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Tambaya 1 Rahoto
A single force which produces the same effect as a set of forces acting together at a point is known as the
Bayanin Amsa
The single force which produces the same effect as a set of forces acting together at a point is known as the "resultant". In other words, the resultant is the net force that results from combining all the individual forces acting on an object. It represents the combined effect of all the forces acting on the object and is the force that would produce the same motion as the original set of forces acting together. Therefore, when solving problems in physics, it is often useful to find the resultant force in order to determine the overall effect of multiple forces on an object.
Tambaya 2 Rahoto
When water is boiling, it
Bayanin Amsa
When water is boiling, it changes from a liquid state to a gaseous state called steam. This happens when the water is heated to its boiling point, which is when it reaches a temperature of 100 degrees Celsius (212 degrees Fahrenheit) at sea level. As the water is heated, it absorbs energy and the molecules start to move faster and faster, eventually reaching a point where they escape into the air as steam. The temperature of the water during boiling does not change, as all the energy is being used to break the bonds between the water molecules rather than increasing the temperature. Therefore, the options "gets hotter," "increase in mass," and "decreases in mass" are not correct when describing what happens when water is boiling.
Tambaya 3 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 4 Rahoto
The part of the human eye that does similar work as the diaphragm of a camera lens is the
Bayanin Amsa
The part of the human eye that does similar work as the diaphragm of a camera lens is the iris. The iris is the colored part of the eye and is responsible for controlling the amount of light that enters the eye. Just like the diaphragm in a camera lens, the iris can adjust its size to allow more or less light into the eye. This helps to regulate the amount of light reaching the retina, which is responsible for sensing light and transmitting the image to the brain.
Tambaya 5 Rahoto
A vibrator causes water ripples to travel across the surface of a tank. The wave travels 50cm in 2s and the distance between successive crests is 5cm. Calculate the frequency of the vibrator
Bayanin Amsa
The frequency of the vibrator can be calculated using the formula: frequency = speed / wavelength where speed is the speed of the wave, and wavelength is the distance between successive crests. In this case, we are given that the wave travels 50cm in 2s, which means the speed of the wave is: speed = distance / time = 50cm / 2s = 25cm/s We are also given that the distance between successive crests is 5cm, which is the wavelength. Therefore, the frequency of the vibrator is: frequency = speed / wavelength = 25cm/s / 5cm = 5Hz So the correct answer is 5Hz.
Tambaya 6 Rahoto
A ray of light passes through the centre of curvature of a concave mirror and strikes the mirror. At what angle is the ray reflected?
Bayanin Amsa
When a light ray passes through the center of curvature of a concave mirror and strikes the mirror, the reflected ray will be reflected back on itself, creating an angle of 0 degrees. Therefore, the correct answer is 0o.
Tambaya 7 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 8 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
Tambaya 9 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 10 Rahoto
The statement 'Heat lost by the hot body equals that gained by the cold one' is assumed when determining specific that heat capacity by the method of mixtures. Which of the following validates the assumption?
I. Lagging the Calorimeter
II. Ensuring that only S.I units are used
III. Weighing the calorimeter, the lid and the stirrer.
Bayanin Amsa
The assumption 'Heat lost by the hot body equals that gained by the cold one' is based on the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred from one system to another. Thus, to validate this assumption, it's important to have a well-designed and insulated calorimeter so that as little heat as possible is lost to the environment. This is accomplished by lagging the calorimeter (Option I). Additionally, using the correct units (Option II) helps ensure that the energy transfer is accurately calculated and reported. Weighing the calorimeter, the lid, and the stirrer (Option III) is important for accurately measuring the amount of heat transferred, but by itself is not enough to validate the assumption. Therefore, the correct answer is "I and III only".
Tambaya 11 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 12 Rahoto
A well 1km deep is filled with a liquid of density 950kg/m3 and g = 10m/s2 , the pressure at the bottom of the well is
Bayanin Amsa
P = Pa + ρgh = (1.00 × 105
) + (950 × 10 × 1000)
P = 105
+ (95 × 105
) = 105
(1 + 95) = 96 × 105
P = 9.6 × 106
N/m2
Tambaya 14 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 15 Rahoto
When two objects A and B are supplied with the same quantity of heat, the temperature change in A is obtained to be twice that of B. The mass of P is half that of Q. The ratio of the specific heat capacity of A to B is
Bayanin Amsa
θA = 2θB ,
mA | = | 12 | mB |
H = MCθ
mA
cA
θA
= mB
cB
θB
( 1/2 mB
)CA
(2θB
) = mB
cB
θB
CA CB | = | 11 |
⇒ 1 : 1
Tambaya 16 Rahoto
A metal rod has a length of 100cm at 200oC . At what temperature will its length be 99.4cm. If the linear expansivity of the material of the rod is 2 × 10−5C−1
Bayanin Amsa
The linear expansivity of a material describes how its length changes with temperature. If the linear expansivity is given as 2 × 10^-5/°C, this means that for every 1°C change in temperature, the length of the material will change by 2 × 10^-5 times its original length. Given that the rod has a length of 100 cm at 200°C, we can use this information to find its length at a different temperature. If we let L be the length of the rod at temperature T, we can write the relationship as follows: L = 100 cm * (1 + 2 × 10^-5 * (T - 200°C)) To find the temperature at which the rod will have a length of 99.4 cm, we can set L equal to 99.4 cm and solve for T: 99.4 cm = 100 cm * (1 + 2 × 10^-5 * (T - 200°C)) 99.4 cm / 100 cm = 1 + 2 × 10^-5 * (T - 200°C) 0.994 = 1 + 2 × 10^-5 * (T - 200°C) -0.006 = 2 × 10^-5 * (T - 200°C) -0.006 / 2 × 10^-5 = T - 200°C -0.006 / (2 × 10^-5) = T - 200°C -0.006 / (2 × 10^-5) + 200°C = T So the temperature at which the rod will have a length of 99.4 cm is approximately equal to -0.006 / (2 × 10^-5) + 200°C, or -100°C. Therefore, the answer is -100°C.
Tambaya 17 Rahoto
The conductivity of gases at low pressure can be termed as
I. hot cathode emission
II. thermo ionic emission
III. cold cathode emission
IV. Field emission
Bayanin Amsa
As conduction of gases is at low pressure and high voltage, called field or cold cathode emission.
Tambaya 18 Rahoto
In a slide wire bridge, the balance is obtained at a point 25cm from one end of wire 1m long. The resistance to be tested is connected to that end and a standard resistance of 3.6Ω is connected to the other end of the wire. Determine the value of the unknown resistance
Bayanin Amsa
R3.6=7525=13
3R = 3.6
R = 1.2Ω
Tambaya 19 Rahoto
An alternating current can induce voltage because it has
Bayanin Amsa
An alternating current can induce voltage because it has a varying magnetic field. An alternating current (AC) is an electrical current that periodically reverses direction, unlike direct current (DC), which flows in one direction. When an AC current flows through a wire, it generates a magnetic field that changes direction with the current. As the current alternates, the magnetic field expands and contracts, inducing an electromotive force (EMF) in any nearby conductor or coil of wire. This phenomenon is known as electromagnetic induction, and it is the basis for the operation of many electrical devices, such as generators and transformers. The induced voltage depends on the strength and rate of change of the magnetic field and the number of turns in the coil. In summary, an alternating current can induce voltage because it creates a varying magnetic field, which in turn generates an electromotive force in nearby conductors or coils of wire, according to the principle of electromagnetic induction.
Tambaya 20 Rahoto
The pitch of a screw jack is 0.45cm and the arm is 60cm long. If the efficiency of the Jack is 75/π %, calculate the mechanical advantage.
Bayanin Amsa
P = 0.45cm, L = 60cm, Eff = 75/π%
VR | (Screw | system) | = | 2πrP | = | 2πLP |
M.A | = | Eff% × VR100 | = | 75π | × | 1100 | × | 2π × 600.45 | = | 75 × 800300 | = | 200 |
Tambaya 21 Rahoto
A straight wire 15cm long, carrying a current of 6.0A is in a uniform field of 0.40T. What is the force on the wire when it is at right angle to the field
Bayanin Amsa
The force on a current-carrying wire in a uniform magnetic field can be calculated using the equation: F = BILsinθ where F is the force in Newtons, B is the magnetic field strength in Tesla, I is the current in Amperes, L is the length of the wire in meters, and θ is the angle between the wire and the magnetic field. In this problem, the wire is 15cm long (0.15m), carrying a current of 6.0A, and the magnetic field is 0.40T. The angle between the wire and the magnetic field is 90 degrees (since the wire is at right angles to the field). Substituting the given values into the equation, we get: F = (0.40T)(6.0A)(0.15m)sin90 sin90 = 1, so we can simplify the equation to: F = (0.40T)(6.0A)(0.15m) F = 0.36N Therefore, the force on the wire is 0.36N. Answer option C is the correct answer.
Tambaya 22 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 23 Rahoto
Calculate the velocity ratio of a screw jack of pitch 0.2cm if the length of the tommy bar is 23cm
Bayanin Amsa
P = 0.2cm, L = r = 23cm
VR | = | 2?rP | = | 2?LP | = | 2?×230.2 | = | 230? |
Tambaya 24 Rahoto
The point at which the molecules of a loaded wire begin to slide across each other resulting in a rapid increase in extension is
Bayanin Amsa
The point at which the molecules of a loaded wire begin to slide across each other resulting in a rapid increase in extension is called the yield point. At this point, the material no longer behaves elastically and becomes permanently deformed. The yield point is an important parameter in material science and engineering as it indicates the maximum stress a material can withstand before it begins to deform plastically. Therefore, the yield point is a critical factor to consider when designing materials for specific applications.
Tambaya 25 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 26 Rahoto
One newton × One meter equals?
Bayanin Amsa
One newton times one meter is equal to one Joule. A newton is the unit of measurement for force, and a meter is the unit of measurement for distance. When force is applied over a distance, work is done, which is measured in Joules. Therefore, one newton multiplied by one meter results in one Joule of work done. The other options listed (one water, one ampere, one kilogram) are not correct units of measurement for this calculation.
Tambaya 27 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 28 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 29 Rahoto
When blue and green colours of light are mixed, the resultant colour is
Tambaya 30 Rahoto
Electrons were discovered by
Bayanin Amsa
Electrons were discovered by J.J. Thompson. In the late 19th century, he performed a series of experiments using cathode ray tubes, which are glass tubes containing low-pressure gas and electrodes. By applying high voltage, he observed a beam of negatively charged particles traveling from the negative electrode to the positive electrode. He concluded that these particles, which he called "corpuscles," were fundamental units of negative charge and later were renamed electrons. This discovery led to the development of the modern understanding of atomic structure and the electron's role in it.
Tambaya 31 Rahoto
A microscope is focused on a mark on a table, when the mark is covered by a plate of glass 2m thick, the microscope has to be raised 0.67cm for the mark to be once more in focus. Calculate the refractive index.
Bayanin Amsa
R = th = 2cm, d = 0.67cm
n | = | RA | = | RR.d | = | 22-0.67 | = | 1.52 |
Tambaya 32 Rahoto
In which of the points labelled A, B, C, D and E on the conductor shown would electric charge tend to concentrate most
Bayanin Amsa
- Charge are mostly concentrated at the outermost part of a hollow conductor
- Charge are also mostly concentrated at the pointed ends or places with high density point.
Tambaya 33 Rahoto
Lamps in domestic lightings are usually in
Bayanin Amsa
Lamps in domestic lighting are usually connected in parallel. This means that each lamp is connected directly to the power supply, rather than being connected in a series or divergent or convergent configuration. In a parallel configuration, each lamp operates independently of the others, and if one lamp fails, the other lamps will continue to function. This is an important feature for domestic lighting, as it ensures that a single lamp failure will not leave the entire room in darkness. Additionally, in a parallel configuration, each lamp can be controlled independently, for example by a switch or dimmer, without affecting the operation of the other lamps. This allows for greater flexibility in lighting design and control. In summary, lamps in domestic lighting are usually connected in parallel because it allows for independent operation of each lamp and ensures that a single lamp failure does not affect the operation of the others.
Tambaya 34 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 35 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 36 Rahoto
Which of the following is consistent with Charles' law?
I
II
III
IV.
Bayanin Amsa
This is the correct graph. The graph is volume against 1/ temperature where temperature is in Celsius.
Tambaya 37 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 38 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 39 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 40 Rahoto
The earth's gravitational field intensity at its surface is about
(G = 6.7 × 10−11 Nm2 /kg2 , mass of the earth is 6 × 1024 kg, radius of the earth is 6.4 × 106 m, g on the earth = 9.8m/s2 )
Bayanin Amsa
The earth's gravitational field intensity at its surface can be calculated using the formula: g = G * M / r^2 where G is the gravitational constant, M is the mass of the earth, r is the radius of the earth, and g is the gravitational field intensity at the surface of the earth. Substituting the given values, we get: g = (6.7 × 10^-11 Nm^2/kg^2) * (6 × 10^24 kg) / (6.4 × 10^6 m)^2 g = 9.8 N/kg (approx.) Therefore, the answer is 9.8N/kg.
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