WAEC SSCE - Physics - 2008
Tambaya 1 Rahoto
The design of the thermostat of an electric iron is based on the
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The design of the thermostat of an electric iron is based on the "increase in the size of metals when heated". The thermostat of an electric iron is designed to maintain a constant temperature of the iron's metal plate. It consists of a bimetallic strip made up of two different metals that have different coefficients of thermal expansion. This means that when the temperature changes, the two metals expand at different rates, causing the strip to bend. The bending of the bimetallic strip is used to control the temperature of the electric iron. When the iron is switched on, the bimetallic strip heats up and bends, which then opens or closes the electrical contacts that control the heating element. As the temperature rises, the strip bends further, eventually causing the contacts to open and turn off the heating element. As the temperature falls, the strip bends in the opposite direction, causing the contacts to close and turn on the heating element again. The reason why the bimetallic strip bends is due to the different coefficients of thermal expansion of the two metals. When heated, one metal expands more than the other, causing the strip to bend. This bending action is used to control the temperature of the electric iron. In summary, the design of the thermostat of an electric iron is based on the "increase in the size of metals when heated". The bimetallic strip used in the thermostat is made up of two metals with different coefficients of thermal expansion, which causes the strip to bend when heated, and is used to control the temperature of the iron.
Tambaya 2 Rahoto
The distance between a concave mirror and an object placed in front of it is 1.0m. If the raduis of curvature of the mirror is 4.0m, the image formed will be
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To determine the position and nature of the image formed by a concave mirror, we can use the mirror formula, which is given by: 1/v + 1/u = 1/f where v is the distance of the image from the mirror, u is the distance of the object from the mirror, and f is the focal length of the mirror. We are given u = -1.0 m (since the object is placed in front of the mirror) and f = -2.0 m (since the mirror is concave). To find v, we need to rearrange the equation: 1/v = 1/f - 1/u 1/v = -1/2 - 1/(-1) 1/v = -1/2 + 1 1/v = 1/2 v = 2.0 m Since the image is formed 2.0 m from the mirror, and the object is placed in front of the mirror, the image is formed behind the mirror. Therefore, the correct option is: - 2.00m behind the mirror.
Tambaya 3 Rahoto
A direct current of 5A flow through a 0.2H inductor. Calculate the energy stored in the inductor
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Tambaya 4 Rahoto
The diagram above illustrates a Hare's apparatus. \(\mathit{e}_1, \mathit{e}_2\) represent densities and \(\mathit{h}_1, \mathit{h}_2\) heights of columns of liquids. Which of the following equations is correct?
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In a Hare's apparatus, a liquid with a higher density is placed in the taller column and a liquid with a lower density is placed in the shorter column. The pressure at the base of each column is the same. Therefore, we can equate the pressures at the base of each column: \(\text{Pressure}=\text{Height} \times \text{Density} \times \text{Gravitational acceleration}\) Thus, we can write: \(\mathit{h}_1 \mathit{e}_1 g = \mathit{h}_2 \mathit{e}_2 g\) Where g is the acceleration due to gravity. We can cancel g from both sides of the equation, which gives: \(\mathit{h}_1 \mathit{e}_1 = \mathit{h}_2 \mathit{e}_2\) Then, we can rearrange this equation to find \(\mathit{h}_1\): \(\mathit{h}_1 = \frac{\mathit{h}_2 \mathit{e}_2}{\mathit{e}_1}\) Therefore, the correct equation is: \(\mathit{h}_1 = \frac{\mathit{h}_2 \mathit{e}_2}{\mathit{e}_1}\)
Tambaya 5 Rahoto
Light is considered as a transverse wave because it travels
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Light is a form of electromagnetic radiation that travels in waves. There are two types of waves: longitudinal and transverse. Longitudinal waves are waves in which the particles of the medium vibrate parallel to the direction of wave propagation, while in transverse waves, the particles vibrate perpendicular to the direction of wave propagation. Light is considered a transverse wave because it travels in a direction perpendicular to the plane containing the electric and magnetic fields. The electric and magnetic fields are at right angles to each other, and both are perpendicular to the direction in which the light is traveling. The oscillation of the electric and magnetic fields is perpendicular to the direction of travel of the wave, which makes it a transverse wave. Moreover, light can travel through space without the need for a medium. Unlike sound waves, which require a medium such as air or water to propagate, light can travel through a vacuum. This is because light is made up of electromagnetic waves, which do not require a medium to travel through. So, the answer is "in a direction perpendicular to the plane containing the electric and magnetic fields".
Tambaya 6 Rahoto
If the air inside a rigid box is heated, the i, average speed of the molecules increases ii. pressure of the air increases iii. average separation of the molecule increases. Which of the statements above are correct?
Bayanin Amsa
The correct statements are i and ii only: i) When the air inside a rigid box is heated, the average speed of the molecules increases. This is because heat energy is transferred to the air molecules, causing them to move faster and increase their kinetic energy. ii) The pressure of the air inside the box increases as well. This is because as the air molecules move faster, they collide more frequently with the walls of the box, causing an increase in the force exerted by the molecules on the walls of the box, and hence an increase in pressure. iii) The average separation of the molecules does not increase. When the temperature of the air inside the box is increased, the volume of the air also increases (assuming the box is not allowed to expand). This means that the same number of molecules now occupy a larger volume, which results in a decrease in the average separation between the molecules. In summary, when the air inside a rigid box is heated, the average speed of the molecules increases, and the pressure of the air increases, while the average separation of the molecules decreases.
Tambaya 7 Rahoto
Nuclear fission is preffered to nuclear fusion in the generation of energy because
Tambaya 8 Rahoto
The diagram above shoes three forces F1, F2 and F3, which of the following ewuations is correct?
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Tambaya 9 Rahoto
A negatively charged rod is held close to an uncharged metal ball on an insulating stand. The ball will
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When a negatively charged rod is held close to an uncharged metal ball on an insulating stand, the electrons in the metal ball will be repelled by the negative charges in the rod, and will move as far away as possible from the rod. This movement of electrons will result in a net positive charge on the side of the ball nearest to the rod, and a net negative charge on the opposite side. Therefore, the ball will have an excess of positive charges on the side nearest the rod. So, the correct option is: the ball will have an excess of positively charged on the side nearest the rod.
Tambaya 10 Rahoto
The diagram above represents the displacement-time graph of the motion of an object thrown upwards. The velocity of the object at the point P is
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The displacement-time graph represents the motion of an object thrown upwards. At point P, the object reaches its maximum height and then starts to fall back down due to gravity. At this point, the velocity of the object is zero, as it momentarily stops moving upwards and starts to fall downwards. Therefore, the correct answer is "zero".
Tambaya 11 Rahoto
A radioactive element has a decay constant of 0.077s-1. Calculate its half-life
Bayanin Amsa
The half-life of a radioactive element is the amount of time it takes for half of the radioactive atoms in a sample to decay. The relationship between the decay constant (λ) and the half-life (t1/2) is given by the equation: t1/2 = ln(2) / λ where ln(2) is the natural logarithm of 2, which is approximately 0.693. Using the above equation and the given decay constant of 0.077s-1, we can calculate the half-life as: t1/2 = ln(2) / λ = 0.693 / 0.077 = 9.0s Therefore, the half-life of the radioactive element is 9.0 seconds. So, the correct option is: 9.0s.
Tambaya 12 Rahoto
a sound wave is produced fron a source ans an echo is heard t seconds afterwards. If d is the distance of the reflecting surface from the source, v the speed, p the wavelength and T the period of the wave, then
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When a sound wave is produced from a source, it travels through a medium (such as air) until it reaches a reflecting surface (such as a wall). At the reflecting surface, the sound wave bounces off and returns back towards the source, producing an echo. Let's assume that the time interval between the production of the sound wave and the reception of its echo is t seconds. The speed of sound in the medium is represented by v, the wavelength of the sound wave is represented by λ, and the period of the sound wave is represented by T. We want to find the distance (d) between the source and the reflecting surface. To do this, we can use the formula: d = vt/2 This formula relates the distance (d) to the speed (v) of the sound wave and the time (t) it takes for the echo to be heard. We can also express the speed (v) of the sound wave in terms of its wavelength (λ) and period (T): v = λ/T Substituting this expression for v into the first formula, we get: d = λt/2T Therefore, the answer is (b) d = \(\frac{\lambda t}{2T}\).
Tambaya 13 Rahoto
An object of mass 0.25kg moves at a height h above the ground with a speed of 4ms-1, if its mechanical energy at this height is 12 J, determine the value of h. [g = 10ms-1]
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Tambaya 15 Rahoto
Which of the following is/are in random motion? i. Pollen grains in water ii. The molecules of hydrogen gas ii. Fine chalk particles floating in air
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The correct answer is (i), (ii), and (iii). All the options given are examples of particles that undergo random motion. In liquids, such as water, particles move around in a random manner known as Brownian motion, caused by collisions with other particles. Pollen grains in water, being solid particles, will exhibit Brownian motion. Similarly, gas molecules such as hydrogen gas are in constant random motion, bouncing off one another in all directions. Lastly, small particles, such as fine chalk particles, suspended in air are also subject to random motion due to collisions with air molecules. Therefore, all the options given represent particles undergoing random motion.
Tambaya 16 Rahoto
A metal rod of length 50cm is heated from 40oC to 80oC. If the linear expansivity of the material is \(\alpha\), calculate the increase in length of the rod (in metres) in terms of \(\alpha\)
Bayanin Amsa
The increase in length of a metal rod due to heating can be calculated using the formula: ΔL = L₀αΔT Where ΔL is the increase in length, L₀ is the original length of the rod, α is the linear expansivity of the material, and ΔT is the change in temperature. In this question, the original length of the rod is 50cm, or 0.5m. The change in temperature is 80oC - 40oC = 40oC. Therefore, the increase in length of the rod is: ΔL = (0.5m)α(40oC) = 20αm So the correct answer is 20\(\alpha\).
Tambaya 17 Rahoto
The mass attached to a string is moving in a circular path. If the speed is doubled, the tension in the string will be
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The tension in the string is related to the centripetal force required to keep the object moving in a circular path. According to centripetal force equation, F = mv²/r, where m is the mass of the object, v is the speed, and r is the radius of the circular path. Since the mass of the object remains constant, doubling the speed would increase the centripetal force required to keep it moving in a circular path by a factor of four (since F = mv²/r). Therefore, the tension in the string would also increase by a factor of four, so the correct answer is "four times as great."
Tambaya 18 Rahoto
Two bodies each carrying a charge of 2.00 x 10-10C are 5cm apart. Calculate the magnitude of the force between the charges.[\(\frac{1}{4\pi \mathit{E}_o} = 9.00 \times 10^9 Nm^2C{-2}\)]
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Tambaya 19 Rahoto
Water of mass 1.5kg is heated from from 30oC to 80oC using an electric kettle which is rated 5A, 230V. Calculate the time taken to reach the final destination [specific heat capacity of water 4200Jkg-1K-1]
Bayanin Amsa
The amount of heat required to raise the temperature of water can be calculated using the formula: Q = mcΔT Where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water and ΔT is the change in temperature. In this case, the mass of water is 1.5 kg, the specific heat capacity of water is 4200 Jkg^-1K^-1, and ΔT is 80 - 30 = 50 K. So, Q = 1.5 kg x 4200 Jkg^-1K^-1 x 50 K = 315000 J The power rating of the electric kettle is 5A x 230V = 1150W = 1150J/s. Therefore, the time taken to heat the water can be calculated using the formula: t = Q/P where t is the time taken, Q is the heat energy, and P is the power rating of the electric kettle. So, t = 315000 J / 1150 J/s = 273.91 s ≈ 4.57 minutes (to 2 decimal places) Therefore, the time taken to heat the water from 30oC to 80oC using the electric kettle is approximately 4.57 minutes. Hence, the correct option is (a) 4.57 minutes.
Tambaya 20 Rahoto
In the day time, it is possible to see under shady areas such as under a tree because light has undergone
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When light shines on an object, it can either be absorbed, transmitted, or reflected. When light is reflected, it can either be reflected in a single direction (like a mirror) or scattered in many different directions. When light hits a smooth surface, like a mirror, it reflects in a single direction. However, when light hits a rough surface, like a tree, the light is scattered in many different directions. This is known as diffuse reflection. Diffuse reflection causes light to scatter in all directions, including in the direction of your eyes. This is why you are able to see under shady areas like under a tree during the day. The leaves and branches of the tree scatter the light, allowing some of it to reach the ground and your eyes. Therefore, the correct answer to the question is diffuse reflection.
Tambaya 21 Rahoto
Both \(\gamma\)-rays and X-rays i. can be produced by thermionic emission ii. have low frequencies iii. have short wavelengths. Which of the statements above is/are correct?
Bayanin Amsa
The third statement, "Both gamma-rays and X-rays have short wavelengths," is correct. Gamma-rays and X-rays are both forms of electromagnetic radiation that have higher frequencies and shorter wavelengths than visible light. Gamma-rays have the highest frequency and shortest wavelength, followed by X-rays. Therefore, statement iii is true. However, the first statement, "Both gamma-rays and X-rays can be produced by thermionic emission," is not necessarily true. While X-rays can be produced by thermionic emission, which is the process of releasing electrons from a heated metal surface, gamma-rays are typically produced by nuclear reactions or radioactive decay. The second statement, "Both gamma-rays and X-rays have low frequencies," is not correct. As mentioned earlier, gamma-rays and X-rays have higher frequencies than visible light, which means they have higher energy and shorter wavelengths. Therefore, statement ii is false. In summary, only statement iii is correct, which is "Both gamma-rays and X-rays have short wavelengths."
Tambaya 22 Rahoto
which of the following statements about Archimedes' principle is correct? The upthrust on a body is a measure of the
Bayanin Amsa
The correct statement about Archimedes' principle is that the upthrust on a body is a measure of the weight of the fluid displaced. This principle states that when an object is immersed in a fluid, it experiences an upward force or buoyant force equal to the weight of the fluid displaced by the object. This means that the weight of the fluid displaced by an object is equal to the upthrust or buoyant force acting on the object. Therefore, option D, "weight of the fluid displaced," is the correct statement about Archimedes' principle.
Tambaya 23 Rahoto
A car travelling with a uniform velocity of 30ms-1 along a horizontal road overcomes a constant frictional force of 600N. Calculate the power of the engine of the car
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Tambaya 24 Rahoto
Which of the following actions will not increase the sensitivity of a moving coil galvanometer?
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Using a light pointer will not increase the sensitivity of a moving coil galvanometer. The sensitivity of a moving coil galvanometer depends on various factors, such as the strength of the magnetic field, the number of turns of the coil, the area of the coil, and the moment of inertia of the coil. A more sensitive galvanometer is one that can detect even a small current flowing through it and produce a large deflection. Using a strong temporary magnet, increasing the area and number of turns of the coil, and using a weak hair spring are all actions that can increase the sensitivity of a moving coil galvanometer. A strong magnet will increase the magnetic field strength and thus increase the torque on the coil. Increasing the area and number of turns of the coil will increase the amount of current induced in the coil for a given magnetic field, and a weak hair spring will reduce the restoring torque and allow the coil to be more easily deflected. However, using a light pointer will not affect the sensitivity of a moving coil galvanometer. A light pointer is simply used to reduce the moment of inertia of the coil, which makes it easier to deflect the coil for a given amount of torque. This does not directly affect the sensitivity of the galvanometer.
Tambaya 25 Rahoto
If the charge on an object is measured as 4.0 x 10-18C, how many excess electrons does the object possess, given that the charge of an electron is 1.6 x 10-18C
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Tambaya 26 Rahoto
Control in nuclear reactors is effected with boron rods because rhe rods have the ability to
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The control in nuclear reactors is very important to ensure the safety and stability of the reactor. One of the methods used for control is the use of boron rods. These rods are inserted or withdrawn from the reactor core to regulate the nuclear reaction taking place. The reason boron rods are used for control in nuclear reactors is that they have the ability to absorb neutrons. Neutrons are the particles that are responsible for the nuclear reaction that takes place in the core of the reactor. When the boron rods are inserted into the reactor core, they absorb the neutrons and prevent them from interacting with the fuel, which slows down or even stops the nuclear reaction. This ability of boron to absorb neutrons makes it a good material for control rods. The concentration of boron in the rods can be adjusted to control the rate of reaction in the reactor. When the boron rods are withdrawn from the reactor core, fewer neutrons are absorbed, and the nuclear reaction can proceed at a faster rate. So, the answer is "absorb neutron".
Tambaya 27 Rahoto
Which of the following devices convert heat energy to electrical energy?
Bayanin Amsa
The device that converts heat energy to electrical energy is a thermocouple. A thermocouple is made up of two different metals that are joined together at one end. When one junction of the two metals is heated, it generates a small voltage. This voltage is a result of the temperature difference between the two junctions, which is known as the Seebeck effect. The voltage generated by the thermocouple can be used to power a circuit or device. Therefore, the correct answer is "thermocouple". A transformer is a device that converts high-voltage, low-current power to low-voltage, high-current power (or vice versa). A dynamo is a device that converts mechanical energy to electrical energy. A thermostat is a device that controls temperature by turning a heating or cooling system on and off. Therefore, these devices do not convert heat energy to electrical energy, and they are not the correct answers.
Tambaya 28 Rahoto
A ball is dropped and it hits the floor at a point A. It rebounds upward to a point B. While moving from A to B its?
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Tambaya 29 Rahoto
Which of the following equations for the efficiency of a machine is correct?
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Tambaya 30 Rahoto
When viewed straight down, a fish in a pond is
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When viewed straight down, a fish in a pond appears nearer than it actually is. This is due to the bending of light as it travels from water to air, which causes the fish to appear closer to the surface than it actually is. This phenomenon is called refraction. As a result, the fish's actual position in the water is farther below than it appears to be when viewed from above.
Tambaya 31 Rahoto
Which of the following defects of vision is as a result of the eye ball being too long?
Bayanin Amsa
The defect of vision that is a result of the eye ball being too long is short sight. Short sight, also known as myopia, is a condition where distant objects appear blurred while near objects are seen clearly. In short-sightedness, the light entering the eye is focused in front of the retina instead of on it, due to the elongated shape of the eyeball. This makes it difficult for the eye to see distant objects clearly. To correct this defect, concave lenses are used to diverge the incoming light rays and bring them to focus on the retina.
Tambaya 32 Rahoto
A wire 1.0m long and with cross-sectional area 2.0 x 10-7 m2 has a resistance of 0.1\(\Omega\). Calculate the electrical conductivity of the wire
Bayanin Amsa
The electrical conductivity of a wire is defined as the inverse of its electrical resistivity, which is given by the formula: resistivity = (resistance x cross-sectional area) / length We can rearrange this formula to solve for the conductivity: conductivity = (length / cross-sectional area) x (1 / resistance) Substituting the given values, we get: conductivity = (1.0 m / 2.0 x 10^-7 m^2) x (1 / 0.1 Ω) = 5.0 x 10^7 Ω^-1 m^-1 Therefore, the electrical conductivity of the wire is 5.0 x 10^7 Ω^-1 m^-1.
Tambaya 33 Rahoto
A very sensitive spring balance was used to determine the weight of an object at the north pole. When the same spring balance was used to measure the weight of the same object at the equator, it was found to reduce. The explanation for this observation is that
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The explanation for this observation is that the acceleration of free fall due to gravity varies with location. At the North Pole, the object experiences a greater acceleration of free fall due to gravity than it does at the Equator. This is because the Earth is not a perfect sphere but rather an oblate spheroid. At the poles, the Earth's rotation axis is perpendicular to the ground, while at the equator, the Earth's rotation axis is parallel to the ground. This difference in orientation results in a difference in the centripetal force on the object due to the Earth's rotation, which affects the object's weight. Since the spring balance measures the force required to support the object's weight, the weight of the object decreases at the Equator due to the reduced gravitational force.
Tambaya 34 Rahoto
The image of an object formed by a convex mirror is
Bayanin Amsa
The image of an object formed by a convex mirror is virtual, erect and diminished. A convex mirror is a spherical mirror that curves outward, and when an object is placed in front of it, the light rays from the object are reflected off the mirror's surface. The reflected rays of light do not converge at a point, but instead appear to diverge from a point behind the mirror. This point is called the virtual focus or focal point. The image formed by a convex mirror is virtual because the reflected rays do not actually converge at a real point, but only appear to diverge from the virtual focus. The image is erect because the object and the image are both oriented in the same direction. Lastly, the image is diminished because the size of the image is smaller than the size of the object.
Tambaya 35 Rahoto
Which of the following sets of quantities is fundamental?
Bayanin Amsa
The set of quantities that is fundamental is "length, mass, and time." These quantities are considered fundamental because they are independent and cannot be defined in terms of one another. All other physical quantities can be derived from these fundamental quantities using mathematical equations. For example, speed can be derived as distance over time, and acceleration can be derived as change in speed over time. Therefore, length, mass, and time are considered the three basic building blocks of physics and are used to describe all physical phenomena.
Tambaya 36 Rahoto
When a radioactive substance undergoes a beta decay, its
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When a radioactive substance undergoes a beta decay, its "atomic number increases by 1". Beta decay is a type of radioactive decay in which a beta particle (an electron) is emitted from the nucleus of an atom. During beta decay, a neutron in the nucleus of the atom is converted into a proton, and an electron is emitted from the nucleus. Since the neutron is converted into a proton, the atomic number of the nucleus increases by 1, since the atomic number is the number of protons in the nucleus. However, the mass number (the sum of protons and neutrons in the nucleus) remains the same, since only one subatomic particle (the neutron) has been converted into another (the proton and electron). In summary, when a radioactive substance undergoes a beta decay, its "atomic number increases by 1". This is because a neutron in the nucleus is converted into a proton during beta decay, increasing the number of protons in the nucleus, which is equal to the atomic number.
Tambaya 37 Rahoto
Three capacitors each of capacitance 1.5\(\mu F\) are connected in series. The total capacitance in the circuit is
Bayanin Amsa
When capacitors are connected in series, the reciprocal of the total capacitance is equal to the sum of the reciprocals of the individual capacitances. Mathematically, we can express it as: \(\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ...\) In this case, we have three capacitors of equal capacitance, so we can simplify the equation to: \(\frac{1}{C_{total}} = \frac{1}{1.5\mu F} + \frac{1}{1.5\mu F} + \frac{1}{1.5\mu F}\) \(\frac{1}{C_{total}} = \frac{3}{1.5\mu F}\) \(\frac{1}{C_{total}} = 2\) \(C_{total} = \frac{1}{2}\mu F\) Therefore, the total capacitance in the circuit is 0.5 \(\mu F\), which is option (D).
Tambaya 38 Rahoto
The cubic expansivity of a certain gas at constant pressure is \(\frac{1}{273} k^{-1}\). If a given mass of the gas is held at constant pressure and its volume at 0oC is 273m3, determine the volume of the gas at 273oC.
Bayanin Amsa
The cubic expansivity of a gas at constant pressure is defined as the fractional increase in volume per degree rise in temperature. Let the volume of the gas at 0°C be V₀ and the temperature at which we need to find the volume of the gas be T₁. The increase in temperature = T₁ - 0 = T₁. The fractional increase in volume = cubic expansivity x increase in temperature = \(\frac{1}{273}\) x T₁. The increase in volume = V₀ x fractional increase in volume = V₀ x \(\frac{1}{273}\) x T₁. The final volume of the gas at T₁ = V₀ + increase in volume = V₀ + V₀ x \(\frac{1}{273}\) x T₁ = V₀(1 + \(\frac{1}{273}\) x T₁). Substituting the given values, we get: Volume at 273oC = 273m3(1 + \(\frac{1}{273}\) x 273) = 546m3. Therefore, the volume of the gas at 273oC is 546m3. The correct option is (b).
Tambaya 39 Rahoto
A converging lens and a screen are placed 20 cm and 80 cm respectively from an object in a straight line so that a sharp image of the object is formed on the screen. If the object is 3cm high. Calculate the height of the image formed
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Tambaya 41 Rahoto
Which of the following is a unit of the time rate of flow of electric charges?
Bayanin Amsa
The unit of the time rate of flow of electric charges is ampere (A). Ampere is the SI unit of electric current, which is defined as the amount of charge flowing through a conductor per unit time. One ampere is equivalent to the flow of one coulomb of electric charge per second. Therefore, amperes are used to measure the rate of flow of electric charges in a circuit. Coulomb is a unit of electric charge, volt is a unit of electric potential or voltage, and watt is a unit of power.
Tambaya 42 Rahoto
A rectangular block of dimensions 2.0m x 1.0m x 0.5m weighs 200N. calculate the maximum pressure exerted by the block on a horizontal floor.
Bayanin Amsa
To calculate the maximum pressure exerted by the rectangular block on a horizontal floor, we need to use the formula: Pressure = Force / Area where Force is the weight of the block, and Area is the area of the block that is in contact with the floor. The weight of the block is given as 200N. To find the area of the block that is in contact with the floor, we need to consider the dimensions of the block. The length and width of the block are 2.0m and 1.0m respectively, which means the area of the base of the block is: Area = Length x Width = 2.0m x 1.0m = 2.0m2 Therefore, the maximum pressure exerted by the block on the floor is: Pressure = Force / Area = 200N / 2.0m2 = 100Nm-2 So, the answer is 100Nm-2.
Tambaya 43 Rahoto
It takes a shorter time for a liquid to boil at the top of a mountain than at the base because at the top, the
Bayanin Amsa
The correct answer is "pressure is lower." When you climb up a mountain, you go to a higher altitude where atmospheric pressure decreases. Since boiling occurs when the vapor pressure of the liquid becomes equal to the atmospheric pressure above it, when the atmospheric pressure is lower, the liquid boils at a lower temperature. This is because less heat is required to convert the liquid into vapor when the atmospheric pressure is low. Therefore, liquids boil at a lower temperature at higher altitudes because the pressure there is lower.
Tambaya 44 Rahoto
The working principle of a pressure cooker is based on the
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The working principle of a pressure cooker is based on the increase in the pressure of the pot. When a liquid is heated, its temperature increases, and it evaporates to form steam. In a normal cooking pot, the steam escapes into the atmosphere, and the pressure inside the pot remains constant. However, in a pressure cooker, the lid is airtight, which prevents the steam from escaping. As the steam is continuously generated, it builds up pressure inside the cooker, which increases the boiling point of water. This allows the food to cook faster and more evenly because the higher temperature causes the food to become tender quickly. In summary, the pressure cooker works by trapping the steam inside and increasing the pressure of the pot.
Tambaya 45 Rahoto
The diagram above illustrates a bar magnet near a coil connected to a galvanometer, When the magnet is rapidly moves towards the coil i. e.m.f. is induced in the coil ii. the galvanometer needle deflects iii. the magnet is attracted by the coil. Which of the statements above is/are correct?
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Tambaya 46 Rahoto
An object of mass 2kg moving with a velocity 3ms-11 collides head-on with another object of mass 1kg moving in the opposite direction with a velocity of 4ms-1. If the objects stick together after collision, calculate their common speed.
Bayanin Amsa
In a head-on collision of two objects, the total momentum before the collision is equal to the total momentum after the collision. The equation for this is: m1v1i + m2v2i = (m1 + m2)vf where m1 and v1i are the mass and initial velocity of the first object, m2 and v2i are the mass and initial velocity of the second object, and vf is the final velocity of the combined objects. Plugging in the values given: (2 kg)(3 m/s) + (1 kg)(-4 m/s) = (2 kg + 1 kg) vf 6 kg·m/s - 4 kg·m/s = 3 kg vf 2 kg·m/s = 3 kg vf vf = 2/3 m/s Therefore, the common speed of the objects after the collision is 0.67 m/s. Answer: 0.67ms-1.
Tambaya 48 Rahoto
A ball bearing falls through a viscous liquid
(a) Using a labelled diagram of a tall vessel, show all the forces acting on it
(b) When will it attain terminal velocity?
Ball bearing falling through a viscous liquid
(a) Labelled diagram showing the forces acting on the ball
Inside a tall vessel of viscous liquid, a spherical ball bearing falling vertically downwards experiences three forces:
- Weight, \(W = mg\), acting vertically downwards through the centre of the ball.
- Upthrust (buoyancy), \(U\), acting vertically upwards, equal to the weight of liquid displaced.
- Viscous drag, \(F\), a fluid-friction force given by Stokes' law \(F = 6\pi\eta r v\), acting vertically upwards, opposing the motion.
(b) When terminal velocity is attained
As the ball accelerates downwards its speed \(v\) increases, and because the viscous drag \(F = 6\pi\eta r v\) grows with speed, the total upward force \((U + F)\) increases too. The ball attains its terminal velocity at the instant the upward forces exactly balance the downward weight, so the resultant force and hence the acceleration become zero:
\[ W = U + F \quad\Rightarrow\quad mg = U + 6\pi\eta r v \]From that moment onwards there is no further acceleration, and the ball continues to fall through the liquid with a constant (maximum, uniform) velocity, which is the terminal velocity.
Bayanin Amsa
Ball bearing falling through a viscous liquid
(a) Labelled diagram showing the forces acting on the ball
Inside a tall vessel of viscous liquid, a spherical ball bearing falling vertically downwards experiences three forces:
- Weight, \(W = mg\), acting vertically downwards through the centre of the ball.
- Upthrust (buoyancy), \(U\), acting vertically upwards, equal to the weight of liquid displaced.
- Viscous drag, \(F\), a fluid-friction force given by Stokes' law \(F = 6\pi\eta r v\), acting vertically upwards, opposing the motion.
(b) When terminal velocity is attained
As the ball accelerates downwards its speed \(v\) increases, and because the viscous drag \(F = 6\pi\eta r v\) grows with speed, the total upward force \((U + F)\) increases too. The ball attains its terminal velocity at the instant the upward forces exactly balance the downward weight, so the resultant force and hence the acceleration become zero:
\[ W = U + F \quad\Rightarrow\quad mg = U + 6\pi\eta r v \]From that moment onwards there is no further acceleration, and the ball continues to fall through the liquid with a constant (maximum, uniform) velocity, which is the terminal velocity.
Tambaya 49 Rahoto
(a) State Heisenberg's Uncertainty Principle.
(b) State one phenomenon that can only be explained in terms of the wave nature of light.
(a) Heisenberg's Uncertainty Principle
It is impossible to determine simultaneously, with perfect accuracy, both the position and momentum of a particle. Mathematically,
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
(b) Interference of light is a phenomenon that can only be explained in terms of the wave nature of light. Diffraction is also acceptable.
Bayanin Amsa
(a) Heisenberg's Uncertainty Principle
It is impossible to determine simultaneously, with perfect accuracy, both the position and momentum of a particle. Mathematically,
\[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \]
(b) Interference of light is a phenomenon that can only be explained in terms of the wave nature of light. Diffraction is also acceptable.
Tambaya 50 Rahoto
An electron of charge 1.60 x 10\(^{-19}\)C is accelerated under a potential difference of 1.0 x 10\(^{5}\) V. Calculate the energy of the electron in joules.
Energy gained by an accelerated electron
When a charge \(q\) is accelerated through a potential difference \(V\), the energy it gains is the work done on it:
\[ E = qV \]Substituting \(q = 1.60\times10^{-19}\,\text{C}\) and \(V = 1.0\times10^{5}\,\text{V}\):
\[ E = (1.60\times10^{-19})(1.0\times10^{5}) = 1.6\times10^{-14}\,\text{J} \]The energy of the electron is \(1.6\times10^{-14}\,\text{J}\).
Bayanin Amsa
Energy gained by an accelerated electron
When a charge \(q\) is accelerated through a potential difference \(V\), the energy it gains is the work done on it:
\[ E = qV \]Substituting \(q = 1.60\times10^{-19}\,\text{C}\) and \(V = 1.0\times10^{5}\,\text{V}\):
\[ E = (1.60\times10^{-19})(1.0\times10^{5}) = 1.6\times10^{-14}\,\text{J} \]The energy of the electron is \(1.6\times10^{-14}\,\text{J}\).
Tambaya 51 Rahoto
The horizontal component of the initial speed of a particle projected at 30° to the horizontal is 50 m\(^{-1}\). If the acceleration of free fall due to gravity is 10 ms\(^{-2}\), determine its;
(a) initial speed;
(b) speed at the maximum height reached.
Projectile launched at 30 degrees with horizontal component 50 ms\(^{-1}\)
(a) Initial speed
The horizontal component is \(v_x = v\cos\theta\), so:
\[ v = \frac{v_x}{\cos\theta} = \frac{50}{\cos 30^\circ} = \frac{50}{0.8660} = 57.7\,\text{ms}^{-1} \](b) Speed at maximum height
At the highest point of the flight the vertical component of velocity is zero; only the horizontal component remains (it is unchanged throughout the motion). Therefore:
\[ v_{\text{top}} = v_x = 50\,\text{ms}^{-1} \]The speed at maximum height is \(50\,\text{ms}^{-1}\).
Bayanin Amsa
Projectile launched at 30 degrees with horizontal component 50 ms\(^{-1}\)
(a) Initial speed
The horizontal component is \(v_x = v\cos\theta\), so:
\[ v = \frac{v_x}{\cos\theta} = \frac{50}{\cos 30^\circ} = \frac{50}{0.8660} = 57.7\,\text{ms}^{-1} \](b) Speed at maximum height
At the highest point of the flight the vertical component of velocity is zero; only the horizontal component remains (it is unchanged throughout the motion). Therefore:
\[ v_{\text{top}} = v_x = 50\,\text{ms}^{-1} \]The speed at maximum height is \(50\,\text{ms}^{-1}\).
Tambaya 52 Rahoto
a) On what principle does lighting in a fluorescent tube operate?
(b) State two factors which determine the colour of light produced in a fluorescent tube.
(a) Principle of the fluorescent tube
A fluorescent tube works on the principle of gas discharge and fluorescence. When a high voltage is applied, electrons flow through the low-pressure mercury vapour in the tube, exciting the mercury atoms; these emit ultraviolet radiation. The ultraviolet light strikes the fluorescent (phosphor) coating on the inside of the tube, which absorbs it and re-emits the energy as visible light (fluorescence).
(b) Two factors determining the colour of the light produced
- The nature (chemical composition) of the phosphor / fluorescent powder coating the inside of the tube.
- The type of gas or vapour (and its pressure) contained in the tube.
Bayanin Amsa
(a) Principle of the fluorescent tube
A fluorescent tube works on the principle of gas discharge and fluorescence. When a high voltage is applied, electrons flow through the low-pressure mercury vapour in the tube, exciting the mercury atoms; these emit ultraviolet radiation. The ultraviolet light strikes the fluorescent (phosphor) coating on the inside of the tube, which absorbs it and re-emits the energy as visible light (fluorescence).
(b) Two factors determining the colour of the light produced
- The nature (chemical composition) of the phosphor / fluorescent powder coating the inside of the tube.
- The type of gas or vapour (and its pressure) contained in the tube.
Tambaya 53 Rahoto
(a) Define electrolysis.
(b) Classify each of the following substances as an electrolyte or a non-electrolyte
(i) sugar solution;
(ii) kerosene;
(iii) alkaline solution
(iv) lemon fruit juice.
(a) Electrolysis is the chemical decomposition of a substance (an electrolyte) brought about by passing an electric current through it when in the molten state or in aqueous solution.
(b) Classification
| Substance | Classification |
|---|---|
| (i) Sugar solution | Non-electrolyte |
| (ii) Kerosene | Non-electrolyte |
| (iii) Alkaline solution | Electrolyte |
| (iv) Lemon fruit juice | Electrolyte |
Electrolytes (alkaline solution and lemon juice, which is acidic) conduct electricity because they contain free mobile ions, whereas sugar solution and kerosene contain no free ions and so do not conduct.
Bayanin Amsa
(a) Electrolysis is the chemical decomposition of a substance (an electrolyte) brought about by passing an electric current through it when in the molten state or in aqueous solution.
(b) Classification
| Substance | Classification |
|---|---|
| (i) Sugar solution | Non-electrolyte |
| (ii) Kerosene | Non-electrolyte |
| (iii) Alkaline solution | Electrolyte |
| (iv) Lemon fruit juice | Electrolyte |
Electrolytes (alkaline solution and lemon juice, which is acidic) conduct electricity because they contain free mobile ions, whereas sugar solution and kerosene contain no free ions and so do not conduct.
Tambaya 54 Rahoto
(a)(i) State Einstein's equation of photoelectric effect
(ii) What conservation principle does the equation represent?
(b) List three applications of photocells
(c) A photo emissive surface has a threshold frequency of 4.02 x 10\(^{14}\)Hz. If the surface is illuminated by light of frequency 5.0 x 10\(^{15}\)Hz, calculate the:
(i) threshold wavelength;
(ii) work function;
(iii) kinetic energy of the emitted photoelectrons. [ c = 3.0 x 10\(^{8}\) ms\(^{-1}\), h = 6.63 x 10\(^{-34}\) Js]
Tambaya 55 Rahoto
(a) What is a magnetic field?
(b) With the aid of a labelled diagram describe an experiment to show that a magnetic field exists around a straight wire carrying current.
(c) A 40 \(\mu\)F capacitor in series with a 40 Q resistor is connected to a 100 V, 50 Hz a.c. supply.
(i) Draw a circuit diagram of the arrangement
(ii) Calculate the: I. impedance in the circuit; II. current in the circuit III. potential difference across the capacitor.
(a) A magnetic field is the region of space around a magnet or a current-carrying conductor within which a magnetic force is experienced by another magnet, a magnetic material, or a moving charge.
(b) Experiment to show that a magnetic field exists around a straight current-carrying wire (Oersted's experiment)
A stiff, straight copper wire is passed vertically through a hole in the centre of a horizontal cardboard sheet. The wire is connected in series with a battery, a switch (key) and a rheostat so that a large current can flow through it. Iron filings are sprinkled evenly over the card, and a small plotting compass is placed on the card near the wire.
When the switch is closed and a strong current flows, the card is gently tapped. The iron filings arrange themselves into a pattern of concentric circles centred on the wire, and the plotting compass needle sets itself tangential to these circles. This shows that a magnetic field exists in the space around the wire. When the direction of the current is reversed, the compass needle turns to point the opposite way, showing that the direction of the field also reverses.
(c) Series R-C circuit on a.c. supply: \(C = 40\,\mu\text{F} = 40\times10^{-6}\,\text{F}\), \(R = 40\,\Omega\), \(V = 100\,\text{V}\), \(f = 50\,\text{Hz}\).
(i) Circuit diagram — the 40 Ω resistor and the 40 µF capacitor connected in series across the 100 V, 50 Hz a.c. supply:
(ii) I. Impedance
Capacitive reactance:
\[ X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi(50)(40\times10^{-6})} = 79.6\,\Omega \]Impedance:
\[ Z = \sqrt{R^2 + X_C^2} = \sqrt{40^2 + 79.6^2} = \sqrt{1600 + 6337} = \sqrt{7937} = 89.1\,\Omega \]II. Current in the circuit
\[ I = \frac{V}{Z} = \frac{100}{89.1} = 1.12\,\text{A} \]III. Potential difference across the capacitor
\[ V_C = I\,X_C = 1.12 \times 79.6 = 89.2\,\text{V} \]Bayanin Amsa
(a) A magnetic field is the region of space around a magnet or a current-carrying conductor within which a magnetic force is experienced by another magnet, a magnetic material, or a moving charge.
(b) Experiment to show that a magnetic field exists around a straight current-carrying wire (Oersted's experiment)
A stiff, straight copper wire is passed vertically through a hole in the centre of a horizontal cardboard sheet. The wire is connected in series with a battery, a switch (key) and a rheostat so that a large current can flow through it. Iron filings are sprinkled evenly over the card, and a small plotting compass is placed on the card near the wire.
When the switch is closed and a strong current flows, the card is gently tapped. The iron filings arrange themselves into a pattern of concentric circles centred on the wire, and the plotting compass needle sets itself tangential to these circles. This shows that a magnetic field exists in the space around the wire. When the direction of the current is reversed, the compass needle turns to point the opposite way, showing that the direction of the field also reverses.
(c) Series R-C circuit on a.c. supply: \(C = 40\,\mu\text{F} = 40\times10^{-6}\,\text{F}\), \(R = 40\,\Omega\), \(V = 100\,\text{V}\), \(f = 50\,\text{Hz}\).
(i) Circuit diagram — the 40 Ω resistor and the 40 µF capacitor connected in series across the 100 V, 50 Hz a.c. supply:
(ii) I. Impedance
Capacitive reactance:
\[ X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi(50)(40\times10^{-6})} = 79.6\,\Omega \]Impedance:
\[ Z = \sqrt{R^2 + X_C^2} = \sqrt{40^2 + 79.6^2} = \sqrt{1600 + 6337} = \sqrt{7937} = 89.1\,\Omega \]II. Current in the circuit
\[ I = \frac{V}{Z} = \frac{100}{89.1} = 1.12\,\text{A} \]III. Potential difference across the capacitor
\[ V_C = I\,X_C = 1.12 \times 79.6 = 89.2\,\text{V} \]Tambaya 56 Rahoto
A wire gradually stretched by loading it until it snaps.
(a) Sketch a load-extension graph for the wire
(b) Indicate on the graph the
(i) elastic limit (E);
(ii) yield point (Y);
(iii) breaking point (B).
```html
Load-extension graph for a wire stretched to breaking
Plot load (force) on the vertical axis against extension on the horizontal axis. The wire extends elastically at first. Beyond the elastic limit, permanent extension begins. At the yield point, the wire extends greatly with little increase in load. Eventually, the wire breaks at point B.
E is the elastic limit: up to this point, the wire returns to its original length when the load is removed.
Y is the yield point: the wire begins to undergo noticeable permanent deformation.
B is the breaking point: the wire snaps.
```Bayanin Amsa
```html
Load-extension graph for a wire stretched to breaking
Plot load (force) on the vertical axis against extension on the horizontal axis. The wire extends elastically at first. Beyond the elastic limit, permanent extension begins. At the yield point, the wire extends greatly with little increase in load. Eventually, the wire breaks at point B.
E is the elastic limit: up to this point, the wire returns to its original length when the load is removed.
Y is the yield point: the wire begins to undergo noticeable permanent deformation.
B is the breaking point: the wire snaps.
```Tambaya 57 Rahoto
Use the diagram to answer question;
Camphor is a chemical that sub-limes and interacts with water reducing the surface tension where it is put. Explain why the toy boat illustrated above would move forward with the camphor placed at the back of it but would be stationary before the placement of the camphor.
What the diagram shows. The pointed toy boat floats on water. Camphor is fixed at the flat rear end. The label \(F_1\) is a leftward (backward) arrow acting at the rear where the camphor is, and \(F_2\) is a rightward (forward) arrow acting at the pointed front. \(F_1\) and \(F_2\) represent the surface-tension pulls of the water on the two ends of the boat.
Key idea: forces due to surface tension. The free surface of water behaves like a stretched elastic skin. Wherever this skin touches the edge of the floating boat, it pulls the boat outward (away from the boat) along the surface. So the water pulls the rear of the boat backward with force \(F_1\) and pulls the front of the boat forward with force \(F_2\). The size of each pull is proportional to the surface tension of the water touching that edge.
Before the camphor is placed (boat stationary). The surface tension of the water is the same all around the boat, so the backward pull equals the forward pull:
\[ F_1 = F_2 \]The two pulls cancel, the resultant horizontal force is zero, and the boat stays at rest.
After the camphor is placed at the back. The camphor sublimes and dissolves into the water immediately behind the boat, and it reduces the surface tension of the water there. The water in front of the boat still has its full (higher) surface tension. Therefore the backward pull becomes small while the forward pull stays large:
\[ F_2 > F_1 \]There is now an unbalanced resultant force acting forward:
\[ F_{net} = F_2 - F_1 \quad (\text{directed forward}) \]By Newton's second law this net force accelerates the boat, so it moves forward (in the direction of \(F_2\)). Equivalently, the higher-tension water in front pulls harder than the weakened-tension water behind, and the boat is dragged toward the region of greater surface tension. As soon as the camphor is used up and the surface tension becomes uniform again, \(F_1\) once more equals \(F_2\) and the boat stops.
Bayanin Amsa
What the diagram shows. The pointed toy boat floats on water. Camphor is fixed at the flat rear end. The label \(F_1\) is a leftward (backward) arrow acting at the rear where the camphor is, and \(F_2\) is a rightward (forward) arrow acting at the pointed front. \(F_1\) and \(F_2\) represent the surface-tension pulls of the water on the two ends of the boat.
Key idea: forces due to surface tension. The free surface of water behaves like a stretched elastic skin. Wherever this skin touches the edge of the floating boat, it pulls the boat outward (away from the boat) along the surface. So the water pulls the rear of the boat backward with force \(F_1\) and pulls the front of the boat forward with force \(F_2\). The size of each pull is proportional to the surface tension of the water touching that edge.
Before the camphor is placed (boat stationary). The surface tension of the water is the same all around the boat, so the backward pull equals the forward pull:
\[ F_1 = F_2 \]The two pulls cancel, the resultant horizontal force is zero, and the boat stays at rest.
After the camphor is placed at the back. The camphor sublimes and dissolves into the water immediately behind the boat, and it reduces the surface tension of the water there. The water in front of the boat still has its full (higher) surface tension. Therefore the backward pull becomes small while the forward pull stays large:
\[ F_2 > F_1 \]There is now an unbalanced resultant force acting forward:
\[ F_{net} = F_2 - F_1 \quad (\text{directed forward}) \]By Newton's second law this net force accelerates the boat, so it moves forward (in the direction of \(F_2\)). Equivalently, the higher-tension water in front pulls harder than the weakened-tension water behind, and the boat is dragged toward the region of greater surface tension. As soon as the camphor is used up and the surface tension becomes uniform again, \(F_1\) once more equals \(F_2\) and the boat stops.
Tambaya 58 Rahoto
(a) The equation y = \(\alpha\) sin (wt - kx) represents a plane wave travelling in a medium along the x-direction, being the displacement at the point x at time t.
(i) Given that x is in metres and t is in seconds, state the units of k and w
(ii) What physical quantity does \(\frac{W}{k}\) represent? Justify your answer
(iii) State whether the wave is travelling in the positive or negative x-direction.
(b)(i) What are beats?
(ii) A sonometer wire has a frequency of 259 Hz. It is sounded alongside a tuning fork of frequency 256 Hz. Calculate the beat frequency
(iii) The sonometer wire in (b)(ii) above is under a tension of 1200 N. If a metre of the wire has a mass of 0.03 kg, calculate the length l of the wire when it is vibrating in the fundamental mode.
(c) List two similarities between the human eye and the photographic camera.
(a)(i) Units in \(y = a\sin(\omega t - kx)\): the argument of sine must be a pure number. Since \(t\) is in seconds, \(\omega\) has unit rad s\(^{-1}\) (per second); since \(x\) is in metres, \(k\) has unit rad m\(^{-1}\) (per metre).
(a)(ii) \(\dfrac{\omega}{k}\) represents the speed (velocity) of the wave. Justification: \(\omega = 2\pi f\) and \(k = \dfrac{2\pi}{\lambda}\), so \(\dfrac{\omega}{k} = \dfrac{2\pi f}{2\pi/\lambda} = f\lambda = v\), which is the wave speed.
(a)(iii) The form \((\omega t - kx)\) represents a wave travelling in the positive x-direction.
(b)(i) Beats are the periodic rise and fall in the loudness (amplitude) of sound heard when two notes of slightly different frequencies are sounded together, caused by their alternate reinforcement and cancellation.
(b)(ii) Beat frequency:
\[ f_{\text{beat}} = |259 - 256| = 3\,\text{Hz} \](b)(iii) Length of the sonometer wire (fundamental mode): the wave speed on the wire is
\[ v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{1200}{0.03}} = \sqrt{40000} = 200\,\text{ms}^{-1} \]For the fundamental, \(f = \dfrac{v}{2l}\), so:
\[ l = \frac{v}{2f} = \frac{200}{2\times 259} = 0.386\,\text{m}\ (\approx 38.6\,\text{cm}) \](c) Two similarities between the human eye and the photographic camera:
- Both use a converging (convex) lens to form a real, inverted, diminished image of the object.
- Both form the image on a light-sensitive surface (the retina in the eye, the film/sensor in the camera) and both control the amount of light entering through an adjustable aperture (the iris/pupil in the eye, the diaphragm in the camera).
Bayanin Amsa
(a)(i) Units in \(y = a\sin(\omega t - kx)\): the argument of sine must be a pure number. Since \(t\) is in seconds, \(\omega\) has unit rad s\(^{-1}\) (per second); since \(x\) is in metres, \(k\) has unit rad m\(^{-1}\) (per metre).
(a)(ii) \(\dfrac{\omega}{k}\) represents the speed (velocity) of the wave. Justification: \(\omega = 2\pi f\) and \(k = \dfrac{2\pi}{\lambda}\), so \(\dfrac{\omega}{k} = \dfrac{2\pi f}{2\pi/\lambda} = f\lambda = v\), which is the wave speed.
(a)(iii) The form \((\omega t - kx)\) represents a wave travelling in the positive x-direction.
(b)(i) Beats are the periodic rise and fall in the loudness (amplitude) of sound heard when two notes of slightly different frequencies are sounded together, caused by their alternate reinforcement and cancellation.
(b)(ii) Beat frequency:
\[ f_{\text{beat}} = |259 - 256| = 3\,\text{Hz} \](b)(iii) Length of the sonometer wire (fundamental mode): the wave speed on the wire is
\[ v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{1200}{0.03}} = \sqrt{40000} = 200\,\text{ms}^{-1} \]For the fundamental, \(f = \dfrac{v}{2l}\), so:
\[ l = \frac{v}{2f} = \frac{200}{2\times 259} = 0.386\,\text{m}\ (\approx 38.6\,\text{cm}) \](c) Two similarities between the human eye and the photographic camera:
- Both use a converging (convex) lens to form a real, inverted, diminished image of the object.
- Both form the image on a light-sensitive surface (the retina in the eye, the film/sensor in the camera) and both control the amount of light entering through an adjustable aperture (the iris/pupil in the eye, the diaphragm in the camera).
Tambaya 59 Rahoto
In an electrolysis experiment, the ammeter records a steady current of 1 A. The mass of copper deposited in 30 minutes is 0.66 g. Calculate the error in the ammeter reading. [ Electrochemical equivalent of copper = 0.00033 g C\(^{-1}\)]
Error in the ammeter reading (electrolysis of copper)
By Faraday's first law, the true charge that passed is found from the mass deposited and the electrochemical equivalent \(z\):
\[ m = zQ \;\Rightarrow\; Q = \frac{m}{z} = \frac{0.66}{0.00033} = 2000\,\text{C} \]The time of the experiment is \(t = 30\,\text{min} = 1800\,\text{s}\), so the true (actual) current is:
\[ I_{\text{true}} = \frac{Q}{t} = \frac{2000}{1800} = 1.11\,\text{A} \]The ammeter reads \(1.0\,\text{A}\), so the error is:
\[ \text{error} = I_{\text{true}} - I_{\text{read}} = 1.11 - 1.00 = 0.11\,\text{A} \]The ammeter reads about \(0.11\,\text{A}\) too low (a percentage error of roughly \(10\%\)).
Bayanin Amsa
Error in the ammeter reading (electrolysis of copper)
By Faraday's first law, the true charge that passed is found from the mass deposited and the electrochemical equivalent \(z\):
\[ m = zQ \;\Rightarrow\; Q = \frac{m}{z} = \frac{0.66}{0.00033} = 2000\,\text{C} \]The time of the experiment is \(t = 30\,\text{min} = 1800\,\text{s}\), so the true (actual) current is:
\[ I_{\text{true}} = \frac{Q}{t} = \frac{2000}{1800} = 1.11\,\text{A} \]The ammeter reads \(1.0\,\text{A}\), so the error is:
\[ \text{error} = I_{\text{true}} - I_{\text{read}} = 1.11 - 1.00 = 0.11\,\text{A} \]The ammeter reads about \(0.11\,\text{A}\) too low (a percentage error of roughly \(10\%\)).
Tambaya 60 Rahoto
Explain plane polarized light
Plane polarized light
Ordinary (unpolarized) light consists of transverse waves whose electric-field vibrations occur in all possible directions perpendicular to the direction of travel. Plane polarized light is light in which these vibrations have been restricted so that they take place in only one plane containing the direction of propagation.
In other words, the electric vector of a plane polarized beam oscillates along a single fixed direction (perpendicular to the direction of travel), rather than in every transverse direction. Such light can be produced by passing ordinary light through a polaroid (or by reflection at the polarizing angle), and its state can be detected using a second polaroid (analyser) which cuts off the light when its axis is crossed with the first. The fact that light can be polarized is direct evidence that light is a transverse wave.
Bayanin Amsa
Plane polarized light
Ordinary (unpolarized) light consists of transverse waves whose electric-field vibrations occur in all possible directions perpendicular to the direction of travel. Plane polarized light is light in which these vibrations have been restricted so that they take place in only one plane containing the direction of propagation.
In other words, the electric vector of a plane polarized beam oscillates along a single fixed direction (perpendicular to the direction of travel), rather than in every transverse direction. Such light can be produced by passing ordinary light through a polaroid (or by reflection at the polarizing angle), and its state can be detected using a second polaroid (analyser) which cuts off the light when its axis is crossed with the first. The fact that light can be polarized is direct evidence that light is a transverse wave.
Tambaya 61 Rahoto
(a) State two differences between boiling and evaporation.
(b) A closed inexpansibie vessel contains air saturated with water vapour at 77°C. The total pressure in the vessel is 1007 mmHg. Calculate the new pressure in the vessel if the temperature is reduced 1.2 27°C. [ The s.v.p. of water at 77°C and 27°C respectively are 314 mmHg and 27 mmHg. Treat the air in the vessel as an ideal gas]
(c) The lengtn of a zinc rod at 23°C is 200 m. Calculate the increase in length of the rod when its temperature rises to 33°C. [ expansivity of zinc = 2.6 x 10\(^{-5}\) K\(^{-1}\)]
(d) Explain with there is no temperature change when a solid being heated changes into liquid at its melting point.
(a) Two differences between boiling and evaporation
| Boiling | Evaporation |
|---|---|
| Occurs at a fixed temperature (the boiling point) | Occurs at all temperatures |
| Takes place throughout the whole liquid (bubbles form within it) | Takes place only at the exposed surface of the liquid |
(b) New pressure in the vessel after cooling
At \(77^\circ\text{C}\ (350\,\text{K})\), total pressure = 1007 mmHg and s.v.p. of water = 314 mmHg, so the partial pressure of the dry air is:
\[ P_{\text{air}} = 1007 - 314 = 693\,\text{mmHg} \]The air (ideal, constant volume) is cooled to \(27^\circ\text{C}\ (300\,\text{K})\):
\[ P'_{\text{air}} = 693\times\frac{300}{350} = 594\,\text{mmHg} \]Adding the s.v.p. of water at \(27^\circ\text{C}\) (27 mmHg):
\[ P_{\text{total}} = 594 + 27 = 621\,\text{mmHg} \](c) Increase in length of the zinc rod
\[ \Delta L = L\,\alpha\,\Delta\theta = 200 \times (2.6\times10^{-5}) \times (33-23) \] \[ \Delta L = 200 \times 2.6\times10^{-5} \times 10 = 0.052\,\text{m} \ (5.2\,\text{cm}) \](d) When a solid melts at its melting point, the heat supplied (the latent heat of fusion) is used entirely to break the bonds/forces holding the particles in the fixed solid lattice, not to increase their kinetic energy. Since temperature depends on the average kinetic energy of the particles, and this does not change while the solid is turning to liquid, the temperature stays constant until melting is complete.
Bayanin Amsa
(a) Two differences between boiling and evaporation
| Boiling | Evaporation |
|---|---|
| Occurs at a fixed temperature (the boiling point) | Occurs at all temperatures |
| Takes place throughout the whole liquid (bubbles form within it) | Takes place only at the exposed surface of the liquid |
(b) New pressure in the vessel after cooling
At \(77^\circ\text{C}\ (350\,\text{K})\), total pressure = 1007 mmHg and s.v.p. of water = 314 mmHg, so the partial pressure of the dry air is:
\[ P_{\text{air}} = 1007 - 314 = 693\,\text{mmHg} \]The air (ideal, constant volume) is cooled to \(27^\circ\text{C}\ (300\,\text{K})\):
\[ P'_{\text{air}} = 693\times\frac{300}{350} = 594\,\text{mmHg} \]Adding the s.v.p. of water at \(27^\circ\text{C}\) (27 mmHg):
\[ P_{\text{total}} = 594 + 27 = 621\,\text{mmHg} \](c) Increase in length of the zinc rod
\[ \Delta L = L\,\alpha\,\Delta\theta = 200 \times (2.6\times10^{-5}) \times (33-23) \] \[ \Delta L = 200 \times 2.6\times10^{-5} \times 10 = 0.052\,\text{m} \ (5.2\,\text{cm}) \](d) When a solid melts at its melting point, the heat supplied (the latent heat of fusion) is used entirely to break the bonds/forces holding the particles in the fixed solid lattice, not to increase their kinetic energy. Since temperature depends on the average kinetic energy of the particles, and this does not change while the solid is turning to liquid, the temperature stays constant until melting is complete.
Tambaya 62 Rahoto
(a) A particle moves on a straight path with an initial speed u and final speed v in time t. Show that the total distance X covered by the particle is given by
\(x = ut + \frac{1}{2}at^2\)
where a is the magnitude of acceleration
(b) State;
(i) Newton's second law of motion
(ii) the principle of conservation of energy
(iii) the law of floatation.
(c) Consider a balloon of mass 0.030 kg being inflated with a gas of density 0.54 kg m\(^{-3}\). What will be the volume of the balloon when it just begins to rise in air of density 1.29 kg m\(^{-3}\)? [ g = 10 ms\(^{-2}\)]
(a) Derivation of \(x = ut + \tfrac{1}{2}at^2\)
For uniform acceleration \(a\), initial speed \(u\), final speed \(v\) after time \(t\), the distance is the area under the velocity-time graph (a trapezium):
\[ x = \left(\frac{u+v}{2}\right)t \]But \(v = u + at\). Substituting:
\[ x = \left(\frac{u + (u+at)}{2}\right)t = \left(\frac{2u + at}{2}\right)t = ut + \tfrac{1}{2}at^2 \quad\text{(shown)} \](b)(i) Newton's second law: The rate of change of momentum of a body is directly proportional to the applied (resultant) force and takes place in the direction of the force; hence \(F = ma\).
(b)(ii) Principle of conservation of energy: Energy can neither be created nor destroyed but only changed from one form to another; the total energy of an isolated system remains constant.
(b)(iii) Law of floatation: A floating body displaces its own weight of the fluid in which it floats.
(c) Volume of the balloon when it just begins to rise
The balloon just begins to rise when the upthrust from the air equals the total weight (of the balloon fabric plus the gas):
\[ \rho_{\text{air}}\,V g = (m + \rho_{\text{gas}}\,V)g \] \[ 1.29\,V = 0.030 + 0.54\,V \] \[ (1.29 - 0.54)V = 0.030 \;\Rightarrow\; 0.75\,V = 0.030 \] \[ V = \frac{0.030}{0.75} = 0.04\,\text{m}^3 \]The volume of the balloon is \(0.04\,\text{m}^3\).
Bayanin Amsa
(a) Derivation of \(x = ut + \tfrac{1}{2}at^2\)
For uniform acceleration \(a\), initial speed \(u\), final speed \(v\) after time \(t\), the distance is the area under the velocity-time graph (a trapezium):
\[ x = \left(\frac{u+v}{2}\right)t \]But \(v = u + at\). Substituting:
\[ x = \left(\frac{u + (u+at)}{2}\right)t = \left(\frac{2u + at}{2}\right)t = ut + \tfrac{1}{2}at^2 \quad\text{(shown)} \](b)(i) Newton's second law: The rate of change of momentum of a body is directly proportional to the applied (resultant) force and takes place in the direction of the force; hence \(F = ma\).
(b)(ii) Principle of conservation of energy: Energy can neither be created nor destroyed but only changed from one form to another; the total energy of an isolated system remains constant.
(b)(iii) Law of floatation: A floating body displaces its own weight of the fluid in which it floats.
(c) Volume of the balloon when it just begins to rise
The balloon just begins to rise when the upthrust from the air equals the total weight (of the balloon fabric plus the gas):
\[ \rho_{\text{air}}\,V g = (m + \rho_{\text{gas}}\,V)g \] \[ 1.29\,V = 0.030 + 0.54\,V \] \[ (1.29 - 0.54)V = 0.030 \;\Rightarrow\; 0.75\,V = 0.030 \] \[ V = \frac{0.030}{0.75} = 0.04\,\text{m}^3 \]The volume of the balloon is \(0.04\,\text{m}^3\).
