WAEC SSCE - Physics - 1994
Ajụjụ 1 Ripọtì
A particle of charge q and mass m moving with a velocity v enters a uniform magnetic field of strength B in the direction of the field. The force on the particle is
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The force on a charged particle moving in a magnetic field is given by the formula F = qvB, where F is the force, q is the charge of the particle, v is its velocity, and B is the strength of the magnetic field. In this case, the particle is moving in the direction of the field, so the angle between its velocity and the magnetic field is zero. Therefore, the force on the particle is simply qvB. The mass of the particle does not affect the force in this case. So, the correct option is "qvB".
Ajụjụ 2 Ripọtì
\(500cm^{-3}\) of a gas is collected at 0°C and at a pressure of 72cm of mercury. What is the volume of the the gas at the same temperature and at a pressure of 76cm of the same mercury?
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The relationship between the pressure and volume of a gas is given by the Boyle's Law, which states that at constant temperature, the pressure and volume of a gas are inversely proportional to each other. This means that if the pressure of a gas increases, its volume decreases and vice versa. Using the Boyle's Law, we can solve the problem as follows: Let V be the volume of the gas at a pressure of 76 cm of mercury. Using Boyle's Law, we have: P₁V₁ = P₂V₂ where P₁ and V₁ are the initial pressure and volume of the gas, respectively, and P₂ and V₂ are the final pressure and volume of the gas, respectively. Substituting the given values, we get: (72 cm)(500 cm³) = (76 cm)(V₂) Solving for V₂, we have: V₂ = (72 cm)(500 cm³) / (76 cm) V₂ = 459.21 cm³ (rounded to two decimal places) Therefore, the volume of the gas at the same temperature and at a pressure of 76 cm of the same mercury is approximately 459.21 cm³. Answer: \(\frac{72\times500}{76}cm^3\)
Ajụjụ 3 Ripọtì
The instrument used in the measurement of atmospheric pressure is?
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The instrument used in the measurement of atmospheric pressure is a barometer. A barometer is an instrument used to measure atmospheric pressure. It consists of a glass tube filled with mercury and inverted into a dish of mercury. Atmospheric pressure exerts a force on the surface of the mercury in the dish, causing it to rise up the tube. The height of the mercury column is directly proportional to the atmospheric pressure. The barometer is calibrated in units of pressure, such as millimeters of mercury (mmHg) or kilopascals (kPa).
Ajụjụ 4 Ripọtì
The density of a body is \(5\times10^3kgm{^-3}\) and it weighs 1N in air. Calculate the apparent weight of the body when totally immersed in water. (Density of water =\(10^3kgm{-3}, g= 10ms^{-2}
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Ajụjụ 5 Ripọtì
A rectangular coil can rotate in a magnetic field. The two ends of the coil are soldered respectively to the two halves of a commutator. Two carbon brushes are made to press lightly against the commutator and when these are connected in a circuit with a battery and a rheostat, the coil rotates. This is a description of
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The given description is of a D.C electric motor. It consists of a rectangular coil that can rotate in a magnetic field. When a current is passed through the coil, it experiences a force due to the interaction with the magnetic field. As a result, the coil rotates. The two ends of the coil are connected to the two halves of a commutator, and carbon brushes are made to press lightly against the commutator. The commutator allows the direction of the current to be reversed each time the coil passes a half-turn, ensuring that the coil continues to rotate in the same direction. This continuous rotation of the coil provides mechanical energy that can be used to perform work.
Ajụjụ 6 Ripọtì
A transformer with 5500 turns in its primary winding is used between a 240V a.c supply and 120V kettle. Calculate the number of turns in the secondary winding
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The voltage ratio of a transformer is equal to the turns ratio. In this case, the primary winding has 5500 turns and is connected to a 240V supply, while the secondary winding is connected to a 120V kettle. Since the voltage is halved from the primary to the secondary, we know that the turns ratio must also be halved. Thus, the number of turns in the secondary winding would be: 5500 / 2 = 2750 Therefore, the answer is 2750.
Ajụjụ 7 Ripọtì
The wave function of a metal is 8.0 x \(10^{-19}\)J. Calculate the wavelength of its threshold frequency. (Speed of light in a vacuum = 3 x \(10^8ms^{-1}, Planck's constant = 6.6 \times 10^{-34}\)Js)
Ajụjụ 8 Ripọtì
Which of the following substances is not an insulator?
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Aluminium is not an insulator because it is a good conductor of electricity. When a substance conducts electricity, it allows electric charges to flow freely through it. In contrast, insulators prevent or limit the flow of electric charges, making them ideal for use in electrical insulation. Therefore, the answer to the question is Aluminium.
Ajụjụ 9 Ripọtì
A gas occupies a certain volume at 27°C. At what temperature will its volume be three times the original volume assuming that its pressure remains constant?
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The problem can be solved using Charles's law, which states that the volume of a gas at a constant pressure is directly proportional to its temperature in kelvins. Let the initial temperature of the gas be T1 = 27°C + 273.15 = 300.15 K, and let its initial volume be V1. According to the problem, when the volume is tripled, the new volume is V2 = 3V1. We want to find the new temperature T2. Using Charles's law, we can write: V1 / T1 = V2 / T2 Substituting the values: V1 / 300.15 = 3V1 / T2 Solving for T2: T2 = (3V1 x 300.15) / V1 = 900.45 K Converting back to Celsius: T2 = 900.45 - 273.15 = 627.3°C (approx) Therefore, the answer is 627°C.
Ajụjụ 10 Ripọtì
Which of the following has the shortest wavelength?
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Gamma rays have the shortest wavelength among the given options. This is because gamma rays have the highest frequency and energy, which corresponds to a shorter wavelength according to the equation E = hf, where E is energy, h is Planck's constant, and f is frequency. Gamma rays have a wavelength of less than 10 picometers, which is even shorter than X-rays. Infrared rays, ultraviolet rays, radio waves, and visible light all have longer wavelengths than gamma rays, in that order.
Ajụjụ 11 Ripọtì
The nucleon number and the proton number of a neutral atom of an element are 238 and 92 respectively. What is the number of neutrons in the atom?
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The number of neutrons in an atom is calculated by subtracting the number of protons from the nucleon number. In this case, the nucleon number is given as 238 and the number of protons is given as 92. Therefore, the number of neutrons is calculated as follows: Number of neutrons = Nucleon number - Proton number Number of neutrons = 238 - 92 Number of neutrons = 146 So, the answer is 146. Option (c) is the correct answer.
Ajụjụ 12 Ripọtì
In which of the following simple machines is the effort applied between the load and the fulcrum?
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Ajụjụ 13 Ripọtì
Which of the following is not a part of an A.C generator?
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The correct option is (c) Commutator. An AC generator (also called an alternator) is a device that converts mechanical energy into electrical energy. It works on the principle of electromagnetic induction. An AC generator consists of two main parts: a stationary part and a rotating part. The stationary part contains the field magnet, and the rotating part contains the armature. The field magnet is a permanent magnet or an electromagnet that produces a magnetic field. The armature is a coil of wire that is rotated inside the magnetic field. When the armature rotates, the magnetic field induces an alternating current (AC) in the coil. The carbon brushes and slip rings are parts of a DC generator, not an AC generator. Carbon brushes are used to transfer the current from the stationary part to the rotating part, while slip rings are used to connect the rotating part to an external circuit. The commutator is also a part of a DC generator, not an AC generator. It is a device that helps to convert the alternating current produced in the armature into direct current (DC) that can be used in an external circuit. Since AC generators produce alternating current, they do not need a commutator to convert the current. Therefore, the correct answer is (c) Commutator.
Ajụjụ 14 Ripọtì
In 24 days a radioactive isotope decreases in mass from 64g to 2g. What is the half life of the radioactive material
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The half-life of a radioactive substance is the time it takes for half of the initial mass of the substance to decay. Let's find the fraction of the initial mass remaining after 24 days: mass remaining/mass at start = (2g/64g) = 1/32 This means that after one half-life, the mass remaining would be half of the initial mass, or (1/2) x 64g = 32g. After two half-lives, the mass remaining would be half of 32g, or 16g. Similarly, after three half-lives, the mass remaining would be 8g, and after four half-lives, the mass remaining would be 4g. Since the mass remaining after 24 days is 1/32 of the initial mass, we know that the number of half-lives that have passed is 5, because 2 raised to the 5th power (2^5) is 32. Therefore, the half-life of the substance is 24 days / 5 = 4.8 days. So the correct option is (d) 4.80 days.
Ajụjụ 15 Ripọtì
A note of frequency of 2000Hz has a velocity of \(400ms^{-1}\). What is the wavelength of the note?
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The formula for calculating wavelength is: wavelength = velocity/frequency In this question, the frequency is given as 2000Hz and the velocity is given as 400m/s. Substituting these values into the formula, we get: wavelength = 400/2000 wavelength = 0.2m Therefore, the wavelength of the note is 0.2 meters.
Ajụjụ 16 Ripọtì
Beta particles are
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Beta particles are electrons that are emitted from the nucleus of an atom during beta decay. When a neutron in the nucleus of an atom changes into a proton, it emits an electron and an antineutrino. This electron is called a beta particle. Beta particles have a negative charge and a relatively small mass, about 1/1836th that of a proton. Beta particles can be easily stopped by a few millimeters of aluminum or a few centimeters of air. Beta particles can be emitted by a variety of radioactive isotopes and can pose a health hazard if ingested or inhaled.
Ajụjụ 18 Ripọtì
which of the following is a scalar quantity?
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A scalar quantity is a physical quantity that has only magnitude, but no direction. Therefore, it can be represented only by a number (positive or negative) and a unit. Distance is a scalar quantity as it only has magnitude and no direction associated with it. The other options - tension, weight, impulse, and upthrust are vector quantities as they have both magnitude and direction associated with them.
Ajụjụ 19 Ripọtì
Calculate the time taken to heat 2kg of water from 50°C to 100°C in an electric kettle taking 5A, from a 210V supply. (Specific heat capacity of water = 4200\(kg^{-1}K{-1}\)
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The formula to use is: energy supplied = mass x specific heat capacity x temperature change First, let's calculate the energy required to heat the water: energy = 2kg x 4200\(kg^{-1}K^{-1}\) x (100°C - 50°C) energy = 420000 J Now, we can calculate the power of the electric kettle: power = voltage x current power = 210V x 5A power = 1050W The time taken to heat the water can be calculated using the formula: time = energy / power Substituting the values we have: time = 420000 J / 1050W time = 400 seconds Converting seconds to minutes: time = 400 seconds ÷ 60 seconds/minute time = 6.7 minutes Therefore, the time taken to heat 2kg of water from 50°C to 100°C in an electric kettle taking 5A, from a 210V supply is 6.7 minutes.
Ajụjụ 20 Ripọtì
Which of the following are transverse waves? I. Ripples on water II. Sound waves in air III. Light waves from the sun
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Ripples on water and light waves from the sun are transverse waves. Transverse waves are characterized by their perpendicular motion to the direction of wave propagation. Ripples on water move up and down perpendicular to the direction of wave propagation. Similarly, light waves consist of electric and magnetic fields that oscillate perpendicular to the direction of the wave. Sound waves in air, on the other hand, are longitudinal waves. In longitudinal waves, the particles of the medium oscillate in the same direction as the wave propagation. In the case of sound waves in air, air particles vibrate back and forth in the same direction that the sound wave is traveling. Therefore, the answer is (B) I and II only.
Ajụjụ 21 Ripọtì
A car fuse is marked 15A and operates normally on a 12V battery. Calculate the resistance of the fuse wire
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The power dissipated by the fuse wire can be calculated using the formula: Power (P) = Current (I) x Voltage (V) The current can be calculated using Ohm's law: V = I x R where V is the voltage, I is the current, and R is the resistance. Therefore, we can rewrite the power formula as: P = (V/R) x V or P = V^2 / R From the given information, the voltage (V) is 12V and the current (I) is 15A. Substituting these values into Ohm's law gives: 12 = 15 x R Solving for R gives: R = 12/15 R = 0.8 Ω Therefore, the correct answer is: 0.8Ω.
Ajụjụ 22 Ripọtì
When two objects P and Q are supplied with the same quantity of heat,the temperature change in P is observed to be twice that in Q. If the masses of P and Q are the same,calculate the ratio of the specific heat capacities of P and Q
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The specific heat capacity of a substance is defined as the amount of heat energy required to raise the temperature of one kilogram of the substance by one degree Celsius. Let the specific heat capacity of object P be Cp and that of Q be Cq, and let their masses be m each. According to the question, the heat supplied to P and Q are the same, let's say Q. So, Heat supplied to P = Heat supplied to Q m × Cp × ΔT_P = m × Cq × ΔT_Q where ΔT_P and ΔT_Q are the temperature changes in P and Q, respectively. We are given that ΔT_P = 2ΔT_Q. Therefore, substituting this into the above equation, we get: m × Cp × 2ΔT_Q = m × Cq × ΔT_Q Cp/Cq = ΔT_Q / ΔT_Q x 2 Cp/Cq = 1/2 Therefore, the ratio of the specific heat capacities of P and Q is 1:2. Hence, the answer is (d) 1:2.
Ajụjụ 23 Ripọtì
The density of water is 1g\(cm^3\) while that of ice is 0.9\(cm^3\), calculate the volume change when 90g of ice is completely melted
Ajụjụ 24 Ripọtì
500g of water is heated so that its temperature rises from 30°C to 72°C in 7 minutes. calculate the supplied per minute. (Specific heat capacity of water is 4200J\(kg^{-1}K^{-1}\)
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The specific heat capacity of water is the amount of energy required to raise the temperature of 1 kg of water by 1°C. It is given as 4200 J/(kg·K). The energy required to heat the water from 30°C to 72°C can be calculated using the formula: energy = mass x specific heat capacity x temperature change Substituting the values given in the question, we have: energy = 500g x 4200 J/(kg·K) x (72°C - 30°C) energy = 500g x 4200 J/(kg·K) x 42°C energy = 8820000 J This is the total energy supplied to the water in 7 minutes. To calculate the energy supplied per minute, we need to divide this value by 7: energy per minute = 8820000 J ÷ 7 energy per minute = 1260000 J/min Therefore, the answer is 12600J.
Ajụjụ 25 Ripọtì
The image of an object placed at the center of the curvature of a concave mirror is
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When an object is placed at the center of curvature of a concave mirror, the image formed will be real, inverted and the same size as the object. So, the correct option is (E) "at the center of curvature". The light rays from the object will fall parallel to the principal axis of the mirror, and after reflection, they will converge at the center of curvature. The image formed will be located at the same distance from the mirror as the object, and it will be inverted and real.
Ajụjụ 26 Ripọtì
A tap supplies water at 30°C while another supplies at 86°C. If a man wishes to bath with water at 44°C, calculate the ratio of mass of hot water to that of cold water required
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The ratio of hot water to cold water can be calculated using the principle of mixture of substances. Let m1 be the mass of cold water required and m2 be the mass of hot water required. The specific heat capacity of water is 4200 J/kg K. The heat gained by cold water, Q1 = m1 x 4200 J/kg K x (44 - 30) K The heat gained by hot water, Q2 = m2 x 4200 J/kg K x (86 - 44) K As per the principle of mixture of substances, the total heat gained by cold water is equal to the total heat lost by hot water. Hence, Q1 = Q2 Therefore, m1 x 4200 J/kg K x (44 - 30) K = m2 x 4200 J/kg K x (86 - 44) K Simplifying this equation gives us: m2/m1 = (44 - 30)/(86 - 44) = 14/42 = 1/3 Therefore, the ratio of mass of hot water to that of cold water required is 1:3. Hence, the answer is (a) 1:3.
Ajụjụ 27 Ripọtì
A light wave travels from air into a medium of refractive index 1.54. If the wavelength of the light in air is \(5.9\times10^{-7}\)m, calculate its wavelength in the medium
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The relationship between the wavelength of light in two different media and their refractive indices is given by the formula: \begin{equation*} \frac{\lambda_1}{\lambda_2} = \frac{n_2}{n_1} \end{equation*} Where: \begin{itemize} \item \(\lambda_1\) is the wavelength of light in the first medium \item \(\lambda_2\) is the wavelength of light in the second medium \item \(n_1\) is the refractive index of the first medium \item \(n_2\) is the refractive index of the second medium \end{itemize} In this case, the light wave travels from air (first medium) into a medium of refractive index 1.54 (second medium). Therefore, we can use the above formula to find the wavelength of the light in the second medium: \begin{align*} \frac{\lambda_1}{\lambda_2} &= \frac{n_2}{n_1} \\ \lambda_2 &= \frac{n_1}{n_2} \lambda_1 \\ &= \frac{1}{1.54} \times 5.9 \times 10^{-7} \mathrm{m} \\ &\approx 3.84 \times 10^{-7} \mathrm{m} \end{align*} Therefore, the wavelength of the light in the medium is approximately \(3.84\times10^{-7}\)m. Answer: \boxed{2}. \(3.8\times10^{-7}\)m.
Ajụjụ 28 Ripọtì
Which of the following scientists postulated that moving particles exhibit wave properties?
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Louis de Broglie postulated that moving particles exhibit wave properties. This postulation is known as the de Broglie hypothesis. In 1924, de Broglie suggested that matter can exhibit both particle and wave-like behavior. He proposed that moving particles, such as electrons, have wave-like properties, and the wavelength of these waves is inversely proportional to the momentum of the particles. This hypothesis was later experimentally confirmed, and it is now known as wave-particle duality.
Ajụjụ 29 Ripọtì
A steam trap is a component of apparatus used in determining the specific latent heat of vaporization of steam. In the steady state, the steam trap
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Ajụjụ 30 Ripọtì
Which of the following is a stringed instrument?
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Out of the given options, the stringed instrument is the piano. A piano is a musical instrument that is played by striking its keys with the fingers, which in turn causes strings inside the piano to vibrate and produce sound. The sound is amplified and projected by the piano's wooden soundboard and metal strings. The other options are not stringed instruments: the flute is a woodwind instrument, the trumpet is a brass instrument, the drum is a percussion instrument, and the organ is a keyboard instrument that uses air to produce sound.
Ajụjụ 31 Ripọtì
A hygrometer is an instrument used to measure
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A hygrometer is an instrument used to measure relative humidity. Relative humidity refers to the amount of moisture or water vapor present in the air, compared to the maximum amount of moisture the air can hold at a particular temperature. A hygrometer works by measuring the amount of moisture in the air and expressing it as a percentage of the maximum amount of moisture the air can hold at a given temperature.
Ajụjụ 32 Ripọtì
A cell of e.m.f. 1.5V and internal resistance of 2.5Ω is connected in series with an ammeter of resistance 0.5W and a resistor of resistance of 7.0Ω. Calculate the current in the circuit.
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To calculate the current in the circuit, we can use Ohm's Law: V = IR where V is the voltage (or electromotive force), I is the current, and R is the resistance. In this case, the voltage of the cell is 1.5V, and the internal resistance of the cell is 2.5Ω. The total resistance in the circuit is the sum of the ammeter resistance (0.5Ω) and the resistor resistance (7.0Ω), which is 7.5Ω. So, we can write the equation: 1.5V = I (2.5Ω + 7.5Ω) Simplifying the equation, we get: 1.5V = I (10Ω) I = 1.5V / 10Ω I = 0.15A Therefore, the current in the circuit is 0.15A. The correct option is (a) 0.15A.
Ajụjụ 33 Ripọtì
A metal rod of length 40cm at 20°C is heated to a temperature of 45°C. If the new length of the rod is 40.05cm. Calculate its linear expansitivity
Ajụjụ 34 Ripọtì
The pressure of air in a tire is \(22.5Nm^{-2}\) at 27°C. If the air in the tyre heats up to 47°C, calculate the new pressure of the air, assuming that no air leaks out and that the change in volume of the air can be neglected.
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The relationship between pressure, volume, and temperature of a gas is given by the ideal gas law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature in Kelvin. Assuming that the volume of the tire remains constant and no air leaks out, we can use the ideal gas law to calculate the new pressure of the air in the tire when the temperature increases from 27°C to 47°C. First, we need to convert the temperatures from Celsius to Kelvin by adding 273 to each temperature: Initial temperature, T1 = 27°C + 273 = 300K Final temperature, T2 = 47°C + 273 = 320K Next, we can set up the following equation using the ideal gas law: P1V = nRT1 Since the volume and the number of moles of air remain constant, we can rearrange this equation to solve for the initial pressure, P1: P1 = (nR / V) * T1 Similarly, we can use the ideal gas law to calculate the final pressure, P2: P2V = nRT2 P2 = (nR / V) * T2 Since the volume and the number of moles of air remain constant, we can simplify this equation to: P2 = P1 * (T2 / T1) Substituting the values we know: P1 = 22.5 Nm^(-2) T1 = 300 K T2 = 320 K P2 = 22.5 * (320 / 300) = 24 Nm^(-2) Therefore, the new pressure of the air in the tire is 24 Nm^(-2) when the temperature increases from 27°C to 47°C. Therefore, the correct option is: - \(24Nm^{-2}\)
Ajụjụ 35 Ripọtì
A body accelerates uniformly from rest at \(2ms^{-2}\). Calculate the velocity after traveling 9m
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We can use the equation of motion: v^2 = u^2 + 2as where v is final velocity, u is initial velocity, a is acceleration, and s is displacement. In this problem, the body starts from rest, so its initial velocity (u) is 0. The acceleration (a) is given as 2\(ms^{-2}\), and the displacement (s) is 9m. Plugging in these values, we get: v^2 = 0^2 + 2(2\(ms^{-2}\))(9m) v^2 = 36\(ms^{-2}\)m^2 Taking the square root of both sides, we get: v = sqrt(36\(ms^{-2}\)m^2) v = 6\(ms^{-1}\) Therefore, the velocity of the body after traveling 9m is 6\(ms^{-1}\). The answer is.
Ajụjụ 36 Ripọtì
The ice and steam points on a mercury in glass thermometer are found to be 90mm apart. What temperature is recorded in degree Celsius when the length of the mercury thread is 33.6mm above the ice point mark?
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We are given that the ice and steam points on the thermometer are 90mm apart. This means that the thermometer has been calibrated to give readings between the ice point (0°C) and the steam point (100°C) when it is used. The length of the mercury thread above the ice point mark is given to be 33.6mm. We can find the corresponding temperature using proportionality. Let x be the temperature in °C corresponding to the length of the mercury thread above the ice point mark. Then, we have: 90 mm = 100°C - 0°C 33.6 mm = x°C - 0°C Using proportionality, we can write: (33.6 mm) / (90 mm) = (x°C - 0°C) / (100°C - 0°C) Simplifying, we get: x/100 = 0.373 x = 37.3°C Therefore, the temperature recorded in °C when the length of the mercury thread is 33.6mm above the ice point mark is 37.3°C. So, the correct option is (c) 37.3°C.
Ajụjụ 37 Ripọtì
When an atom loses or gains a charge, it becomes an
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When an atom loses or gains a charge, it becomes an ion. An ion is an atom or molecule that has gained or lost one or more electrons, resulting in a net positive or negative electrical charge. If an atom loses electrons, it becomes positively charged and is called a cation. On the other hand, if an atom gains electrons, it becomes negatively charged and is called an anion. Ions are important in chemical reactions, as they can form bonds with other ions or molecules to form compounds.
Ajụjụ 38 Ripọtì
A battery of e.m.f 10V an internal resistance 2Ω is connected to a resistance of 6Ω. Calculate the p.d
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The potential difference (p.d) across a resistor in a circuit can be calculated using Ohm's law which states that the p.d is equal to the product of the current flowing through the resistor and the resistance of the resistor. In this circuit, the battery has an e.m.f of 10V and an internal resistance of 2Ω. Therefore, the p.d across the battery terminals is 10V. Since the resistor of 6Ω is connected in the same series with the battery, the current flowing in the circuit will cause a voltage drop across the internal resistance of the battery. This means that the p.d across the resistor will be less than the 10V. The current flowing in the circuit can be calculated using the formula I = E/(R + r), where E is the e.m.f of the battery, R is the resistance of the external resistor, and r is the internal resistance of the battery. Substituting the given values, we get: I = 10V/(6Ω + 2Ω) = 1.25A The p.d across the resistor can then be calculated using Ohm's law: p.d = IR = 1.25A x 6Ω = 7.5V Therefore, the answer is 7.50V.
Ajụjụ 39 Ripọtì
A ball is dropped from a height of 45m above the ground. Calculate the velocity of the ball just before it strikes the ground. (Neglect air resistance and take g as \(10ms^{-2}\)
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The velocity of an object just before it strikes the ground can be found using the formula: v = \(\sqrt{2gh}\) where v is the velocity, g is the acceleration due to gravity, and h is the height from which the object was dropped. In this question, the height from which the ball was dropped is given as 45m, and the acceleration due to gravity is given as \(10ms^{-2}\). Substituting these values in the formula, we get: v = \(\sqrt{2 \times 10 \times 45}\) v = \(\sqrt{900}\) v = 30\(ms^{-1}\) Therefore, the velocity of the ball just before it strikes the ground is 30\(ms^{-1}\). So, the correct answer is the second option: 30\(ms^{-1}\).
Ajụjụ 40 Ripọtì
A simple magnifying glass is used to view an object. At what distance from the lens must the object be placed so that an image 5 times the size of the object is produced 20cm from the lens?
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Ajụjụ 41 Ripọtì
The electrical energy supplied by a Leclanche cell is obtained from the
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The electrical energy supplied by a Leclanche cell is obtained from the chemical energy stored within the cell. Specifically, it is a type of battery that converts chemical energy into electrical energy through a redox reaction. The Leclanche cell contains a zinc anode and a manganese dioxide cathode separated by a paste of ammonium chloride and manganese dioxide. When the cell is connected to a circuit, a chemical reaction takes place that causes electrons to flow through the circuit, generating an electric current. The chemical reaction within the cell continues until one or both of the electrodes are consumed, at which point the cell needs to be replaced or recharged.
Ajụjụ 42 Ripọtì
An electric kettle rated 1500W boils away 0.3kg of a liquid at its boiling point every 300s. Calculate the specific latent heat of the vaporisation of the liquid
Akọwa Nkọwa
The specific latent heat of vaporization is the amount of energy required to change the state of a unit mass of a substance from liquid to gas, without a change in temperature. To find the specific latent heat of vaporization, we can use the formula: specific latent heat = power x time / mass where power is the power rating of the electric kettle, time is the time taken to boil away the liquid, and mass is the mass of the liquid boiled away. Substituting the given values: specific latent heat = 1500W x 300s / 0.3kg = 1.5 x \(10^6Jkg^-1\) Therefore, the specific latent heat of vaporization of the liquid is 1.5 x \(10^6Jkg^-1\). Answer: 1.5 x \(10^6Jkg^-1\)
Ajụjụ 43 Ripọtì
A lamp marked 100W, 250V is lit for 10 hours. If it operates normally and 1KWh of electrical energy costs 2k, what is the cost of lighting the lamp?
Akọwa Nkọwa
The power rating of the lamp is given as 100W, and the voltage as 250V. Using the formula, Power = Voltage × Current, we can calculate the current as: Current = Power / Voltage = 100 / 250 = 0.4A The energy consumed by the lamp can be calculated using the formula Energy = Power × Time, where the power is in kilowatts (kW) and the time is in hours. Converting the power to kilowatts and using the given time of 10 hours, we get: Energy = (100 / 1000) × 10 = 1kWh From the question, we are told that 1kWh of electrical energy costs 2k. Therefore, for 1kWh of energy consumed by the lamp, the cost is 2k. Thus, for 1kWh of energy consumed by the lamp, the cost is 2k, and for the energy consumed by the lamp (which is 1kWh), the cost is: Cost = 1 × 2 = 2k Therefore, the cost of lighting the lamp for 10 hours is 2k. Answer: 2k.
Ajụjụ 44 Ripọtì
(a) Describe an experiment to show how the frequency of the note emitted by a vibrating string depends on the tension in the string
(b) Draw diagrams showing a vibrating string fixed at both ends emitting (i) fundamental frequency (ii) second overtone indicate the nodes and antinodes on the diagrams
(c) With the aid of a ray diagram show how a virtual image of an object is formed by a (i) concave mirror (ii) converging lens
(a) Experiment: how the frequency of the note depends on the tension
Apparatus: a sonometer (a hollow wooden box carrying a thin, uniform stretched wire), two movable bridges, a fixed peg, a frictionless pulley at one end, a scale-pan carrying known masses, a set of tuning forks of known frequency, and a small paper rider.
Procedure: The wire is stretched over the two bridges so that the vibrating (effective) length \(l\) between them is kept constant throughout, and the same uniform wire is used so that the mass per unit length \(\mu\) is constant. One end is fixed to the peg; the other passes over the pulley and carries the scale-pan. A known mass \(m\) is placed on the pan, giving a tension \(T = mg\) in the wire. A tuning fork of known frequency \(f\) is struck and its stem pressed on the box. The mass (tension) is adjusted until the wire vibrates in unison with the fork in its fundamental mode; at resonance a small paper rider placed at the middle of the wire is violently thrown off. The tension \(T\) and the fork frequency \(f\) are recorded. The experiment is repeated with four more forks of different frequency, each time altering the mass to obtain resonance, while \(l\) and \(\mu\) are held fixed.
Sample results:
| Tension T / N | √T / N1/2 | Frequency f / Hz |
|---|---|---|
| 10 | 3.16 | 100 |
| 40 | 6.32 | 200 |
| 90 | 9.49 | 300 |
| 160 | 12.65 | 400 |
| 250 | 15.81 | 500 |
A graph of \(f\) against \(\sqrt{T}\) is plotted:
The graph is a straight line passing through the origin of slope \(\approx 31.6\ \text{Hz N}^{-1/2}\). This shows that the frequency is directly proportional to the square root of the tension, \( f \propto \sqrt{T}\), in agreement with \( f = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}\).
(b) Vibrating string fixed at both ends
(i) Fundamental frequency (first harmonic): the string vibrates in a single loop, with a node (N) at each fixed end and one antinode (A) at the middle.
(ii) Second overtone (third harmonic): the string vibrates in three loops, giving four nodes (N) and three antinodes (A).
(c) Formation of a virtual image (ray diagrams)
(i) Concave mirror: with the object O placed between the pole P and the principal focus F, the two reflected rays diverge. One incident ray parallel to the axis reflects through F; a second ray directed to the pole P reflects symmetrically about the axis. Producing the reflected rays backwards behind the mirror (broken lines) locates a virtual, erect and magnified image I.
(ii) Converging lens: with the object O placed between the lens and its principal focus F (the magnifying-glass position), the refracted rays diverge. One incident ray parallel to the axis refracts through the far focus F'; a second ray through the optical centre passes straight on. Producing the refracted rays backwards on the same side as the object (broken lines) locates a virtual, erect and magnified image I.
Akọwa Nkọwa
(a) Experiment: how the frequency of the note depends on the tension
Apparatus: a sonometer (a hollow wooden box carrying a thin, uniform stretched wire), two movable bridges, a fixed peg, a frictionless pulley at one end, a scale-pan carrying known masses, a set of tuning forks of known frequency, and a small paper rider.
Procedure: The wire is stretched over the two bridges so that the vibrating (effective) length \(l\) between them is kept constant throughout, and the same uniform wire is used so that the mass per unit length \(\mu\) is constant. One end is fixed to the peg; the other passes over the pulley and carries the scale-pan. A known mass \(m\) is placed on the pan, giving a tension \(T = mg\) in the wire. A tuning fork of known frequency \(f\) is struck and its stem pressed on the box. The mass (tension) is adjusted until the wire vibrates in unison with the fork in its fundamental mode; at resonance a small paper rider placed at the middle of the wire is violently thrown off. The tension \(T\) and the fork frequency \(f\) are recorded. The experiment is repeated with four more forks of different frequency, each time altering the mass to obtain resonance, while \(l\) and \(\mu\) are held fixed.
Sample results:
| Tension T / N | √T / N1/2 | Frequency f / Hz |
|---|---|---|
| 10 | 3.16 | 100 |
| 40 | 6.32 | 200 |
| 90 | 9.49 | 300 |
| 160 | 12.65 | 400 |
| 250 | 15.81 | 500 |
A graph of \(f\) against \(\sqrt{T}\) is plotted:
The graph is a straight line passing through the origin of slope \(\approx 31.6\ \text{Hz N}^{-1/2}\). This shows that the frequency is directly proportional to the square root of the tension, \( f \propto \sqrt{T}\), in agreement with \( f = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}\).
(b) Vibrating string fixed at both ends
(i) Fundamental frequency (first harmonic): the string vibrates in a single loop, with a node (N) at each fixed end and one antinode (A) at the middle.
(ii) Second overtone (third harmonic): the string vibrates in three loops, giving four nodes (N) and three antinodes (A).
(c) Formation of a virtual image (ray diagrams)
(i) Concave mirror: with the object O placed between the pole P and the principal focus F, the two reflected rays diverge. One incident ray parallel to the axis reflects through F; a second ray directed to the pole P reflects symmetrically about the axis. Producing the reflected rays backwards behind the mirror (broken lines) locates a virtual, erect and magnified image I.
(ii) Converging lens: with the object O placed between the lens and its principal focus F (the magnifying-glass position), the refracted rays diverge. One incident ray parallel to the axis refracts through the far focus F'; a second ray through the optical centre passes straight on. Producing the refracted rays backwards on the same side as the object (broken lines) locates a virtual, erect and magnified image I.
Ajụjụ 45 Ripọtì
(a) State the laws of electromagnetic induction
(b) Explain how one of the laws illustrates the principle of conservation of energy
(c)(i) Draw a labelled diagram of a simple d.c. electric motor and explain how it works.
(ii) State two reasons why the efficiency of an electric motor is less than 100%.
(a) Laws of electromagnetic induction
Faraday's law: Whenever there is a change in the magnetic flux linking a circuit, an e.m.f. is induced in the circuit, and the magnitude of the induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage, \(E \propto -\dfrac{d\Phi}{dt}\).
Lenz's law: The induced current (or e.m.f.) always flows in such a direction as to oppose the change producing it.
(b) How Lenz's law illustrates conservation of energy
Lenz's law states that the induced current opposes the change that causes it. Therefore, to keep the flux changing (for example, to keep pushing a bar magnet into a coil) an external agent must do work against the opposing force set up by the induced current. The mechanical (kinetic) energy supplied by the external agent is exactly converted into the electrical energy of the induced current (which finally appears as heat in the resistance of the circuit). If the induced current instead aided the motion, the system would accelerate itself and create energy from nothing. Hence Lenz's law is simply a statement of the principle of conservation of energy.
(c)(i) Simple d.c. electric motor
It consists of a rectangular coil of insulated wire (the armature) mounted on an axle between the poles of a permanent magnet, a split-ring commutator fixed to the coil, and two carbon brushes that press against the commutator and feed current from a d.c. supply (battery).
Action: When current flows through the coil, side AB and side CD each lie in the magnetic field and carry current, so by the motor rule (Fleming's left-hand rule) each side experiences a force. On side AB the force acts upward and on side CD it acts downward (the two forces are equal and opposite), producing a couple (turning moment) that rotates the coil. As the coil passes the vertical position the split-ring commutator reverses the direction of current in the coil, so the forces on the two sides continue to turn the coil in the same direction, giving continuous rotation.
(c)(ii) Two reasons why efficiency is less than 100%
- Some energy is wasted as heat in the coil windings because of their electrical resistance (\(I^{2}R\) heating).
- Energy is lost through friction at the bearings and between the brushes and commutator (and through air resistance), so the useful mechanical output is always less than the electrical input.
Akọwa Nkọwa
(a) Laws of electromagnetic induction
Faraday's law: Whenever there is a change in the magnetic flux linking a circuit, an e.m.f. is induced in the circuit, and the magnitude of the induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage, \(E \propto -\dfrac{d\Phi}{dt}\).
Lenz's law: The induced current (or e.m.f.) always flows in such a direction as to oppose the change producing it.
(b) How Lenz's law illustrates conservation of energy
Lenz's law states that the induced current opposes the change that causes it. Therefore, to keep the flux changing (for example, to keep pushing a bar magnet into a coil) an external agent must do work against the opposing force set up by the induced current. The mechanical (kinetic) energy supplied by the external agent is exactly converted into the electrical energy of the induced current (which finally appears as heat in the resistance of the circuit). If the induced current instead aided the motion, the system would accelerate itself and create energy from nothing. Hence Lenz's law is simply a statement of the principle of conservation of energy.
(c)(i) Simple d.c. electric motor
It consists of a rectangular coil of insulated wire (the armature) mounted on an axle between the poles of a permanent magnet, a split-ring commutator fixed to the coil, and two carbon brushes that press against the commutator and feed current from a d.c. supply (battery).
Action: When current flows through the coil, side AB and side CD each lie in the magnetic field and carry current, so by the motor rule (Fleming's left-hand rule) each side experiences a force. On side AB the force acts upward and on side CD it acts downward (the two forces are equal and opposite), producing a couple (turning moment) that rotates the coil. As the coil passes the vertical position the split-ring commutator reverses the direction of current in the coil, so the forces on the two sides continue to turn the coil in the same direction, giving continuous rotation.
(c)(ii) Two reasons why efficiency is less than 100%
- Some energy is wasted as heat in the coil windings because of their electrical resistance (\(I^{2}R\) heating).
- Energy is lost through friction at the bearings and between the brushes and commutator (and through air resistance), so the useful mechanical output is always less than the electrical input.
Ajụjụ 46 Ripọtì
(a) Explain what is meant by a magnetic field
(b)(i) Describe an experiment to show that a magnetic field exists around a straight wire carrying current
(ii) Draw a labelled diagram showing the pattern and direction of the magnetic field rroduced around the wire. (Neglect the earth's magnetic field).
(c) Sketch the magnetic field due to two straight parallel wires carrying current in the same direction. Indicate the neutral point in the field
(d) Explain, with the aid of a labelled diagram, how a delicate magnetic material could be protected from the earth's magnetic field.
(a) Magnetic field. A magnetic field is the region of space around a magnet or a current-carrying conductor within which a magnetic force is experienced (for example by a magnetic material, a moving charge, or another magnet). It is a vector quantity, represented by lines of magnetic flux whose direction at any point is the direction in which a free north pole would move.
(b)(i) Experiment to show a field around a straight wire (Oersted's experiment). A thick straight copper wire is passed vertically through a small hole in the middle of a horizontal piece of stiff cardboard. Several small plotting compasses are placed on the card around the wire, and the wire is connected through a switch to a battery. When a large current is switched on and the card is tapped gently, the compass needles swing round and settle so that they all point along circles centred on the wire. Iron filings sprinkled on the card set into the same pattern of concentric circles. This shows that a magnetic field exists around the wire. When the current is reversed, every compass needle reverses its direction, showing that the field direction depends on the current direction.
(b)(ii) Pattern and direction of the field. The lines of force are concentric circles centred on the wire and lying in the plane at right angles to it. Their direction is given by the right-hand grip rule: gripping the wire with the right hand so that the thumb points along the conventional current, the curled fingers give the field direction. With the current coming out of the page (shown by the central dot), the field lines run anticlockwise, as drawn below.
(c) Two parallel wires carrying current in the same direction. Each wire produces its own set of concentric circular field lines. At the midpoint of the line joining the two wires the two fields are equal in magnitude but opposite in direction, so they cancel exactly, giving a neutral point N. (Outside the pair the fields reinforce, which is why two such wires attract each other.)
(d) Protecting a delicate magnetic material (magnetic shielding). The instrument is enclosed in a thick ring or box of soft iron, which has a very high magnetic permeability. Because the flux prefers the easy path through the iron, the earth's field lines are drawn into the shield and channelled round its walls instead of crossing the enclosed space. The cavity inside is therefore left almost field-free, screening the delicate material from the earth's magnetic field.
Akọwa Nkọwa
(a) Magnetic field. A magnetic field is the region of space around a magnet or a current-carrying conductor within which a magnetic force is experienced (for example by a magnetic material, a moving charge, or another magnet). It is a vector quantity, represented by lines of magnetic flux whose direction at any point is the direction in which a free north pole would move.
(b)(i) Experiment to show a field around a straight wire (Oersted's experiment). A thick straight copper wire is passed vertically through a small hole in the middle of a horizontal piece of stiff cardboard. Several small plotting compasses are placed on the card around the wire, and the wire is connected through a switch to a battery. When a large current is switched on and the card is tapped gently, the compass needles swing round and settle so that they all point along circles centred on the wire. Iron filings sprinkled on the card set into the same pattern of concentric circles. This shows that a magnetic field exists around the wire. When the current is reversed, every compass needle reverses its direction, showing that the field direction depends on the current direction.
(b)(ii) Pattern and direction of the field. The lines of force are concentric circles centred on the wire and lying in the plane at right angles to it. Their direction is given by the right-hand grip rule: gripping the wire with the right hand so that the thumb points along the conventional current, the curled fingers give the field direction. With the current coming out of the page (shown by the central dot), the field lines run anticlockwise, as drawn below.
(c) Two parallel wires carrying current in the same direction. Each wire produces its own set of concentric circular field lines. At the midpoint of the line joining the two wires the two fields are equal in magnitude but opposite in direction, so they cancel exactly, giving a neutral point N. (Outside the pair the fields reinforce, which is why two such wires attract each other.)
(d) Protecting a delicate magnetic material (magnetic shielding). The instrument is enclosed in a thick ring or box of soft iron, which has a very high magnetic permeability. Because the flux prefers the easy path through the iron, the earth's field lines are drawn into the shield and channelled round its walls instead of crossing the enclosed space. The cavity inside is therefore left almost field-free, screening the delicate material from the earth's magnetic field.
Ajụjụ 47 Ripọtì
(a) What is meant by the statement: The specific heat capacity of copper is \(400 J kg^{-1}K^{-1}\)?
(b)(i) Describe an experiment to determine the specific heat capacity of copper using a copper ball.
(ii) State two precautions necessary to obtain accurate results
(iii) A piece of copper ball of mass 20 g at 200°C is placed in a copper calorimeter of mass 60 g containing 50 g of water at 30°C, ignoring heat losses, calculate the final steady temperature of the mixture (Specific heat capacity of water = \(4.2 J g^{-1}K^{-1}\)) (Specific heat capacity of copper = \(0.4 Jg^{-1}K^{-1}\)).
(a) "The specific heat capacity of copper is \(400\ \text{J kg}^{-1}\text{K}^{-1}\)" means that 400 joules of heat energy are required to raise the temperature of 1 kilogram of copper by 1 kelvin (1 °C).
(b)(i) Experiment (method of mixtures).
- Weigh the copper ball to get its mass \(m_c\). Heat it in boiling water for some minutes so it reaches a known steady temperature \(\theta_1\) (about 100 °C, read from a thermometer in the water).
- Weigh a calorimeter (mass \(m_{cal}\)), pour in a known mass of water \(m_w\), and record the initial temperature \(\theta_2\) of the calorimeter and water.
- Quickly transfer the hot copper ball into the calorimeter, stir gently, and record the highest (final steady) temperature \(\theta_3\).
- Apply the principle of mixtures: heat lost by copper ball = heat gained by water + heat gained by calorimeter.
Every quantity except the specific heat capacity of copper \(c_c\) is known or measured, so \(c_c\) is calculated.
(b)(ii) Two precautions.
- Transfer the hot ball quickly and lag (insulate) the calorimeter to reduce heat loss to the surroundings.
- Stir the water gently before taking the final temperature so the mixture is uniform.
(b)(iii) Calculation. Let the final steady temperature be \(\theta\).
Heat lost by copper ball \(= 20\times0.4\times(200-\theta)=8(200-\theta)\)
Heat gained by water \(= 50\times4.2\times(\theta-30)=210(\theta-30)\)
Heat gained by calorimeter \(= 60\times0.4\times(\theta-30)=24(\theta-30)\)
\[ 8(200-\theta) = (210+24)(\theta-30) \] \[ 1600 - 8\theta = 234\theta - 7020 \] \[ 8620 = 242\theta \] \[ \theta = 35.6\ ^{\circ}\text{C} \]The final steady temperature of the mixture is about 35.6 °C.
Akọwa Nkọwa
(a) "The specific heat capacity of copper is \(400\ \text{J kg}^{-1}\text{K}^{-1}\)" means that 400 joules of heat energy are required to raise the temperature of 1 kilogram of copper by 1 kelvin (1 °C).
(b)(i) Experiment (method of mixtures).
- Weigh the copper ball to get its mass \(m_c\). Heat it in boiling water for some minutes so it reaches a known steady temperature \(\theta_1\) (about 100 °C, read from a thermometer in the water).
- Weigh a calorimeter (mass \(m_{cal}\)), pour in a known mass of water \(m_w\), and record the initial temperature \(\theta_2\) of the calorimeter and water.
- Quickly transfer the hot copper ball into the calorimeter, stir gently, and record the highest (final steady) temperature \(\theta_3\).
- Apply the principle of mixtures: heat lost by copper ball = heat gained by water + heat gained by calorimeter.
Every quantity except the specific heat capacity of copper \(c_c\) is known or measured, so \(c_c\) is calculated.
(b)(ii) Two precautions.
- Transfer the hot ball quickly and lag (insulate) the calorimeter to reduce heat loss to the surroundings.
- Stir the water gently before taking the final temperature so the mixture is uniform.
(b)(iii) Calculation. Let the final steady temperature be \(\theta\).
Heat lost by copper ball \(= 20\times0.4\times(200-\theta)=8(200-\theta)\)
Heat gained by water \(= 50\times4.2\times(\theta-30)=210(\theta-30)\)
Heat gained by calorimeter \(= 60\times0.4\times(\theta-30)=24(\theta-30)\)
\[ 8(200-\theta) = (210+24)(\theta-30) \] \[ 1600 - 8\theta = 234\theta - 7020 \] \[ 8620 = 242\theta \] \[ \theta = 35.6\ ^{\circ}\text{C} \]The final steady temperature of the mixture is about 35.6 °C.
