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Tambaya 1 Rahoto
Which of the following bodies, each with centre of gravity G, lying on a horizontal table, is/are in unstable equilibrium?
Bayanin Amsa
- I and II are in neutral equilibrium. They will roll continuously on the table
- III is a body with high centre of gravity (unstable)
- IV is a body with high centre of gravity (stable)
Tambaya 3 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 4 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 5 Rahoto
Any line or section taken through an advancing wave in which all the particles are in the same phase is called the
Bayanin Amsa
The answer is: wave front. A wave front is any imaginary line or surface that connects all points of a wave that are in the same phase, meaning they are at the same point in their cycle. In other words, it is a line or surface that separates the points of a wave that are in-phase from those that are out-of-phase. For example, consider the ripples on the surface of a pond when a stone is thrown in. The wave fronts are the concentric circles that emanate from the point where the stone entered the water. All points along a given circle are in-phase, meaning the water molecules at those points are at the same point in their oscillation cycle. In summary, a wave front is a line or surface that separates points in a wave that are in-phase from those that are out-of-phase.
Tambaya 6 Rahoto
When blue and green colours of light are mixed, the resultant colour is
Tambaya 7 Rahoto
A car moving at 20m/s with its horn blowing (f = 1200Hz) is chasing another car going at 15m/s. What is the apparent frequency of the horn as heard by the driver being chased?
Bayanin Amsa
f1 | = | f(v - vo )v - vs | = | 1200(340 - 15)340 - 20 | = | 1.22KHz |
Tambaya 8 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 9 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 10 Rahoto
If the attraction of the sun is suddenly ceased, the earth would continue to move in a straight line making a tangent with the original orbit. This statement is derived from Neutron's
Bayanin Amsa
The correct answer is the First law of motion. The First law of motion, also known as the law of inertia, states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. In this case, the earth is moving in its orbit around the sun because of the force of gravity between the two objects. If the force of gravity suddenly ceased, the earth would no longer be acted upon by an external force and would continue to move in a straight line, making a tangent with its original orbit. This idea is attributed to Sir Isaac Newton, who developed the laws of motion and the law of universal gravitation. However, the specific statement mentioned in the question is derived from the First law of motion.
Tambaya 11 Rahoto
The limiting frictional force between two surface 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 surface
Bayanin Amsa
The correct answer is "I and IV only". The limiting frictional force between two surfaces depends on the normal reaction between the surfaces (I) and the nature of the surface (IV). The normal reaction is the force that the surfaces exert on each other perpendicular to the plane of contact. The greater the normal reaction, the greater the frictional force that can be applied before motion occurs. The nature of the surface is determined by factors such as roughness, hardness, and texture, which can affect the frictional force. The area of surface in contact (II) does not directly affect the limiting frictional force, although it can affect the force required to initiate motion. For example, if the area of contact is small, the pressure between the surfaces will be higher, making it harder to initiate motion. The relative velocity between the surfaces (III) also does not directly affect the limiting frictional force, although it can affect the force required to maintain motion. If the surfaces are already in motion, a lower force may be required to keep them moving than to initiate motion. In summary, the limiting frictional force between two surfaces depends primarily on the normal reaction and the nature of the surface, and is not directly affected by the area of contact or the relative velocity between the surfaces.
Tambaya 13 Rahoto
During the transformation of matter from the solid to the liquid state, the heat supplied does not produce temperature increase because
Bayanin Amsa
When a solid is heated to its melting point, the heat supplied is used to overcome the intermolecular forces holding the molecules in a fixed position, resulting in the breaking of these bonds. As a result, the solid transforms into a liquid without any change in temperature. This is because the heat energy supplied is used in breaking the bonds between molecules rather than increasing the kinetic energy of the molecules, which is what causes an increase in temperature. Therefore, the correct option is: "all the heat is used to break the bonds holding the molecules of the solid together."
Tambaya 14 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 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
The pin-hole camera produces a less sharply defined image when the
Bayanin Amsa
The pin-hole camera produces a less sharply defined image when the pin-hole is larger. A pin-hole camera works by allowing light to pass through a small hole (the pin-hole) and project an inverted image of the outside world onto a screen or surface located behind the hole. The smaller the pin-hole, the sharper the resulting image, as light passing through a smaller hole produces less diffraction or spreading out of the light. When the pin-hole is larger, more light enters the camera, but the light rays also become more scattered, resulting in a less well-defined image. This is because the larger opening allows more light rays to enter at different angles, creating a wider range of paths that the light can take as it travels through the camera and onto the screen. As a result, the image is less clear and less defined, with less sharp edges and more blurring. is the correct answer because it correctly identifies the effect of a larger pin-hole on the image produced by the pin-hole camera. less illumination, would actually produce a dimmer image, but it would not affect the sharpness or definition of the image. the distance of the screen from the pin-hole, and the distance of the object from the pin-hole, would affect the size of the image and the scale of the objects, but they would not affect the sharpness or definition of the image.
Tambaya 17 Rahoto
A copper rod, 5m long when heated through 20c, expands by 1mm. If a second copper rod, 2.5m long is heated through 5c, by how much will it expand?
Bayanin Amsa
l1
= 5m, ΔT = 10c, l2
- l1
= 1mm
l1
= 2.5m, ΔT = 5c, l2
- l1
= ?
using | α | = | l2 - l1 l1 ΔT |
15(10) | = | l2 - l1 2.5(5) |
l2 | - | l2 | = | 2.5(5)5(10) | = | 14 | = | 0.25mm |
Tambaya 18 Rahoto
A train has an initial velocity of 44m/s and an acceleration of -4m/s2 . Calculate its velocity after 10 seconds
Bayanin Amsa
The velocity of the train after 10 seconds can be calculated using the formula: v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Substituting the given values, we get: v = 44 m/s + (-4 m/s^2) x 10 s v = 44 m/s - 40 m/s v = 4 m/s Therefore, the velocity of the train after 10 seconds is 4m/s. Answer option D is correct. Explanation: The train has an initial velocity of 44 m/s and an acceleration of -4 m/s^2. The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which means that the train is slowing down. After 10 seconds, the train's velocity decreases by 40 m/s (4 m/s^2 x 10 s) to reach a final velocity of 4 m/s.
Tambaya 19 Rahoto
The following are some units
I. Ns
II. Non
III. Nm−2
IV. J°K−1
V. JKj−1
What are the units of latent heat?
Bayanin Amsa
Latent heat or specific latent heat = L
Heat | energy | = | mL | or | L | = | Hm | = | energymass |
Tambaya 20 Rahoto
The mass of a nucleus is the
Bayanin Amsa
The mass of a nucleus is the total number of its protons and neutrons. The protons and neutrons are the subatomic particles that make up the nucleus of an atom. The mass of an atom is mostly concentrated in its nucleus, and the electrons orbiting the nucleus have a much smaller mass. Therefore, the mass of an atom is mostly determined by the number of protons and neutrons in its nucleus. The number of protons determines the element, and the number of neutrons can vary, resulting in isotopes of that element.
Tambaya 21 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 22 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 23 Rahoto
Which of the following characteristics of a wave is used in the measurement of the depth of the Sea?
Bayanin Amsa
Depth of sea can be measured by echo, a reflected sound waves.
Tambaya 24 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 25 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 26 Rahoto
A thermocouple thermometer is connected to a millivoltmeter which can read up to 10mV. When one junction is in ice at 0°C and the other is steam at 100°C, the millivoltmeter reads 4mV. What is the maximum temperature which this arrangement can measure
Bayanin Amsa
The maximum temperature which this arrangement can measure is 250°C. A thermocouple thermometer works by using the thermoelectric effect, which is the phenomenon that occurs when two dissimilar metals are joined together to form a loop and a temperature difference is established between the two junctions. This temperature difference generates a small electrical voltage, which can be measured using a millivoltmeter. The voltage generated is proportional to the temperature difference between the two junctions. In the case of the thermocouple thermometer described, one junction is in ice at 0°C and the other is steam at 100°C, and the millivoltmeter reads 4mV. This means that the voltage generated by the thermocouple is 4 millivolts, which corresponds to a temperature difference of 100°C. However, the millivoltmeter can only read up to 10mV, so the maximum temperature difference it can measure is 10mV / 4mV/°C = 250°C. This means that the maximum temperature which this arrangement can measure is 250°C.
Tambaya 27 Rahoto
Which of the following statements is/are correct for a freely falling body?
I. the total is entirely kinetic
II. the ratio of potential energy to kinetic energy is constant
III. the sum of potential and kinetic energy is constant
Bayanin Amsa
The correct answer is "III only". A freely falling body is one that is falling under the influence of gravity and experiences no other force or constraint. In this situation, the total energy of the body is conserved, meaning that the sum of its potential and kinetic energy remains constant. The potential energy of a body is directly proportional to its height above the ground, and its kinetic energy is directly proportional to its velocity. As the body falls, its potential energy decreases and its kinetic energy increases, but the total energy remains constant. Statement III is correct because the sum of potential and kinetic energy is indeed constant for a freely falling body. Statement I is incorrect because the body has both potential and kinetic energy, so the total energy is not entirely kinetic. Statement II is incorrect because the ratio of potential energy to kinetic energy is not constant for a freely falling body, as both are changing as the body falls.
Tambaya 28 Rahoto
The height at which the atmosphere cases to exist is about 80km. If the atmospheric pressure on the ground level is 760mmHg, the pressure at a height of 20km above the ground level is
(ρm = 13.6g/cm3 ρ = 0.00013g/cm3 )
Bayanin Amsa
ρm
hm
= ρa
ha
13.68(760 - p) × 10−3
= 13 × 10−5
(20 × 103
)
760 | - | p | = | 13 × 10−5 × 20 × 103 13.68 × 10−3 | = | 19.00 | × | 101 |
760 - p = 190
p = 760 - 190 = 570mmHg
Tambaya 29 Rahoto
Aluminium is sometimes used as the leaf of an electroscope because it
Bayanin Amsa
- Aluminium can be made in thin sheet like Gold.
- the leaf is a thin material that can be diverged easily.
Tambaya 30 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 31 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 32 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 33 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 34 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 35 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 36 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 37 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 38 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 39 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 40 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.
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