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Tambaya 1 Rahoto
A body was slightly displaced from its equilibrium position. Which one of the following is a condition for its stable equilibrium
Bayanin Amsa
The condition for stable equilibrium of a body that has been slightly displaced from its equilibrium position is "an increase in the potential energy of the body." When an object is at its equilibrium position, it has a minimum potential energy. When the object is displaced from its equilibrium position, it has a higher potential energy. For the object to be in stable equilibrium, it must be able to return to its equilibrium position after it has been displaced. If the potential energy of the object increases as it is displaced, it means that the equilibrium position is a point of stable equilibrium. This is because the object will experience a restoring force that will push it back towards its equilibrium position, as the potential energy decreases. Therefore, an increase in potential energy is a condition for a body to be in stable equilibrium after it has been slightly displaced from its equilibrium position. An increase in kinetic energy or height does not necessarily indicate stability, as it depends on the specific situation and other factors at play.
Tambaya 2 Rahoto
A mixture of blue and red pigment when illuminated by white light will appear
Bayanin Amsa
A mixture of blue and red pigment when illuminated by white light will appear purple. This is because when white light shines on a surface, it contains all the colors of the visible spectrum. When blue and red pigments are mixed together, they absorb all the other colors in the spectrum except for blue and red. Therefore, when white light shines on this mixture, the blue pigment absorbs all the colors except blue, while the red pigment absorbs all the colors except red. The result of this is that the blue and red pigments reflect only blue and red light, which then combines to form purple. Therefore, the mixture of blue and red pigments appears purple when illuminated by white light.
Tambaya 3 Rahoto
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)
Bayanin Amsa
n = 200, S = 132 rev/min, v = 350m/s2
| f | = | ns | = | 200 | × | 132 | revmin | × | 1min60s | = | 440Hz |
| λ | = | vf | = | 350440 | = | 0.875m |
Tambaya 4 Rahoto
The diagram shows four positions of the bob of a simple pendulum. At which of these positions does the bob have maximum kinetic energy and minimum potential energy
Bayanin Amsa
At position 1, the bob of the simple pendulum has the maximum potential energy and zero kinetic energy. At position 4, the bob has the maximum kinetic energy and minimum potential energy. To understand this, we need to know that the energy of a simple pendulum is converted back and forth between kinetic energy and potential energy as it swings back and forth. When the bob is at its highest point (position 1), it has the maximum potential energy because it is farthest from the ground and has the most potential to move downward. At this point, the bob has zero kinetic energy because it is momentarily at rest. As the bob swings downward towards the equilibrium point, it gains speed and its potential energy is converted to kinetic energy. At the equilibrium point (position 2), the bob has equal amounts of kinetic and potential energy. As the bob continues to move downward, its potential energy decreases and its kinetic energy increases. At position 3, the bob has minimum potential energy and some amount of kinetic energy. At the lowest point of its swing (position 4), the bob has maximum kinetic energy because it is moving at its fastest speed. At this point, the bob has minimum potential energy because it is closest to the ground and has the least amount of potential to move downward. So, to summarize, the bob has maximum potential energy at position 1, equal amounts of kinetic and potential energy at position 2, minimum potential energy at position 3, and maximum kinetic energy at position 4.
Tambaya 5 Rahoto
Neutrons were discovered by
Bayanin Amsa
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.
Tambaya 6 Rahoto
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.
Bayanin Amsa
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%.
Tambaya 7 Rahoto
The lead-acid accumulator consists of
Bayanin Amsa
- the positive pole is lead peroxide (PbO2
)
- the negative pole is head
- the electrolyte is H2
SO4
Tambaya 8 Rahoto
The volume of 0.354g of helium at 273°C and 114cm of mercury pressure is 2667cm3 . Calculate the volume
Bayanin Amsa
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 |
Tambaya 9 Rahoto
Radio waves belongs to the class of ware whose velocity is about
Bayanin Amsa
Radio waves belong to the class of waves whose velocity is approximately 3 x 10^8 m/s. This velocity is commonly denoted as the speed of light, which is the speed at which all electromagnetic waves, including radio waves, travel in a vacuum. This constant velocity is one of the fundamental principles of physics and is important in understanding the behavior and properties of light and other electromagnetic waves. The speed of light is incredibly fast, and it's difficult for us to imagine just how fast it is. To put it into perspective, light can travel around the Earth's equator almost 7.5 times in just one second. This high speed is essential for radio communication, as it enables radio waves to travel long distances in a short amount of time, allowing us to communicate with people and devices far away from us.
Tambaya 10 Rahoto
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.
Bayanin Amsa
From two parts 20m apart
a = 10m, x = 6m, A = 5
V = ω√A2−X2
= 5√102−62
= 40m/s
Tambaya 11 Rahoto
Workdone on an object to bring it to a certain point in space is called
Bayanin Amsa
The work done on an object to bring it to a certain point in space is called "Potential Energy". Potential energy is a form of energy that an object possesses due to its position relative to other objects. When an object is lifted or moved to a higher point against gravity, work is done on it, and this work is stored as potential energy. The potential energy of an object is directly proportional to its height and mass. It can be converted into other forms of energy, such as kinetic energy, when the object is released or allowed to move freely. Therefore, potential energy is a type of stored energy that an object has due to its position, and it can be released to do work.
Tambaya 12 Rahoto
Which of the following statements are correct of the production and propagation of waves?
I. vibration produces waves
II. waves transmit energy along the medium
III. the medium through which the wave travels does not travel with the wave
IV. waves do not require any medium for transmission
Bayanin Amsa
The correct statement is: I and II and III only. Explanation: - Statement I is correct because the production of waves involves some kind of disturbance that creates a vibration in the medium, which then propagates as a wave. - Statement II is correct because waves carry energy along the medium as they propagate. This is why waves can be used to transmit information or power over long distances. - Statement III is correct because the medium through which a wave travels does not move with the wave. Instead, the wave passes through the medium, causing it to oscillate or vibrate, but not to move along with the wave. - Statement IV is incorrect because most waves require a medium through which to propagate. For example, sound waves require air, water waves require water, and seismic waves require the Earth's crust. There are some types of waves, such as electromagnetic waves, that can propagate through a vacuum, but this is not true for all waves.
Tambaya 13 Rahoto
The volume of a stone having an irregular shape can be determined using?
Bayanin Amsa
The volume of a stone with an irregular shape can be determined using a measuring cylinder. A measuring cylinder is a glass or plastic container with a narrow cylindrical shape and markings on the side to indicate the volume it contains. To determine the volume of an irregularly shaped stone, you would fill the measuring cylinder with water, carefully lower the stone into the water, and note the increase in the volume of the water. The difference in the volume of the water before and after the stone was added is equal to the volume of the stone. The meter rule, vernier calliper, and micrometer screw gauge are all measuring instruments, but they are not designed to measure the volume of irregularly shaped objects. The meter rule is a measuring tool used for measuring length. The vernier calliper is used for measuring the diameter of objects, and the micrometer screw gauge is used for precise measurements of small distances.
Tambaya 14 Rahoto
Which of the following media allow the transmission of sound waves through them?
I. air
II. liquid
III. solids
Bayanin Amsa
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".
Tambaya 15 Rahoto
Which of the following readings cannot be determined with a meter rule?
Bayanin Amsa
Meter rule has a reading accuracy of 0.5mm or 0.05cm, thus measurement is M ± 0.05cm i.e 2.00, 2.05, 2.50, 2.55 etc.
The reading that cannot be read is 2.56cm.
Tambaya 16 Rahoto
When blue and green colours of light are mixed, the resultant colour is
Tambaya 17 Rahoto
- angle of dip is zero at the magnetic equator
- angle of variation is the same as angle of declination.
Bayanin Amsa
- angle of dip is zero at the magnetic equator
- angle of variation is the same as angle of declination.
Tambaya 18 Rahoto
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.
Bayanin Amsa
Devices with different liquids
d1
h1
= d2
h2
1 × 20 = 0.8 × h
| h | = | 200.8 | = | 25cm |
Tambaya 19 Rahoto
The following are parts of the eye
I. Retina
II. Pupil
III. Iris
The correct equivalent in the camera in the same order are
Bayanin Amsa
- retina is similar to film
- pupil is similar to aperture
- iris is similar to diaphragm
Tambaya 20 Rahoto
In the molecular explanation of conduction, heat is transferred by the
Bayanin Amsa
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."
Tambaya 21 Rahoto
A supply of 400V is connected across capacitors of 3μf and 6μf in series. Calculate the charge
Bayanin Amsa
| CT | = | C1 × C2 C1 + C2 |
| = | 3 × 63 + 6 |
= 189
= 2μf
Q = CV
⇒ 2 × 10−6
× 400
⇒ 800 × 10−6
C = 8 × 10−4
C
Tambaya 23 Rahoto
Efficiency of conduction in liquids and gases compared to solids is
Bayanin Amsa
The efficiency of conduction in liquids and gases compared to solids is generally less efficient. This means that solids are better conductors of heat and electricity than liquids and gases. This is because the particles in solids are closely packed and are tightly bound to one another, allowing heat and electricity to flow easily through the material. On the other hand, the particles in liquids and gases are more spread out and less tightly bound, making it more difficult for heat and electricity to flow through these materials. However, it is important to note that the efficiency of conduction can vary depending on the specific liquid or gas and the specific solid being compared. Some liquids and gases may have properties that make them better conductors than certain solids, but this is not a general rule.
Tambaya 24 Rahoto
If a body moves with a constant speed and at the same time undergoes an acceleration, its motion is said to be
Bayanin Amsa
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.
Tambaya 25 Rahoto
The limiting frictional force between two surfaces depends on
I. the normal reaction between the surfaces
II. the area of surface in contact
III. the relative velocity between the surfaces
IV. the nature of the surfaces
Bayanin Amsa
- Friction depends on the nature of the surfaces in contact
- Solid friction is independent of the area of the surfaces in contact and the relative velocity between the surfaces.
Tambaya 26 Rahoto
Ripple in a power supply unit is caused by
Bayanin Amsa
The correct option is "Using a zener diode" as fluctuation of d.c signal results from the rectification of a.c to d.c.
Tambaya 27 Rahoto
According to kinetic molecular model, in gases
Bayanin Amsa
According to the kinetic molecular model, in gases, the molecules are very fast apart and occupy all the space made available. This means that gas molecules are in constant random motion and they move freely in all directions without any regular arrangement. They collide with each other and with the walls of the container, exerting pressure. The temperature of the gas is related to the average kinetic energy of the gas molecules. The higher the temperature, the faster the gas molecules move, and the higher the kinetic energy.
Tambaya 28 Rahoto
The resultant capacitance in the figure above is
Bayanin Amsa
For the parallel arrangement = 2 + 4 = 6μf
| For | the | series | arrangement | = | 1CT | = | 12 | + | 13 | + | 16 | + | 14 |
| 1CT | = | 1512 |
| CT | = | 1215 | = | 0.8μf |
Tambaya 29 Rahoto
The equilibrium position of objects in any field corresponds to situation of
Bayanin Amsa
The equilibrium position of an object in any field corresponds to the situation of minimum potential energy. This means that at the equilibrium position, the object has the lowest possible potential energy within the field. In other words, the forces acting on the object are balanced, and the object is not being pushed or pulled in any direction. Therefore, the object will remain at rest at the equilibrium position unless it is acted upon by an external force. Of the options given, the correct answer is "minimum potential energy".
Tambaya 31 Rahoto
The momentum of a car moving at a constant speed in a circular track
Bayanin Amsa
Movement of an object in a circle with an acceleration towards its center is provided by change in velocity and centripetal force a α V α Fc
Tambaya 32 Rahoto
Which of the following statement about the electromagnet shown above is correct?
Bayanin Amsa
A - B = S - N.
Also, starting end of the current is south while terminating end is North.
Tambaya 33 Rahoto
The Earth's magnetic equator passes through Jos in Nigeria. At Jos, the
Bayanin Amsa
The Earth has a magnetic field that is generated by the movement of molten iron in its core. The magnetic field has different properties at different locations on the Earth's surface. The magnetic equator is an imaginary line on the Earth's surface where the inclination or tilt of the Earth's magnetic field is zero, meaning that the magnetic field lines are parallel to the Earth's surface. At Jos, Nigeria, the Earth's magnetic equator passes through, which means that the angle of inclination or dip of the Earth's magnetic field is zero. Therefore, the correct answer is that the angle of dip is zero. This means that a magnetic needle suspended by a thread or placed on a horizontal surface would remain horizontal and not point downwards or upwards, as it would at other locations on the Earth's surface. This makes Jos an important location for studying the Earth's magnetic field and for conducting experiments related to magnetism.
Tambaya 34 Rahoto
Heat may be transferred by conduction, convention and radiation. By which of these methods does heat travel through vacuum?
Bayanin Amsa
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".
Tambaya 35 Rahoto
A coil X is moved quickly away from the end Y of a stationary metal bar and a current then flows in X as shown above.
Then
Bayanin Amsa
N - S magnet is moved towards a coil production clockwise direction of current in the coil.
- This is the same as a coil moved away from S-N (Y - North pole)
Tambaya 36 Rahoto
A rectangular solid black has length 10cm, breadth 5cm and height 2cm. If it lies on a horizontal surface, and has density 100kg/m3 , calculate the pressure it exerts on the surface.
Bayanin Amsa
To calculate the pressure that the rectangular solid exerts on the surface, we need to use the formula for pressure: Pressure = Force / Area In this case, the force is the weight of the rectangular solid, which we can calculate using the formula: Weight = Mass x Gravity The mass of the rectangular solid can be calculated using its density and volume: Mass = Density x Volume The volume of the rectangular solid is simply its length x breadth x height: Volume = Length x Breadth x Height = 10 cm x 5 cm x 2 cm = 100 cm3 We need to convert this volume to cubic meters to use the density given in kg/m3: Volume = 100 cm3 = 0.0001 m3 Now we can calculate the mass: Mass = Density x Volume = 100 kg/m3 x 0.0001 m3 = 0.01 kg The gravity is the acceleration due to gravity, which we can assume to be 9.81 m/s2. Therefore, the weight is: Weight = Mass x Gravity = 0.01 kg x 9.81 m/s2 = 0.0981 N Now we can use this weight to calculate the pressure on the surface. The surface area in contact with the rectangular solid is simply its length x breadth: Area = Length x Breadth = 10 cm x 5 cm = 50 cm2 We need to convert this area to square meters: Area = 50 cm2 = 0.005 m2 Therefore, the pressure is: Pressure = Force / Area = 0.0981 N / 0.005 m2 = 19.62 N/m2 We can convert this to units of N/cm2 or N/mm2 if desired. This is equivalent to: Pressure = 0.1962 N/cm2 = 0.0001962 N/mm2 So the pressure that the rectangular solid exerts on the surface is 19.62 N/m2, which is approximately 20 N/m2. Therefore, the answer is 200 N/m2.
Tambaya 37 Rahoto
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
Bayanin Amsa
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.
Tambaya 38 Rahoto
When the temperature of a liquid is increased, its surface tension
Bayanin Amsa
Surface tension or elasticity of a fluid decreases with increased in temperature
Tambaya 39 Rahoto
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
Bayanin Amsa
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
Tambaya 40 Rahoto
The distance between an object and its real image in a convex lens is 40cm. If the magnification of the image is 3, calculate the focal length of the lens
Bayanin Amsa
u + v = 40
vu = 3
v = 3u
u + 3u = 40
4u = 40
u = 10cm
v = 3u = 30cm
f = uvu+v=10(30)10+30=30040
= 7.5 cm
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