JAMB UTME - Physics - 2002
Ajụjụ 1 Ripọtì
Calculate the refractive index of the material for the glass prism in the diagram above
Akọwa Nkọwa
To calculate the refractive index of the material for the glass prism in the diagram, we need to use Snell's law. Snell's law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two media.
In this case, we can assume that the light is traveling from air into the glass prism. Let's call the angle of incidence θ1 and the angle of refraction θ2. We know that θ1 = 45°, and we can measure θ2 from the diagram.
Now we can use Snell's law:
sin(θ1)/sin(θ2) = n1/n2
where n1 is the refractive index of air, which is approximately 1, and n2 is the refractive index of the glass prism, which we want to find.
Rearranging the equation, we get:
n2 = n1*sin(θ1)/sin(θ2)
Plugging in the values we know, we get:
n2 = 1*sin(45°)/sin(θ2)
We can use trigonometry to find sin(θ2) based on the values in the diagram:
sin(θ2) = (1/2)*sin(90°-θ2) = (1/2)*sin(θ1) = (1/2)*(√2/2) = √2/4
Plugging this back into the equation, we get:
n2 = 1*sin(45°)/(√2/4) = 2/√2 = √2
Therefore, the refractive index of the material for the glass prism is √2.
Ajụjụ 2 Ripọtì
an electric cell with nominal voltage E has a resistance of 3Ω connected across it. If the voltage falls to 0.6E, the internal resistance of the cell is
Akọwa Nkọwa
I = V/R = 0.6E/3 = 0.2E
Again, I = E / (R+ϒ) = E / (3+ϒ)
Divide eqn (2) by eqn (1)
I/I = E / (3+ϒ) x I / 0.2E
=> 0.2(3+ϒ) = 1
0.6 x 0.2ϒ = 1
∴0.2ϒ = -0.6
=> ϒ = 0.6 / 0.2 = 3Ω
ϒ = 3Ω
Again, I = E / (R+ϒ) = E / (3+ϒ)
Divide eqn (2) by eqn (1)
I/I = E / (3+ϒ) x I / 0.2E
=> 0.2(3+ϒ) = 1
0.6 x 0.2ϒ = 1
∴0.2ϒ = -0.6
=> ϒ = 0.6 / 0.2 = 3Ω
ϒ = 3Ω
Ajụjụ 3 Ripọtì
A concave mirror of radius of curvature 40 cm forms a real image twice as large as the object.
The object distance is
Akọwa Nkọwa
if r = 40 cm , then f = r/2 = 40/2 = 20 cm
then for a real image by a concave mirror,
magnification M = f/u-f => 2 = 20/U-20
therefore 2(u-20) = 20
2u - 40 = 20
2u = 40 + 20
2u = 60/2 = 30 cm
then for a real image by a concave mirror,
magnification M = f/u-f => 2 = 20/U-20
therefore 2(u-20) = 20
2u - 40 = 20
2u = 40 + 20
2u = 60/2 = 30 cm
Ajụjụ 4 Ripọtì
a ray of light which strikes a glass slab from air at normal incidence passes through the slab
Akọwa Nkọwa
When a ray of light travels from air into a denser medium, such as a glass slab, it slows down and changes direction, which is called refraction. However, when the incident ray is perpendicular to the surface of the glass slab, it is called normal incidence, and the ray does not change direction, but only changes speed as it passes through the glass. Therefore, the answer is that the ray of light will be undeviated and undisplaced at a lower speed. So, the correct option is undeviated and undisplaced at a lower speed.
Ajụjụ 6 Ripọtì
The electromagnetic wave that can produce a heating effect on the environment is
Akọwa Nkọwa
Ajụjụ 7 Ripọtì
A wheel and axle is used to raise a load of 500 N by the application of an effort of 250 N. If the radii of the wheel and the axle are 0.4 cm and 0.1 cm respectively, the efficiency of the machine is
Akọwa Nkọwa
A wheel and axle is a simple machine consisting of two circular objects of different sizes. The wheel is the larger circular object and the axle is the smaller one that is connected to the wheel. The load is attached to the axle, and the effort is applied to the wheel to raise the load.
The efficiency of a machine is defined as the ratio of output work to input work, expressed as a percentage. In this case, the output work is the work done on the load, which is equal to the product of the load lifted and the height it is lifted to. The input work is the work done by the effort, which is equal to the product of the effort applied and the distance moved by the effort.
To calculate the output work, we need to determine the distance moved by the load when it is raised by one revolution of the wheel. This distance is equal to the circumference of the axle, which is given by 2πr, where r is the radius of the axle. Therefore, the distance moved by the load in one revolution of the wheel is 2π x 0.1 cm = 0.628 cm.
To calculate the input work, we need to determine the distance moved by the effort when the wheel makes one revolution. This distance is equal to the circumference of the wheel, which is given by 2πr, where r is the radius of the wheel. Therefore, the distance moved by the effort in one revolution of the wheel is 2π x 0.4 cm = 2.512 cm.
The load lifted is 500 N, and the effort applied is 250 N. The height to which the load is lifted is equal to the distance moved by the load in one revolution of the wheel, which is 0.628 cm. Therefore, the output work is 500 N x 0.628 cm = 314 J.
The work done by the effort is equal to the product of the effort applied and the distance moved by the effort, which is 250 N x 2.512 cm = 628 J.
The efficiency of the machine is the ratio of output work to input work, expressed as a percentage. Therefore, the efficiency of the machine is (314 J / 628 J) x 100% = 50%.
Therefore, the closest answer option to the efficiency of the machine is 50%.
Ajụjụ 8 Ripọtì
A body weighing 80N stands in an elevator that is about to move. The force exerted by floor on the body as the elevator moves upward with an acceleration of 5 ms-2 is
[g = 10ms-2]
Akọwa Nkọwa
Let M = mass of elevator
m = mass of the body
g = acceleration of free fall
a = upward acceleration of the elevator (5m/ss)
when the elevator is at rest, the force it exerts on the body is equal to the weight of the body. And from Newtons third law, Mg = mg
However as the elevator ascends the net upward force F = Mg - mg = (M + m)a.
Thus, if the weight of the body = 80 N, then 88 its mass, m = 80/10 = 8kg.
Therefore M x 10 - 8 x 10 = (M + 8) 5
10M - 80 = 5M + 40
10M - 5M = 40 x 80
5M = 120 => M = 120/5 = 24kg.
Thus, net upward force = force exerted by the elevator floor on the body = (M x m)a
= (24 x 8) x5
= 32 x 5
= 160 N
m = mass of the body
g = acceleration of free fall
a = upward acceleration of the elevator (5m/ss)
when the elevator is at rest, the force it exerts on the body is equal to the weight of the body. And from Newtons third law, Mg = mg
However as the elevator ascends the net upward force F = Mg - mg = (M + m)a.
Thus, if the weight of the body = 80 N, then 88 its mass, m = 80/10 = 8kg.
Therefore M x 10 - 8 x 10 = (M + 8) 5
10M - 80 = 5M + 40
10M - 5M = 40 x 80
5M = 120 => M = 120/5 = 24kg.
Thus, net upward force = force exerted by the elevator floor on the body = (M x m)a
= (24 x 8) x5
= 32 x 5
= 160 N
Ajụjụ 9 Ripọtì
The radiator of a motor car is cooled by
Akọwa Nkọwa
The radiator of a motor car is cooled by convection. When the engine is running, it produces heat that is transferred to the coolant fluid circulating through the engine block. This heated fluid then flows through the radiator, which is designed to dissipate the heat through the process of convection. In this process, the hot coolant transfers its heat to the metal fins of the radiator, which are in contact with the outside air. As the air flows over the fins, it absorbs the heat from the coolant and carries it away, cooling the engine. Radiation and conduction can also contribute to cooling, but the primary cooling mechanism for a car radiator is convection.
Ajụjụ 10 Ripọtì
The energy associated with the photon of a radio transmission at 3 x 105 Hz is
[h = 6.63 x 10-34Js]
Akọwa Nkọwa
A photon of energy E = hf
= 6.63 x 10-34 x 3 x 105
= 19.89 x 10-29
= 2.0 x 10-28J
= 6.63 x 10-34 x 3 x 105
= 19.89 x 10-29
= 2.0 x 10-28J
Ajụjụ 11 Ripọtì
Which of the following eye defects can be corrected using a cylindrical lens.
Akọwa Nkọwa
A cylindrical lens is typically used to correct astigmatism, which is a common eye condition that results in blurred vision due to an irregularly shaped cornea or lens. Unlike a spherical lens that has a uniform curvature, a cylindrical lens has different curvatures in two perpendicular planes, allowing it to bend light more in one direction than the other. This helps to compensate for the irregularities in the cornea or lens and to focus light more precisely on the retina, thereby improving vision clarity.
Myopia, also known as nearsightedness, is a condition where distant objects appear blurred, while nearby objects are clear. This condition can be corrected using a spherical lens that is concave (thinner in the center), which helps to diverge the light and focus it more accurately on the retina.
Presbyopia, on the other hand, is an age-related condition where the lens loses its flexibility, making it difficult to focus on nearby objects. This condition can be corrected using a bifocal or progressive lens that has different sections for distance and near vision.
Chromatic aberration is a condition where different colors of light are refracted differently, causing a blurred or distorted image. This condition can be corrected using a lens that has multiple refractive indices, such as an achromatic lens.
Therefore, the eye defect that can be corrected using a cylindrical lens is astigmatism.
Ajụjụ 12 Ripọtì
A coin place below a rectangular glass block of thickness 9cm and refractive index 1.5 is viewed vertically above the block. The apparent displacement of the coin is
Akọwa Nkọwa
The apparent displacement of the coin is 3 cm. The rectangular glass block with a thickness of 9 cm and a refractive index of 1.5 will refract light passing through it. When the coin is placed beneath the block and viewed vertically from above, the light from the coin will refract as it passes through the glass block, making the coin appear to be displaced from its original position. The amount of displacement will depend on the thickness of the glass block and the refractive index of the material. Using the formula for the apparent depth of an object submerged in a material with a different refractive index, we can calculate the displacement: Apparent depth = Actual depth / Refractive index In this case, the actual depth of the coin is the thickness of the glass block, which is 9 cm. The refractive index of the glass block is 1.5. Therefore, the apparent depth of the coin will be: Apparent depth = 9 cm / 1.5 = 6 cm However, this only calculates the apparent depth of the coin, not its apparent displacement. To find the displacement, we need to subtract the actual depth of the coin (which is 0 cm) from the apparent depth: Displacement = Apparent depth - Actual depth Displacement = 6 cm - 0 cm Displacement = 6 cm So the coin will appear to be displaced by 6 cm. However, since the coin is placed at the center of the block, the displacement will only be half of this value, or 3 cm, in any direction. Therefore, the correct answer is 3 cm.
Ajụjụ 13 Ripọtì
Pure silicon can be converted to a p-type by adding a controlled amount of
Akọwa Nkọwa
Pure silicon can be converted into a p-type semiconductor by adding a controlled amount of trivalent atoms.
A semiconductor can be classified into two types: p-type and n-type. A p-type semiconductor is created by adding impurities to pure silicon such that some of the silicon atoms are replaced by impurity atoms that have one less valence electron than silicon. These impurities are called acceptors since they accept electrons in the crystal structure.
The trivalent impurities, such as boron, aluminum, and indium, have three valence electrons, and when they are added to the pure silicon crystal, they create "holes" or vacant spaces where an electron is missing. These holes act as positive charge carriers in the semiconductor, and they are attracted to the negatively charged regions of the crystal.
By adding a controlled amount of trivalent impurities, the p-type semiconductor can be created with a certain level of conductivity. The impurities are added in such a way that the concentration of holes is much greater than the concentration of electrons, which results in the p-type semiconductor being a majority carrier hole semiconductor.
In contrast, the n-type semiconductor is created by adding pentavalent impurities, such as phosphorus, arsenic, or antimony, to the pure silicon crystal, which have one more valence electron than silicon. These impurities are called donors since they donate extra electrons to the crystal structure, which then act as negative charge carriers.
Therefore, the answer to the given question is option (A) trivalent atoms.
Ajụjụ 14 Ripọtì
blowing air over a liquid aids evaporation by
Akọwa Nkọwa
When a liquid evaporates, its molecules at the surface gain enough energy to escape the attractive forces of the liquid and become vapor. This process requires energy and is influenced by several factors, including temperature, pressure, and surface area. Blowing air over a liquid aids evaporation by increasing its surface area.
When air is blown over the surface of a liquid, it creates a flow of air that disturbs the boundary layer of the liquid, which is the layer of still air that surrounds the liquid surface. This movement of air creates turbulence at the surface, which in turn increases the surface area of the liquid that is exposed to the air. The increased surface area allows more molecules to escape from the liquid and become vapor, which increases the rate of evaporation.
The other options given in the question are not true for the reason why blowing air over a liquid aids evaporation. Blowing air over a liquid does not decrease its vapor pressure, which is the pressure exerted by the vapor of a substance in equilibrium with its liquid phase at a given temperature. Increasing the temperature of a liquid can increase the rate of evaporation, but it is not related to the process of blowing air over the liquid. Similarly, decreasing the density of a liquid does not aid evaporation.
Ajụjụ 15 Ripọtì
Which of the following is a characteristics of stationary waves?
Akọwa Nkọwa
Stationary waves are formed by the superposition of two waves of the same frequency and amplitude travelling in opposite directions. This means that the first characteristic listed is correct.
Stationary waves can be transverse or longitudinal, which is the second characteristic listed.
The distance between two successive nodes in a stationary wave is equal to half of the wavelength, not one wavelength as stated in the third option.
Finally, the antinode is a point of maximum displacement, not minimum displacement as stated in the fourth option.
In summary, the correct characteristics of stationary waves are: they are formed by two identical waves travelling in opposite directions, they can be transverse or longitudinal, the distance between two successive nodes is half of the wavelength, and the antinode is a point of maximum displacement.
Ajụjụ 16 Ripọtì
At what frequency would a capacitor of 2.5μF
Used in radio circuit have a resistance of 250Ω
Akọwa Nkọwa
| Capacitive Reactance = | ___1 |
| 2π + C |
| => 250 = | ____1 |
| 2 x π x f x 2.5 x 106 |
∴ 250/1 = 106/5.0xf
| => F = | ___106 |
| 250 x 5.0 x π |
| = | 106 |
| 1250π |
800/π
Ajụjụ 17 Ripọtì
If the distance between two suspended masses 10kg each is tripled, the gravitational force of attraction between them is reduced by
Akọwa Nkọwa
The gravitational force of attraction between two masses depends on their masses and the distance between them. According to Newton's law of gravitation, the force of attraction is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.
So, if the distance between two suspended masses 10kg each is tripled, the force of attraction between them will be reduced by a factor of 1/9 (1/3 squared). This means the gravitational force of attraction will decrease by 1/9th of its original value.
To understand this, imagine two magnets. The force of attraction between them decreases as they move away from each other. If you triple the distance between them, the force of attraction will be reduced to 1/9th of its original value. Similarly, the gravitational force of attraction between two masses follows the same principle, and the force decreases as the distance between them increases.
Ajụjụ 18 Ripọtì
The particle that is responsible for the nuclear fission in a nuclear reactor is
Akọwa Nkọwa
Nuclear fission is a process in which the nucleus of an atom is split into two or more smaller nuclei, along with the release of a large amount of energy. This process is typically initiated by the absorption of a neutron by the nucleus of an atom.
In a nuclear reactor, the fuel consists of heavy atoms, such as uranium-235 or plutonium-239, which are capable of undergoing nuclear fission when struck by a neutron. When a neutron is absorbed by the nucleus of an atom, the nucleus becomes unstable and splits into two smaller nuclei, along with the release of two or three neutrons and a large amount of energy.
These released neutrons can then go on to collide with other atoms and cause further fission reactions, leading to a chain reaction that releases a large amount of energy. In a nuclear reactor, the chain reaction is carefully controlled to ensure that the energy is released at a steady rate and that the reactor does not overheat or become unstable.
Therefore, the particle that is responsible for the nuclear fission in a nuclear reactor is a neutron.
Ajụjụ 19 Ripọtì
When connected to a mains of 250V, the fuse rating in the plug of an electric device of 1 KW is
Akọwa Nkọwa
The fuse rating in the plug of an electric device of 1 KW when connected to a mains of 250V can be calculated using the formula P = V x I, where P is the power in watts, V is the voltage in volts, and I is the current in amperes.
In this case, the power of the device is 1 KW (or 1000 watts), and the voltage is 250V. So we can rearrange the formula to find the current:
I = P / V = 1000 / 250 = 4 amps
Therefore, the correct fuse rating for this electric device would be 4 amps. This is because the fuse should be rated slightly higher than the expected current, to ensure that it can handle any momentary surges or spikes in current without blowing.
So, option (C) 4 A is the correct answer.
Ajụjụ 20 Ripọtì
The percentage of the original nuclei of the sample of a radioactive substance left after 5 half - lives is
Akọwa Nkọwa
The percentage of the original nuclei of a sample of a radioactive substance left after 5 half-lives can be calculated using the radioactive decay formula. For each half-life that passes, the amount of radioactive substance remaining is halved. After five half-lives, the amount of radioactive substance remaining is 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32 of the original amount.
To convert this to a percentage, we multiply by 100, giving 3.125%, which is closest to 3%. Therefore, the correct option is (c) 3%.
Ajụjụ 21 Ripọtì
The diagram above show s the force-extension curve of a piece of wire. The energy stored when the wire is stretched from E to F is
Akọwa Nkọwa
Energy store is equal to the work done in stretching the wire from E to F given by the area of the shaded trapezium under the graph.
= 1/2(0.1 x 0.2) x (0.1 - 0.05)
= 1/2(0.3) x 0.05
= 0.15 x 0.05
= 0.075
= 7.5 x 10-1J
= 1/2(0.1 x 0.2) x (0.1 - 0.05)
= 1/2(0.3) x 0.05
= 0.15 x 0.05
= 0.075
= 7.5 x 10-1J
Ajụjụ 22 Ripọtì
A particle in circular motion performs 30 oscillation in 6 seconds. Its angular velocity is
Akọwa Nkọwa
The frequency of oscillation is given by the number of oscillations divided by time taken for the oscillations, which is: f = number of oscillations / time In this case, the particle performs 30 oscillations in 6 seconds, so the frequency is: f = 30 / 6 = 5 Hz The angular velocity (ω) is related to frequency (f) by the formula: ω = 2πf Substituting the value of frequency (f) into the formula, we get: ω = 2π × 5 = 10π rad s^-1 Therefore, the correct option is: 10π rad s-1
Ajụjụ 23 Ripọtì
The electric field between two - point charges is shown in the diagram above. What types of charges are at X and Y?
Akọwa Nkọwa
Ajụjụ 24 Ripọtì
A bead travelling on a straight line wire is brought to rest at 0.2 by friction. If the mass of the bead is 0.01kg and the coefficient of friction between the bead and the wire is 0.1, determine the work one by the friction. [g = 10ms2]
Akọwa Nkọwa
Work done by friction = Friction force ×
displacement from the relation, f = μR
where μ = coefficient of friction ; R = normal reaction
=> F = μ mg (r = mg)
= 0.01 × 0.01 × 10
∴ work done = F × displacement
= 0.1 × 0.01 × 10 × 0.02
= 0.002
= 2 × 10-3
where μ = coefficient of friction ; R = normal reaction
=> F = μ mg (r = mg)
= 0.01 × 0.01 × 10
∴ work done = F × displacement
= 0.1 × 0.01 × 10 × 0.02
= 0.002
= 2 × 10-3
Ajụjụ 25 Ripọtì
the pressure of 3 moles of an ideal gas at a temperature of 27°C having a volume of 10-3m3 is
Akọwa Nkọwa
From the relation PV = nRT
V = 10-33; n = 3 moles; T = 27 + 273 = 300k
R = 8.3 j mol-1 K-1
=> P = nRT/V = 3 X 8.3 X 300/10-3 = 7.47 X 106 Nm-2
V = 10-33; n = 3 moles; T = 27 + 273 = 300k
R = 8.3 j mol-1 K-1
=> P = nRT/V = 3 X 8.3 X 300/10-3 = 7.47 X 106 Nm-2
Ajụjụ 27 Ripọtì
A bead traveling on a straight wire is brought to rest at 0.2m by friction. If the mass of the bead is 0.01kg and the coefficient of friction between the bead and the wire is 0.1 determine the work done by the friction
Akọwa Nkọwa
The work done by friction can be found by calculating the force of friction and multiplying it by the distance the bead traveled.
The force of friction can be found using the formula:
force of friction = coefficient of friction * normal force
where the normal force is the force exerted by the wire on the bead, which is equal and opposite to the weight of the bead.
normal force = weight of the bead = mass * gravity
where gravity is the acceleration due to gravity, approximately equal to 9.81 m/s^2.
So, the normal force is:
normal force = 0.01 kg * 9.81 m/s^2 = 0.0981 N
And the force of friction is:
force of friction = 0.1 * 0.0981 N = 0.00981 N
Now, to find the work done by friction, we multiply the force of friction by the distance traveled:
work done by friction = force of friction * distance traveled
In this case, the distance traveled is 0.2 m. So:
work done by friction = 0.00981 N * 0.2 m = 0.001962 J
Therefore, the work done by friction is 2 × 10^-3 J.
Ajụjụ 28 Ripọtì
If tension is maintained on a stretched string of length 0.6m, such that its fundamental frequency of 220Hz is excited, determine the velocity of the transverse wave in the string.
Akọwa Nkọwa
The velocity of a wave traveling along a stretched string can be calculated using the formula:
v = fλ
where v is the velocity, f is the frequency of the wave, and λ is the wavelength.
For a stretched string, the fundamental frequency is given by:
f1 = (1/2L)√(T/μ)
where L is the length of the string, T is the tension in the string, and μ is the linear density of the string.
We can rearrange this equation to solve for the linear density:
μ = (T/4)(L/f1)^2
Now we can substitute this expression for μ into the wavelength equation:
λ = 2L/n
where n is the harmonic number (in this case, n = 1 for the fundamental frequency).
Substituting for μ and λ in the velocity equation, we get:
v = f1(2L/1) = 2f1L
So we just need to plug in the given values:
L = 0.6 m
f1 = 220 Hz
v = 2(220 Hz)(0.6 m) = 264 m/s
Therefore, the velocity of the transverse wave in the string is 264 m/s.
is the correct answer.
Ajụjụ 29 Ripọtì
From the diagram above, determine the value of the resistance X
Akọwa Nkọwa
Let the combined parallel resistance of 3 Ω and x = Q Ω
∴ Total resistance in the circuit = 2 + Q + 1.5 = 3.5 + Q Ω
Total current in the circuit = 2A
P.D a cross the circuit 12 volts
∴ From Ohm’s law, V = IR
12 = 2(3.5 + Q)
= 7.0 + 2Q
∴ 2Q = 12 – 7
= 5
∴ Q = 5/2 = 2.5 Ω
Thus from parallel formula: I/Q = 1/3 + 1/x
1/2.5 = 1/3 + 1/x
∴ 1/x = 1/2.5 - 1/3 = 3-2.5/7.5 = 0.5/7.5
X = 7.5/0.5 = 75/5 = 15 Ω
∴ X = 15 Ω
∴ Total resistance in the circuit = 2 + Q + 1.5 = 3.5 + Q Ω
Total current in the circuit = 2A
P.D a cross the circuit 12 volts
∴ From Ohm’s law, V = IR
12 = 2(3.5 + Q)
= 7.0 + 2Q
∴ 2Q = 12 – 7
= 5
∴ Q = 5/2 = 2.5 Ω
Thus from parallel formula: I/Q = 1/3 + 1/x
1/2.5 = 1/3 + 1/x
∴ 1/x = 1/2.5 - 1/3 = 3-2.5/7.5 = 0.5/7.5
X = 7.5/0.5 = 75/5 = 15 Ω
∴ X = 15 Ω
Ajụjụ 30 Ripọtì
The energy stored in a capacitor of capacitance 10μF carrying a charge of 100μC
Akọwa Nkọwa
The energy stored in a capacitor is given by the formula:
E = 1/2 * C * V^2
where E is the energy stored, C is the capacitance of the capacitor, and V is the voltage across the capacitor.
Since the capacitor is carrying a charge of 100μC, the voltage across the capacitor can be calculated using the formula:
V = Q/C
where Q is the charge on the capacitor and C is the capacitance.
Substituting the given values, we get:
V = Q/C = 100μC/10μF = 10V
Now substituting the values of C and V in the energy formula, we get:
E = 1/2 * C * V^2 = 1/2 * 10μF * (10V)^2 = 500μJ
So the energy stored in the capacitor is 500μJ, which is equivalent to 5 x 10^-4 J (in scientific notation). Therefore, the answer is option (D) 5 x 10^-4 J.
Ajụjụ 31 Ripọtì
To produce an enlarged and erect image with a concave mirror,the object must be positioned
Akọwa Nkọwa
To produce an enlarged and erect image with a concave mirror, the object must be positioned between the principal focus and the pole of the mirror.
A concave mirror is a curved mirror that has a reflecting surface that curves inward like a spoon. It is also called a converging mirror because it converges the parallel incident rays of light to a single point known as the focus.
When an object is placed in front of a concave mirror, the light rays coming from the object reflect off the mirror and converge to form an image. Depending on the position of the object relative to the mirror, the image can be real or virtual, erect or inverted, and magnified or reduced.
If the object is placed at the principal focus of the mirror, the reflected rays of light will be parallel to each other, and the image will be formed at infinity. This image is real, inverted, and highly diminished.
If the object is placed beyond the center of curvature, the image formed will be real, inverted, and diminished.
If the object is placed between the principal focus and the center of curvature, the image formed will be real, inverted, and magnified.
Finally, if the object is placed between the principal focus and the pole of the mirror, the image formed will be virtual, erect, and magnified.
Thus, to produce an enlarged and erect image with a concave mirror, the object must be positioned between the principal focus and the pole of the mirror. Therefore, the answer to the given question is option (D) between the principal focus and the pole.
Ajụjụ 32 Ripọtì
An electric iron is rated 1000W, 230V. What is the resistance of its elements?
Akọwa Nkọwa
The formula for calculating resistance is R = V/I, where R is the resistance in ohms, V is the voltage in volts, and I is the current in amperes.
First, we need to find the current flowing through the iron. We can use the formula P = VI, where P is the power in watts. We know that the power of the iron is 1000W and the voltage is 230V, so we can rearrange the formula to find the current:
I = P/V = 1000/230 = 4.35A
Now that we have the current, we can use the formula R = V/I to find the resistance. We know the voltage is 230V, so:
R = V/I = 230/4.35 = 52.9Ω
Therefore, the resistance of the iron's heating element is 52.9Ω. Answer is correct.
Ajụjụ 33 Ripọtì
A transistor is used in the amplification of signals because it
Akọwa Nkọwa
A transistor is a semiconductor device that is used in the amplification and switching of electronic signals. It is used in electronic circuits to control the flow of current, and it can be thought of as an electrically controlled switch.
The reason why a transistor is used in the amplification of signals is that it can control the amount of current flowing through it based on a small input signal. This allows a weak input signal to be amplified to a larger output signal. The input signal is used to control the amount of current flowing through the transistor, which in turn controls the output signal. This makes the transistor a key component in many electronic devices, including radios, televisions, and computers.
The other options given in the question are not true for the reason why a transistor is used in the amplification of signals. A transistor does not consume a lot of power, but instead can be designed to operate at low power levels. Doping is a process used in the manufacture of transistors to create n-type and p-type semiconductors, but it is not the reason why a transistor is used in the amplification of signals. Finally, while a transistor does contain electron and hole carriers, this is not the reason why it is used in the amplification of signals.
Ajụjụ 34 Ripọtì
I. Temperature.
II. Density of air molecules
III. Pressure
IV. Pitch
Which of the above will affect the velocity of sound in the air
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
A copper cube weighs 0.25 N in air, 0.17N when completely immersed in paraffin oil and 0.15 N when completely immersed in water.The ratio of upthrust in oil to upthrust in water is
Akọwa Nkọwa
The upthrust on a submerged object is equal to the weight of fluid displaced by the object. The weight of an object in a fluid is given by the difference between the weight of the object in air and the upthrust on the object in that fluid.
Let's call the weight of the copper cube in air W. Then, the upthrust in paraffin oil is (W - 0.17 N), and the upthrust in water is (W - 0.15 N).
Now, we can use the fact that the ratio of upthrust in oil to upthrust in water is equal to the ratio of the weight of water displaced to the weight of oil displaced. Let's call the weight of water displaced Ww and the weight of oil displaced Wo.
Then, we have:
(W - 0.17 N)/(W - 0.15 N) = Ww/Wo
Simplifying this expression by cross-multiplying, we get:
(W - 0.17 N) * Wo = (W - 0.15 N) * Ww
Expanding the brackets, we get:
Wo * W - 0.17 N * Wo = Ww * W - 0.15 N * Ww
Rearranging this expression, we get:
(Ww/Wo) = (W - 0.17 N)/(W - 0.15 N)
Substituting the given values, we get:
(Ww/Wo) = (0.25 N - 0.17 N)/(0.25 N - 0.15 N) = 0.08 N/0.10 N = 4/5
Therefore, the ratio of upthrust in oil to upthrust in water is 4:5, which is option (B).
Ajụjụ 36 Ripọtì
The diagram above shows the force (F) acting on an object through a distance (x). The work done on this object is expressed as
Akọwa Nkọwa
The work done on this object is given by the area of the triangle, which is equal to 1/2 fx
Ajụjụ 37 Ripọtì
A current of 0.5 A flowing 3h Deposits 2g of metal during electrolysis. The quantity of the same metal that would be deposited by the currents of 1.5A flowing in 1h is
Akọwa Nkọwa
From Faradays Law, mass deposited during electrolysis is given by
M = ItZ
=> Z = M/(It)
For a current of 1.5A flowing for 1h, the mass M, deposited is given by
M= ZiT
= 2 / (0.5x3x60x60) x 0.5x1x60x60 / 1
= 2.0g
M = 2.0g
M = ItZ
=> Z = M/(It)
| = | __2 |
| 0.5x3x60x60 |
For a current of 1.5A flowing for 1h, the mass M, deposited is given by
M= ZiT
= 2 / (0.5x3x60x60) x 0.5x1x60x60 / 1
= 2.0g
M = 2.0g
Ajụjụ 38 Ripọtì
Two force each of 10N acts on a body, one towards the north and the other towards the east.
The magnitude and the direction of the resultant force are
Akọwa Nkọwa
The problem gives us two forces acting on a body: one 10N force towards the north and the other 10N force towards the east. To find the magnitude and direction of the resultant force, we can use vector addition. We can draw a diagram where the two forces are represented as arrows with their tails at the same point. Using the Pythagorean theorem, we can find the magnitude of the resultant force by finding the hypotenuse of the right triangle formed by the two forces. $R = \sqrt{(10N)^2 + (10N)^2} = 10\sqrt{2}N$ To find the direction of the resultant force, we can use trigonometry. We can find the angle between the resultant force and the x-axis (east) by taking the inverse tangent of the ratio of the y-component (north) to the x-component (east) of the resultant force. $\theta = \tan^{-1}(\frac{10N}{10N}) = 45°$ Therefore, the magnitude and direction of the resultant force are 10√2N and 45°E. The answer is: 10√2N, 45°E.
Ajụjụ 40 Ripọtì
Which of the following statements is true of the properties of a moving coil galvanometer?
Akọwa Nkọwa
Ajụjụ 41 Ripọtì
| 23 | Na + X → | 20 | F + | 4 | He | |
| 11 | 9 | 2 |
What particle is X in the reaction above?
Akọwa Nkọwa
The particle in the reaction is neutron
| 23 | Na + | 1 | n → | 20 | F + | 4 | He |
| 11 | 0 | 9 | 2 |
Ajụjụ 42 Ripọtì
The carbon-granules microphone works on the principle of change in
Akọwa Nkọwa
The carbon-granules microphone works on the principle of change in resistance. When sound waves enter the microphone, they cause the carbon granules to vibrate, which changes the resistance of the granules. This change in resistance produces an electrical signal that corresponds to the sound wave, which can then be amplified and transmitted through a speaker.
Ajụjụ 43 Ripọtì
Vibration in a stretched spring cannot be polarized because they are
Akọwa Nkọwa
Polarization is a property that applies to transverse waves, where the direction of wave oscillation is perpendicular to the direction of wave propagation. Longitudinal waves, such as vibrations in a stretched spring, have oscillations that are parallel to the direction of wave propagation and therefore cannot be polarized. Thus, vibrations in a stretched spring cannot be polarized because they are longitudinal waves.
Ajụjụ 44 Ripọtì
The hydrostatic blood pressure difference between the head and the feet of a boy standing straight is 1.65 x 104 Nm-2. Find the height of the boy.
Akọwa Nkọwa
[ Density of blood = 1.1 x 103 kgm-3, g = 10ms-2]
Pressure = pgh, where p = density, h = the height, g = acceleration of free fall due to gravity
1.6.5 x 104 = 1.1 x 103 x 10 x h
therefore h = 1.65 x 104/1.1 x 103 x 10 = 1.5m
Pressure = pgh, where p = density, h = the height, g = acceleration of free fall due to gravity
1.6.5 x 104 = 1.1 x 103 x 10 x h
therefore h = 1.65 x 104/1.1 x 103 x 10 = 1.5m
Ajụjụ 45 Ripọtì
the property that is propagated in a traveling wave is
Akọwa Nkọwa
The property that is propagated in a traveling wave is energy.
A traveling wave is a disturbance that propagates through space and time, carrying energy from one place to another without transporting any matter. Waves are characterized by several properties, such as amplitude, frequency, wavelength, and velocity.
Amplitude refers to the maximum displacement of the particles in the medium from their equilibrium position. Frequency is the number of oscillations or cycles of the wave that occur in one second. Wavelength is the distance between two consecutive points on the wave that are in phase.
However, the most fundamental property of a wave is its energy. Waves transport energy from one point to another by transferring the energy of their oscillations to the particles in the medium. This energy is propagated along the direction of the wave and can be transformed into different forms, such as mechanical, electrical, or electromagnetic energy.
For example, when a sound wave travels through the air, it transfers energy from the vibrating source to the air particles, causing them to vibrate and transmit the sound energy to our ears. Similarly, when an electromagnetic wave, such as light, travels through space, it carries energy from the emitting source and can be transformed into electrical energy when it interacts with matter.
Therefore, the property that is propagated in a traveling wave is energy. Hence, the answer to the given question is option (C) energy.
Ajụjụ 46 Ripọtì
The eye controls the amount of light reaching the retina by adjusting the
Akọwa Nkọwa
The sentence is stating that there is a part of the eye responsible for controlling the amount of light that reaches the retina. This part must be able to adjust and regulate the amount of light that passes through the eye. The options provided are: cornea, iris, retina, and optic nerve. Based on this information, the correct option is iris, because the iris is the part of the eye that controls the size of the pupil and regulates the amount of light entering the eye.
Ajụjụ 47 Ripọtì
The effect of a particle in a fluid attaining its terminal velocity is that the
Akọwa Nkọwa
When a particle falls through a fluid, it experiences a resistance force due to the friction between the particle and the fluid. This resistance force increases as the velocity of the particle increases, until it reaches a point where the resistance force is equal to the force of gravity on the particle.
At this point, the particle stops accelerating and falls at a constant velocity called the terminal velocity. Therefore, when a particle attains its terminal velocity in a fluid, the acceleration becomes zero, and the weight of the particle is balanced by the retarding force due to the viscosity of the fluid.
This means that "weight is equal to the retarding force," is the correct answer. "acceleration is maximum," is incorrect because the particle has stopped accelerating. "buoyancy force is equal to the viscous retarding force," is not necessarily true, as the buoyancy force may be negligible in some situations. "buoyancy force is more than the weight of the fluid displaced," is also not necessarily true, as the buoyancy force may be less than the weight of the fluid displaced in some situations.
Ajụjụ 49 Ripọtì
If the total force acting on a particle is zero, the linear momentum will
Akọwa Nkọwa
The correct option is "Be Constant."
The sentence states that if the total force acting on a particle is zero, the linear momentum of the particle will remain constant. Linear momentum is a physical quantity that describes the motion of an object and is equal to the product of its mass and velocity. According to Newton's first law of motion, an object will remain in a state of rest or uniform motion in a straight line unless acted upon by an external force.
When the total force acting on a particle is zero, there is no external force acting on the particle. Therefore, the particle will continue to move at a constant velocity in a straight line, and its linear momentum will remain constant. This means that the momentum will not increase, decrease, or change in any way. Thus, the correct option is "Be Constant."
Ajụjụ 50 Ripọtì
The phenomenon that makes sound persist when its source has been removed is known as
Akọwa Nkọwa
When a sound is produced in a room or an enclosed space, it travels and bounces off the walls, ceiling, and floor, creating multiple reflections. These reflections of sound waves mix with the original sound waves, creating a series of echoes that our ears perceive as reverberation. Therefore, the correct answer is "reverberation."
