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Ajụjụ 1 Ripọtì
During the transformation of matter from the solid to the liquid state, the heat supplied does not produce temperature increase because
Akọwa Nkọwa
When a solid is heated to its melting point, the heat supplied is used to overcome the intermolecular forces holding the molecules in a fixed position, resulting in the breaking of these bonds. As a result, the solid transforms into a liquid without any change in temperature. This is because the heat energy supplied is used in breaking the bonds between molecules rather than increasing the kinetic energy of the molecules, which is what causes an increase in temperature. Therefore, the correct option is: "all the heat is used to break the bonds holding the molecules of the solid together."
Ajụjụ 2 Ripọtì
The limiting frictional force between two surfaces depends on
I. the normal reaction between the surfaces
II. the area of surface in contact
III. the relative velocity between the surfaces
IV. the nature of the surfaces
Akọwa Nkọwa
- Friction depends on the nature of the surfaces in contact
- Solid friction is independent of the area of the surfaces in contact and the relative velocity between the surfaces.
Ajụjụ 3 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 4 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 5 Ripọtì
The resultant capacitance in the figure above is
Akọwa Nkọwa
For the parallel arrangement = 2 + 4 = 6μf
| For | the | series | arrangement | = | 1CT | = | 12 | + | 13 | + | 16 | + | 14 |
| 1CT | = | 1512 |
| CT | = | 1215 | = | 0.8μf |
Ajụjụ 6 Ripọtì
Which of the following characteristics of a wave is used in the measurement of the depth of the Sea?
Akọwa Nkọwa
Depth of sea can be measured by echo, a reflected sound waves.
Ajụjụ 7 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 8 Ripọtì
- angle of dip is zero at the magnetic equator
- angle of variation is the same as angle of declination.
Akọwa Nkọwa
- angle of dip is zero at the magnetic equator
- angle of variation is the same as angle of declination.
Ajụjụ 9 Ripọtì
A train has an initial velocity of 44m/s and an acceleration of -4m/s2 . Calculate its velocity after 10 seconds
Akọwa Nkọwa
The velocity of the train after 10 seconds can be calculated using the formula: v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Substituting the given values, we get: v = 44 m/s + (-4 m/s^2) x 10 s v = 44 m/s - 40 m/s v = 4 m/s Therefore, the velocity of the train after 10 seconds is 4m/s. Answer option D is correct. Explanation: The train has an initial velocity of 44 m/s and an acceleration of -4 m/s^2. The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which means that the train is slowing down. After 10 seconds, the train's velocity decreases by 40 m/s (4 m/s^2 x 10 s) to reach a final velocity of 4 m/s.
Ajụjụ 10 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 but at the same time undergoes an acceleration, its motion is called rectilinear motion. This means that the body moves in a straight line and its speed changes at a constant rate, causing an acceleration. It is different from oscillation, circular and rotational motions which involve changes in direction, as well as changes in speed.
Ajụjụ 11 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 12 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
Ajụjụ 13 Ripọtì
The diagram shows four positions of the bob of a simple pendulum. At which of these positions does the bob have maximum kinetic energy and minimum potential energy
Akọwa Nkọwa
At position 1, the bob of the simple pendulum has the maximum potential energy and zero kinetic energy. At position 4, the bob has the maximum kinetic energy and minimum potential energy. To understand this, we need to know that the energy of a simple pendulum is converted back and forth between kinetic energy and potential energy as it swings back and forth. When the bob is at its highest point (position 1), it has the maximum potential energy because it is farthest from the ground and has the most potential to move downward. At this point, the bob has zero kinetic energy because it is momentarily at rest. As the bob swings downward towards the equilibrium point, it gains speed and its potential energy is converted to kinetic energy. At the equilibrium point (position 2), the bob has equal amounts of kinetic and potential energy. As the bob continues to move downward, its potential energy decreases and its kinetic energy increases. At position 3, the bob has minimum potential energy and some amount of kinetic energy. At the lowest point of its swing (position 4), the bob has maximum kinetic energy because it is moving at its fastest speed. At this point, the bob has minimum potential energy because it is closest to the ground and has the least amount of potential to move downward. So, to summarize, the bob has maximum potential energy at position 1, equal amounts of kinetic and potential energy at position 2, minimum potential energy at position 3, and maximum kinetic energy at position 4.
Ajụjụ 14 Ripọtì
A mass of 0.5kg is whirled in a vertical circle of radius 2m at a steady rate of 2 rev/s. Calculate the centripetal force
Akọwa Nkọwa
The centripetal force is the force that acts towards the center and keeps an object moving in a circular path. To calculate the centripetal force, we can use the following formula: f = m * v^2 / r where: - f = centripetal force - m = mass of the object (0.5 kg) - v = velocity of the object (2 rev/s * 2 * pi m/rev = 12.57 m/s) - r = radius of the circle (2 m) Plugging in the values, we get: f = 0.5 kg * 12.57 m/s^2 / 2 m f = 31.43 N Rounding to the nearest whole number, the centripetal force is 31 N. So, the closest answer from the options is 160N.
Ajụjụ 15 Ripọtì
The equilibrium position of objects in any field corresponds to situation of
Akọwa Nkọwa
The equilibrium position of an object in any field corresponds to the situation of minimum potential energy. This means that at the equilibrium position, the object has the lowest possible potential energy within the field. In other words, the forces acting on the object are balanced, and the object is not being pushed or pulled in any direction. Therefore, the object will remain at rest at the equilibrium position unless it is acted upon by an external force. Of the options given, the correct answer is "minimum potential energy".
Ajụjụ 16 Ripọtì
A siren having a ring of 200 hole makes 132 rev/min. A jet of air is directed on the set of holes. Calculate the frequency and wavelength in air of the note produced (take v = 350m/s)
Akọwa Nkọwa
n = 200, S = 132 rev/min, v = 350m/s2
| f | = | ns | = | 200 | × | 132 | revmin | × | 1min60s | = | 440Hz |
| λ | = | vf | = | 350440 | = | 0.875m |
Ajụjụ 17 Ripọtì
The diagram above represents the stress-strain graph of a loaded wire. Which of these statements is correct?
Akọwa Nkọwa
- I is the elastic limit
- the end of the constant part J is the yield point
- L is the break point.
Ajụjụ 18 Ripọtì
A microscope is focused on a mark on a table, when the mark is covered by a plate of glass 2m thick, the microscope has to be raised 0.67cm for the mark to be once more in focus. Calculate the refractive index.
Akọwa Nkọwa
R = th = 2cm, d = 0.67cm
| n | = | RA | = | RR.d | = | 22-0.67 | = | 1.52 |
Ajụjụ 19 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 20 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 21 Ripọtì
When two objects A and B are supplied with the same quantity of heat, the temperature change in A is obtained to be twice that of B. The mass of P is half that of Q. The ratio of the specific heat capacity of A to B is
Akọwa Nkọwa
θA = 2θB ,
| mA | = | 12 | mB |
H = MCθ
mA
cA
θA
= mB
cB
θB
( 1/2 mB
)CA
(2θB
) = mB
cB
θB
| CA CB | = | 11 |
⇒ 1 : 1
Ajụjụ 22 Ripọtì
In the molecular explanation of conduction, heat is transferred by the
Akọwa Nkọwa
In the molecular explanation of conduction, heat is transferred by the Free electrons. In metals, free electrons move randomly and collide with other particles as they gain kinetic energy. These free electrons transfer the energy to the adjacent particles, which in turn gain kinetic energy and transmit it to other adjacent particles, thus transferring heat energy from one part of the material to another. This process of heat transfer by free electrons is called conduction. Therefore, the correct option is "Free electrons."
Ajụjụ 23 Ripọtì
Gases conduct electricity under
Akọwa Nkọwa
Gases conduct electricity under low pressure and high voltage
Ajụjụ 24 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 25 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 26 Ripọtì
When blue and green colours of light are mixed, the resultant colour is
Ajụjụ 27 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 29 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 30 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 31 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 32 Ripọtì
Three resistors with resistance 200Ω, 500Ω and 1kΩ are connected in series. A 6v battery is connected to either end of the combination. Calculate the potential difference between the ends of 200Ω resistance.
Akọwa Nkọwa
To calculate the potential difference between the ends of the 200Ω resistance, we need to use Ohm's Law, which states that the potential difference (V) across a resistor is equal to the current (I) flowing through the resistor multiplied by the resistance (R) of the resistor. First, we need to find the total resistance of the series combination of resistors. We add up the individual resistances: Total resistance = 200Ω + 500Ω + 1kΩ = 1.7kΩ Next, we can use Ohm's Law to find the current flowing through the circuit. We know that the battery voltage is 6V, and the total resistance is 1.7kΩ: I = V / R = 6V / 1.7kΩ = 0.0035A Now we can use Ohm's Law again to find the potential difference across the 200Ω resistor: V = IR = 0.0035A * 200Ω = 0.7V Therefore, the potential difference between the ends of the 200Ω resistance is 0.7V. The correct answer is option B.
Ajụjụ 33 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 34 Ripọtì
The limiting frictional force between two surface depends on
I. the normal reaction between the surfaces
II. the area of surface in contact
III. the relative velocity between the surfaces
IV. the nature of the surface
Akọwa Nkọwa
The correct answer is "I and IV only". The limiting frictional force between two surfaces depends on the normal reaction between the surfaces (I) and the nature of the surface (IV). The normal reaction is the force that the surfaces exert on each other perpendicular to the plane of contact. The greater the normal reaction, the greater the frictional force that can be applied before motion occurs. The nature of the surface is determined by factors such as roughness, hardness, and texture, which can affect the frictional force. The area of surface in contact (II) does not directly affect the limiting frictional force, although it can affect the force required to initiate motion. For example, if the area of contact is small, the pressure between the surfaces will be higher, making it harder to initiate motion. The relative velocity between the surfaces (III) also does not directly affect the limiting frictional force, although it can affect the force required to maintain motion. If the surfaces are already in motion, a lower force may be required to keep them moving than to initiate motion. In summary, the limiting frictional force between two surfaces depends primarily on the normal reaction and the nature of the surface, and is not directly affected by the area of contact or the relative velocity between the surfaces.
Ajụjụ 35 Ripọtì
A metal rod has a length of 100cm at 200oC . At what temperature will its length be 99.4cm. If the linear expansivity of the material of the rod is 2 × 10−5C−1
Akọwa Nkọwa
The linear expansivity of a material describes how its length changes with temperature. If the linear expansivity is given as 2 × 10^-5/°C, this means that for every 1°C change in temperature, the length of the material will change by 2 × 10^-5 times its original length. Given that the rod has a length of 100 cm at 200°C, we can use this information to find its length at a different temperature. If we let L be the length of the rod at temperature T, we can write the relationship as follows: L = 100 cm * (1 + 2 × 10^-5 * (T - 200°C)) To find the temperature at which the rod will have a length of 99.4 cm, we can set L equal to 99.4 cm and solve for T: 99.4 cm = 100 cm * (1 + 2 × 10^-5 * (T - 200°C)) 99.4 cm / 100 cm = 1 + 2 × 10^-5 * (T - 200°C) 0.994 = 1 + 2 × 10^-5 * (T - 200°C) -0.006 = 2 × 10^-5 * (T - 200°C) -0.006 / 2 × 10^-5 = T - 200°C -0.006 / (2 × 10^-5) = T - 200°C -0.006 / (2 × 10^-5) + 200°C = T So the temperature at which the rod will have a length of 99.4 cm is approximately equal to -0.006 / (2 × 10^-5) + 200°C, or -100°C. Therefore, the answer is -100°C.
Ajụjụ 36 Ripọtì
In a slide wire bridge, the balance is obtained at a point 25cm from one end of wire 1m long. The resistance to be tested is connected to that end and a standard resistance of 3.6Ω is connected to the other end of the wire. Determine the value of the unknown resistance
Akọwa Nkọwa
R3.6=7525=13
3R = 3.6
R = 1.2Ω
Ajụjụ 37 Ripọtì
According to kinetic molecular model, in gases
Akọwa Nkọwa
According to the kinetic molecular model, in gases, the molecules are very fast apart and occupy all the space made available. This means that gas molecules are in constant random motion and they move freely in all directions without any regular arrangement. They collide with each other and with the walls of the container, exerting pressure. The temperature of the gas is related to the average kinetic energy of the gas molecules. The higher the temperature, the faster the gas molecules move, and the higher the kinetic energy.
Ajụjụ 38 Ripọtì
An alternating current can induce voltage because it has
Akọwa Nkọwa
An alternating current can induce voltage because it has a varying magnetic field. An alternating current (AC) is an electrical current that periodically reverses direction, unlike direct current (DC), which flows in one direction. When an AC current flows through a wire, it generates a magnetic field that changes direction with the current. As the current alternates, the magnetic field expands and contracts, inducing an electromotive force (EMF) in any nearby conductor or coil of wire. This phenomenon is known as electromagnetic induction, and it is the basis for the operation of many electrical devices, such as generators and transformers. The induced voltage depends on the strength and rate of change of the magnetic field and the number of turns in the coil. In summary, an alternating current can induce voltage because it creates a varying magnetic field, which in turn generates an electromotive force in nearby conductors or coils of wire, according to the principle of electromagnetic induction.
Ajụjụ 39 Ripọtì
Which of the following statement about the electromagnet shown above is correct?
Akọwa Nkọwa
A - B = S - N.
Also, starting end of the current is south while terminating end is North.
Ajụjụ 40 Ripọtì
A cone is in unstable equilibrium has its potential energy
Akọwa Nkọwa
In unstable equilibrium, potential energy decreases as the height decreases.
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