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Ajụjụ 1 Ripọtì
A vibrator causes water ripples to travel across the surface of a tank. The wave travels 50cm in 2s and the distance between successive crests is 5cm. Calculate the frequency of the vibrator
Akọwa Nkọwa
The frequency of the vibrator can be calculated using the formula: frequency = speed / wavelength where speed is the speed of the wave, and wavelength is the distance between successive crests. In this case, we are given that the wave travels 50cm in 2s, which means the speed of the wave is: speed = distance / time = 50cm / 2s = 25cm/s We are also given that the distance between successive crests is 5cm, which is the wavelength. Therefore, the frequency of the vibrator is: frequency = speed / wavelength = 25cm/s / 5cm = 5Hz So the correct answer is 5Hz.
Ajụjụ 2 Ripọtì
The diagram shows a uniform meter rule AB which balances horizontally at the 90cm mark when a mass of 0.2kg is suspended from B. Calculate the mass of the meter rule.
Akọwa Nkọwa
Mr
(90 - 50) = 0.2(100 - 90)
40Mr
= 0.2 × 10
Mr
= 240
= 0.05kg
Ajụjụ 3 Ripọtì
Which of the following equations is the correct definition of the reactance of an indicator L?
Akọwa Nkọwa
The correct definition of the reactance of an inductor L is: Reactance = (Amplitude of voltage) ÷ (Amplitude of current) The reactance of an inductor is a measure of the opposition offered by the inductor to the flow of alternating current (AC). It is denoted by the symbol Xl and is measured in ohms. When AC flows through an inductor, a magnetic field is generated around the inductor, which opposes any changes in the current flowing through it. This opposition to the flow of current is called reactance. The reactance of an inductor depends on its inductance, frequency of the AC signal, and the amplitude of the AC signal. However, the reactance of an inductor is directly proportional to the frequency of the AC signal and the inductance of the inductor. The reactance of an inductor is also affected by the amplitude of the AC signal, but this effect is not as significant as the other two factors. is the correct definition of the reactance of an inductor, as it expresses the ratio of the amplitude of voltage to the amplitude of current, which is a common way to define reactance. is incorrect, as it represents the power delivered by the AC signal, not the reactance. and are also incorrect, as they involve squaring either the amplitude of current or the amplitude of voltage, which is not a valid method of calculating reactance. Therefore, the correct option is.
Ajụjụ 4 Ripọtì
Workdone on an object to bring it to a certain point in space is called
Akọwa Nkọwa
The work done on an object to bring it to a certain point in space is called "Potential Energy". Potential energy is a form of energy that an object possesses due to its position relative to other objects. When an object is lifted or moved to a higher point against gravity, work is done on it, and this work is stored as potential energy. The potential energy of an object is directly proportional to its height and mass. It can be converted into other forms of energy, such as kinetic energy, when the object is released or allowed to move freely. Therefore, potential energy is a type of stored energy that an object has due to its position, and it can be released to do work.
Ajụjụ 6 Ripọtì
A copper rod, 5m long when heated through 20c, expands by 1mm. If a second copper rod, 2.5m long is heated through 5c, by how much will it expand?
Akọwa Nkọwa
l1
= 5m, ΔT = 10c, l2
- l1
= 1mm
l1
= 2.5m, ΔT = 5c, l2
- l1
= ?
| using | α | = | l2 - l1 l1 ΔT |
| 15(10) | = | l2 - l1 2.5(5) |
| l2 | - | l2 | = | 2.5(5)5(10) | = | 14 | = | 0.25mm |
Ajụjụ 8 Ripọtì
A single force which produces the same effect as a set of forces acting together at a point is known as the
Akọwa Nkọwa
The single force which produces the same effect as a set of forces acting together at a point is known as the "resultant". In other words, the resultant is the net force that results from combining all the individual forces acting on an object. It represents the combined effect of all the forces acting on the object and is the force that would produce the same motion as the original set of forces acting together. Therefore, when solving problems in physics, it is often useful to find the resultant force in order to determine the overall effect of multiple forces on an object.
Ajụjụ 9 Ripọtì
If a body moves with a constant speed and at the same time undergoes an acceleration, its motion is said to be
Akọwa Nkọwa
If a body moves with a constant speed and at the same time undergoes an acceleration, its motion is said to be rectilinear. When an object moves with constant speed, it means that it covers the same distance in equal time intervals. On the other hand, acceleration is the rate of change of velocity with time. If an object undergoes acceleration, its velocity changes with time. Therefore, if a body moves with constant speed and undergoes an acceleration, it means that its direction of motion changes while it covers equal distances in equal time intervals. This type of motion is called rectilinear motion, where the object moves in a straight line, but its velocity changes due to the acceleration. In contrast, circular motion is when an object moves in a circular path with a constant speed, while oscillatory motion is when an object moves back and forth around a fixed point. Rotational motion is when an object rotates around an axis. None of these descriptions fit the scenario of a body moving with constant speed and undergoing acceleration, so the answer is rectilinear motion.
Ajụjụ 10 Ripọtì
A ray of light passes through the centre of curvature of a concave mirror and strikes the mirror. At what angle is the ray reflected?
Akọwa Nkọwa
When a light ray passes through the center of curvature of a concave mirror and strikes the mirror, the reflected ray will be reflected back on itself, creating an angle of 0 degrees. Therefore, the correct answer is 0o.
Ajụjụ 11 Ripọtì
The mass of water vapour in a given volume of air is 0.05g at 20°C, while the mass of water vapour required to saturate it at the same temperature is 0.15g. Calculate the relative humidity of the air.
Akọwa Nkọwa
Relative humidity is a measure of how much water vapor the air is holding compared to the maximum amount it could hold at a given temperature. It is expressed as a percentage. To calculate the relative humidity of the air in this problem, we need to use the formula: Relative humidity = (mass of water vapor in air / mass of water vapor required for saturation) x 100% We are given that the mass of water vapor in the air is 0.05g and the mass of water vapor required for saturation at the same temperature is 0.15g. Plugging these values into the formula, we get: Relative humidity = (0.05 / 0.15) x 100% = 33.33% Therefore, the relative humidity of the air is 33.33%. So the answer is 33.33%.
Ajụjụ 12 Ripọtì
Which of the following statements is/are correct for a freely falling body?
I. the total is entirely kinetic
II. the ratio of potential energy to kinetic energy is constant
III. the sum of potential and kinetic energy is constant
Akọwa Nkọwa
The correct answer is "III only". A freely falling body is one that is falling under the influence of gravity and experiences no other force or constraint. In this situation, the total energy of the body is conserved, meaning that the sum of its potential and kinetic energy remains constant. The potential energy of a body is directly proportional to its height above the ground, and its kinetic energy is directly proportional to its velocity. As the body falls, its potential energy decreases and its kinetic energy increases, but the total energy remains constant. Statement III is correct because the sum of potential and kinetic energy is indeed constant for a freely falling body. Statement I is incorrect because the body has both potential and kinetic energy, so the total energy is not entirely kinetic. Statement II is incorrect because the ratio of potential energy to kinetic energy is not constant for a freely falling body, as both are changing as the body falls.
Ajụjụ 13 Ripọtì
Which of the following readings cannot be determined with a meter rule?
Akọwa Nkọwa
Meter rule has a reading accuracy of 0.5mm or 0.05cm, thus measurement is M ± 0.05cm i.e 2.00, 2.05, 2.50, 2.55 etc.
The reading that cannot be read is 2.56cm.
Ajụjụ 14 Ripọtì
In which of the points labelled A, B, C, D and E on the conductor shown would electric charge tend to concentrate most
Akọwa Nkọwa
- Charge are mostly concentrated at the outermost part of a hollow conductor
- Charge are also mostly concentrated at the pointed ends or places with high density point.
Ajụjụ 15 Ripọtì
The volume of 0.354g of helium at 273°C and 114cm of mercury pressure is 2667cm3 . Calculate the volume
Akọwa Nkọwa
m = 0.354g, T1
= 273°C = 273 + 273 = 576K
P1
= 114cmHg, V1
= 2667cm3
at STP
T2
= 273K, P2
= 76cmHg, V2
= ?
| P1 V1 T1 | = | P2 V2 T1 |
| V2 | = | 114 × 2667 × 27376 × 576 | = | 2000.25cm3 |
Ajụjụ 16 Ripọtì
A man on a bench will exert the greatest pressure on the bench when he
Akọwa Nkọwa
The man on the bench will exert the greatest pressure when he stands on the toes of one foot. This is because when he stands on one foot, all his weight is concentrated on a smaller surface area of the bench, resulting in more pressure. The pressure he exerts is calculated by dividing his weight by the surface area in contact with the bench. When he stands on one foot, the surface area is smaller, which means the pressure exerted is greater. In comparison, when he lies flat on his back or belly, or when he stands on both feet, his weight is distributed over a larger surface area, resulting in less pressure.
Ajụjụ 17 Ripọtì
A rectangular solid black has length 10cm, breadth 5cm and height 2cm. If it lies on a horizontal surface, and has density 100kg/m3 , calculate the pressure it exerts on the surface.
Akọwa Nkọwa
To calculate the pressure that the rectangular solid exerts on the surface, we need to use the formula for pressure: Pressure = Force / Area In this case, the force is the weight of the rectangular solid, which we can calculate using the formula: Weight = Mass x Gravity The mass of the rectangular solid can be calculated using its density and volume: Mass = Density x Volume The volume of the rectangular solid is simply its length x breadth x height: Volume = Length x Breadth x Height = 10 cm x 5 cm x 2 cm = 100 cm3 We need to convert this volume to cubic meters to use the density given in kg/m3: Volume = 100 cm3 = 0.0001 m3 Now we can calculate the mass: Mass = Density x Volume = 100 kg/m3 x 0.0001 m3 = 0.01 kg The gravity is the acceleration due to gravity, which we can assume to be 9.81 m/s2. Therefore, the weight is: Weight = Mass x Gravity = 0.01 kg x 9.81 m/s2 = 0.0981 N Now we can use this weight to calculate the pressure on the surface. The surface area in contact with the rectangular solid is simply its length x breadth: Area = Length x Breadth = 10 cm x 5 cm = 50 cm2 We need to convert this area to square meters: Area = 50 cm2 = 0.005 m2 Therefore, the pressure is: Pressure = Force / Area = 0.0981 N / 0.005 m2 = 19.62 N/m2 We can convert this to units of N/cm2 or N/mm2 if desired. This is equivalent to: Pressure = 0.1962 N/cm2 = 0.0001962 N/mm2 So the pressure that the rectangular solid exerts on the surface is 19.62 N/m2, which is approximately 20 N/m2. Therefore, the answer is 200 N/m2.
Ajụjụ 18 Ripọtì
The statement 'Heat lost by the hot body equals that gained by the cold one' is assumed when determining specific that heat capacity by the method of mixtures. Which of the following validates the assumption?
I. Lagging the Calorimeter
II. Ensuring that only S.I units are used
III. Weighing the calorimeter, the lid and the stirrer.
Akọwa Nkọwa
The assumption 'Heat lost by the hot body equals that gained by the cold one' is based on the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred from one system to another. Thus, to validate this assumption, it's important to have a well-designed and insulated calorimeter so that as little heat as possible is lost to the environment. This is accomplished by lagging the calorimeter (Option I). Additionally, using the correct units (Option II) helps ensure that the energy transfer is accurately calculated and reported. Weighing the calorimeter, the lid, and the stirrer (Option III) is important for accurately measuring the amount of heat transferred, but by itself is not enough to validate the assumption. Therefore, the correct answer is "I and III only".
Ajụjụ 19 Ripọtì
If the time of flight is 96seconds, calculate the horizontal range through the point of projection.
Akọwa Nkọwa
Time of flight, T = 96s
R = (Ucosθ) *time* T = 640 × 96 = 61,440m
Ajụjụ 20 Ripọtì
The earth's gravitational field intensity at its surface is about
(G = 6.7 × 10−11 Nm2 /kg2 , mass of the earth is 6 × 1024 kg, radius of the earth is 6.4 × 106 m, g on the earth = 9.8m/s2 )
Akọwa Nkọwa
The earth's gravitational field intensity at its surface can be calculated using the formula: g = G * M / r^2 where G is the gravitational constant, M is the mass of the earth, r is the radius of the earth, and g is the gravitational field intensity at the surface of the earth. Substituting the given values, we get: g = (6.7 × 10^-11 Nm^2/kg^2) * (6 × 10^24 kg) / (6.4 × 10^6 m)^2 g = 9.8 N/kg (approx.) Therefore, the answer is 9.8N/kg.
Ajụjụ 21 Ripọtì
The lead-acid accumulator consists of
Akọwa Nkọwa
- the positive pole is lead peroxide (PbO2
)
- the negative pole is head
- the electrolyte is H2
SO4
Ajụjụ 22 Ripọtì
When a girl moves towards a plane mirror at a speed of 4.0m/s, the distance between the girl and her image reduces a speed of
Akọwa Nkọwa
| v | = | dt | or | v | α | d |
d = x, v = 4m/s
d = 2x, v = ? (girl and image)
| v | = | 2 × 4x | = | 8 | ms |
Ajụjụ 23 Ripọtì
The momentum of a car moving at a constant speed in a circular track
Akọwa Nkọwa
Movement of an object in a circle with an acceleration towards its center is provided by change in velocity and centripetal force a α V α Fc
Ajụjụ 25 Ripọtì
A well 1km deep is filled with a liquid of density 950kg/m3 and g = 10m/s2 , the pressure at the bottom of the well is
Akọwa Nkọwa
P = Pa + ρgh = (1.00 × 105
) + (950 × 10 × 1000)
P = 105
+ (95 × 105
) = 105
(1 + 95) = 96 × 105
P = 9.6 × 106
N/m2
Ajụjụ 26 Ripọtì
Which of the following media allow the transmission of sound waves through them?
I. air
II. liquid
III. solids
Akọwa Nkọwa
Sound waves are disturbances in a medium that propagate through the medium and transfer energy from one point to another. The transmission of sound waves depends on the physical properties of the medium, including its elasticity and density. Air (Option I) is a gas that is compressible and has a relatively low density, which makes it an excellent medium for transmitting sound waves. Liquids (Option II) are also able to transmit sound waves, although the speed of sound in liquids is slower than in gases because liquids are more dense and less compressible. Solids (Option III) are able to transmit sound waves as well, but their density and elasticity make them more rigid, which means that sound waves in solids tend to be transmitted as elastic waves or mechanical waves, rather than as acoustic waves. Therefore, the correct answer is "I, II, and III".
Ajụjụ 27 Ripọtì
Which of the following is/are the limitations to the Rutherford's atomic models?
I. It is applicable when energy is radiated as electrons are revolving
II. It is applicable when energy is radiated in a continuous mode
III. It is applicable to an atom with only one electron in the other shell
Akọwa Nkọwa
Rutherford assumed that (I) energy is radiated when electrons are revolving (II) energy is radiated in a continuous mode. These are limitations of Rutherford's model
Ajụjụ 28 Ripọtì
The distance between an object and its real image in a convex lens is 40cm. If the magnification of the image is 3, calculate the focal length of the lens
Akọwa Nkọwa
u + v = 40
vu = 3
v = 3u
u + 3u = 40
4u = 40
u = 10cm
v = 3u = 30cm
f = uvu+v=10(30)10+30=30040
= 7.5 cm
Ajụjụ 29 Ripọtì
Efficiency of conduction in liquids and gases compared to solids is
Akọwa Nkọwa
The efficiency of conduction in liquids and gases compared to solids is generally less efficient. This means that solids are better conductors of heat and electricity than liquids and gases. This is because the particles in solids are closely packed and are tightly bound to one another, allowing heat and electricity to flow easily through the material. On the other hand, the particles in liquids and gases are more spread out and less tightly bound, making it more difficult for heat and electricity to flow through these materials. However, it is important to note that the efficiency of conduction can vary depending on the specific liquid or gas and the specific solid being compared. Some liquids and gases may have properties that make them better conductors than certain solids, but this is not a general rule.
Ajụjụ 30 Ripọtì
Lamps in domestic lightings are usually in
Akọwa Nkọwa
Lamps in domestic lighting are usually connected in parallel. This means that each lamp is connected directly to the power supply, rather than being connected in a series or divergent or convergent configuration. In a parallel configuration, each lamp operates independently of the others, and if one lamp fails, the other lamps will continue to function. This is an important feature for domestic lighting, as it ensures that a single lamp failure will not leave the entire room in darkness. Additionally, in a parallel configuration, each lamp can be controlled independently, for example by a switch or dimmer, without affecting the operation of the other lamps. This allows for greater flexibility in lighting design and control. In summary, lamps in domestic lighting are usually connected in parallel because it allows for independent operation of each lamp and ensures that a single lamp failure does not affect the operation of the others.
Ajụjụ 31 Ripọtì
The mass of a nucleus is the
Akọwa Nkọwa
The mass of a nucleus is the total number of its protons and neutrons. The protons and neutrons are the subatomic particles that make up the nucleus of an atom. The mass of an atom is mostly concentrated in its nucleus, and the electrons orbiting the nucleus have a much smaller mass. Therefore, the mass of an atom is mostly determined by the number of protons and neutrons in its nucleus. The number of protons determines the element, and the number of neutrons can vary, resulting in isotopes of that element.
Ajụjụ 32 Ripọtì
In semi-conductor, the carriers of current at room temperature are
Akọwa Nkọwa
In a semiconductor, the carriers of current at room temperature are both electrons and holes. Semiconductors are materials with properties that are in between those of conductors (e.g. metals) and insulators (e.g. rubber). At room temperature, a semiconductor crystal contains both free electrons and positively charged vacancies called holes. When a voltage is applied across the semiconductor, the electrons move towards the positive end of the circuit and the holes move towards the negative end. This movement of charge carriers constitutes an electric current. In summary, both electrons and holes can carry current in a semiconductor at room temperature, making the correct answer.
Ajụjụ 33 Ripọtì
Aluminium is sometimes used as the leaf of an electroscope because it
Akọwa Nkọwa
- Aluminium can be made in thin sheet like Gold.
- the leaf is a thin material that can be diverged easily.
Ajụjụ 34 Ripọtì
Which of the following statements are correct of the production and propagation of waves?
I. vibration produces waves
II. waves transmit energy along the medium
III. the medium through which the wave travels does not travel with the wave
IV. waves do not require any medium for transmission
Akọwa Nkọwa
The correct statement is: I and II and III only. Explanation: - Statement I is correct because the production of waves involves some kind of disturbance that creates a vibration in the medium, which then propagates as a wave. - Statement II is correct because waves carry energy along the medium as they propagate. This is why waves can be used to transmit information or power over long distances. - Statement III is correct because the medium through which a wave travels does not move with the wave. Instead, the wave passes through the medium, causing it to oscillate or vibrate, but not to move along with the wave. - Statement IV is incorrect because most waves require a medium through which to propagate. For example, sound waves require air, water waves require water, and seismic waves require the Earth's crust. There are some types of waves, such as electromagnetic waves, that can propagate through a vacuum, but this is not true for all waves.
Ajụjụ 35 Ripọtì
If the attraction of the sun is suddenly ceased, the earth would continue to move in a straight line making a tangent with the original orbit. This statement is derived from Neutron's
Akọwa Nkọwa
The correct answer is the First law of motion. The First law of motion, also known as the law of inertia, states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. In this case, the earth is moving in its orbit around the sun because of the force of gravity between the two objects. If the force of gravity suddenly ceased, the earth would no longer be acted upon by an external force and would continue to move in a straight line, making a tangent with its original orbit. This idea is attributed to Sir Isaac Newton, who developed the laws of motion and the law of universal gravitation. However, the specific statement mentioned in the question is derived from the First law of motion.
Ajụjụ 36 Ripọtì
Radio waves belongs to the class of ware whose velocity is about
Akọwa Nkọwa
Radio waves belong to the class of waves whose velocity is approximately 3 x 10^8 m/s. This velocity is commonly denoted as the speed of light, which is the speed at which all electromagnetic waves, including radio waves, travel in a vacuum. This constant velocity is one of the fundamental principles of physics and is important in understanding the behavior and properties of light and other electromagnetic waves. The speed of light is incredibly fast, and it's difficult for us to imagine just how fast it is. To put it into perspective, light can travel around the Earth's equator almost 7.5 times in just one second. This high speed is essential for radio communication, as it enables radio waves to travel long distances in a short amount of time, allowing us to communicate with people and devices far away from us.
Ajụjụ 37 Ripọtì
The point at which the molecules of a loaded wire begin to slide across each other resulting in a rapid increase in extension is
Akọwa Nkọwa
The point at which the molecules of a loaded wire begin to slide across each other resulting in a rapid increase in extension is called the yield point. At this point, the material no longer behaves elastically and becomes permanently deformed. The yield point is an important parameter in material science and engineering as it indicates the maximum stress a material can withstand before it begins to deform plastically. Therefore, the yield point is a critical factor to consider when designing materials for specific applications.
Ajụjụ 38 Ripọtì
Neutrons were discovered by
Akọwa Nkọwa
Neutrons were discovered by James Chadwick. In 1932, he conducted an experiment in which he bombarded a thin sheet of beryllium with alpha particles. He observed that a new type of radiation was emitted that was not affected by electric or magnetic fields. He concluded that this radiation was composed of particles that were neutral and had a mass similar to that of a proton. He called these particles "neutrons," and his discovery revolutionized our understanding of atomic structure and led to the development of nuclear energy.
Ajụjụ 39 Ripọtì
The pin-hole camera produces a less sharply defined image when the
Akọwa Nkọwa
The pin-hole camera produces a less sharply defined image when the pin-hole is larger. A pin-hole camera works by allowing light to pass through a small hole (the pin-hole) and project an inverted image of the outside world onto a screen or surface located behind the hole. The smaller the pin-hole, the sharper the resulting image, as light passing through a smaller hole produces less diffraction or spreading out of the light. When the pin-hole is larger, more light enters the camera, but the light rays also become more scattered, resulting in a less well-defined image. This is because the larger opening allows more light rays to enter at different angles, creating a wider range of paths that the light can take as it travels through the camera and onto the screen. As a result, the image is less clear and less defined, with less sharp edges and more blurring. is the correct answer because it correctly identifies the effect of a larger pin-hole on the image produced by the pin-hole camera. less illumination, would actually produce a dimmer image, but it would not affect the sharpness or definition of the image. the distance of the screen from the pin-hole, and the distance of the object from the pin-hole, would affect the size of the image and the scale of the objects, but they would not affect the sharpness or definition of the image.
Ajụjụ 40 Ripọtì
A body moves in SHM between two point 20m on the straight line Joining the points. If the angular speed of the body is 5 rad/s. Calculate its speed when it is 6m from the center of the motion.
Akọwa Nkọwa
From two parts 20m apart
a = 10m, x = 6m, A = 5
V = ω√A2−X2
= 5√102−62
= 40m/s
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