up of a series of thermocouples, which are devices that generate a voltage when there is a difference in temperature between their ends. When the thermopile absorbs radiant energy, it causes a temperature difference between the ends of the thermocouples, which in turn generates a voltage. Therefore, the correct option is: radiant energy.
Tambaya 43 Rahoto
A metal ball of weight W falls through a column of glycerine of viscosity V. If the ball experiences an upthrust U and terminal velocity is attained, then
Bayanin Amsa
When the metal ball falls through the column of glycerine, it experiences both weight and upthrust, which opposes the motion of the ball. Eventually, the forces will be balanced and the velocity of the ball becomes constant, which is called terminal velocity. At this point, the forces on the ball are in equilibrium, which means that the weight of the ball (W) is equal to the upthrust (U) plus the viscous drag force (V). Therefore, the correct answer is: W = U + V Option (B)
Tambaya 44 Rahoto
Which of the following radiations emitted in radioactive decay has momentum, a fairly high penetrating power and is deflected by a magnet?
Bayanin Amsa
The radiation emitted in radioactive decay that has momentum, a fairly high penetrating power, and is deflected by a magnet is the beta particle. Beta particles are fast-moving electrons or positrons emitted from the nucleus of an atom during radioactive decay. They have a negative charge, a mass of about 1/1836th of a proton, and can penetrate several millimeters of matter. Beta particles can be deflected by a magnetic field, which is a characteristic of charged particles in motion. In contrast, alpha particles are heavy particles consisting of two protons and two neutrons, which have a high ionizing power but are not deflected by a magnetic field. Gamma radiation and X-radiation are both electromagnetic radiation and do not have charge or mass, so they are not deflected by a magnetic field.
Tambaya 45 Rahoto
Which of the following statements about X-rays is correct? X-rays
Bayanin Amsa
X-rays are produced when a metal target is used to block fast moving electrons. When high-speed electrons are fired at a metal target, they collide with the atoms in the target and produce X-rays. X-rays have short wavelengths and are highly penetrating, which makes them useful in medical imaging and materials analysis. Therefore, the correct statement among the options is "are produced when a metal target is used to block fast moving electrons."
Tambaya 46 Rahoto
Which of the following modes is the most economical method of transmitting electrical power over long distances?
Bayanin Amsa
The most economical method of transmitting electrical power over long distances is alternating current at high voltage and low current. This is because the power loss in a transmission line is given by the formula P=I²R, where P is the power loss, I is the current and R is the resistance of the transmission line. By transmitting power at high voltage and low current, the current can be reduced, resulting in a lower power loss due to the lower value of I in the formula. This means that more power can be transmitted over longer distances with less loss, resulting in a more efficient and economical method of power transmission.
Tambaya 47 Rahoto
When two like charges are brought together, the potential energy of the system formed will
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Tambaya 48 Rahoto
(a) State the principle of conservation of linear momentum.
(b) Explain the mode of action of a propelled rocket.
(c) During a training session, two footballers pass a ball repeatedly between themselves. Give two reasons why the to and fro motion of the ball is not simple harmonic.
(d) A ball is dropped from a height, at the same time as another ball is projected horizontally from the same height.
(i) Would the balls hit the ground at the same time?
(ii) Explain your answer in (i).
(e) A ball of mass 0.10 kg is projected horizontally onto a vertical wall with a speed of 17 ms\(^{-1}\). The ball makes contact with the wall for 0.15 s and rebounds horizontally with a speed of 13 ms\(^{-1}\).
Calculate the:
(i) change in momentum of the ball;
(ii) average force exerted on the ball during its collision with the wall.
Tambaya 49 Rahoto
TEST OF PRACTICAL KNOWLEDGE QUESTION
You are provided with a potentiometer, an ammeter, a voltmeter, a standard resistor, and other necessary apparatus. Using the circuit diagram above as a guide carry out the following instructions.
- Set up a circuit as illustrated in the diagram above.
- Close the key, K.
- Read and record the ammeter reading l\(_{o}\) and the voltmeter reading V\(_{o}\) when jockey J is not making contact with the potentiometer wire OQ.
- Using J, make a contact with the potentiometer wire OQ at a point P such that OP = 1Ocm.
- Read and record the current and the corresponding value of the voltage V.
- Repeat the procedure for other values of OP= 20cm, 30cm, 40cm, 50cm, and 60cm.
- Tabulate your readings.
- Plot a graph with V on the vertical axis and I on the horizontal axis, starting both axes from the origin (0, 0).
- Determine the slope, s, of the graph.
- Determine the value of V when I = 0.
- State two precautions taken to obtain accurate results.
(b) i. State two advantages of a lead-acid accumulator over a dry Leclanche cell.
ii. A cell of emf 2V and internal resistance of 1\(\Omega\) passes current through an external load of 9\(\Omega\). Calculate the potential drop across the cell.
Bayanin Amsa
None
Tambaya 50 Rahoto
(a) Define magnetic line of force.
(b) A wire of length 10 cm carrying a current of 4.0 A is placed between the poles of a powerful electromagnet of magnetic flux density 2.0 T. Calculate the:
(i) force on the wire when it is parallel to the field;
(ii) maximum force on the wire;
(iii) force on the wire when it makes an angle of 60° with the field.
(c) Describe how keepers can be used to preserve the magnetic strength of permanent bar magnets.
(d) A sailor observes that his mariners' compass reads N 10° W at a place where the angle of declination is N15° W. Calculate the true bearing of the place.
Bayanin Amsa
None
Tambaya 51 Rahoto
a) Define surface tension.
(b) State two methods by which the surface tension of a liquid can be reduced.
Tambaya 52 Rahoto
Explain why it is desirable to install an air conditioner near the ceiling of a room and not close to the floor.
Bayanin Amsa
None
Tambaya 53 Rahoto
(a) On what principle does lighting in a fluorescent tube operate?
(b) State two factors which determine the colour of light produced in a fluorescent tube.
Bayanin Amsa
None
Tambaya 54 Rahoto
(a) Explain wave-particle duality of light.
(b) Illustrate your answer in (a) with observable phenomena.
Bayanin Amsa
None
Tambaya 55 Rahoto
(a) Explain the term critical angle.
(b) List two factors which determine the deviation of a ray of light by a triangular glass prism.
(c) The angle of refraction (r) of a ray of white light from air through a triangular glass prism of refractive index 1.5 is 29.0°. Calculate the angle through which the ray is least deviated.
(d) Study the ray diagram below and use it to answer the questions that follow.
Calculate the:
(i) values of angles P,Q and R;
(ii) refractive index n of the glass prism;
(iii) value of e;
(iv) total deviation D.
Bayanin Amsa
None
Tambaya 56 Rahoto
Explain why sound waves cannot be plane Polarized.
Tambaya 57 Rahoto
A ray of light is incident on an air-glass boundary at an angle \(\theta\). If the angle between the partially reflected ray and the refracted ray is 90°, calculate \(\theta\), given that the refractive index of glass is 1.50.
Bayanin Amsa
None
Tambaya 58 Rahoto
TEST OF PRACTICAL KNOWLEDGE QUESTION
Study the diagrams above and use them as guides in carrying out the following instructions.
- Using the spring balance provided, determine the weight of the object of mass M= 50.0g in air. Record this weight as W\(_{1}\).
- Determine the weight of the object when it is completely immersed in water contained in a beaker as shown in the diagram above. Record the weight as W\(_{2}\).
- Determine the weight of the object when it is completely immersed in a liquid labeled L. Record the weight as W\(_{3}\).
- Evaluate U = (W\(_{1}\) - W\(_{2}\)) and V = (W\(_{1}\) -W\(_{3}\)).
- Repeat the procedure with the objects of masses M= 100g, 150g, 200g, and 250g
- In each case, evaluate U = (W\(_{1}\) - W\(_{2}\)) and V = (W\(_{1}\) -W\(_{3}\)).
- Tabulate your readings.
- Plot a graph with V on the vertical axis and U on the horizontal axis.
- Determine the slope, s, of the horizontal graph.
- State two precautions taken to ensure accurate results.
(b)i. State Archimedes' principle.
ii. A piece of brass of mass 20.0g is hung on a spring balance from a rigid support and completely immersed in kerosene from of density 8.0 x 10\(^{2}\)kgm\(^{-3}\). Determine the readings of the spring balance (g= 10ms\(^{-2}\), density of brass 8.0 x 10\(^{3}\)kgm\(^{-3}\))
Bayanin Amsa
None
Tambaya 59 Rahoto
(a) (i) Explain why x-rays can be used to produce photographs of fractures in bones.
(ii) List four uses of x-rays other than in medicine.
(b) State the energy transformations which takes place during the operation of an x-ray tube.
(c) (i) Explain three named dangers to which human beings may be exposed when subjected to large doses of x-rays.
(ii) State two precautions that must be taken by persons working with x-rays.
(d) In an x-ray tube, an electron is accelerated from rest towards a tungsten target biased at a potential of 33 kV. Calculate, for the electron, the
(i) kinetic energy;
(ii) velocity. [h = 6.6 x 10\(^{-14}\) Js; Me = 9.1 x10\(^{-31}\) kg; c = 3.0 x 10\(^4\) ms\(^{-1}\); e = 1.6 x 10\(^{-19}\) C.]
Tambaya 60 Rahoto
The diagram above illustrates an arrangement of a cathode ray from an electron gun and a bar magnet placed perpendicularly to the direction of the ray. Will the ray bend downward or upward? Explain.
Bayanin Amsa
None
Tambaya 61 Rahoto
TEST OF PRACTICAL KNOWLEDGE QUESTION
You are provided with a syringe, a petri-dish firmly attached to the base of the movable piston (plunger) of the syringe, a Set of weights, and other necessary apparatus.
- Pull the piston of the syringe upward until it can no longer move. Read and record this position of the piston on the graduated mark on the syringe as V\(_{o}\).
- Clamp the syringe and ensure that it is vertical.
- Place a mass M= 500g gently at the center of the petri-dish.
- Read and record the new position of the piston as V.
- Evaluate V\(^{1}\).
- Repeat the procedure for four other values of M= 1000g, 1500g, 2000g, and 2500g.
- Tabulate your readings.
- Plot a graph with V\(^{-1}\) on the vertical axis and M on the horizontal axis, starting both axes from the origin (0,0).
- Determine the slope, s, of the graph.
- Evaluate k = s\(^{-1}\).
- State two precautions taken to ensure accurate results.
(b)i. When a weight is placed on the petri-dish, which quantities of the gas in the syringe (\(\Omega\)) increases; (\(\beta\)) decrease?
ii. What is responsible for the pressure exerted by a gas in a closed vessel?
Tambaya 62 Rahoto
(a) Explain the term electrodes in electric cells.
b) An electric current passing through an electrolyte for 2 minutes deposited 200 g of a substance. If the electrochemical equivalent of the substance is 8.33 x 10\(^{-4}\)g C\(^{-1}\), calculate the current passed.
Bayanin Amsa
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