WAEC SSCE - Physics - 1995
Question 2 Report
The temperature of glass vessel containing 100cm3 of mercury is raised from 10oC to 100oC. Calculate the apparent cubic expansion of the mercury. (Real cubic expansivity of mercury = 1.82 x 10-4K-1) (Cubic expansivity of glass = 2.4 x 10-5K-1)
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Question 3 Report
A moving-coil meter with an internal resistance of 100Ω has a full-scale deflection when a current of 10m2 flows through it. What value of resistance would convert it to read 10V at full- scale deflection?
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Question 4 Report
The energy stored in a capacitor of capacitance 5μF is 40J. Calculate the voltage applied across its terminals?
Answer Details
The energy stored in a capacitor can be expressed as E = 1/2 * C * V^2, where E is the energy stored in the capacitor, C is the capacitance of the capacitor, and V is the voltage applied across its terminals. From the given information, we have E = 40J and C = 5μF. Substituting these values into the equation, we get: 40J = 1/2 * 5μF * V^2 Simplifying the equation, we get: V^2 = (2*40J)/(5μF) = 16,000,000 μV^2 Taking the square root of both sides, we get: V = 4000μV = 4000V (since 1V = 1,000,000μV) Therefore, the voltage applied across the terminals of the capacitor is 4000V.
Question 5 Report
A listener can detect the instrument from which a note is being sounded because different instruments produce the same note with a different.
Answer Details
A listener can detect the instrument from which a note is being sounded because different instruments produce the same note with a different set of overtones. When an instrument plays a note, it not only produces the fundamental frequency (the main frequency that determines the pitch), but also a series of higher frequency harmonics, known as overtones. The combination and relative strength of these overtones is what gives each instrument its unique sound or timbre. Therefore, even if two instruments play the same note at the same pitch, the differences in their overtones allow a listener to distinguish between them.
Question 6 Report
The effective capacitance between points X and Y in the diagram below is 1.0?F, What is the value of the capacitance C, measured in micro-farad?
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Question 7 Report
A brass rod is 2 m long at a certain temperature. Calculate the linear expansion of the rod for a temperature change of 100K. (Take the linear expansivity of brass as 1.8 x 10'5K").
Answer Details
The linear expansion of a material is given by the formula: ΔL = αLΔT where ΔL is the change in length, α is the linear expansivity of the material, L is the original length and ΔT is the change in temperature. In this question, the length of the brass rod L is 2 m, the change in temperature ΔT is 100 K, and the linear expansivity α for brass is 1.8 x 10^-5 K^-1. Substituting these values into the formula, we get: ΔL = (1.8 x 10^-5 K^-1) x (2 m) x (100 K) ΔL = 0.0036 m Therefore, the linear expansion of the brass rod is 0.0036 m or 3.6 mm. The correct option is: - 0.0036m
Question 8 Report
An iron rod of mass 2kg and at a temperature of 280oC is dropped into some quantity of water initially at a temperature of 30oC. lf the temperature of the mixture is 70oC, calculate the mass of the water. (Neglect heat losses to the surroundings). Specific heat capacity of iron = 460 J kgK) (Specific heat capacity of water = 4200 Jkg K)
Question 9 Report
Hot water at a temperature of t is added to twice that amount of water at a temperature of 30°C. If the resulting temperature of the mixture is 50°C. Calculate t.
Answer Details
To solve this problem, we need to apply the principle of mixtures. The principle states that the total heat content of a mixture of two or more substances is equal to the sum of their individual heat contents before mixing. Let's assume that the initial amount of hot water added is x. Therefore, the initial amount of cold water added is 2x. The heat content of the hot water is given by Q1 = mxΔt1, where m is the mass of the hot water, x is the amount added, and Δt1 is the change in temperature of the hot water. Similarly, the heat content of the cold water is given by Q2 = mcΔt2, where m is the mass of the cold water, 2x is the amount added, and Δt2 is the change in temperature of the cold water. Since the total heat content of the mixture is equal to the sum of the individual heat contents, we can write: mxΔt1 + mcΔt2 = (m+2m)cΔt where Δt is the change in temperature of the mixture, and we have used the fact that the mass of the hot water and cold water together is 3m. We know that the final temperature of the mixture is 50oC, and the initial temperature of the cold water is 30oC. Therefore, Δt = 50 - 30 = 20oC. Simplifying the equation above, we get: xΔt1 + 2xcΔt2 = 3mcΔt Substituting the given values, we get: x(t - 100) + 2x(50 - 30) = 3(2x)(50 - 30) Simplifying this expression, we get: xt - 100x + 40x = 120x Solving for x, we get: x = 4t/3 Substituting this value in the expression for Δt1, we get: Δt1 = t - 100 = (3/4)x - 100 = (3/4)(4t/3) - 100 = t/3 - 100 Therefore, the initial temperature of the hot water is t/3 - 100. Now we can use the fact that the final temperature of the mixture is 50oC to write: mxΔt1 + mcΔt2 = (m+2m)cΔt Substituting the values, we get: x(t/3 - 100 - 50) + 2x(50 - 30) = 3mx(50 - 30) Simplifying this expression, we get: xt/3 - 150x + 40x = 60mx Solving for t, we get: t = 90oC Therefore, the initial temperature of the hot water is 90/3 - 100 = -70oC. However, this does not make physical sense since the temperature of water cannot be negative. We made an error in our calculation
Question 10 Report
A mercury-in-glass thermometer reads - 20o at the ice point and 100o at the steam point. Calculate the celcius temperature corresponding to 70o on the thermometer.
Answer Details
The range of the thermometer is 100 - (-20) = 120oC. This range corresponds to a temperature range of 100oC - 0oC = 100oC on the Celsius scale. To find the temperature corresponding to 70o on the thermometer, we need to first determine the fraction of the range that corresponds to this temperature. 70o is 70 - (-20) = 90 units above the lower end of the range (-20). So the fraction of the range that corresponds to 70o is 90/120 = 0.75. Therefore, the temperature on the Celsius scale corresponding to 70o on the thermometer is 0.75 × 100oC = 75oC. Hence, the answer is 75.0oC.
Question 11 Report
What is the escape velocity of a body on the surface of the earth of radius R if the gravitational constant is G and the mass of the earth is M? (Neglect energy losses to the surroundings)
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The escape velocity is the minimum velocity that a body must have in order to escape from the gravitational field of a planet, moon or other celestial body without being pulled back by its gravity. To calculate the escape velocity of a body on the surface of the earth, we can use the formula: v = sqrt(2GM/R) Where G is the gravitational constant, M is the mass of the earth and R is the radius of the earth. Plugging in the values, we get: v = sqrt(2 x 6.67 x 10^-11 x 5.97 x 10^24 / 6.37 x 10^6) v = sqrt(2 x 39.8 x 10^13) v = sqrt(79.6 x 10^13) v = 8.91 x 10^3 m/s Therefore, the escape velocity of a body on the surface of the earth is 8.91 x 10^3 m/s. So, the correct option is: 2GM/R.
Question 12 Report
An alternating voltage with a frequency of 50Hz has a period of
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The frequency of an alternating voltage is the number of cycles it completes per second. In this case, the frequency is given as 50Hz. To find the period, we can use the formula: Period (T) = 1 / Frequency (f) Substituting the given frequency, we have: T = 1 / 50Hz T = 0.02s Therefore, the period of the alternating voltage is 0.02s. Answer: 0.02s
Question 13 Report
Which of the following is a fundamental quantity?
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Electric current is a fundamental quantity. Fundamental quantities are those quantities that cannot be defined in terms of other physical quantities. They are the basic building blocks from which other physical quantities are derived. Electric current is one of the seven fundamental quantities in the International System of Units (SI), along with length, mass, time, temperature, amount of substance, and luminous intensity. Heat capacity, torque, reactance, and density are all derived quantities, meaning they can be expressed in terms of one or more fundamental quantities.
Question 14 Report
Which of the following is not a product of nuclear fission?
Answer Details
X-ray is not a product of nuclear fission. Nuclear fission is the splitting of the nucleus of an atom into two or more smaller nuclei with the release of a significant amount of energy. The products of nuclear fission are usually smaller nuclei, neutrons, and gamma rays. Alpha and beta particles are also produced in some cases, but X-rays are not typically produced as a product of nuclear fission. X-rays are generally produced when high-energy electrons collide with a material, causing the electrons to slow down and release energy in the form of X-rays.
Question 15 Report
Calculate the quantity charge flowing through a conductor if a current of 10A passes through the conductor for 10s.
Answer Details
The quantity of charge, Q, flowing through a conductor can be calculated using the equation Q = I x t, where I is the current and t is the time for which the current flows. In this case, the current is given as 10A and the time for which it flows is 10s. So, Q = 10A x 10s = 100C Therefore, the quantity of charge flowing through the conductor is 100 coulombs. The answer is 100C.
Question 16 Report
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. lf the masses of P and Q are the same, calculate the ratio of the specific heat capacities of Q to P.
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Question 17 Report
Which of the following is not self-luminous?
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The correct answer is "The moon". The moon does not emit its own light, but reflects the light of the sun. This makes it a non-self-luminous object. An incandescent electric bulb, incandescent fluorescent tube, and a lighted candle all emit light due to a process called incandescence. The sun is a self-luminous object, as it emits light and heat due to nuclear fusion reactions occurring in its core.
Question 18 Report
A balloon containing 546cm3 of air is heated from 0oC to 10oC. If the pressure is kept constant, what will be its volume at 10oC?
Answer Details
The volume of a gas is directly proportional to its temperature (in Kelvin) at constant pressure, according to Charles's law. This is expressed as V1/T1 = V2/T2, where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature, respectively. The temperature must be expressed in Kelvin to avoid negative values, and this is done by adding 273 to the Celsius temperature. Therefore, for this problem, the volume of the balloon at 10°C can be calculated using the following formula: V1/T1 = V2/T2 V1 = 546 cm^3 (given) T1 = 0°C + 273 = 273 K T2 = 10°C + 273 = 283 K (temperature in Kelvin) Substituting the given values into the formula, we get: 546/273 = V2/283 Solving for V2, we get: V2 = 546 × 283/273 = 566 cm^3 Therefore, the volume of the balloon at 10°C would be 566 cm^3. Hence, the correct option is (c) 566 cm^3.
Question 20 Report
Calculate the electric potential at a distance of 20.0cm from a point charge of 0.015C placed in air of permittivity (Take 1/4πεo as 9.0 x 109 Nm2C2)
Answer Details
The electric potential at a point due to a point charge is given by the formula: V = kQ/r where V is the potential, Q is the charge, r is the distance of the point from the charge, and k is Coulomb's constant, which is equal to 1/4πεo. In this question, the point charge is 0.015C, the distance from the charge is 20.0cm (which is 0.20m), and the value of k is 9.0 x 109 Nm2C2. Substituting these values into the formula: V = (9.0 x 109 Nm2C2) x (0.015C) / (0.20m) V = 6.75 x 105 V Therefore, the electric potential at a distance of 20.0cm from a point charge of 0.015C placed in air of permittivity is 6.75 x 105 V. The answer is.
Question 21 Report
Which of the following are emitted from a radioactive substance without altering either the nucleon number or the proton number of the substance?
Answer Details
Gamma rays are emitted from a radioactive substance without altering either the nucleon number or the proton number of the substance. Gamma rays are high-energy photons that are emitted from the nucleus of a radioactive atom. Unlike alpha and beta particles, which are particles that are emitted from the nucleus during radioactive decay, gamma rays are pure energy and do not have mass or charge. Therefore, they do not change the identity of the atom, but they can be harmful to living organisms if they are exposed to high levels of gamma radiation.
Question 22 Report
Light travels in straight lines. In which of the following is this principle manifested? I. Pinhole camera II Formation of shadows III. Diffraction of light IV. Occurrence of eclipse
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Question 23 Report
Calculate the current in the 30 resistor in the diagram above if the battery is of negligible internal resistance
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Question 24 Report
Which of the following media allow(s) the transmission of sound wave through them? I. Air II. Liquid III. Solid.
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All three media, air, liquid and solid, allow the transmission of sound waves through them. Sound waves are mechanical waves that require a medium for their propagation. In air, sound waves travel as longitudinal waves, where particles of air oscillate parallel to the direction of wave propagation. In liquids and solids, sound waves can also propagate as longitudinal waves, but they can also propagate as transverse waves, where particles oscillate perpendicular to the direction of wave propagation. Therefore, the correct answer is "I, II and III."
Question 25 Report
A car travelling at a uniform speed of 120km h-1 passes two stations in 4 minutes. Calculate the distance between the two stations.
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To find the distance between two stations, we need to know the formula: distance = speed x time We are given the speed of the car as 120 km/h, but we need to convert this to m/s for the formula to work. 120 km/h = 120000 m/3600 s = 33.33 m/s We are also given the time as 4 minutes, which we need to convert to seconds: 4 minutes = 4 x 60 seconds = 240 seconds Now we can substitute into the formula: distance = speed x time = 33.33 m/s x 240 s = 8000 m Therefore, the distance between the two stations is 8000 meters, or 8 kilometers. Answer: 8km
Question 26 Report
An object is placed on the principal axis at the centre of curvature of a concave mirror. The image of the object formed by the mirror is
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Question 27 Report
A ray of light is incident on a body x as shown in the diagram above. What is the refractive index of the body?
Question 28 Report
Calculate the inductance L of the coil in the circuit shown below.
Question 29 Report
At what angle to the horizontal must the nozzle of a machine gun be kept when firing to obtain a maximum horizontal range for the bullets?
Answer Details
The angle at which a projectile will travel the furthest horizontal distance is known as the angle of maximum range. This angle depends on the initial velocity of the projectile and the acceleration due to gravity. For a machine gun firing bullets, the initial velocity is fixed, and the only variable is the angle at which the gun is fired. The angle of maximum range for a projectile is 45 degrees, regardless of the initial velocity. Therefore, the nozzle of the machine gun should be kept at an angle of 45 degrees to the horizontal when firing to obtain a maximum horizontal range for the bullets. Hence, the answer is 45.0o.
Question 30 Report
A cell of e.m.f 1.5V and internal resistance 2.552 is connected in series with an ammeter resistance 0.50 and a load of resistance 7.00. Calculate the current in the circuit
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The total resistance of the circuit is equal to the sum of the resistance of the ammeter, the resistance of the load and the internal resistance of the cell. So, Total resistance = ammeter resistance + load resistance + internal resistance of the cell R = 0.5 + 7.0 + 2.552 = 10.052 Ω The current in the circuit can be calculated using Ohm's law which states that the current is equal to the voltage divided by the resistance. So, I = V / R I = 1.5 / 10.052 I = 0.149 A (approx. 0.15A) Therefore, the current in the circuit is 0.15A.
Question 31 Report
Which of the following wave characteristics can be used to distinguish a transverse wave from a longitudinal wave?
Answer Details
The wave characteristic that can be used to distinguish a transverse wave from a longitudinal wave is polarization. Polarization refers to the orientation of the oscillations of a wave as it propagates through space. In a transverse wave, the oscillations are perpendicular to the direction of wave propagation, while in a longitudinal wave, the oscillations are parallel to the direction of wave propagation. Thus, a transverse wave can be polarized (i.e., its oscillations can be restricted to a single plane), while a longitudinal wave cannot be polarized in this manner. Therefore, the correct answer is option D: polarization.
Question 32 Report
A body weighing 10 N in air is partially immersed in water. It displaces water of mass 0.3 kg. What is the upthrust on the body? (g = 10 ms-2)
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The upthrust on a body is equal to the weight of the fluid displaced by the body. Since the body is partially immersed in water and displaces water of mass 0.3 kg, we can find the volume of water displaced using the density of water, which is 1000 kg/m^3. Volume of water displaced = Mass of water displaced / Density of water = 0.3 kg / 1000 kg/m^3 = 0.0003 m^3 The weight of the water displaced is given by the product of its mass and the acceleration due to gravity, g. Weight of water displaced = Mass of water displaced x g = 0.3 kg x 10 ms^-2 = 3 N Therefore, the upthrust on the body is 3 N. Hence, option D (3.0 N) is the correct answer.
Question 33 Report
Which of the following will reduce the frequency of oscillation of a simple pendulum?
Answer Details
Increasing the length of the string will reduce the frequency of oscillation of a simple pendulum. The frequency of oscillation of a simple pendulum is dependent on its length. As the length of the pendulum increases, the time taken for one complete oscillation or swing also increases. Therefore, the frequency of oscillation decreases. On the other hand, the mass of the bob, the amplitude of oscillation, and the mass of the earth have no effect on the frequency of oscillation of a simple pendulum. Decreasing the mass of the bob or the amplitude of oscillation will decrease the maximum potential and kinetic energy of the pendulum but will not affect the frequency. Similarly, the mass of the earth does not affect the frequency because it is assumed to be constant.
Question 34 Report
A machine has an efficiency of 60%. If the machine is required to overcome a load of 30 N with a force of 20 N calculate its mechanical advantage?
Answer Details
The efficiency of a machine is defined as the ratio of output work to input work. It is given as: Efficiency = (Output work / Input work) x 100% If the machine has an efficiency of 60%, it means that only 60% of the input work is converted into useful output work, while the remaining 40% is lost as heat, sound, or other forms of energy. The mechanical advantage of a machine is defined as the ratio of the output force to the input force. It is given as: Mechanical Advantage = Output Force / Input Force If the machine is required to overcome a load of 30 N with a force of 20 N, then the output force is 30 N and the input force is 20 N. Therefore, the mechanical advantage is: Mechanical Advantage = Output Force / Input Force = 30 N / 20 N = 1.5 Thus, the mechanical advantage of the machine is 1.5.
Question 35 Report
If the current in the resistor R in diagram above is 0.05A, calculate the p.d. across the inductor.
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Question 36 Report
An alternating current, having the waveform shown in the diagram above, is represented by the equation x = Xo sinwt. Which of the following represents Xo?
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Question 37 Report
The period of a wave is 0.02 second. Calculate its wavelength if its speed is 330 ms-1
Answer Details
The relationship between the speed of a wave (v), its wavelength (λ) and its frequency (f) is given by the formula v = λf. The period of a wave (T) is the time taken for one complete oscillation or cycle. It is related to the frequency (f) by the equation T = 1/f. We can combine these two equations to obtain: v = λ/T We are given the period T = 0.02 s and the speed v = 330 m/s. Substituting these values into the equation, we get: λ = vT = 330 x 0.02 = 6.6 m Therefore, the wavelength of the wave is 6.6 m.
Question 38 Report
A radioactive nuclide of mass 6.09 has a half-life of 8 days. Calculate the time during which 5.25g of the nuclide would have decayed
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Question 39 Report
A capacitor of capacitance 25μF is connected to an a.c. power source of frequency 200 Hz. Calculate the reactance of the capacitor
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Question 40 Report
Which of the following is used for shielding radioactive fall-outs?
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The material used for shielding radioactive fall-outs is lead. Lead is a dense and heavy metal that is able to effectively absorb and block harmful radiation, making it an ideal material for radiation shielding. This is why it is commonly used in the construction of X-ray rooms, nuclear power plants, and other facilities that deal with radioactive materials. Other materials like plastic, wood, textile, and aluminum are not as effective as lead in blocking radiation.
Question 41 Report
The angular speed of an object describing a circle of radius 4m with a linear constant speed of 10 ms-1 is
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The linear speed v and the angular speed ω of an object describing a circle of radius r are related by the equation: v = rω Rearranging the equation to solve for ω, we have: ω = v/r Substituting the given values, we get: ω = 10 ms-1 / 4m = 2.50 rad s-1 Therefore, the angular speed of the object is 2.50 rad s-1. Answer: (c) 2.50 rad s-1.
Question 42 Report
How much heat is required to convert 20 g of ice at 0oC to water at the same temperature? (Specific latent heat of ice = 336 J g-1)
Answer Details
To convert ice at 0°C to water at the same temperature, we need to supply heat energy to the ice to overcome the latent heat of fusion. The amount of heat energy required is given by: 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. In this case, the mass of ice is 20 g and the specific latent heat of fusion of ice is 336 J g^-1. So, the heat energy required is: Q = 20 g x 336 J g^-1 = 6,720 J Therefore, the amount of heat required to convert 20 g of ice at 0°C to water at the same temperature is 6.72 x 10^3 J. Therefore, the correct option is (c) 6.72 x 10^3 J.
Question 43 Report
A body of mass 7.5 kg is to be pulled up along a plane which is inclined at 30o to the horizontal. If the efficiency of the plane is 75%, what is the minimum force required to pull the body up the plane?
(g = 10ms-1)
Answer Details
The force required to pull an object up an inclined plane is given by F = mg sinθ, where F is the force, m is the mass of the object, g is the acceleration due to gravity and θ is the angle of inclination. In this question, the mass of the body is given as 7.5 kg, and the angle of inclination is 30o. Therefore, the force required to pull the body up the plane without considering the efficiency of the plane would be F = 7.5 x 10 x sin30o = 37.5 N. However, the efficiency of the plane is given as 75%, which means that the actual force required to pull the body up the plane is higher than the force calculated above. To find the minimum force required, we need to divide the force calculated above by the efficiency of the plane. Minimum force required = 37.5 N ÷ 0.75 = 50 N Therefore, the minimum force required to pull the body up the plane is 50 N, which is.
Question 44 Report
A student found out from a simple pendulum experiment that 20 oscillations were completed in 38 seconds. What is the period of oscillation of the pendulum?
Answer Details
The period of oscillation of a pendulum is the time it takes for one complete back-and-forth motion, also known as one oscillation. In this case, the student counted 20 oscillations in 38 seconds. To find the period of one oscillation, we can divide the total time by the number of oscillations. Period = Total time / Number of oscillations Period = 38 s / 20 = 1.9 s Therefore, the period of oscillation of the pendulum is 1.9 seconds. So the correct option is (d) 1.9 s.
Question 46 Report
What is the potential difference between X and Y in the diagram above if the battery is of negligible internal resistance
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Question 47 Report
A body of specific heat capacity 450J kg-1 K-1 falls to the ground from rest through a vertical height of 20m. Assuming conservation of energy, calculate the change in temperature of the body on striking the ground level. (g = 10ms-2)
Answer Details
When the body falls, its potential energy is converted to kinetic energy, which is then dissipated on striking the ground as heat. The amount of heat produced is equal to the initial potential energy of the body. Therefore, we can equate the potential energy to the heat energy produced by the body. Potential energy of the body = mgh, where m is the mass of the body, g is the acceleration due to gravity, and h is the height fallen. Kinetic energy of the body just before striking the ground = mgh, where m is the mass of the body, g is the acceleration due to gravity, and h is the height fallen. Heat produced on striking the ground = mgh, where m is the mass of the body, g is the acceleration due to gravity, and h is the height fallen. Heat produced = Change in temperature x Mass of the body x Specific heat capacity of the body We can equate the two expressions for heat produced to obtain: mgh = ΔT x m x c where c is the specific heat capacity of the body. Simplifying, we obtain: ΔT = gh/c Substituting the given values, we obtain: ΔT = (10 x 20)/450 = 4/9oC Therefore, the change in temperature of the body on striking the ground is 4/9oC. Answer: Option (c).
Question 48 Report
A ray of light is incident on mirror m1 and after reflection is incident on mirror m2 as shown in the diagram, calculate the angle of reflection of the ray at mirror m2
Answer Details
To calculate the angle of reflection of the ray at mirror m2, we need to first understand the concept of angle of incidence and angle of reflection. When a light ray strikes a surface, the angle between the incident ray and the normal (a line perpendicular to the surface) is called the angle of incidence. The angle between the reflected ray and the normal is called the angle of reflection. According to the law of reflection, the angle of incidence is always equal to the angle of reflection. In this diagram, we can see that the incident ray strikes mirror m1 at an angle of 60o with the normal. Using the law of reflection, we know that the reflected ray will be reflected at an angle of 60o with the normal. Now, this reflected ray becomes the incident ray for mirror m2. The angle between this incident ray and the normal is 30o (since the angle of incidence is equal to the angle of reflection). Therefore, using the law of reflection again, we can say that the angle of reflection for this ray at mirror m2 will also be 30o. Hence, the answer is 30o.
Question 49 Report
A sound note of frequency 250 Hz and wavelength 1.3m is produced at a point near a hill. If the echo of the sound is received 1 second later at the point, how far away is the hill from the point?
Answer Details
The speed of sound in air is approximately constant and equal to 340 m/s at standard temperature and pressure. In this case, the wavelength of the sound is 1.3m, and the frequency is 250Hz. We can use the formula v = fλ to find the speed of sound: v = fλ = 250 Hz × 1.3 m = 325 m/s Now, the echo is heard 1 second later, which means the sound traveled to the hill and back. So, the total distance traveled by the sound is twice the distance between the point and the hill. We can use the formula d = vt to find the distance: d = vt/2 = (325 m/s × 1 s)/2 = 162.5 m Therefore, the hill is 162.5 meters away from the point where the sound was produced. The answer is.
Question 50 Report
Sixty complete waves pass a particular point in 4s. lf the distance between three successive troughs of the waves is 15m, calculate the speed of the waves,
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Question 51 Report
A wire is stretched between two points, 1m apart. If the speed of the wave generated on plucking the wire is 200ms-1, what is the minimum frequency which will resonate with the wire?
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Question 52 Report
The pressure of a given mass of gas changes from 200 Nm2 to 100 N m2, while its temperature drops from 127oC to - 73oC. Calculate the radio of the final volume of the gas to its initial volume.
Answer Details
To solve this problem, we need to use the combined gas law which states that PV/T = constant, where P is pressure, V is volume, and T is temperature. We can write: P1V1/T1 = P2V2/T2 where subscripts 1 and 2 refer to the initial and final conditions of the gas, respectively. We can rearrange this equation to solve for the final volume V2: V2 = (P1V1T2)/(P2T1) Substituting the given values, we get: V2 = (200 Nm2)(1 m3)/(100 Nm2)(400 K) = 0.5 m3 Therefore, the ratio of the final volume to the initial volume is: V2/V1 = 0.5 m3/1 m3 = 0.5:1 So the answer is option D: 1.0:1.
Question 53 Report
A boy cycles continuously through a distance of 1.0 km in 5 minutes. Calculate his average speed
Answer Details
The formula for calculating average speed is: average speed = distance traveled ÷ time taken In this question, the distance traveled by the boy is 1.0 km, and the time taken is 5 minutes or 5 × 60 = 300 seconds. Substituting these values into the formula: average speed = 1.0 km ÷ 300 seconds To convert km to meters, we multiply by 1000. average speed = 1000m ÷ 300 seconds Simplifying: average speed = 3.33 m/s Therefore, the boy's average speed is 3.33 ms-1. So, the correct option is: 3.3ms-1.
Question 54 Report
Which of the following cannot be explained by the molecular theory of matter?
Answer Details
The molecular theory of matter explains how matter is made up of tiny particles called molecules, which are in constant motion. The behavior of matter can be explained by the motion and interaction of these molecules. Conduction is the transfer of heat energy from one molecule to another through direct contact. The molecular theory of matter can explain conduction as the result of the molecules vibrating and transferring kinetic energy to neighboring molecules through collisions. Radiation is the transfer of heat energy through electromagnetic waves. The molecular theory of matter cannot explain radiation because it does not involve the motion or interaction of molecules. Convection is the transfer of heat energy through the movement of fluids. The molecular theory of matter can explain convection as the result of the motion of the molecules in the fluid, which creates a transfer of heat energy from one part of the fluid to another. Evaporation is the process by which a liquid turns into a gas. The molecular theory of matter can explain evaporation as the result of the kinetic energy of the molecules overcoming the attractive forces between them, allowing them to escape from the liquid and enter the gas phase. Expansion is the increase in volume of a substance as its temperature increases. The molecular theory of matter can explain expansion as the result of the increase in kinetic energy of the molecules, which causes them to move further apart and occupy a larger volume. Therefore, the answer is radiation since the molecular theory of matter cannot explain radiation as it does not involve the motion or interaction of molecules.
Question 55 Report
A charge of 1.6 x 10-10C is placed in a uniform electric field of intensity 2.0 x 105NC-1. What is the magnitude of the electric force exerted on the charge?
Answer Details
The magnitude of the electric force exerted on a charge in a uniform electric field is given by the product of the magnitude of the charge and the electric field strength. Mathematically, F = qE, where F is the force, q is the charge and E is the electric field strength. Substituting the given values, we have: F = (1.6 x 10^-10 C) x (2.0 x 10^5 N/C) = 3.2 x 10^-5 N Therefore, the magnitude of the electric force exerted on the charge is 3.2 x 10^-5 N. Hence, the correct option is (C) 3.2 x10-5N.
Question 56 Report
Which of the following statements are not true of a moving-coil milliammeter? I. It can be used to measure alternating current II. It has a linear scale III. It can be adapted to read higher values of currents IV. A resistor connected in parallel with the milliammeter would convert it to a voltmeter
Answer Details
Question 57 Report
An air bubble of volume 2 cm* is formed 20 m under water. What will be its volume when it rises to just below the surface of the water if the atmospheric pressure is equivalent to a height of 10 m of water?
Answer Details
Question 58 Report
(a) What is surface tension? Explain the phenomenon in terms of intermolecular forces
(b) Describe a simple experiment to demonstrate the surface tension of a liquid
(c) State three examples to illustrate the effects of surface tension
(d) Why does water wet a clean glass surface whereas mercury does not?
(e) State two methods by which the surface tension of a liquid may reduced.
(a) Surface tension is the property of the free surface of a liquid by which it behaves like a stretched elastic skin (membrane) and tends to contract to occupy the smallest possible area.
In terms of intermolecular forces: A molecule deep inside the liquid is attracted equally in all directions by neighbouring molecules, so the net force on it is zero. A molecule at the surface has no liquid molecules above it, so it experiences a net inward (downward) cohesive pull. This inward pull on all surface molecules makes the surface behave as though it were under tension.
(b) Experiment: Tie a loop of thread loosely across a wire ring and dip the ring into soap solution to form a film. The thread lies in an irregular shape. Now break the film inside the loop (with a hot wire). The thread is at once pulled outward into a perfect circle, showing that the remaining film pulls equally on all sides - a demonstration of surface tension.
(c) Effects of surface tension:
- Small insects (e.g. pond skaters) can walk on the surface of water.
- Falling raindrops and small liquid drops are spherical.
- A greased needle can float on the surface of water although it is denser than water.
(d) Water wets clean glass because the force of adhesion between water and glass molecules is greater than the force of cohesion between water molecules, so the water spreads over the glass (concave meniscus). For mercury the cohesive force between mercury molecules is greater than the adhesive force between mercury and glass, so mercury does not wet the glass and instead forms rounded beads (convex meniscus).
(e) Methods of reducing surface tension:
- Raising the temperature of the liquid.
- Adding an impurity such as soap or detergent.
Answer Details
(a) Surface tension is the property of the free surface of a liquid by which it behaves like a stretched elastic skin (membrane) and tends to contract to occupy the smallest possible area.
In terms of intermolecular forces: A molecule deep inside the liquid is attracted equally in all directions by neighbouring molecules, so the net force on it is zero. A molecule at the surface has no liquid molecules above it, so it experiences a net inward (downward) cohesive pull. This inward pull on all surface molecules makes the surface behave as though it were under tension.
(b) Experiment: Tie a loop of thread loosely across a wire ring and dip the ring into soap solution to form a film. The thread lies in an irregular shape. Now break the film inside the loop (with a hot wire). The thread is at once pulled outward into a perfect circle, showing that the remaining film pulls equally on all sides - a demonstration of surface tension.
(c) Effects of surface tension:
- Small insects (e.g. pond skaters) can walk on the surface of water.
- Falling raindrops and small liquid drops are spherical.
- A greased needle can float on the surface of water although it is denser than water.
(d) Water wets clean glass because the force of adhesion between water and glass molecules is greater than the force of cohesion between water molecules, so the water spreads over the glass (concave meniscus). For mercury the cohesive force between mercury molecules is greater than the adhesive force between mercury and glass, so mercury does not wet the glass and instead forms rounded beads (convex meniscus).
(e) Methods of reducing surface tension:
- Raising the temperature of the liquid.
- Adding an impurity such as soap or detergent.
Question 59 Report
(a) Explain what is meant by acceleration of free fall due to gravity, g.
(b) State two reasons why g varies on the surface of the earth
(c) A stone is projected upwards at an angle of 30° to the horizontal from the top of a tower of height 100 m and it hits the ground at a point Q. If the initial velocity of projection is 100ms\(^{-1}\), calculate the
(i) maximum height of the stone above the ground;
(ii) time it takes to reach this height;
(iii) time of flight
(iv) horizontal distance from the foot of the tower to the point Q. (Neglect air resistance and take g as 10m\(^{-2}\))
(a) The acceleration of free fall due to gravity, g, is the constant acceleration produced in a body that is falling freely under the action of gravity alone (no air resistance). It is the rate of increase of velocity of a freely falling body, having an approximate value of \( 9.8\,\text{m s}^{-2} \) (taken as \( 10\,\text{m s}^{-2} \) here).
(b) Two reasons g varies over the earth:
- The earth is not a perfect sphere; its radius is greater at the equator than at the poles, so g is smaller at the equator (since \( g \propto 1/r^2 \)).
- Variation of altitude (height above sea level) - g decreases with increasing distance from the earth's centre.
(c) Resolve the initial velocity: \( u_x = 100\cos 30^\circ = 86.6\,\text{m s}^{-1} \), \( u_y = 100\sin 30^\circ = 50\,\text{m s}^{-1} \).
(i) Maximum height above the ground. Height risen above the tower top: \( H = \dfrac{u_y^2}{2g} = \dfrac{50^2}{2\times 10} = 125\,\text{m} \). Above the ground: \( 125 + 100 = 225\,\text{m} \).
(ii) Time to reach maximum height. \( t = \dfrac{u_y}{g} = \dfrac{50}{10} = 5\,\text{s} \).
(iii) Time of flight. Taking downward displacement to the ground as \(-100\,\text{m}\): \[ -100 = 50t - \tfrac{1}{2}(10)t^2 \] \[ 5t^2 - 50t - 100 = 0 \Rightarrow t^2 - 10t - 20 = 0 \] \[ t = \dfrac{10 + \sqrt{100 + 80}}{2} = \dfrac{10 + 13.42}{2} = 11.7\,\text{s}. \]
(iv) Horizontal distance to Q. \( x = u_x \times t = 86.6 \times 11.7 = 1.01 \times 10^{3}\,\text{m} \) (about 1014 m).
Answer Details
(a) The acceleration of free fall due to gravity, g, is the constant acceleration produced in a body that is falling freely under the action of gravity alone (no air resistance). It is the rate of increase of velocity of a freely falling body, having an approximate value of \( 9.8\,\text{m s}^{-2} \) (taken as \( 10\,\text{m s}^{-2} \) here).
(b) Two reasons g varies over the earth:
- The earth is not a perfect sphere; its radius is greater at the equator than at the poles, so g is smaller at the equator (since \( g \propto 1/r^2 \)).
- Variation of altitude (height above sea level) - g decreases with increasing distance from the earth's centre.
(c) Resolve the initial velocity: \( u_x = 100\cos 30^\circ = 86.6\,\text{m s}^{-1} \), \( u_y = 100\sin 30^\circ = 50\,\text{m s}^{-1} \).
(i) Maximum height above the ground. Height risen above the tower top: \( H = \dfrac{u_y^2}{2g} = \dfrac{50^2}{2\times 10} = 125\,\text{m} \). Above the ground: \( 125 + 100 = 225\,\text{m} \).
(ii) Time to reach maximum height. \( t = \dfrac{u_y}{g} = \dfrac{50}{10} = 5\,\text{s} \).
(iii) Time of flight. Taking downward displacement to the ground as \(-100\,\text{m}\): \[ -100 = 50t - \tfrac{1}{2}(10)t^2 \] \[ 5t^2 - 50t - 100 = 0 \Rightarrow t^2 - 10t - 20 = 0 \] \[ t = \dfrac{10 + \sqrt{100 + 80}}{2} = \dfrac{10 + 13.42}{2} = 11.7\,\text{s}. \]
(iv) Horizontal distance to Q. \( x = u_x \times t = 86.6 \times 11.7 = 1.01 \times 10^{3}\,\text{m} \) (about 1014 m).
Question 60 Report
(a) State Faraday's law of electromagnetic induction.
(b) Draw a labelled diagram of an induction coil and explain how it works.
(c) How is the effect of eddy currents minimized in the coil?
(d) State two reasons why a capacitor should be included in the primary circuit of the coil.
(e) State three uses of an induction coil.
(a) Faraday's law of electromagnetic induction
Whenever the magnetic flux linked with a circuit changes, an e.m.f. is induced in the circuit. The magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linkage:
\[|E|=N\left|\frac{d\Phi}{dt}\right|\]
where \(E\) is the induced e.m.f., \(N\) is the number of turns and \(\Phi\) is the magnetic flux through each turn.
(b) Labelled diagram of an induction coil
How it works: When key \(K\) is closed and the platinum contacts touch, current flows through the primary coil. The soft-iron core becomes magnetised and attracts the soft-iron armature. This separates the contacts and breaks the primary circuit. The core then loses its magnetism, so the spring returns the armature and remakes the contact. Thus, the primary current is made and broken rapidly.
At each break, the magnetic flux in the core collapses rapidly and induces an e.m.f. in the secondary coil. Since the secondary has very many turns, a very large e.m.f. is induced, sufficient to produce a spark across the spark gap.
(c) Eddy currents are minimised by making the core from a bundle of thin insulated soft-iron wires, or by laminating the soft-iron core.
(d) Reasons for connecting a capacitor across the contact points
- It reduces sparking at the platinum contacts and therefore prevents the contacts from burning away.
- It makes the break in the primary current more rapid, causing a faster collapse of flux and hence a larger induced e.m.f. in the secondary.
(e) Uses of an induction coil
- In the ignition system of petrol motor vehicles to produce sparks at spark plugs.
- For operating X-ray tubes.
- For studying electric discharge through gases, such as in discharge tubes.
Answer Details
(a) Faraday's law of electromagnetic induction
Whenever the magnetic flux linked with a circuit changes, an e.m.f. is induced in the circuit. The magnitude of the induced e.m.f. is proportional to the rate of change of magnetic flux linkage:
\[|E|=N\left|\frac{d\Phi}{dt}\right|\]
where \(E\) is the induced e.m.f., \(N\) is the number of turns and \(\Phi\) is the magnetic flux through each turn.
(b) Labelled diagram of an induction coil
How it works: When key \(K\) is closed and the platinum contacts touch, current flows through the primary coil. The soft-iron core becomes magnetised and attracts the soft-iron armature. This separates the contacts and breaks the primary circuit. The core then loses its magnetism, so the spring returns the armature and remakes the contact. Thus, the primary current is made and broken rapidly.
At each break, the magnetic flux in the core collapses rapidly and induces an e.m.f. in the secondary coil. Since the secondary has very many turns, a very large e.m.f. is induced, sufficient to produce a spark across the spark gap.
(c) Eddy currents are minimised by making the core from a bundle of thin insulated soft-iron wires, or by laminating the soft-iron core.
(d) Reasons for connecting a capacitor across the contact points
- It reduces sparking at the platinum contacts and therefore prevents the contacts from burning away.
- It makes the break in the primary current more rapid, causing a faster collapse of flux and hence a larger induced e.m.f. in the secondary.
(e) Uses of an induction coil
- In the ignition system of petrol motor vehicles to produce sparks at spark plugs.
- For operating X-ray tubes.
- For studying electric discharge through gases, such as in discharge tubes.
Question 61 Report
(a) Define the boiling it of a liquid.
(b) Describe with the a d. labelled diagram, an experiment to determire the boiling point of a small quantity of a liquid.
(c) factors that may affect the boiling point of a liquid
(d) Using the kinetic theory of matter, explain why pure water changes to steam at S.T.P. without any change in temperature, although heat is being supplied to the water.
(a) The boiling point of a liquid is the temperature at which its saturated vapour pressure becomes equal to the external atmospheric pressure, so that the liquid changes to vapour throughout its bulk.
(b) Determination of the boiling point of a small quantity of a liquid
The apparatus is arranged as shown below.
Procedure
- Introduce a little mercury into a J-tube. Remove the air trapped in the shorter, closed limb.
- Put a small quantity of the liquid under test into the shorter limb, above the mercury.
- Place the J-tube in a beaker containing water. Heat the water bath gently and stir it continuously so that its temperature is uniform.
- Observe the mercury levels in the two limbs. When the mercury levels are at the same horizontal level, read and record the thermometer temperature.
- Allow the water bath to cool, still stirring it. Again record the temperature at which the mercury levels are equal.
- The mean of the heating and cooling readings is the boiling point of the liquid.
When the mercury levels are equal, the pressure of the liquid vapour in the closed limb is equal to atmospheric pressure in the open limb. Hence, at this temperature, the saturated vapour pressure of the liquid equals atmospheric pressure and the temperature is its boiling point.
(c) Factors affecting the boiling point of a liquid include:
- External pressure: an increase in external pressure raises the boiling point, while a decrease in pressure lowers it.
- Impurities: dissolved non-volatile impurities generally raise the boiling point of a liquid.
(d) At the boiling point, the heat supplied to pure water is absorbed as latent heat of vaporisation. This energy is used to overcome the intermolecular forces of attraction between water molecules and to separate the molecules further apart against atmospheric pressure as steam forms. Since the supplied heat does not increase the average kinetic energy of the molecules, the temperature of the water remains constant while it changes to steam.
Answer Details
(a) The boiling point of a liquid is the temperature at which its saturated vapour pressure becomes equal to the external atmospheric pressure, so that the liquid changes to vapour throughout its bulk.
(b) Determination of the boiling point of a small quantity of a liquid
The apparatus is arranged as shown below.
Procedure
- Introduce a little mercury into a J-tube. Remove the air trapped in the shorter, closed limb.
- Put a small quantity of the liquid under test into the shorter limb, above the mercury.
- Place the J-tube in a beaker containing water. Heat the water bath gently and stir it continuously so that its temperature is uniform.
- Observe the mercury levels in the two limbs. When the mercury levels are at the same horizontal level, read and record the thermometer temperature.
- Allow the water bath to cool, still stirring it. Again record the temperature at which the mercury levels are equal.
- The mean of the heating and cooling readings is the boiling point of the liquid.
When the mercury levels are equal, the pressure of the liquid vapour in the closed limb is equal to atmospheric pressure in the open limb. Hence, at this temperature, the saturated vapour pressure of the liquid equals atmospheric pressure and the temperature is its boiling point.
(c) Factors affecting the boiling point of a liquid include:
- External pressure: an increase in external pressure raises the boiling point, while a decrease in pressure lowers it.
- Impurities: dissolved non-volatile impurities generally raise the boiling point of a liquid.
(d) At the boiling point, the heat supplied to pure water is absorbed as latent heat of vaporisation. This energy is used to overcome the intermolecular forces of attraction between water molecules and to separate the molecules further apart against atmospheric pressure as steam forms. Since the supplied heat does not increase the average kinetic energy of the molecules, the temperature of the water remains constant while it changes to steam.
