WAEC SSCE - Physics - 1999
Swali 1 Ripoti
Iron at 30oC feels colder to the hand than wood at 28oC because
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The correct answer is "iron conducts heat more quickly from the hand than wood." The sensation of temperature is related to heat flow, which is the transfer of thermal energy from one object to another due to a temperature difference. When we touch an object, heat flows from our hand to the object if the object is cooler than our hand, or from the object to our hand if it is warmer. In this case, the iron feels colder to the hand than wood because it conducts heat more quickly from the hand than wood. Iron is a good conductor of heat, whereas wood is a poor conductor. Therefore, when we touch a piece of iron and a piece of wood at the same temperature, more heat flows from our hand to the iron, making it feel colder, while less heat flows to the wood, making it feel warmer.
Swali 2 Ripoti
The diagram above illustrates the energy transitions of five electrons of an atom. Which of the transitions will produce the emission of longest wavelength?
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Swali 3 Ripoti
The force between molecules of the same substance is termed
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The force between molecules of the same substance is termed cohesive force. Cohesive force refers to the attractive force between molecules of the same substance. It is responsible for the ability of liquid molecules to stick together and form a distinct surface. Cohesive forces arise due to the dipole moments of the molecules, which create a net attractive force between them. The strength of the cohesive force depends on various factors, including the nature of the molecules, the temperature, and the pressure. In liquids, cohesive forces dominate over the kinetic energy of the molecules, which results in a definite shape and volume of the liquid. Therefore, the force between molecules of the same substance is termed cohesive force.
Swali 4 Ripoti
Using the diagram above, calculate the effective capacitance of the circuit
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Swali 5 Ripoti
The half-life of a radioactive substance is
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The half-life of a radioactive substance is the length of time during which the activity of the substance reduces by 50%. In other words, it is the time required for half of the initial number of radioactive atoms to decay. This means that after one half-life, half of the initial amount of the radioactive substance will have decayed, and after two half-lives, one quarter of the initial amount will remain. The concept of half-life is important in radioactivity, as it can be used to predict the decay rate and the amount of time it will take for a given amount of the substance to decay.
Swali 6 Ripoti
A metal has work function of 4.375 eV Calculate its threshold frequency. (h = 6.6 x 10-34Js; 1eV = 1.6 x 10-19J)
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The threshold frequency of a metal is the minimum frequency of light required to eject an electron from its surface. The formula for calculating the threshold frequency is given by: Threshold frequency = Work function / Planck's constant We are given the work function of the metal as 4.375 eV. We need to convert this into joules using the conversion factor 1 eV = 1.6 x 10^-19 J: Work function = 4.375 x 1.6 x 10^-19 J/eV = 7 x 10^-19 J We are also given Planck's constant as 6.6 x 10^-34 J s. Now, substituting the values into the formula, we have: Threshold frequency = Work function / Planck's constant Threshold frequency = (7 x 10^-19 J) / (6.6 x 10^-34 J s) Threshold frequency = 1.06 x 10^15 Hz Therefore, the threshold frequency of the metal is 1.06 x 10^15 Hz. Option (ii) is the correct answer.
Swali 7 Ripoti
A motorcyclist, passing a road junction, moves due west for 8 s at a uniform speed of 5ms-1. He then moves due north for another 6 s with the same speed. At the end of the 6s his displacement from the road junction is 50m in the direction of
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The motorcyclist travels west for 8 seconds at a speed of 5 m/s, so his distance traveled is: distance = speed × time = 5 m/s × 8 s = 40 m Next, he turns and travels north for 6 seconds at the same speed of 5 m/s. His distance traveled in the north direction is: distance = speed × time = 5 m/s × 6 s = 30 m To find the displacement from the road junction, we need to use the Pythagorean theorem because the displacement is the straight-line distance between the starting point and the ending point. The displacement is the hypotenuse of a right-angled triangle with the westward distance as one side and the northward distance as the other side. Using the Pythagorean theorem: displacement2 = (40 m)2 + (30 m)2 = 1600 m2 + 900 m2 = 2500 m2 displacement = √(2500 m2) = 50 m Thus, the motorcyclist's displacement from the road junction is 50 m. The displacement is in the direction of the hypotenuse of the right-angled triangle, which can be found using trigonometry. The angle is given by: tan θ = opposite / adjacent = 40 m / 30 m θ = tan-1(40/30) = 53.13° Since the motorcyclist moved north after moving west, the direction of displacement from the road junction is N53oE. Therefore, the answer is (a) N53oE.
Swali 8 Ripoti
Using the data on the diagram above, calculate the potential difference across the 20- ? resistor. (Neglect the internal resistance of the cell)
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Swali 9 Ripoti
An iron rod of length 50m and at a temperature of 60oC is heated to 70oC. Calculate its new length. (Linear expansivity of iron = 1.2 x 10-5K-1)
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The formula for linear thermal expansion is: ΔL = αLΔT Where: ΔL = change in length α = coefficient of linear expansion L = original length ΔT = change in temperature Using the values given: L = 50 m α = 1.2 x 10^-5 K^-1 ΔT = 70°C - 60°C = 10°C Substituting the values in the formula: ΔL = (1.2 x 10^-5 K^-1)(50 m)(10°C) ΔL = 0.006 m The new length of the iron rod is the sum of the original length and the change in length: New length = 50 m + 0.006 m = 50.006 m Therefore, the correct answer is: 50.006m.
Swali 10 Ripoti
A resistance wire of length 2 m and of uniform cross-sectional area 5.0 x 10-7 m2 has a resistance of 2.2 Ω Calculate its resistivity
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The formula for the resistance of a wire is given by: R = (ρ × L) / A Where R is the resistance, ρ is the resistivity, L is the length of the wire, and A is the cross-sectional area of the wire. Rearranging this equation, we get: ρ = (R × A) / L Substituting the given values, we get: ρ = (2.2 Ω × 5.0 x 10-7 m2) / 2 m Simplifying this expression, we get: ρ = 5.5 x 10-7 Ωm Therefore, the resistivity of the wire is 5.5 x 10-7 Ωm. Answer: 5.5 x 10-7Ωm
Swali 11 Ripoti
Calculate the heat energy required to change 0.1kg of ice at 0oC to water boiling at 100oC. (Specific heat capacity of water = 4200 J kg-1K-1) (Specific latent heat of fusion of ice = 336,000 J kg-1)
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To calculate the heat energy required, we need to consider the different stages involved in the process: 1. Heat energy required to melt the ice: This is given by the formula Q = mL, where Q is the heat energy, m is the mass of ice and L is the specific latent heat of fusion of ice. So, Q = 0.1kg x 336,000 J kg-1 = 33,600 J. 2. Heat energy required to raise the temperature of the melted ice (now water) from 0oC to 100oC: This is given by the formula Q = mcΔθ, where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water, and Δθ is the change in temperature. So, Q = 0.1kg x 4200 J kg-1K-1 x (100oC - 0oC) = 42,000 J. Therefore, the total heat energy required is the sum of the heat energy required for melting the ice and for raising the temperature of the water: 33,600 J + 42,000 J = 75,600 J. Therefore, the answer is 75,600J.
Swali 12 Ripoti
The refractive index of glass for yellow light is greater than that for red light because yellow light
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Swali 13 Ripoti
A net force of magnitude 0.6N acts on a body of mass 40g, initially at rest. Calculate the magnitude of tire resulting acceleration.
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The magnitude of the resulting acceleration can be calculated using Newton's second law of motion which states that the acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass. Mathematically, this can be represented as: a = F/m where a is the acceleration of the object, F is the net force acting on the object, and m is the mass of the object. Substituting the given values, we get: a = 0.6N / 0.04kg a = 15 ms^-2 Therefore, the magnitude of the resulting acceleration is 15ms^-2. The correct answer is: 15ms-2.
Swali 15 Ripoti
The amplitude of a wave is the
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The amplitude of a wave is the maximum displacement of a wave particle from its equilibrium position. In simpler terms, it is the distance from the middle or equilibrium point of a wave to the highest point (crest) or lowest point (trough) of the wave. It represents the amount of energy that is being carried by the wave, with higher amplitude waves carrying more energy than lower amplitude waves. The amplitude of a wave can be measured in various units, such as meters (m), centimeters (cm), or even millimeters (mm), depending on the type of wave being considered.
Swali 16 Ripoti
Calculate the power rating of an immersion heater used for 10 minutes to increase the temperature of 10kg of water by 15K. (Specific heat capacity of water = 4200 J l-1K-1)
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Swali 17 Ripoti
Bohr’s atomic model proves most successful for the explanation of the. I. structure of the hydrogen atom; II. line spectra of the hydrogen atom; III. Multi-electron atoms. Which of the statements above is/are correct?
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Bohr’s atomic model is most successful for explaining the structure of the hydrogen atom and the line spectra of the hydrogen atom. Therefore, statements I and II are correct. However, the model is not as successful in explaining multi-electron atoms, so statement III is incorrect. The correct answer is (c) I and II only.
Swali 18 Ripoti
A 15kg mass, suspended from a ceiling, is pulled aside with a horizontal force, F, as shown in the diagram above. Calculate the value of the tension, T. (g = 10ms-2)
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Swali 19 Ripoti
If the position of resonance in a resonance tube is 16.50cm from open end of the tube, calculate the distance from the open end to the next position where resonance occurs. (Neglect end-correction)
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Swali 20 Ripoti
The temperature of a piece of metal of mass 9 g is raised from 10oC to 110oC when it absorbs 108J of heat energy. Determine the specific heat capacity of the metal in J kg-1K-1
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The specific heat capacity of a material is the amount of heat energy required to raise the temperature of 1 kg of the material by 1 degree Celsius. The formula for calculating the specific heat capacity of a material is: specific heat capacity = (heat energy absorbed) / (mass of material × change in temperature) In this question, the mass of the metal is 9 g, which is 0.009 kg. The temperature of the metal has been raised by 100 degrees Celsius, i.e. a change of 100 degrees Celsius. Substituting the given values into the formula, we get: specific heat capacity = (108 J) / (0.009 kg × 100 °C) specific heat capacity = 1200 J kg-1K-1 Therefore, the specific heat capacity of the metal is 1200 J kg-1K-1. The answer is 120.0.
Swali 21 Ripoti
ln order to obtain a sound note of a high pitch from a wire stretched by a constant tension, the wire must be
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To obtain a sound note of a high pitch from a wire stretched by a constant tension, the wire must be short and thin. The pitch of a sound is determined by its frequency, which is the number of vibrations (or cycles) per second. In the case of a wire that is stretched under tension, its frequency of vibration is determined by the tension on the wire, its length, and its mass per unit length (also known as linear density). The tension on the wire remains constant, so to increase the frequency (and therefore the pitch) of the note, the wire must be made thinner or shorter. This is because a thinner wire has a lower mass per unit length, which allows it to vibrate more quickly. Similarly, a shorter wire vibrates over a shorter distance, so it can complete more vibrations per second. Therefore, to obtain a sound note of a high pitch from a wire stretched by a constant tension, the wire must be short and thin.
Swali 22 Ripoti
The voltage and current in the primary of a transformer are 200V and 2Arespectively. lf the transformer is used to light ten 12 V, 30 W bulbs, calculate its efficiency.
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The efficiency of a transformer is given by the ratio of the output power to the input power. In this case, the input power is given by the product of the voltage and current in the primary coil of the transformer: Input power = Voltage × Current = 200V × 2A = 400W The output power is given by the total power of the bulbs, which is the product of their voltage and current ratings, multiplied by the number of bulbs: Output power = Number of bulbs × Voltage × Current = 10 × 12V × 2.5A = 300W The efficiency of the transformer is therefore: Efficiency = Output power / Input power = 300W / 400W = 0.75 or 75% Therefore, the correct option is (c) 75%.
Swali 24 Ripoti
The air column in a resonance tube is set into resonance by a vibrating tuning fork. The resultant waves in the air column will be
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When a vibrating tuning fork is placed near the open end of a resonance tube, it creates sound waves that travel down the tube. If the length of the tube is such that a standing wave is set up, the air column in the tube will resonate at a certain frequency. In a resonance tube, the waves created by the tuning fork will be longitudinal waves, which means that the vibrations of the air particles will be parallel to the direction of wave propagation. This is because sound waves are always longitudinal waves in air and other fluids. When the waves reach the closed end of the tube, they are reflected back and interfere with the incoming waves to form standing waves. These standing waves are stationary, which means that the air particles at each point in the tube are oscillating back and forth, but not moving in any particular direction. Therefore, the resultant waves in the air column of a resonance tube set into resonance by a vibrating tuning fork are stationary and longitudinal.
Swali 25 Ripoti
Which of the following are the essential parts of an atomic bomb?
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Uranium and neutrons are the essential parts of an atomic bomb. An atomic bomb works by initiating a chain reaction of nuclear fission in a highly enriched uranium-235 core. This is achieved by firing a neutron into the nucleus of a uranium-235 atom, which causes it to split into two smaller atoms, releasing energy and more neutrons in the process. These released neutrons then go on to collide with other uranium-235 atoms, continuing the chain reaction and releasing a tremendous amount of energy in the form of an explosion. Therefore, is the correct answer.
Swali 26 Ripoti
If the force of attraction between the sun and the 1 planets is removed; the planets will
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Swali 27 Ripoti
Which of the following is not a safety precaution in a nuclear station?
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Swali 28 Ripoti
Which of the transitions will produce the emission of highest frequency?
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Swali 29 Ripoti
A sheet of paper is placed in the path of the following radiations: I. Alpha particles II. Beta particles III. Gamma rays. Which of the radiations will pass through the sheet of paper?
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Swali 30 Ripoti
The half-life of a radioactive element is 5 Calculate its decay constant
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The half-life of a radioactive element is the time it takes for half of the original atoms to decay. In this case, the half-life is 5. The decay constant (represented by the symbol λ) is a measure of how quickly the radioactive atoms decay. It is defined as the probability of decay per unit time. To calculate the decay constant, we can use the formula: λ = ln(2) / T1/2 where ln(2) is the natural logarithm of 2, and T1/2 is the half-life. Substituting the given values, we get: λ = ln(2) / 5 Calculating this expression gives: λ ≈ 0.139 s-1 Therefore, the decay constant of the radioactive element is approximately 0.139 s-1.
Swali 31 Ripoti
Which of the following statements is not correct about long sight?
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Swali 32 Ripoti
Magnetic flux density is defined as the
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Magnetic flux density is defined as the number of magnetic lines of force per unit area that are normal (perpendicular) to the magnetic field. It is also known as magnetic induction or simply magnetic field strength. It is usually denoted by the symbol B and is measured in tesla (T) or gauss (G). Magnetic flux density is a vector quantity, which means it has both magnitude and direction.
Swali 33 Ripoti
A 10- Ω coil takes 21 s to melt 10g of ice at 0oC. Assuming no heat losses, determine the current in the coil. (Specific latent heat of fusion of ice = 336 Jg-1)
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The amount of heat required to melt 10g of ice is given by: Q = mL where Q is the heat required, m is the mass of ice and L is the specific latent heat of fusion of ice. Therefore, Q = (10 g) x (336 Jg^-1) = 3360 J. The time taken to melt the ice is 21 s. The power output of the coil is given by: P = Q/t where P is the power output, Q is the heat required and t is the time taken. Therefore, P = (3360 J)/(21 s) = 160 W. Using Ohm's law, we can find the current in the coil: V = IR where V is the voltage, I is the current and R is the resistance. The resistance of the coil is given as 10 Ω, and the voltage is not given directly in the question. However, we can use the power output to find the voltage: P = IV Therefore, V = P/I = (160 W)/(I). Substituting this value of V in the equation V = IR, we get: (I)R = (160 W)/(I) Solving for I, we get: I^2 = (160 W)/(10 Ω) = 16 A Taking the square root of both sides, we get: I = 4 A. Therefore, the current in the coil is 4A. Answer: 4A.
Swali 34 Ripoti
The real image of an object formed by a converging lens of focal length 15cm, is three times the size of the object. Calculate the object distance
Maelezo ya Majibu
The magnification produced by a lens is given by the ratio of the size of the image to the size of the object: magnification = size of image / size of object In this case, we are given that the size of the image is three times the size of the object, so the magnification is: magnification = 3/1 = 3 We can also use the formula for the magnification produced by a converging lens: magnification = -image distance / object distance where the negative sign indicates that the image is inverted. We know the focal length of the lens is 15 cm. We can substitute in the known values and solve for the object distance: 3 = -image distance / object distance image distance = -3 object distance 1/f = 1/image distance + 1/object distance 1/0.15 = 1/-3x + 1/x x = 20 cm Therefore, the object distance is 20 cm. Answer: 20cm
Swali 35 Ripoti
A body weighing 10N moves with a speed of 5ms-1 in a horizontal circular path of radius 5m. Calculate the . magnitude of the centripetal force acting on the body. (g = 10ms-2)
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Swali 36 Ripoti
In a uniform electric field, the magnitude of the force on a charge of 0.2 C is 4N. Calculate the electric field intensity
Maelezo ya Majibu
The electric force, F, on a charge, q, in an electric field, E, is given by the formula: F = qE Here, the force, F, is given as 4 N and the charge, q, is 0.2 C. We can rearrange the formula to solve for the electric field intensity, E: E = F/q Substituting the values we have: E = 4 N / 0.2 C E = 20 N C-1 Therefore, the electric field intensity is 20 N C-1, which is closest to 20 N C-1. In simpler terms, the electric field intensity is the force experienced by a unit charge in an electric field. In this case, a charge of 0.2 C experiences a force of 4 N, so the electric field intensity is the force per unit charge, which is 4 N divided by 0.2 C, giving 20 N C-1.
Swali 37 Ripoti
The image of an object is located 6cm behind a convex mirror. If its magnification is 0.6, calculate the focal length of the mirror
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The magnification of a mirror is given by the ratio of the height of the image to the height of the object. magnification (m) = height of image (h') / height of object (h) Also, for a convex mirror, the focal length (f) is negative, and the relationship between the focal length, image distance (v) and object distance (u) is given by the mirror equation: 1/f = 1/v + 1/u where u is the distance of the object from the mirror, and v is the distance of the image from the mirror. Given that the magnification of the mirror is 0.6, we know that: m = h'/h = 0.6 We also know that v = -6 cm (since the image is behind the mirror), and we can assume that the object is far enough away from the mirror that u can be considered infinite. Using the magnification equation, we can write: 0.6 = h'/h = -v/u Solving for u, we get: u = -v/h' = -6 / 0.6 = -10 cm Substituting these values into the mirror equation, we get: 1/f = 1/v + 1/u = -1/6 - 1/(-10) = -5/30 + 3/30 = -2/30 Simplifying, we get: f = -15 cm Therefore, the focal length of the convex mirror is 15 cm, which corresponds to.
Swali 38 Ripoti
The time rate of change of displacement is known as
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The time rate of change of displacement is known as velocity. Velocity is a vector quantity that gives the speed and direction of an object's motion. It is calculated by dividing the change in displacement by the time interval during which the change occurred. Therefore, velocity takes into account both the speed of an object and the direction in which it is moving.
Swali 39 Ripoti
The ice point of an ungraduated mercury-in-glass thermometer is X, while its steam point is 90o. This thermometer reads 60o when the true temperature is 40oC Calculate the value of X.
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Swali 40 Ripoti
Which of the following statements about evaporation is not correct?
Maelezo ya Majibu
The statement "Evaporation takes place inside a liquid" is not correct. Evaporation is a process in which a liquid changes into a gas or vapor state at its surface, without reaching its boiling point. It occurs when the faster molecules escape from the surface of a liquid, leaving behind the slower ones, which results in a decrease in temperature of the remaining liquid. The rate of evaporation depends on various factors, including temperature, humidity, surface area, and air flow. However, evaporation does not take place inside a liquid. The process occurs only at the surface of the liquid, where the faster molecules have enough energy to break the attractive forces of the liquid and escape into the surrounding space as vapor. The remaining liquid remains unaffected and does not evaporate from the inside. Therefore, the statement "Evaporation takes place inside a liquid" is not correct.
Swali 41 Ripoti
A pendulum bob executing simple harmonic motion has 2cm and 12Hz as amplitude and frequency respectively. Calculate the period of the motion.
Maelezo ya Majibu
The period of a pendulum is the time taken for one complete oscillation or swing. The relationship between the period, T, and frequency, f, of an object in simple harmonic motion is given by: T = 1/f In the question, the frequency of the pendulum is given as 12 Hz. Therefore, we can calculate the period as: T = 1/f = 1/12 Hz = 0.0833 s Rounding off to two significant figures, the period is 0.08s. Therefore, the correct option is (c) 0.08s.
Swali 42 Ripoti
Thunder is usually heard some seconds after lightning is observed because
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Thunder is usually heard some seconds after lightning is observed because light travels faster than sound. When lightning occurs, it produces a sudden burst of light which travels at the speed of light, about 300,000 kilometers per second. On the other hand, sound travels at a much slower speed, about 343 meters per second. As a result, the sound waves produced by the lightning take a longer time to reach our ears compared to the light waves. Therefore, we see the lightning before we hear the thunder, which is why there is usually a time delay between the two.
Swali 43 Ripoti
I. All elements emit and absorb characteristic spectra. II. Spectral analysis is an important method of identifying environmental pollutants. III. The chemical composition of stars could be determined using spectral analysis. Which of the statements above are correct?
Maelezo ya Majibu
The correct answer is (A) I, II, and III. I. All elements emit and absorb characteristic spectra: This is a fundamental principle of spectroscopy. When atoms or molecules absorb or emit light, they do so at specific wavelengths that are unique to that particular element or compound. This allows scientists to identify the chemical composition of a sample. II. Spectral analysis is an important method of identifying environmental pollutants: Spectral analysis is widely used to identify pollutants in the environment. By analyzing the absorption or emission spectra of a sample, scientists can determine what compounds are present and in what concentrations. III. The chemical composition of stars could be determined using spectral analysis: Astronomers use spectral analysis to study the chemical composition of stars. When light from a star passes through a prism or diffraction grating, it is separated into its component wavelengths, creating a spectrum. By analyzing the spectrum, astronomers can determine what elements are present in the star. Therefore, all the statements are correct, and the correct answer is (A) I, II, and III.
Swali 44 Ripoti
A stone of mass 0.7kg is projected vertically , upwards with a speed of 5ms-1. Calculate the maximum height reached. (Take g as 10ms-2 and neglect air resistance)
Maelezo ya Majibu
The problem involves a projectile motion where a stone is projected vertically upwards with a speed of 5 m/s. We are given the mass of the stone as 0.7 kg, the initial velocity as 5 m/s, the acceleration due to gravity as 10 m/s\(^2\), and we need to find the maximum height reached by the stone. In projectile motion, the vertical motion of the stone can be analyzed independently of the horizontal motion. The initial vertical velocity of the stone is 5 m/s, and the acceleration due to gravity is 10 m/s\(^2\), acting in the opposite direction to the motion of the stone. Using the kinematic equation of motion for vertical motion, we can calculate the maximum height reached by the stone. The equation is: v\(^2\) = u\(^2\) + 2as where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and s is the displacement. At the maximum height, the final velocity of the stone is zero, and the displacement is the maximum height reached, h. Substituting the given values, we get: 0\(^2\) = 5\(^2\) - 2 × 10 × h Simplifying this equation, we get: h = 1.25 m Therefore, the maximum height reached by the stone is 1.25 m.
Swali 45 Ripoti
The parts of a bar magnet at which the magnetic effect is strongest are called the
Maelezo ya Majibu
The correct answer is "poles". A bar magnet has two poles, a north pole and a south pole. These poles are the points on the magnet where the magnetic field is strongest. When a bar magnet is freely suspended, it will point in a north-south direction, aligning itself with the Earth's magnetic field. This is because the north pole of the magnet is attracted to the Earth's south magnetic pole, and vice versa. The strength of the magnetic force between two magnets or a magnet and a magnetic material depends on the strength of the magnetic field, which is greatest at the poles.
Swali 46 Ripoti
Palm oil from a bottle flows out more easily after it as been heated because the
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Swali 47 Ripoti
A point charge of magnitude 2μC is moved through a distance of 0.02 m against uniform field of intensity 25 Vm-1 Calculate the work done on the charge
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The work done on a point charge moving through a distance against an electric field is given by the formula: W = qEd where: - W is the work done - q is the magnitude of the point charge - E is the intensity of the electric field - d is the distance moved against the electric field In this case, the point charge has a magnitude of 2μC, the electric field has an intensity of 25 Vm-1, and the distance moved against the electric field is 0.02 m. Substituting these values into the formula, we get: W = (2 x 10-6 C) x (25 Vm-1) x (0.02 m) W = 1 x 10-6 J Therefore, the work done on the charge is 1.0 x 10-6J.
Swali 48 Ripoti
A trolley of mass 4kg moving on a smooth horizontal platform with a speed of 1.0ms-1 collides perfectly with a stationary trolley of the same mass on the same platform. Calculate the total momentum of the two trolleys immediately after the collision.
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Swali 49 Ripoti
A ray of light is incident on 'a plane mirror at an angle of 20o. This mirror is rotated through, twice this angle, In this new position the angle between the incident ray and the reflected ray is
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Swali 50 Ripoti
(a) Explain: (i) Fusion. (ii) Fission.
(b) State three advantages of fusion over fission in the generation of power
(c) Calculate in joules, the binding energy for
\(\frac{59}{27}\) Co
(Atomic mass of \(^{59}_{2} Co\) = 58.9332 u)
(Mass of proton = 1.00783u)
(Mass of neutron = 1.00867 u)
(Unified atomic mass unit, U = 931 MeV)
(1 eV = 1.6 x 10\(^{-19}\) J)
(a)(i) Fusion: the joining together of two light nuclei to form a single heavier nucleus, with the release of a large amount of energy.
(ii) Fission: the splitting of a heavy nucleus into two lighter nuclei of comparable mass (together with a few neutrons), with the release of a large amount of energy.
(b) Three advantages of fusion over fission for power generation:
- Fusion releases more energy per unit mass of fuel than fission.
- Its fuel (hydrogen isotopes from water) is very cheap and practically unlimited.
- Its products are non-radioactive (or far less radioactive), so there is little dangerous nuclear waste, and there is no risk of an uncontrolled chain reaction.
(c) Binding energy of \( ^{59}_{27}\text{Co} \): it has \(27\) protons and \(59 - 27 = 32\) neutrons.
Total mass of the separate nucleons:
\[ = 27(1.00783) + 32(1.00867) = 27.2114 + 32.2774 = 59.4888\,\text{u} \]Mass defect:
\[ \Delta m = 59.4888 - 58.9332 = 0.5556\,\text{u} \]Binding energy (using \(1\,\text{u} = 931\,\text{MeV}\)):
\[ E = 0.5556 \times 931 = 517.3\,\text{MeV} \]Converting to joules \((1\,\text{MeV} = 10^{6}\times1.6\times10^{-19}\,\text{J}):\)
\[ E = 517.3\times10^{6} \times 1.6\times10^{-19} = 8.28\times10^{-11}\,\text{J} \]Answer: the binding energy is about \(8.28\times10^{-11}\,\text{J}\).
Maelezo ya Majibu
(a)(i) Fusion: the joining together of two light nuclei to form a single heavier nucleus, with the release of a large amount of energy.
(ii) Fission: the splitting of a heavy nucleus into two lighter nuclei of comparable mass (together with a few neutrons), with the release of a large amount of energy.
(b) Three advantages of fusion over fission for power generation:
- Fusion releases more energy per unit mass of fuel than fission.
- Its fuel (hydrogen isotopes from water) is very cheap and practically unlimited.
- Its products are non-radioactive (or far less radioactive), so there is little dangerous nuclear waste, and there is no risk of an uncontrolled chain reaction.
(c) Binding energy of \( ^{59}_{27}\text{Co} \): it has \(27\) protons and \(59 - 27 = 32\) neutrons.
Total mass of the separate nucleons:
\[ = 27(1.00783) + 32(1.00867) = 27.2114 + 32.2774 = 59.4888\,\text{u} \]Mass defect:
\[ \Delta m = 59.4888 - 58.9332 = 0.5556\,\text{u} \]Binding energy (using \(1\,\text{u} = 931\,\text{MeV}\)):
\[ E = 0.5556 \times 931 = 517.3\,\text{MeV} \]Converting to joules \((1\,\text{MeV} = 10^{6}\times1.6\times10^{-19}\,\text{J}):\)
\[ E = 517.3\times10^{6} \times 1.6\times10^{-19} = 8.28\times10^{-11}\,\text{J} \]Answer: the binding energy is about \(8.28\times10^{-11}\,\text{J}\).
Swali 51 Ripoti
(a) State Heisenberg's uncertainty principle.
(b) Mention two phenomena that can only be explained in terms of the particulate nature of light.
(a) Heisenberg's uncertainty principle: it is impossible to determine simultaneously, with perfect accuracy, both the position and the momentum of a particle. The more accurately one is known, the less accurately the other can be known. In symbols:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]where \(\Delta x\) is the uncertainty in position, \(\Delta p\) the uncertainty in momentum and \(h\) Planck's constant.
(b) Two phenomena explained only by the particulate (quantum) nature of light:
- The photoelectric effect.
- The Compton effect (scattering of X-rays by electrons). Emission of line/black-body spectra is also acceptable.
Maelezo ya Majibu
(a) Heisenberg's uncertainty principle: it is impossible to determine simultaneously, with perfect accuracy, both the position and the momentum of a particle. The more accurately one is known, the less accurately the other can be known. In symbols:
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]where \(\Delta x\) is the uncertainty in position, \(\Delta p\) the uncertainty in momentum and \(h\) Planck's constant.
(b) Two phenomena explained only by the particulate (quantum) nature of light:
- The photoelectric effect.
- The Compton effect (scattering of X-rays by electrons). Emission of line/black-body spectra is also acceptable.
Swali 52 Ripoti
(a) What is surface tension?
(b) State two methods by which the surface tension of a liquid can be reduced.
(a) Surface tension is the force per unit length acting along (and at right angles to) an imaginary line drawn in the free surface of a liquid, tending to make the surface behave like a stretched elastic skin and to contract to the smallest possible area. Its unit is \(\text{N m}^{-1}\).
(b) Two methods of reducing the surface tension of a liquid:
- Raising the temperature of the liquid.
- Adding impurities such as soap, detergent or oil (surfactants) to the liquid.
Maelezo ya Majibu
(a) Surface tension is the force per unit length acting along (and at right angles to) an imaginary line drawn in the free surface of a liquid, tending to make the surface behave like a stretched elastic skin and to contract to the smallest possible area. Its unit is \(\text{N m}^{-1}\).
(b) Two methods of reducing the surface tension of a liquid:
- Raising the temperature of the liquid.
- Adding impurities such as soap, detergent or oil (surfactants) to the liquid.
Swali 53 Ripoti
(a) State the conditions for the equilibrium of a rigid body acted upon by parallel forces
(b)(i) Describe an experiment, using the principle of moments, to determine the mass of a metre rule
(ii) State two precautions necessary to ensure accurate results.
(c) A bullet of mass 120g is fired horizontally into a fixed wooden block with a speed of 20ms\(^{-1}\). The bullet is brought to rest in the block in 0.1s by a constant resistance. Calculate the:
(i) magnitude of the resistance;
(ii) distance moved by the bullet in the wood.
(a) Conditions for equilibrium of a rigid body under parallel forces:
- The algebraic sum of the forces is zero, i.e. the total upward force equals the total downward force (no net force).
- The algebraic sum of the moments of the forces about any point is zero, i.e. the total clockwise moment equals the total anticlockwise moment (principle of moments).
(b)(i) Experiment to find the mass of a metre rule (principle of moments):
- Balance the metre rule alone on a knife-edge and note the balance point; this is its centre of gravity \(G\) (near the 50 cm mark).
- Now shift the knife-edge to one side of \(G\) and suspend a known mass \(m\) from the short arm; adjust its position until the rule balances horizontally.
- Measure the distance \(a\) of the known mass from the knife-edge and the distance \(b\) of \(G\) from the knife-edge.
- By the principle of moments, \( m \times a = M \times b \), so the mass of the rule \( M = \dfrac{m\,a}{b} \).
(ii) Two precautions:
- Read the balance positions with the eye directly above the scale to avoid parallax error.
- Ensure the knife-edge is sharp and the rule is horizontal (balanced) before taking readings.
(c) Bullet: \( m = 120\,\text{g} = 0.12\,\text{kg}, \; u = 20\,\text{m s}^{-1}, \; v = 0, \; t = 0.1\,\text{s}. \)
(i) Resistance (magnitude of the constant force):
\[ F = ma = m\frac{(u - v)}{t} = 0.12 \times \frac{20}{0.1} = 0.12 \times 200 = 24\,\text{N} \](ii) Distance moved in the wood:
\[ s = \left(\frac{u + v}{2}\right)t = \left(\frac{20 + 0}{2}\right)(0.1) = 10 \times 0.1 = 1.0\,\text{m} \]Answers: resistance \(= 24\,\text{N}\); distance \(= 1.0\,\text{m}\).
Maelezo ya Majibu
(a) Conditions for equilibrium of a rigid body under parallel forces:
- The algebraic sum of the forces is zero, i.e. the total upward force equals the total downward force (no net force).
- The algebraic sum of the moments of the forces about any point is zero, i.e. the total clockwise moment equals the total anticlockwise moment (principle of moments).
(b)(i) Experiment to find the mass of a metre rule (principle of moments):
- Balance the metre rule alone on a knife-edge and note the balance point; this is its centre of gravity \(G\) (near the 50 cm mark).
- Now shift the knife-edge to one side of \(G\) and suspend a known mass \(m\) from the short arm; adjust its position until the rule balances horizontally.
- Measure the distance \(a\) of the known mass from the knife-edge and the distance \(b\) of \(G\) from the knife-edge.
- By the principle of moments, \( m \times a = M \times b \), so the mass of the rule \( M = \dfrac{m\,a}{b} \).
(ii) Two precautions:
- Read the balance positions with the eye directly above the scale to avoid parallax error.
- Ensure the knife-edge is sharp and the rule is horizontal (balanced) before taking readings.
(c) Bullet: \( m = 120\,\text{g} = 0.12\,\text{kg}, \; u = 20\,\text{m s}^{-1}, \; v = 0, \; t = 0.1\,\text{s}. \)
(i) Resistance (magnitude of the constant force):
\[ F = ma = m\frac{(u - v)}{t} = 0.12 \times \frac{20}{0.1} = 0.12 \times 200 = 24\,\text{N} \](ii) Distance moved in the wood:
\[ s = \left(\frac{u + v}{2}\right)t = \left(\frac{20 + 0}{2}\right)(0.1) = 10 \times 0.1 = 1.0\,\text{m} \]Answers: resistance \(= 24\,\text{N}\); distance \(= 1.0\,\text{m}\).
Swali 54 Ripoti
(a) Define Young's modulus.
(b) State the physical quantities one has to measure in order to determine the Young's modulus of a wire.
(a) Young's modulus of a material is the ratio of the tensile stress to the tensile strain, provided the elastic limit is not exceeded:
\[ Y = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/L} = \frac{FL}{Ae} \]Its unit is \(\text{N m}^{-2}\) (Pa).
(b) Physical quantities to measure for a wire:
- The stretching force (load) \(F\) applied to the wire.
- The original (unstretched) length \(L\) of the wire.
- The extension \(e\) produced by the load.
- The diameter of the wire (to obtain the cross-sectional area \(A\)).
Maelezo ya Majibu
(a) Young's modulus of a material is the ratio of the tensile stress to the tensile strain, provided the elastic limit is not exceeded:
\[ Y = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/L} = \frac{FL}{Ae} \]Its unit is \(\text{N m}^{-2}\) (Pa).
(b) Physical quantities to measure for a wire:
- The stretching force (load) \(F\) applied to the wire.
- The original (unstretched) length \(L\) of the wire.
- The extension \(e\) produced by the load.
- The diameter of the wire (to obtain the cross-sectional area \(A\)).
Swali 55 Ripoti
A stone is projected horizontally from the top of a tower with a speed of 5ms-1. It lands on the ground level at a horizontal distance of 20m from the foot of the tower. Calculate the height of the tower. (g = 10ms\(^{-2}\))
Horizontal projectile. The horizontal and vertical motions are independent. The horizontal velocity stays constant while the vertical motion is free fall from rest.
Step 1: Find the time of flight from the horizontal motion. There is no horizontal acceleration, so
\[ x = ut \implies 20 = 5 \times t \implies t = 4\,\text{s} \]Step 2: Use the time in the vertical motion. Vertically the stone starts with zero vertical velocity and falls a height \(h\):
\[ h = ut + \tfrac{1}{2}gt^2 = 0 + \tfrac{1}{2}(10)(4)^2 = \tfrac{1}{2}(10)(16) \] \[ h = 80\,\text{m} \]Answer: the height of the tower is \(80\,\text{m}\).
Maelezo ya Majibu
Horizontal projectile. The horizontal and vertical motions are independent. The horizontal velocity stays constant while the vertical motion is free fall from rest.
Step 1: Find the time of flight from the horizontal motion. There is no horizontal acceleration, so
\[ x = ut \implies 20 = 5 \times t \implies t = 4\,\text{s} \]Step 2: Use the time in the vertical motion. Vertically the stone starts with zero vertical velocity and falls a height \(h\):
\[ h = ut + \tfrac{1}{2}gt^2 = 0 + \tfrac{1}{2}(10)(4)^2 = \tfrac{1}{2}(10)(16) \] \[ h = 80\,\text{m} \]Answer: the height of the tower is \(80\,\text{m}\).
Swali 56 Ripoti
(a) List two types of waves, apart from light, that can be plane polarised
(b) State two applications of plane polarised light.
(a) Two types of waves, apart from light, that can be plane polarised. Only transverse waves can be polarised. Examples:
- Radio waves (electromagnetic).
- Microwaves (electromagnetic). (Other acceptable answers: X-rays, or transverse mechanical waves on a stretched string.)
(b) Two applications of plane polarised light:
- Polaroid sunglasses and camera filters, used to cut out glare from reflected light.
- Measuring the concentration of optically active solutions such as sugar (polarimetry / saccharimetry); also used in stress analysis of materials (photoelasticity) and in LCD screens.
Maelezo ya Majibu
(a) Two types of waves, apart from light, that can be plane polarised. Only transverse waves can be polarised. Examples:
- Radio waves (electromagnetic).
- Microwaves (electromagnetic). (Other acceptable answers: X-rays, or transverse mechanical waves on a stretched string.)
(b) Two applications of plane polarised light:
- Polaroid sunglasses and camera filters, used to cut out glare from reflected light.
- Measuring the concentration of optically active solutions such as sugar (polarimetry / saccharimetry); also used in stress analysis of materials (photoelasticity) and in LCD screens.
Swali 57 Ripoti
(a) State three properties of waves
(b)(i) Describe, with the aid of a labelled diagram, an experiment to show how the frequency of the note emitted by a vibrating string depends on the tension in the string
(ii) State two precautions necessary to ensure accurate results.
(c) Draw a ray diagram showing how a virtual image of an object is formed by a concave mirror.
(a) Three properties of waves are:
- Reflection
- Refraction
- Diffraction
(b)(i) Experiment to show the relation between frequency and tension
A sonometer is set up as shown below. A uniform wire is stretched over two knife-edge bridges, A and B. One end of the wire passes over a smooth pulley and supports a scale pan. The distance AB is kept constant throughout the experiment.
- Place a small paper rider at the middle of the wire between A and B.
- Put a known total mass, including the scale pan, on the hanger. The tension in the wire is then \(T=mg\).
- Strike a tuning fork of known frequency and hold its stem on the sonometer box. Pluck the wire gently.
- Adjust the load until resonance occurs. At resonance the vibration of the wire is large and the paper rider is thrown off.
- Record the total suspended mass and the frequency of the tuning fork.
- Repeat the procedure with other tuning forks, keeping the vibrating length AB unchanged.
A set of readings is:
| Frequency, \(f\) (Hz) | Total suspended mass, \(m\) (kg) | \(\sqrt{m}\) (kg1/2) |
|---|---|---|
| 256 | 1.00 | 1.00 |
| 320 | 1.5625 | 1.25 |
| 384 | 2.25 | 1.50 |
| 512 | 4.00 | 2.00 |
The graph of \(f\) against \(\sqrt{m}\) is:
The straight line passes through the origin. Its gradient is
\[\frac{512-256}{2.00-1.00}=256\ \text{Hz kg}^{-1/2}.\]
Hence \(f=256\sqrt{m}\). Since \(T=mg\), \(\sqrt{m}=\sqrt{T/g}\), and therefore
\[f\propto \sqrt{T}.\]
(b)(ii) Precautions
- Keep the distance between the knife-edge bridges fixed, and ensure that the wire is taut and firmly held at the bridges.
- Strike the tuning fork gently on a rubber pad and identify resonance only when the paper rider is thrown off, so that the frequency of the fork is not altered or damped.
(c) Virtual image formed by a concave mirror
When the object is between the pole \(P\) and principal focus \(F\), the reflected rays diverge. Their backward extensions meet behind the mirror at a virtual, erect and magnified image.
Maelezo ya Majibu
(a) Three properties of waves are:
- Reflection
- Refraction
- Diffraction
(b)(i) Experiment to show the relation between frequency and tension
A sonometer is set up as shown below. A uniform wire is stretched over two knife-edge bridges, A and B. One end of the wire passes over a smooth pulley and supports a scale pan. The distance AB is kept constant throughout the experiment.
- Place a small paper rider at the middle of the wire between A and B.
- Put a known total mass, including the scale pan, on the hanger. The tension in the wire is then \(T=mg\).
- Strike a tuning fork of known frequency and hold its stem on the sonometer box. Pluck the wire gently.
- Adjust the load until resonance occurs. At resonance the vibration of the wire is large and the paper rider is thrown off.
- Record the total suspended mass and the frequency of the tuning fork.
- Repeat the procedure with other tuning forks, keeping the vibrating length AB unchanged.
A set of readings is:
| Frequency, \(f\) (Hz) | Total suspended mass, \(m\) (kg) | \(\sqrt{m}\) (kg1/2) |
|---|---|---|
| 256 | 1.00 | 1.00 |
| 320 | 1.5625 | 1.25 |
| 384 | 2.25 | 1.50 |
| 512 | 4.00 | 2.00 |
The graph of \(f\) against \(\sqrt{m}\) is:
The straight line passes through the origin. Its gradient is
\[\frac{512-256}{2.00-1.00}=256\ \text{Hz kg}^{-1/2}.\]
Hence \(f=256\sqrt{m}\). Since \(T=mg\), \(\sqrt{m}=\sqrt{T/g}\), and therefore
\[f\propto \sqrt{T}.\]
(b)(ii) Precautions
- Keep the distance between the knife-edge bridges fixed, and ensure that the wire is taut and firmly held at the bridges.
- Strike the tuning fork gently on a rubber pad and identify resonance only when the paper rider is thrown off, so that the frequency of the fork is not altered or damped.
(c) Virtual image formed by a concave mirror
When the object is between the pole \(P\) and principal focus \(F\), the reflected rays diverge. Their backward extensions meet behind the mirror at a virtual, erect and magnified image.
Swali 58 Ripoti
(a) What is the principle upon which the lighting in fluorescent tubes operate?
(b) State two factors on which the colour of light from a fluorescent tube depends.
(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 ultraviolet radiation. This ultraviolet light strikes the fluorescent (phosphor) powder coating the inside of the tube, which absorbs the invisible ultraviolet and re-emits the energy as visible light.
(b) Two factors on which the colour of the light depends:
- The nature (chemical composition) of the fluorescent powder (phosphor) coating the tube.
- The nature of the gas or vapour (and the pressure) inside the tube.
Maelezo ya Majibu
(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 ultraviolet radiation. This ultraviolet light strikes the fluorescent (phosphor) powder coating the inside of the tube, which absorbs the invisible ultraviolet and re-emits the energy as visible light.
(b) Two factors on which the colour of the light depends:
- The nature (chemical composition) of the fluorescent powder (phosphor) coating the tube.
- The nature of the gas or vapour (and the pressure) inside the tube.
Swali 59 Ripoti
(a) State Hooke's law of elasticity
(b) A spiral spring, loaded with a piece of metal, extends by 10.5cm hair. When the metal is fully submerged in water, the spring extends by 6.8cm. Calculate the relative density of the metal. (Assume Hooke's law is obeyed).
(a) Hooke's law: provided the elastic limit is not exceeded, the extension of an elastic material (spring) is directly proportional to the stretching force (load) applied to it, i.e. \( F = ke \).
(b) Because the spring obeys Hooke's law, its extension is directly proportional to the force stretching it.
In air: the extension is caused by the full weight \(W\) of the metal, so \( W \propto 10.5 \).
In water: the metal experiences an upthrust, so the spring is stretched only by the apparent weight \(W'\), where \( W' \propto 6.8 \).
The upthrust equals the loss in weight (weight of water displaced):
\[ U = W - W' \propto (10.5 - 6.8) = 3.7 \]Relative density:
\[ \text{R.D.} = \frac{\text{weight of metal in air}}{\text{upthrust in water}} = \frac{W}{W - W'} = \frac{10.5}{3.7} \] \[ \text{R.D.} = 2.84 \]Answer: the relative density of the metal is \(2.84\).
Maelezo ya Majibu
(a) Hooke's law: provided the elastic limit is not exceeded, the extension of an elastic material (spring) is directly proportional to the stretching force (load) applied to it, i.e. \( F = ke \).
(b) Because the spring obeys Hooke's law, its extension is directly proportional to the force stretching it.
In air: the extension is caused by the full weight \(W\) of the metal, so \( W \propto 10.5 \).
In water: the metal experiences an upthrust, so the spring is stretched only by the apparent weight \(W'\), where \( W' \propto 6.8 \).
The upthrust equals the loss in weight (weight of water displaced):
\[ U = W - W' \propto (10.5 - 6.8) = 3.7 \]Relative density:
\[ \text{R.D.} = \frac{\text{weight of metal in air}}{\text{upthrust in water}} = \frac{W}{W - W'} = \frac{10.5}{3.7} \] \[ \text{R.D.} = 2.84 \]Answer: the relative density of the metal is \(2.84\).
Swali 60 Ripoti
(a) Sketch the magnetic flux pattern around a long; straight, current-carrying wire
(b) State two methods by which the sensitivity of a moving-coil galvanometer can be increased.
(c) A series RLC circuit comprises a 100-\(\Omega\) resistor, a 3-H inductor and a 4-\(\mu\)f capacitor. The a.c source of tile circuit has an e.m.f of 100V and a frequency of 160 Hz.
(i) Draw the circuit diagram of the arrangement. Calculate the:
(ii) capacitive reactance;
(iii) inductive reactance:
(iv) impedance of the circuit;
(v) current in the circuit;
(vi) average power dissipated in the circuit.
(a) Magnetic flux pattern around a long straight current-carrying wire
The wire carries current out of the plane of the paper, as indicated by the dot. The magnetic field lines are concentric circles centred on the wire. Their direction is anticlockwise, given by the right-hand grip rule.
(b) The sensitivity of a moving-coil galvanometer can be increased by:
- using a weaker control spring, so that the restoring torque is smaller;
- using a stronger magnet to increase the magnetic flux density.
(c) Given: \(R=100\,\Omega\), \(L=3\,\text{H}\), \(C=4\,\mu\text{F}=4\times10^{-6}\,\text{F}\), \(V=100\,\text{V}\), and \(f=\dfrac{160}{\pi}\,\text{Hz}\).
(i) Series RLC circuit
(ii) Capacitive reactance
\[X_C=\frac{1}{2\pi fC}=\frac{1}{2\pi\left(\dfrac{160}{\pi}\right)\left(4\times10^{-6}\right)}=781.25\,\Omega\]
\[\boxed{X_C\approx781\,\Omega}\]
(iii) Inductive reactance
\[X_L=2\pi fL=2\pi\left(\dfrac{160}{\pi}\right)(3)=960\,\Omega\]
\[\boxed{X_L=960\,\Omega}\]
(iv) Impedance
\[Z=\sqrt{R^2+(X_L-X_C)^2}\]
\[Z=\sqrt{100^2+(960-781.25)^2}=204.8\,\Omega\]
\[\boxed{Z\approx205\,\Omega}\]
(v) Current in the circuit
\[I=\frac{V}{Z}=\frac{100}{204.8}=0.488\,\text{A}\]
\[\boxed{I\approx0.49\,\text{A}}\]
(vi) Average power dissipated
Only the resistor dissipates average power:
\[P=I^2R=(0.488)^2(100)=23.8\,\text{W}\]
\[\boxed{P\approx24\,\text{W}}\]
Maelezo ya Majibu
(a) Magnetic flux pattern around a long straight current-carrying wire
The wire carries current out of the plane of the paper, as indicated by the dot. The magnetic field lines are concentric circles centred on the wire. Their direction is anticlockwise, given by the right-hand grip rule.
(b) The sensitivity of a moving-coil galvanometer can be increased by:
- using a weaker control spring, so that the restoring torque is smaller;
- using a stronger magnet to increase the magnetic flux density.
(c) Given: \(R=100\,\Omega\), \(L=3\,\text{H}\), \(C=4\,\mu\text{F}=4\times10^{-6}\,\text{F}\), \(V=100\,\text{V}\), and \(f=\dfrac{160}{\pi}\,\text{Hz}\).
(i) Series RLC circuit
(ii) Capacitive reactance
\[X_C=\frac{1}{2\pi fC}=\frac{1}{2\pi\left(\dfrac{160}{\pi}\right)\left(4\times10^{-6}\right)}=781.25\,\Omega\]
\[\boxed{X_C\approx781\,\Omega}\]
(iii) Inductive reactance
\[X_L=2\pi fL=2\pi\left(\dfrac{160}{\pi}\right)(3)=960\,\Omega\]
\[\boxed{X_L=960\,\Omega}\]
(iv) Impedance
\[Z=\sqrt{R^2+(X_L-X_C)^2}\]
\[Z=\sqrt{100^2+(960-781.25)^2}=204.8\,\Omega\]
\[\boxed{Z\approx205\,\Omega}\]
(v) Current in the circuit
\[I=\frac{V}{Z}=\frac{100}{204.8}=0.488\,\text{A}\]
\[\boxed{I\approx0.49\,\text{A}}\]
(vi) Average power dissipated
Only the resistor dissipates average power:
\[P=I^2R=(0.488)^2(100)=23.8\,\text{W}\]
\[\boxed{P\approx24\,\text{W}}\]
Swali 61 Ripoti
In an electrolysis experiment, the ammeter records a steady current of 1A. The mass of copper deposited is 0.66g in 30 minutes. Calculate the error in the ammeter reading (Electrochemical equivalent of copper = 0.00033g C\(^{-1}\)).
Faraday's first law of electrolysis: \( m = zIt \), where \(z\) is the electrochemical equivalent.
Step 1: Convert the time. \( t = 30 \times 60 = 1800\,\text{s} \).
Step 2: Find the true (actual) current that must have flowed to deposit the copper.
\[ I = \frac{m}{zt} = \frac{0.66}{0.00033 \times 1800} = \frac{0.66}{0.594} = 1.11\,\text{A} \]Step 3: Compare with the ammeter reading. The ammeter shows \(1.0\,\text{A}\), but the mass of copper proves the true current was \(1.11\,\text{A}\).
\[ \text{Error} = 1.11 - 1.00 = 0.11\,\text{A} \]Answer: the ammeter under-reads by \(0.11\,\text{A}\). (As a percentage of the true value this is \(\tfrac{0.11}{1.11}\times100\% \approx 10\%\).)
Maelezo ya Majibu
Faraday's first law of electrolysis: \( m = zIt \), where \(z\) is the electrochemical equivalent.
Step 1: Convert the time. \( t = 30 \times 60 = 1800\,\text{s} \).
Step 2: Find the true (actual) current that must have flowed to deposit the copper.
\[ I = \frac{m}{zt} = \frac{0.66}{0.00033 \times 1800} = \frac{0.66}{0.594} = 1.11\,\text{A} \]Step 3: Compare with the ammeter reading. The ammeter shows \(1.0\,\text{A}\), but the mass of copper proves the true current was \(1.11\,\text{A}\).
\[ \text{Error} = 1.11 - 1.00 = 0.11\,\text{A} \]Answer: the ammeter under-reads by \(0.11\,\text{A}\). (As a percentage of the true value this is \(\tfrac{0.11}{1.11}\times100\% \approx 10\%\).)
Swali 62 Ripoti
Explain plane polarisation of light
Plane polarisation of light. Ordinary (unpolarised) light is a transverse wave whose vibrations of the electric field take place in all directions (all planes) perpendicular to the direction of travel of the wave.
Plane polarisation is the process by which these vibrations are restricted so that they occur in only one plane containing the direction of travel. Light in which the vibrations are confined to a single plane is called plane-polarised light.
When ordinary light passes through a polariser (for example a Polaroid sheet or by reflection at a suitable angle, or through a Nicol prism), only the component of vibration in one particular plane is transmitted; the rest are absorbed or removed. If this plane-polarised light then meets a second polariser (the analyser) whose axis is at right angles to the first, no light is transmitted, confirming that the light is polarised. Only transverse waves can be polarised, so polarisation is proof that light is a transverse wave.
Maelezo ya Majibu
Plane polarisation of light. Ordinary (unpolarised) light is a transverse wave whose vibrations of the electric field take place in all directions (all planes) perpendicular to the direction of travel of the wave.
Plane polarisation is the process by which these vibrations are restricted so that they occur in only one plane containing the direction of travel. Light in which the vibrations are confined to a single plane is called plane-polarised light.
When ordinary light passes through a polariser (for example a Polaroid sheet or by reflection at a suitable angle, or through a Nicol prism), only the component of vibration in one particular plane is transmitted; the rest are absorbed or removed. If this plane-polarised light then meets a second polariser (the analyser) whose axis is at right angles to the first, no light is transmitted, confirming that the light is polarised. Only transverse waves can be polarised, so polarisation is proof that light is a transverse wave.
Swali 63 Ripoti
(a) Explain electrolysis
(b) Classify the following substances as electrolytes and non-electrolytes: Sugar solution: paraffin, suit solution and grape fruit juice.
(a) Electrolysis is the chemical decomposition of an electrolyte, in its molten state or in aqueous solution, brought about by the passage of a direct electric current through it. During the process the positive ions (cations) move to the cathode and the negative ions (anions) move to the anode, where they are discharged and the substance is broken down into its components.
(b) Classification. A substance is an electrolyte only if, when molten or dissolved, it contains free-moving ions that can carry the current. Sugar dissolves as neutral molecules and paraffin is a covalent non-conductor, so neither provides ions. Common salt dissolves to give \(\text{Na}^{+}\) and \(\text{Cl}^{-}\) ions, and grape fruit juice contains dissolved acids (such as citric and ascorbic acids) that release ions, so both conduct.
| Substance | Classification |
|---|---|
| Sugar solution | Non-electrolyte |
| Paraffin | Non-electrolyte |
| Salt solution | Electrolyte |
| Grape fruit juice | Electrolyte |
Common misconception: a solution is not an electrolyte simply because it is a liquid or tastes of something dissolved. Sugar solution is a liquid and conducts almost no current because sugar stays as whole neutral molecules; it is a non-electrolyte. What matters is whether free ions are present.
Maelezo ya Majibu
(a) Electrolysis is the chemical decomposition of an electrolyte, in its molten state or in aqueous solution, brought about by the passage of a direct electric current through it. During the process the positive ions (cations) move to the cathode and the negative ions (anions) move to the anode, where they are discharged and the substance is broken down into its components.
(b) Classification. A substance is an electrolyte only if, when molten or dissolved, it contains free-moving ions that can carry the current. Sugar dissolves as neutral molecules and paraffin is a covalent non-conductor, so neither provides ions. Common salt dissolves to give \(\text{Na}^{+}\) and \(\text{Cl}^{-}\) ions, and grape fruit juice contains dissolved acids (such as citric and ascorbic acids) that release ions, so both conduct.
| Substance | Classification |
|---|---|
| Sugar solution | Non-electrolyte |
| Paraffin | Non-electrolyte |
| Salt solution | Electrolyte |
| Grape fruit juice | Electrolyte |
Common misconception: a solution is not an electrolyte simply because it is a liquid or tastes of something dissolved. Sugar solution is a liquid and conducts almost no current because sugar stays as whole neutral molecules; it is a non-electrolyte. What matters is whether free ions are present.
Swali 64 Ripoti
(a) Explain (i) work: (ii) power
(b) Show that the efficiency E, the force ratio M.A. and the velocity ratio V.R. of a machine are related by the equation: E = M.A. x 100% V.R.
(c) An inclined plane of angle 15° is used raise a load of 4500N through a height of 2m. If the plane is 75% efficient calculate:
(i) velocity ratio of the plane;
(ii) work done on the load
(d) Explain Charles' law using the Kinetic theory of matter.
