JAMB UTME - Physics - 2018
Tambaya 1 Rahoto
A man hears his echo from a nearby hill 2s after he shouted. If the frequency of his voice is 260Hz and the wavelength is 1.29m, how far away is the hill
Bayanin Amsa
The speed of sound in air is approximately 343 meters per second at room temperature. The formula for the speed of sound is:
Speed of sound = Frequency × Wavelength
In this problem, we are given the frequency (260 Hz) and the wavelength (1.29 m) of the sound wave. We can use these values to calculate the speed of sound:
Speed of sound = 260 Hz × 1.29 m = 335.4 m/s
Next, we need to use the fact that the man hears his echo 2 seconds after he shouted. Since the sound wave traveled from the man to the hill and then back to the man, the total distance traveled by the sound wave is twice the distance from the man to the hill. We can use the formula:
Distance = Speed × Time
to calculate the distance from the man to the hill:
Distance = (335.4 m/s) × (2 s/2) = 335.4 m
Therefore, the hill is 335.4 meters away from the man. The answer is option (B), 335.4m.
Tambaya 2 Rahoto
A car traveled at a uniform speed of 100km/h, spends 15m moving from point A to point B along its route. The distance between A and B is
Bayanin Amsa
To calculate the distance between point A and point B, we can use the formula:
Distance = Speed x Time
where the speed is given as 100 km/h and the time is given as 15 minutes, which we need to convert to hours.
1 hour = 60 minutes, so 15 minutes = 15/60 hours = 0.25 hours.
Now, we can substitute these values into the formula:
Distance = 100 km/h x 0.25 h = 25 km
Therefore, the distance between point A and point B is 25 km.
is the correct answer.
Tambaya 3 Rahoto
A lens of focal length 15cm forms on erect image which is three times the size of the object. The distance between the object and the image is ___.
Bayanin Amsa
We can use the lens formula, 1/f = 1/v - 1/u, where f is the focal length of the lens, v is the distance between the lens and the image, and u is the distance between the lens and the object.
From the problem, we know that the focal length of the lens is 15 cm, and the image is erect and three times the size of the object. This means that the image distance v is positive and the object distance u is negative (since the object is in front of the lens).
Let's assume that the object distance u is -x cm, where x is a positive number. Then, the image distance v is +3x cm, since the image is three times the size of the object. Substituting these values into the lens formula, we get:
1/15 = 1/(+3x) - 1/(-x)
Simplifying the right-hand side, we get:
1/15 = (1 + 3)/3x
Multiplying both sides by 3x, we get:
3x/15 = 4
Simplifying, we get:
x = 20
Therefore, the distance between the object and the lens is -20 cm (since it is in front of the lens), and the distance between the image and the lens is +60 cm (since it is behind the lens). The distance between the object and the image is the sum of these distances, which is:
(-20) + (+60) = 40 cm
Therefore, the answer is 40cm.
Tambaya 4 Rahoto
A 40W instrument has a resistance 90 Ohms. On what voltage should it be operated normally
Bayanin Amsa
To calculate the voltage needed for a 40W instrument with a resistance of 90 Ohms, we can use the formula:
Voltage = √(Power x Resistance)
Plugging in the given values, we get:
Voltage = √(40W x 90Ω)
Voltage = √(3600)
Voltage = 60V
Therefore, the instrument should be operated at 60V to generate 40W of power with a resistance of 90 Ohms. The correct answer is, 60V.
Tambaya 5 Rahoto
Which of the following does NOT describe the Image formed by a plane minor?
Bayanin Amsa
The option that does NOT describe the image formed by a plane mirror is "Magnified".
When an object is placed in front of a plane mirror, the image formed is:
1. Erect: The orientation of the object in the mirror is the same as the orientation of the object in real life. For example, if you raise your right hand in front of a plane mirror, the image in the mirror will also show your right hand raised.
2. Laterally inverted: The image formed in the mirror is flipped horizontally, which means that the left side of the object appears on the right side of the image and vice versa. For example, if you wear a shirt with the letter "H" on it and look at it in a plane mirror, the image will show the letter "H" flipped horizontally.
3. Same distance from the mirror as object: The image formed in the mirror is located behind the mirror at the same distance as the object is located in front of the mirror. For example, if you stand 1 meter away from a plane mirror, the image of yourself will also be located 1 meter away from the mirror, behind the mirror.
4. NOT magnified: The image formed in the plane mirror is of the same size as the object, which means that there is no magnification or reduction in the size of the image. For example, if you stand in front of a plane mirror with a height of 1 meter, the image of yourself in the mirror will also have a height of 1 meter.
Therefore, the correct answer is "Magnified", as the image formed by a plane mirror is not magnified.
Tambaya 6 Rahoto
A 20kg mass is to be pulled up a slope inclined at 300 to the horizontal. If efficiency of the plane is 75%. The force required to pull the load up the plane is J [g=10ms\(^{-2}\)]
Bayanin Amsa
The force required to pull a 20kg mass up a slope inclined at 300 can be calculated using the formula:
force = mass * gravity * sin(angle)
where mass is 20kg, gravity is 10 m/s^2 and angle is 300.
The formula for efficiency is:
efficiency = output force / input force
where output force is the force required to pull the mass up the slope and input force is the force applied to the rope.
Since the efficiency of the plane is 75%, the input force is 4 times the output force.
So, the output force can be calculated as:
output force = input force / 4
input force = mass * gravity * sin(angle) / efficiency
input force = 20 * 10 * sin(300) / 0.75
input force = 533.2 N
And the output force can be calculated as:
output force = input force / 4
output force = 533.2 / 4
output force = 133.3 N
So, the force required to pull the load up the plane is 133.3 N.
Tambaya 7 Rahoto
In the diagram given if the atmospheric pressure is 760mm, the pressure in the chamber G Is
Bayanin Amsa
Tambaya 8 Rahoto
Which of the following best describes the energy changes which take place when a steam engine drives a generator which lights a lamp?
Bayanin Amsa
The energy changes that take place when a steam engine drives a generator which lights a lamp can be described as:
Heat energy from burning fuel is used to create steam in the boiler of the steam engine. This steam is then used to drive the turbine, which generates kinetic energy as it spins. The kinetic energy is transferred to the generator, which converts it into electrical energy (electricity). The electricity then flows through the wires to the lamp, where it is converted back into light energy, which is what we see.
Therefore, the correct option would be:
Heat ----> Kinetic ----> Electricity ----> Light
Tambaya 9 Rahoto
Which of the following may be used to explain a mirage?
I. Layers of air near the road surface have varying refractive indices in hot weather
II. Road surfaces sometimes become good reflectors in hot weather
III. Light from the sky can be reflected upwards after coming close to the road surface.
Bayanin Amsa
The phenomenon of a mirage can be explained by options I and III. A mirage is an optical illusion that occurs when light rays passing through a medium with varying refractive indices create a false image of distant objects or even the sky. In hot weather, the air near the road surface becomes hotter and less dense than the air above, causing the light passing through it to bend and create a reflection of the sky or objects in the distance. This effect is known as a temperature inversion. Additionally, light from the sky can be reflected upwards after coming close to the road surface, adding to the illusion of a reflected object or the sky. Option II, which suggests that road surfaces become good reflectors in hot weather, is not a valid explanation for a mirage. Therefore, the correct answer is: I and III only.
Tambaya 10 Rahoto
A body of mass 100g moving with a velocity if 10ms-1 collides with a wall. If after collision it moves with a velocity of 2.0ms−1 in the opposite direction, calculate the change in momentum.
Bayanin Amsa
Momentum = Mass x Velocity
Change in Momentum = Mass x (change in velocity)
Change in Momentum = Mass x (v - u)
Mass = 100g = 0.1kg
U = 2ms−1
V = 10ms−1
Change in momentum = m (v − u)
= 0.1 (10 − 2)
= 0.1 (8)
= 0.8Ns
Change in Momentum = Mass x (change in velocity)
Change in Momentum = Mass x (v - u)
Mass = 100g = 0.1kg
U = 2ms−1
V = 10ms−1
Change in momentum = m (v − u)
= 0.1 (10 − 2)
= 0.1 (8)
= 0.8Ns
Tambaya 11 Rahoto
The photo cell works on the principle of the
Bayanin Amsa
The photocell works on the principle of the emission of electrons by incident radiation. In simple terms, a photocell is a device that converts light energy into electrical energy. It does this by using a material (such as silicon) that releases electrons when it is exposed to light. These electrons can then be collected and used to produce a current, which can be used to power an electrical device. The more light that hits the photocell, the more electrons are released and the greater the electrical current.
Tambaya 12 Rahoto
If the fraction of the atoms of a radioactive material left after 120years is \( \frac{1}{64} \), what is the half-life of the material? \( \frac{1}{64} \)
Bayanin Amsa
The half-life of a radioactive material is the time it takes for half of the atoms in a sample to decay. The fraction of atoms left after a certain number of half-lives can be calculated using the formula:
fraction left = (1/2)^(number of half-lives)
Let's use this formula to solve the problem. We know that the fraction of atoms left after 120 years is 1/64, which means that:
(1/2)^(number of half-lives) = 1/64
To solve for the number of half-lives, we can take the logarithm of both sides:
log[(1/2)^(number of half-lives)] = log(1/64)
Using the rule that log(a^b) = b*log(a), we can simplify the left side of the equation:
number of half-lives * log(1/2) = log(1/64)
Dividing both sides by log(1/2), we get:
number of half-lives = log(1/64) / log(1/2)
Using a calculator or the change of base formula, we can evaluate this expression:
number of half-lives = 6
Therefore, the half-life of the material is 120/6 = 20 years.
Tambaya 13 Rahoto
A boy receives the echo of his clap reflected by a nearby hill 0.8s later. If the speed of sound in air is 3.40ms\(^{-1}\), how far is he from the hill?
Bayanin Amsa
The speed of sound in air is given as 3.40 m/s. The echo of the clap is heard 0.8 seconds after it was produced, which means it traveled to the hill and back. The time taken for sound to travel to the hill is half of the time taken for it to travel to and from the hill. Therefore, the time taken for sound to travel to the hill is 0.8/2 = 0.4 seconds.
We can use the formula:
distance = speed x time
to calculate the distance between the boy and the hill.
distance = speed of sound x time taken for sound to travel to the hill
distance = 3.40 m/s x 0.4 s
distance = 1.36 m
Therefore, the distance between the boy and the hill is 1.36 meters.
However, the answer options provided are in meters and are significantly larger than 1.36 meters. It is possible that the speed of sound provided in the question is incorrect or the answer options are incorrect.
Tambaya 14 Rahoto
The electrochemical equivalent of silver is 0.0012g/C. If 36.0g of silver is to be deposited by electrolysis on a surface by passing a steady current for 5mins, the current must be?
Bayanin Amsa
The electrochemical equivalent of silver is a measure of the amount of silver that is deposited on a surface per unit of charge. In this case, the electrochemical equivalent of silver is 0.0012 grams per Coulomb of charge. To deposit 36.0 grams of silver by electrolysis, we need to know the amount of charge that must be passed through the solution.
The amount of charge is given by:
Q = m/z
where m is the mass of silver to be deposited, 0.0012 is the electrochemical equivalent of silver, and z is the charge on one mole of electrons (z = 1 for a single electron).
So, the amount of charge required is:
Q = 36.0 g / 0.0012 g/C = 30000 C
The current, I, is given by:
I = Q / t
where t is the time for which the current is flowing. In this case, t = 5 minutes.
So, the current required is:
I = 30000 C / (5 x 60 s) = 100 A
Therefore, the current must be 100 Amperes.
Tambaya 15 Rahoto
Which of the following is not a product of nuclear fusion?
Bayanin Amsa
Neutron is not a product of nuclear fusion. Nuclear fusion is the process by which two or more atomic nuclei come together to form a heavier nucleus, releasing a large amount of energy in the process. In most fusion reactions, the resulting products are alpha particles (helium nuclei) and energy in the form of gamma rays.
X-rays and gamma rays are both forms of high-energy electromagnetic radiation that can be produced by nuclear reactions, including nuclear fusion. Alpha particles are also a common product of nuclear fusion, especially in the fusion reactions that power the sun.
However, neutrons are not typically produced in fusion reactions. In fact, one of the major challenges in developing fusion as a practical energy source is finding ways to produce and control the high-energy neutrons that are generated in the process. Neutrons can be produced in some types of fusion reactions, but they are not a primary product.
In summary, neutron is not a product of nuclear fusion, while X-rays, Y-rays (assuming this is a valid form of radiation), and alpha particles are common products of this process.
Tambaya 16 Rahoto
A particle of mass M initially at rest splits into two. If one of the particles of mass M1 moves with velocity V1 , the second particle moves with velocity
Bayanin Amsa
When a particle of mass M splits into two, the total mass is conserved, and so the sum of the masses of the two resulting particles must be equal to M. If one of the particles of mass M1 moves with velocity V1, we can use the law of conservation of momentum to determine the velocity of the second particle.
The law of conservation of momentum states that the total momentum of a system of particles remains constant if no external forces act on the system. In this case, the initial momentum of the system is zero, since the particle was initially at rest. After the particle splits, the momentum of the system is the sum of the momenta of the two resulting particles.
Let's use the subscript 1 to represent the first particle of mass M1 and the subscript 2 to represent the second particle of mass M-M1. By conservation of momentum, we have:
0 = M1*V1 + (M - M1)*V2
Solving for V2, we get:
V2 = -M1/M*(V1)
Therefore, the second particle moves in the opposite direction with velocity -M1/M*(V1). This means that the two particles move in opposite directions, with the ratio of their velocities determined by the ratio of their masses. Option (D) in the table shows the correct answer, which is -M1/M*(V1).
Tambaya 17 Rahoto
An electric heating coil rated at 1KW is used to heat 2kg of water for 2 minutes. The initial water temperature is 30\(^{o}\)C. Taking the specific heat of the water as 4,000Jkg\(^{-1}\) and neglecting that of the container, the final water temperature is
Bayanin Amsa
To determine the final temperature of the water, we can use the formula:
Q = mcΔT
where Q is the heat transferred, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
We know that the power of the electric heating coil is 1KW, which means it transfers 1000 Joules of energy per second. In 2 minutes, or 120 seconds, it transfers 120,000 Joules of energy to the water.
The mass of the water is given as 2kg and the specific heat capacity of water is 4000 J/kg°C. We can assume that the initial temperature of the water is 30°C.
Using the formula, we can solve for the change in temperature:
120,000 J = (2 kg)(4000 J/kg°C)(ΔT)
ΔT = 15°C
Therefore, the final temperature of the water is 30°C + 15°C = 45°C.
So, the final water temperature is 45.0oC.
Tambaya 18 Rahoto
The friction which exist between two layers of liquid in relative motion is called
Bayanin Amsa
The correct answer is "Viscosity". Viscosity is the property of a fluid that describes its resistance to flow. When two layers of liquid are in relative motion, the viscosity of the liquid causes friction between the layers. This friction creates a resistance to the movement of one layer past the other. The greater the viscosity of the liquid, the greater the friction and the more difficult it is for the layers to move past each other. This property is important in many industrial and natural processes, such as the flow of oil in pipelines or the movement of blood through the human body.
Tambaya 19 Rahoto
The process whereby a liquid turns spontaneously into vapour is called
Bayanin Amsa
The process whereby a liquid turns spontaneously into vapor is called evaporation.
Evaporation is the process by which a liquid changes into a gas at a temperature below its boiling point. This happens when the molecules of the liquid gain enough energy to escape from the surface of the liquid into the air as a gas. The rate of evaporation depends on factors such as the temperature, the humidity of the air, and the surface area of the liquid. For example, a shallow pool of water will evaporate faster than a deep one because it has a larger surface area.
Boiling, on the other hand, is the process by which a liquid changes into a gas at its boiling point. This happens when the pressure of the gas generated by the boiling liquid is equal to the atmospheric pressure. The temperature remains constant during boiling.
Regelation and sublimation are different processes altogether. Regelation is the process by which a solid changes into a liquid when it is subjected to pressure. Sublimation is the process by which a solid changes directly into a gas, bypassing the liquid state.
Tambaya 20 Rahoto
Shadows and eclipses result from the
Bayanin Amsa
The rectilinear propagation of light means that light travels in straight lines as a wave. This can be observed in the well-defined shadows formed when an object blocks a light source and through the use of a pinhole camera.
According to Sudipa Sarkar, the formation of shadows with sharp edges demonstrates the rectilinear propagation of light, i.e. The fact that light travels in straight line. When an opaque obstacle is placed between a source of light and a screen, a shadow of the obstacle is formed on the screen. The kind of shadow depends on the size of the source of light. If it is a point source (light from a small hole), the shadow obtained is a region of total darkness, called umbra.
If an extended source of light, e.g. a bulb, is used, the umbra is surrounded by a region of partial darkness, called penumbra. The moon is seen because it reflects the sun's light. An eclipse of the moon (lunar eclipse) occurs when the earth comes between the sun and the moon and prevents some of the light from the sun from reaching the moon. In other words, the earth casts its shadow on the moon. The solar eclipse occurs when the moon comes between the sun and the earth.
According to Sudipa Sarkar, the formation of shadows with sharp edges demonstrates the rectilinear propagation of light, i.e. The fact that light travels in straight line. When an opaque obstacle is placed between a source of light and a screen, a shadow of the obstacle is formed on the screen. The kind of shadow depends on the size of the source of light. If it is a point source (light from a small hole), the shadow obtained is a region of total darkness, called umbra.
If an extended source of light, e.g. a bulb, is used, the umbra is surrounded by a region of partial darkness, called penumbra. The moon is seen because it reflects the sun's light. An eclipse of the moon (lunar eclipse) occurs when the earth comes between the sun and the moon and prevents some of the light from the sun from reaching the moon. In other words, the earth casts its shadow on the moon. The solar eclipse occurs when the moon comes between the sun and the earth.
Tambaya 21 Rahoto
In homes, electrical appliances and lamps are connected in parallel because
Bayanin Amsa
Tambaya 22 Rahoto
A force of 5N acts at a point Y on a rod XYZ as shown in this diagram. If XYZ is 2m, what is the moment of the force about point X?
Bayanin Amsa
To find the moment of the force about point X, we need to first understand what moment is. Moment is the turning effect of a force around a pivot point. It is calculated by multiplying the force by the perpendicular distance between the pivot point and the line of action of the force.
In this case, the force of 5N is acting on the rod at point Y. To find the moment of this force about point X, we need to find the perpendicular distance between point X and the line of action of the force.
From the diagram, we can see that the perpendicular distance between point X and the line of action of the force is 2m (the length of the rod). So, the moment of the force about point X is:
Moment = force x perpendicular distance
= 5N x 2m
= 10Nm
Therefore, the correct answer is: 10Nm.
Tambaya 23 Rahoto
The resistance of a 40W car head lamp, drawing current from a 12V battery is ____.
Bayanin Amsa
The resistance of a 40W car headlamp can be calculated using Ohm's Law, which states that the current (I) flowing through a conductor between two points is directly proportional to the voltage (V) across the two points, and inversely proportional to the resistance (R) of the conductor. The equation can be written as V = IR.
Since the power (P) of the headlamp is given as 40W and the voltage is 12V, we can calculate the current using the equation P = IV. Substituting I = P/V, we get I = 40/12 = 3.33A.
Finally, using Ohm's Law, we can calculate the resistance as R = V/I = 12/3.33 = 3.6Ω.
So, the resistance of the 40W car headlamp, drawing current from a 12V battery, is 3.6Ω.
Tambaya 24 Rahoto
Cathode rays are
Bayanin Amsa
Cathode rays are streams of electrons. They were first discovered by scientists experimenting with vacuum tubes, and they observed that a glowing beam of particles traveled from the negatively charged electrode (the cathode) to the positively charged electrode (the anode). These particles were found to have a negative charge, which was later identified as electrons. Cathode rays played an important role in the development of electronics and the understanding of atomic structure.
Tambaya 25 Rahoto
The product of force and time is?
Bayanin Amsa
The product of force and time is known as impulse.
Impulse can be defined as the change in momentum that an object experiences as a result of a force being applied to it over a period of time. In simpler terms, impulse is the "push" that an object receives from a force acting on it for a certain amount of time. The more force applied, or the longer the time the force is applied, the greater the impulse and the greater the change in momentum of the object.
It's important to note that impulse is a vector quantity, meaning it has both magnitude and direction. Impulse is a measure of the ability of a force to cause an object to change its velocity, and can be used to explain many phenomena in physics, such as why a heavy object is harder to stop than a lighter one, or why a soccer ball changes direction when it is kicked.
Tambaya 26 Rahoto
The process by which protons are converted into helium atoms with a tremendous release of energy is called?
Bayanin Amsa
The process by which protons are converted into helium atoms with a tremendous release of energy is called "thermonuclear fusion".
In this process, two light atomic nuclei combine to form a heavier nucleus, releasing a huge amount of energy in the form of light and heat. This is the same process that powers the sun and other stars. The high temperatures and pressures required for fusion to occur can only be achieved in stars or in controlled environments such as fusion reactors.
Thermonuclear fusion is different from nuclear fission, which is the process of splitting a heavy nucleus into lighter nuclei with the release of energy.
Thermionic emission and photoelectric emission are different processes that involve the emission of electrons from a material due to heating or exposure to light, respectively.
Tambaya 27 Rahoto
A room is heated by means of a charcoal fire, an occupants of the room standing away from the fire is warmed mainly by
Bayanin Amsa
The main way that the occupants of a room standing away from a charcoal fire are warmed is by radiation. Radiation is the transfer of heat energy through electromagnetic waves, and it can travel through empty space.
In this scenario, the charcoal fire emits radiation in the form of infrared waves, which travel through the air and warm up the objects (including the occupants) in the room. Convection, on the other hand, is the transfer of heat through the movement of fluids (such as air), but in this case, the air in the room is not being actively circulated by a fan or other mechanism. Conduction involves the transfer of heat through direct contact between two objects, but the occupants are not in direct contact with the fire. Reflection refers to the bouncing of radiation off a surface, but it is not a significant factor in this scenario as most of the radiation is absorbed by the objects in the room.
Tambaya 28 Rahoto
Palm oil from a bottle flows out more easily after it has been heated because the
Bayanin Amsa
Molecules cannot be given energy during the heating and the molecules of oil cannot force each other out
Tambaya 29 Rahoto
If the focal length of a camera is 20cm, the distance from the film at which the lens must be set to produce a sharp image of 100cm away is
Bayanin Amsa
F = 20cm
V = 100cm
U = ?
1U + 1V = 1F
120 + 1100 = 1F
5+1100 = 1F
F = 1006
= 16.7cm
= 17cm
V = 100cm
U = ?
1U + 1V = 1F
120 + 1100 = 1F
5+1100 = 1F
F = 1006
= 16.7cm
= 17cm
Tambaya 30 Rahoto
The inner diameter of a small test tube can be measured accurately using a
Bayanin Amsa
A vernier caliper is a measuring device used to precisely measure linear dimensions. It is a very useful tool to use when measuring the diameter of a round objects like cylinders because the measuring jaws can be secured on either side of the circumference.
Vernier calipers have both a fixed main scale and a moving vernier scale. The main scale is graduated in either millimetres or tenths of an inch. The vernier scale allows much more precise readings to be taken (usually to the nearest 0.02mm or 0.001 inch) in comparison to a standard ruler (which only measures to th nearest 1mm or 0.25 inch).
The vernier scale was invented by French mathematician Pierre Vernier in 1631. As part of the vernier caliper, it is used together with the main scale, and helps to provide very precise measurements. Vernier calipers usually show either imperial or metric measurements, but some measure in both.
Vernier calipers have both a fixed main scale and a moving vernier scale. The main scale is graduated in either millimetres or tenths of an inch. The vernier scale allows much more precise readings to be taken (usually to the nearest 0.02mm or 0.001 inch) in comparison to a standard ruler (which only measures to th nearest 1mm or 0.25 inch).
The vernier scale was invented by French mathematician Pierre Vernier in 1631. As part of the vernier caliper, it is used together with the main scale, and helps to provide very precise measurements. Vernier calipers usually show either imperial or metric measurements, but some measure in both.
Tambaya 31 Rahoto
The density of 400cm3 of palm oil was 0.9gcm-3 before frying. If the density of the oil was 0.6gcm-3 after frying, assuming no loss of oil due to spilling, its new volume was?
Bayanin Amsa
The density of a substance is defined as its mass per unit volume. Therefore, the mass of the palm oil before frying was:
Mass = Density x Volume = 0.9 g/cm³ x 400 cm³ = 360 g
After frying, the mass of the palm oil remains the same, but its density changes to 0.6 g/cm³. Therefore, the new volume of the palm oil can be calculated by rearranging the density formula:
Volume = Mass / Density = 360 g / 0.6 g/cm³ = 600 cm³
So the new volume of the palm oil after frying is 600 cm³. is the correct answer.
Tambaya 32 Rahoto
Natural radioactivity consists of the emission of
Bayanin Amsa
Radioactive decay releases different types of energetic emissions. The three most common types of radioactive emissions are alpha particles, beta particles, and gamma rays.
Tambaya 33 Rahoto
Musical instruments playing the same note can be distinguished from one another owing to the difference in their
Bayanin Amsa
Different musical instruments playing the same note can be distinguished from one another due to the difference in their "timbre" or "tone color." Timbre refers to the unique character or quality of a sound that allows us to distinguish it from other sounds even when they have the same pitch and loudness. For example, a piano and a guitar playing the same note will sound different due to the differences in their timbre. This is why we can tell the difference between different instruments and why some instruments are better suited to certain styles of music than others.
Tambaya 34 Rahoto
Which of the following obeys Ohm's law?
Bayanin Amsa
Ohm's law states that the current passing through a conductor is directly proportional to the voltage applied across it, given the temperature and other physical conditions remain constant.
Among the given options, only "all metals" obey Ohm's law. This is because metals have a linear relationship between their resistance and the applied voltage, meaning that the resistance of a metal remains constant regardless of the voltage applied. As a result, the current passing through a metal is directly proportional to the voltage applied, following Ohm's law.
On the other hand, a diode, all electrolytes, and glass do not obey Ohm's law. A diode is a semiconductor that has a non-linear current-voltage relationship, and its resistance is not constant. Similarly, electrolytes and glass are non-metallic substances that do not have a linear relationship between their resistance and the applied voltage. Their resistance can change significantly with the voltage applied, and hence they do not follow Ohm's law.
Tambaya 35 Rahoto
Temperature is the property of a body which is proportional to the ____.
Bayanin Amsa
Temperature is proportional to the average kinetic energy of the molecules in a body. This means that as the average kinetic energy of the molecules increases, so does the temperature.
Think about it like this: the hotter an object, the more energy its molecules have. This energy is what makes the molecules move faster, and therefore, the temperature of the object increases.
The average kinetic energy of the molecules is a better measure of temperature than the maximum speed of the molecules because temperature is a measure of the overall energy distribution, not just the energy of a single molecule.
Tambaya 36 Rahoto
An object moves in a circular path of radius 0.5m with a speed of \(1ms^{-1}\). What is its angular velocity?
Bayanin Amsa
Angular velocity is a measure of how fast an object is rotating around a center point. It's usually measured in radians per second (rad/s).
To calculate angular velocity, we use the formula: angular velocity = linear velocity / radius. In this case, the linear velocity is 1 m/s, and the radius is 0.5 m.
So, the angular velocity would be:
1 m/s / 0.5 m = 2 rad/s
Therefore, the answer is 2 rad/s or 2rads^-1
Tambaya 37 Rahoto
The critical angle for light travelling from a transparent medium to air s measured as 340. The refractive index of the medium is
Bayanin Amsa
The critical angle is the angle of incidence at which light is just able to pass through the interface between two media and not reflect back. When light travels from a medium with a higher refractive index to a medium with a lower refractive index, it slows down and bends towards the normal. If the angle of incidence is increased, the light will eventually reach a point where it will not be able to escape the higher index medium and will be totally reflected back. This is the critical angle.
The formula for the critical angle can be expressed as follows:
sin(θc) = n2/n1
Where θc is the critical angle, n1 is the refractive index of the first medium, and n2 is the refractive index of the second medium. In this case, the first medium is the transparent medium and the second medium is air, which has a refractive index of approximately 1.
By substituting the value of sin(θc) with the value of 340, and n2 with 1, we can solve for n1.
sin(340) = n1/1
n1 = 1/sin(340)
The value of n1 calculated using this formula is approximately 1.79, which means that the refractive index of the transparent medium is 1.79.
Tambaya 38 Rahoto
What is the resultant resistance of the circuit in the image shown?
Bayanin Amsa
Formulae resistance in parallel
= 1/R = 1/R1 +1/R2
1/R = 1/2 +1/2 = 1
Resistance are now in series
R = 1 + 3 + 4
= 8 ohms
= 1/R = 1/R1 +1/R2
1/R = 1/2 +1/2 = 1
Resistance are now in series
R = 1 + 3 + 4
= 8 ohms
Tambaya 40 Rahoto
The diagram shown represents a block-and-tackle pulley system on which an effort of W Newtons supports a load of 120.0N. If the efficiency of the machine is 40, then the value of W is?
Bayanin Amsa
Let the total number of pulleys used in both the blocks be n
.
In a block-and-tackle pulley system, the velocity ratio is equal to n.
Efficiency = MAVR×100%
MA=LE,VR=n
Efficiency = LE×1n×100%
E=LEff.×n×100%
E=120N40%×6×100%
E=50N
In a block-and-tackle pulley system, the velocity ratio is equal to n.
Efficiency = MAVR×100%
MA=LE,VR=n
Efficiency = LE×1n×100%
E=LEff.×n×100%
E=120N40%×6×100%
E=50N
