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Ajụjụ 1 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 2 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 3 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 4 Ripọtì
A straight wire 15cm long, carrying a current of 6.0A is in a uniform field of 0.40T. What is the force on the wire when it is at right angle to the field
Akọwa Nkọwa
The force on a current-carrying wire in a uniform magnetic field can be calculated using the equation: F = BILsinθ where F is the force in Newtons, B is the magnetic field strength in Tesla, I is the current in Amperes, L is the length of the wire in meters, and θ is the angle between the wire and the magnetic field. In this problem, the wire is 15cm long (0.15m), carrying a current of 6.0A, and the magnetic field is 0.40T. The angle between the wire and the magnetic field is 90 degrees (since the wire is at right angles to the field). Substituting the given values into the equation, we get: F = (0.40T)(6.0A)(0.15m)sin90 sin90 = 1, so we can simplify the equation to: F = (0.40T)(6.0A)(0.15m) F = 0.36N Therefore, the force on the wire is 0.36N. Answer option C is the correct answer.
Ajụjụ 5 Ripọtì
The conductivity of gases at low pressure can be termed as
I. hot cathode emission
II. thermo ionic emission
III. cold cathode emission
IV. Field emission
Akọwa Nkọwa
As conduction of gases is at low pressure and high voltage, called field or cold cathode emission.
Ajụjụ 7 Ripọtì
Which of the following bodies, each with centre of gravity G, lying on a horizontal table, is/are in unstable equilibrium?
Akọwa Nkọwa
- I and II are in neutral equilibrium. They will roll continuously on the table
- III is a body with high centre of gravity (unstable)
- IV is a body with high centre of gravity (stable)
Ajụjụ 8 Ripọtì
The pitch of a screw jack is 0.45cm and the arm is 60cm long. If the efficiency of the Jack is 75/π %, calculate the mechanical advantage.
Akọwa Nkọwa
P = 0.45cm, L = 60cm, Eff = 75/π%
| VR | (Screw | system) | = | 2πrP | = | 2πLP |
| M.A | = | Eff% × VR100 | = | 75π | × | 1100 | × | 2π × 600.45 | = | 75 × 800300 | = | 200 |
Ajụjụ 9 Ripọtì
Which of the following is consistent with Charles' law?
I
II
III
IV.
Akọwa Nkọwa
This is the correct graph. The graph is volume against 1/ temperature where temperature is in Celsius.
Ajụjụ 10 Ripọtì
A boy pushes a 500kg box along a floor with a force of 2000N. If the velocity of the box is uniform, the co-efficient of friction between the box and the floor is
Akọwa Nkọwa
The coefficient of friction is a measure of the amount of friction between two surfaces. It is represented by the symbol "μ" and is a dimensionless quantity. The coefficient of friction between two surfaces depends on the nature of the surfaces in contact and the force pressing them together. In this problem, the boy is pushing the box with a force of 2000N. If the box is moving with a uniform velocity, then the force of friction acting on the box is equal and opposite to the pushing force applied by the boy. We can calculate the force of friction using the formula: frictional force = coefficient of friction x normal force where the normal force is the force exerted by the floor on the box in a direction perpendicular to the floor. Since the box is not moving up or down, the normal force is equal to the weight of the box. The weight of the box can be calculated using the formula: weight = mass x gravity where mass is the mass of the box and gravity is the acceleration due to gravity (9.8 m/s^2). So, the weight of the box is: weight = 500 kg x 9.8 m/s^2 = 4900 N The force of friction is equal to the pushing force of 2000N, so we can set these two equal to each other and solve for the coefficient of friction: frictional force = 2000N coefficient of friction x normal force = 2000N coefficient of friction x 4900N = 2000N coefficient of friction = 2000N / 4900N = 0.408 So, the coefficient of friction between the box and the floor is approximately 0.4. Therefore, the correct answer is 0.4.
Ajụjụ 11 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 12 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 13 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 14 Ripọtì
Water and Kerosine are drawn respectively into the two limbs of a Hare's apparatus. The destiny of water is 1.0gcm−3 and the density of kerosine is 0.80gcm−3 . If the height of the water column is 20.0cm, calculate the height of the kerosine column.
Akọwa Nkọwa
Devices with different liquids
d1
h1
= d2
h2
1 × 20 = 0.8 × h
| h | = | 200.8 | = | 25cm |
Ajụjụ 15 Ripọtì
Gases conduct electricity under
Akọwa Nkọwa
Gases conduct electricity under low pressure and high voltage
Ajụjụ 16 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 17 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 18 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 19 Ripọtì
When blue and green colours of light are mixed, the resultant colour is
Ajụjụ 20 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 21 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 22 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 23 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 25 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 26 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 27 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 28 Ripọtì
One newton × One meter equals?
Akọwa Nkọwa
One newton times one meter is equal to one Joule. A newton is the unit of measurement for force, and a meter is the unit of measurement for distance. When force is applied over a distance, work is done, which is measured in Joules. Therefore, one newton multiplied by one meter results in one Joule of work done. The other options listed (one water, one ampere, one kilogram) are not correct units of measurement for this calculation.
Ajụjụ 29 Ripọtì
A supply of 400V is connected across capacitors of 3μf and 6μf in series. Calculate the charge
Akọwa Nkọwa
| CT | = | C1 × C2 C1 + C2 |
| = | 3 × 63 + 6 |
= 189
= 2μf
Q = CV
⇒ 2 × 10−6
× 400
⇒ 800 × 10−6
C = 8 × 10−4
C
Ajụjụ 30 Ripọtì
The part of the human eye that does similar work as the diaphragm of a camera lens is the
Akọwa Nkọwa
The part of the human eye that does similar work as the diaphragm of a camera lens is the iris. The iris is the colored part of the eye and is responsible for controlling the amount of light that enters the eye. Just like the diaphragm in a camera lens, the iris can adjust its size to allow more or less light into the eye. This helps to regulate the amount of light reaching the retina, which is responsible for sensing light and transmitting the image to the brain.
Ajụjụ 31 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 32 Ripọtì
During the transformation of matter from the solid to the liquid state, the heat supplied does not produce a temperature increase because
Akọwa Nkọwa
During the transformation of matter from the solid to the liquid state, the heat supplied does not produce a temperature increase because all the heat is used to break the bonds holding the molecules of the solid together
Ajụjụ 33 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 34 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 35 Ripọtì
Heat may be transferred by conduction, convention and radiation. By which of these methods does heat travel through vacuum?
Akọwa Nkọwa
Heat can be transferred by conduction, convection, and radiation. Conduction is the transfer of heat through a material by the movement of heat-carrying particles, such as atoms or molecules, from one part of the material to another. This method of heat transfer is not possible in a vacuum, as there are no particles present to carry heat. Convection is the transfer of heat by the movement of a fluid, such as air or water. This method of heat transfer is also not possible in a vacuum, as there are no fluids present to carry heat. Radiation is the transfer of heat through electromagnetic waves, such as light or infrared radiation. This method of heat transfer does not require any material or fluid medium, and can therefore occur in a vacuum. Therefore, the answer is "Radiation only".
Ajụjụ 36 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 37 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 38 Ripọtì
Calculate the velocity ratio of a screw jack of pitch 0.2cm if the length of the tommy bar is 23cm
Akọwa Nkọwa
P = 0.2cm, L = r = 23cm
| VR | = | 2?rP | = | 2?LP | = | 2?×230.2 | = | 230? |
Ajụjụ 39 Ripọtì
The angular dispersion of a prism depends on
Akọwa Nkọwa
Dispersion is due to different refractive indices speeds and wavelengths.
Ajụjụ 40 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.
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