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Tambaya 2 Rahoto
Two objects of masses 80kg and 50kg are separated by a distance of 0.2m. If the gravitational constant is 6.6 x 10-11Nm kg, calculate the gravitational attraction between them.
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Tambaya 3 Rahoto
A team trap is a component of the apparatus used in determining the specific latent heat of vaporisation of steam. In the steady steady state, the steam trap
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In the apparatus used to determine the specific latent heat of vaporisation of steam, the steam trap is used to ensure that only dry steam gets into the calorimeter. The steam trap works by allowing the steam to pass through it, but trapping any condensed steam or water droplets that may be present in the steam. This is important because if any water droplets or condensed steam get into the calorimeter, they will add to the heat energy in the system and affect the accuracy of the measurement. By trapping any condensed steam, the steam trap ensures that only dry steam gets into the calorimeter, which allows for a more accurate measurement of the specific latent heat of vaporisation of steam. Therefore, the correct option is: ensures that only dry steam gets into the calorimeter.
Tambaya 4 Rahoto
Which of the following statements about wave is not correct? i. A wavefront is a line which contains all particles whose vibration are in phase. ii. The direction of propagation of a wave is the link drawn parallel to the wavefront. iii. A wavefront is a circle which is common to particles that are in the same state of disturbance
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The incorrect statement about waves is iii. A wavefront is not a circle which is common to particles that are in the same state of disturbance. Rather, a wavefront is an imaginary surface that connects all points in space that are reached by a wave at the same time during its propagation. Hence, the correct answer is (e) ii and iii only. (i) A wavefront is a line that connects all particles in the medium that are vibrating in phase, meaning that their displacement from equilibrium is the same at any given moment in time. This means that they reach their maximum and minimum amplitudes at the same time. (ii) The direction of propagation of a wave is always perpendicular to the wavefront. This is because each point on the wavefront is at the same phase of vibration and so has the same displacement from equilibrium at any given moment in time. Therefore, the wavefront itself is a surface of constant phase, and the direction of propagation is the normal to this surface. (iii) As mentioned earlier, a wavefront is not a circle which is common to particles that are in the same state of disturbance. However, it can be a circle in the case of a spherical wave, which emanates from a point source and spreads out uniformly in all directions.
Tambaya 5 Rahoto
A man standing 300m away from the wall sounds a whistle. The echo from the wall reaches him 1.8sec. later, calculate the velocity of sound in air
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The velocity of sound in air can be calculated using the formula: velocity = distance / time In this case, the distance travelled by the sound is twice the distance from the man to the wall, since the sound has to travel to the wall and back. So, the distance travelled by the sound is: 2 x 300m = 600m The time taken for the sound to travel this distance (to the wall and back) is: 1.8s So, using the formula: velocity = distance / time velocity = 600m / 1.8s velocity = 333.3ms-1 Therefore, the velocity of sound in air is 333.3ms-1. Hence, the correct option is ii. 333.3ms-1.
Tambaya 6 Rahoto
the diagram above shows the position of a simple pendulum set in motion. At which of the positions does the pendulum have maximum kinetic energy?
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Tambaya 7 Rahoto
A body moves with a constant speed but has an acceleration. this process if it
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A body that moves with a constant speed but has an acceleration is moving in a circular path. Acceleration is the rate at which the velocity of an object changes. When an object moves in a circular path, it is constantly changing direction, even if it is moving at a constant speed. This means that its velocity is changing, and therefore, it has an acceleration. The direction of the acceleration is always towards the center of the circular path, and its magnitude is given by the formula a = v^2/r, where a is the acceleration, v is the speed of the object, and r is the radius of the circular path. Therefore, the correct option is: moves in a circle.
Tambaya 8 Rahoto
The unit of stress is
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The unit of stress is Nm-2 (Newtons per square meter), also known as Pascal (Pa). Stress is defined as force per unit area and is a measure of the internal forces acting on a material. When a force is applied to a material, it creates internal forces that tend to resist the deformation or change in shape. Stress is the measure of these internal forces per unit area. The unit Nm represents work or energy, while N represents force, and Nm2 does not have a physical meaning.
Tambaya 9 Rahoto
A water fall is 420m high. Calculate the difference in temperature of the water between the top and bottom of the water fall. Neglect heat loses(g = 10.0m-2 specific heat capacity of water = 4.20 x 10-3 JKg-1 K-1)
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Tambaya 10 Rahoto
The set-up illustrated above shows a capillary tube of uniform cross-sectional area in two different arrangements. Using the data in the diagrams. Calculate the pressure of the atmosphere
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From the two diagrams, we can see that the heights of the mercury columns in both the open and closed ends of the capillary tube are different. This difference in height is due to the atmospheric pressure acting on the surface of the mercury in the dish. Let h1 and h2 be the heights of the mercury columns in the two arrangements, and let d be the density of mercury. For the first arrangement, we have: h1 = 7 cm h2 = 6 cm The pressure difference between the two ends of the capillary tube is given by the difference in height of the mercury columns: ΔP = dgh where g is the acceleration due to gravity. ΔP = (13.6 x 103 kg/m3) x (10.0 m/s2) x (0.01 m) x (h1 - h2) ΔP = 1.36 x 103 (h1 - h2) Pa For the second arrangement, we have: h1 = 75 cm h2 = 74 cm Using the same formula, we get: ΔP = 1.36 x 103 (h1 - h2) Pa ΔP = 1.36 x 103 (75 - 74) Pa ΔP = 1.36 x 103 Pa To find the pressure of the atmosphere, we need to add this pressure difference to the atmospheric pressure: Patm = P0 + ΔP where P0 is the atmospheric pressure at the surface of the mercury in the dish. Since the height of the mercury column in the dish is 76 cm, we have: P0 = dgh P0 = (13.6 x 103 kg/m3) x (10.0 m/s2) x (0.01 m) x (76 cm) P0 = 1.032 x 105 Pa Therefore, the atmospheric pressure is: Patm = P0 + ΔP Patm = 1.032 x 105 Pa + 1.36 x 103 Pa Patm = 1.046 x 105 Pa Converting this to cm of Hg, we get: Patm = 1.046 x 105 Pa x (1 cm/133.322 Pa) Patm = 784.8 cm of Hg Therefore, the pressure of the atmosphere is approximately 784.8 cm of Hg. The answer is closest to 75 cm
Tambaya 12 Rahoto
Which of the following statements above viscosity are correct? when a ball falls through a viscous liquid i. viscosity opposes the gravitational force on the ball. ii. viscosity opposes the upthrust on the ball. iii. viscosity is in the same direction as the upthrust on the ball. iv. as the ball falls faster the more viscous the liquid is
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Tambaya 13 Rahoto
A fixed mass of gas of volume 600cm3 at a temperature of 27oC is cooled at constant pressure to a temperature of 0oC. What is the change in volume?
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We can use the combined gas law to solve this problem. The combined gas law states that the product of pressure and volume is directly proportional to the product of temperature and the number of moles of gas, assuming constant mass. Mathematically, this can be written as: P1V1/T1 = P2V2/T2 where P is pressure, V is volume, T is temperature, and the subscripts 1 and 2 refer to the initial and final states, respectively. We are told that the initial volume V1 is 600 cm3 and the initial temperature T1 is 27oC = 300K. We also know that the pressure P remains constant throughout the process, and that the final temperature T2 is 0oC = 273K. We can solve for the final volume V2: P1V1/T1 = P2V2/T2 V2 = (P1/P2) * (T2/T1) * V1 Since the pressure is constant, we can cancel it out, and substitute in the given values: V2 = (273K/300K) * 600 cm3 = 546 cm3 Therefore, the change in volume is 600 cm3 - 546 cm3 = 54 cm3. So, the correct answer is 54 cm3.
Tambaya 14 Rahoto
Two forces, whose resultant is 100N are perpendicular to each other. If one of them makes an angle of 60o with the resultant, calculate the magnitude. (Sin 60o = 0.8660, Cos 60o = 0.500)
Tambaya 15 Rahoto
A wire, 20m long is heated from a temperature of 5oC to 55oC. If the change in length is 0.020m, calculate the linear expansivity of the wire
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Linear expansivity is a measure of how much a material changes in length when its temperature changes. It is denoted by the Greek letter alpha (α) and is expressed in units of K-1. The formula for linear expansivity is: α = ΔL / (L_i * ΔT) Where: ΔL = change in length L_i = initial length ΔT = change in temperature Using the values given in the question, we have: ΔL = 0.020 m L_i = 20 m ΔT = (55 - 5) = 50 K Substituting these values into the formula, we get: α = 0.020 / (20 * 50) = 2.0 x 10-5 K-1 Therefore, the linear expansivity of the wire is 2.0 x 10-5 K-1. Answer choice (C) is the correct answer.
Tambaya 16 Rahoto
An electron charge 1.6 x 10-19C is accelerated in vacuum from rest at zero volt towards a plate at 40k V. Calculate the kinetic energy of the electron
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The work done in accelerating the electron is given by W = qV where q is the charge of the electron and V is the potential difference. The potential difference is given as 40 kV, which is equivalent to 40 x 10^3 V. Therefore, the work done in accelerating the electron is: W = (1.6 x 10^-19 C) x (40 x 10^3 V) = 6.4 x 10^-15 J This work done is equal to the kinetic energy gained by the electron since it started from rest. Therefore, the kinetic energy of the electron is: K.E = 6.4 x 10^-15 J Hence, the correct option is: 3.2 x 10^-15 J.
Tambaya 17 Rahoto
Calculate the refractive index of the material of the glass block shown in the diagram if YZ = 4cm
Tambaya 18 Rahoto
In the circuit above R is a resistor whose resistance increases with increases with increase in temperature L1 and L2 are identical lamps. If the temperature of R increases
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Tambaya 19 Rahoto
A ray of light is incident at an angle of 30o on a glass prism of refractive index 1.5. Calculate the angle through which the ray is minimally deviated in the prism. (The medium surrounding the prism is air)
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Tambaya 20 Rahoto
The sound from a bell in an enclosed jar gradually faints away while the jar is being evacuated. Which of the following explains this observation?
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The observation that the sound from a bell in an enclosed jar gradually faints away while the jar is being evacuated can be explained by the fact that there is no more material medium for the sound waves to propagate through. Sound waves are vibrations that propagate through a medium such as air, water, or solids. When the jar is evacuated, the air is pumped out, resulting in a decrease in pressure inside the jar. As the pressure decreases, the density of the air inside the jar decreases as well. Sound waves require a medium to travel through, and in this case, there is less and less air inside the jar, resulting in fewer particles available to transmit the sound waves. Eventually, when the jar is fully evacuated, there is no more air left inside the jar to transmit the sound waves. Therefore, the correct option is: there is no more material medium.
Tambaya 21 Rahoto
A 90W immersion heater is used to supply energy for 5 minutes. The energy supplied is used to completely melt 160g of a solid at its melting point. Calculate the specific latent heat of the solid.
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To calculate the specific latent heat of the solid, we need to use the formula: Q = mL where Q is the amount of energy supplied by the immersion heater, m is the mass of the solid, and L is the specific latent heat of the solid. First, we need to calculate the amount of energy supplied by the immersion heater. We know that the heater has a power output of 90W and it is used for 5 minutes, so the total energy supplied is: Q = Pt = 90W x 5min x 60s/min = 27,000 J Next, we can use this value of Q and the mass of the solid, which is 160g, to calculate the specific latent heat L: L = Q/m = 27,000 J / 160g = 168.75 Jg-1 Therefore, the specific latent heat of the solid is 168.75 Jg-1. Option (D) 168.75 Jg-1 is the correct answer.
Tambaya 22 Rahoto
The boiling point of a liquid depends on the following except the
Tambaya 23 Rahoto
The graph above shows the expansion of water as the temperature increases from 0oC. Which of the following deductions from the graph are true? i. Water has its maximum density at Q. ii. The volume of water is greater at 0oC than at 4oC. iii. The volume of water decreases uniformly when cooled from 100oC. iv. When water solidifies, its volume increases
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Tambaya 24 Rahoto
plane as 63cm of HG while a ground observer records a reading of 75cm of Hg with his barometer. assuming that the density of air is constant, calculate the height of the plane above the ground. (Take the relative densities of air and mecury as 0.00136 and 13.6 respectively
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To solve this problem, we can use the formula: h = (ρm/ρa) * (P0 - P) where: h = height of the plane above the ground ρm = density of mercury (13.6 g/cm^3) ρa = density of air (0.00136 g/cm^3) P0 = atmospheric pressure at ground level (75 cm of Hg) P = atmospheric pressure at the height of the plane (63 cm of Hg) First, we need to convert the units of the densities from g/cm^3 to kg/m^3 to be consistent with the units of the atmospheric pressure in meters of mercury (m of Hg): ρm = 13.6 * 1000 kg/m^3 = 13600 kg/m^3 ρa = 0.00136 * 1000 kg/m^3 = 1.36 kg/m^3 Now, we can plug in the values and solve for h: h = (13600/1.36) * (75 - 63) h = 10000 * 12 h = 120000 meters Therefore, the height of the plane above the ground is 120,000 meters. Note: This answer seems unreasonable as it is much higher than the typical cruising altitude of commercial airplanes. This may be due to a mistake in the given information or in the calculations.
Tambaya 25 Rahoto
In the arrangement illustrated above, Y and B are yellow and blue transparent light filters respectively through the filters is
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Tambaya 26 Rahoto
Two capacitors C1 and C2 are connected as shown in the diagram. The capacitance C2 is twice C1 when the key is opened the energy stored up in C1 is W. If the key is later closed and the system is allowed to attain electrical equilibrium, the total energy stored in the system will be
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When the key is open, the two capacitors are not connected and the charge on each of them is zero. The energy stored in capacitor C1 is given by the formula: W = (1/2) * C1 * V^2, where V is the voltage across capacitor C1. When the key is closed, the two capacitors are connected in parallel, and a charge Q will flow from one plate of C1 to the other plate of C2 until the voltage across both capacitors is the same. Let the voltage across both capacitors be V. Then, the charge on C1 will be Q/2, and the charge on C2 will be Q. The total energy stored in the system can be calculated as follows: W(total) = (1/2) * C1 * V^2 + (1/2) * C2 * V^2 Substituting C2 = 2C1 and Q = CV, we get: W(total) = (1/2) * C1 * V^2 + (1/2) * 2C1 * V^2 = (1/2) * 3C1 * V^2 = (3/2) * W Therefore, the total energy stored in the system when the key is closed and the system is allowed to attain electrical equilibrium is (3/2) times the energy stored in C1 when the key is open. Hence, the correct option is 3W.
Tambaya 27 Rahoto
If the frequency of the e.m.f. source in the a.c. circuit illustrated above is \(\frac{500}{\pi}\)Hz. What is the reactance of the inductor?
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Tambaya 28 Rahoto
Which of the following has the highest surface tension?
Tambaya 29 Rahoto
The saturation vapour of a liquid depends on its
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The saturation vapor pressure of a liquid depends on its temperature. When a liquid evaporates, it turns into a gas or vapor. This process of evaporation depends on the energy of the liquid molecules. The higher the temperature, the more energy the molecules have, and the more likely they are to escape from the surface of the liquid and become a gas. As the concentration of vapor molecules increases above a liquid surface, some of them may condense back into the liquid phase, and this creates an equilibrium between the liquid and vapor phases, called saturation. The saturation vapor pressure is the pressure exerted by the vapor in equilibrium with its liquid phase at a given temperature. So, as the temperature of the liquid increases, the saturation vapor pressure increases, meaning more molecules are in the gas phase. On the other hand, as the temperature decreases, the saturation vapor pressure decreases, indicating fewer molecules in the gas phase. Therefore, the saturation vapor pressure of a liquid is determined by its temperature and is independent of the volume, mass, density, or pressure of the liquid.
Tambaya 30 Rahoto
The half-life of a radioactive substance is 2 seconds. Calculate the decay constant
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Tambaya 31 Rahoto
Which of the following are true of plane-polarised right? i. Plane polarisation of light is the formation of hydrogen bubbles on the particle of light. ii. Plane components. iii. A plane polarised light vibrates in one plane. iv. Polarisation of light is characteristic of transverse vibration
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Tambaya 33 Rahoto
Melting ice cools an orange drink far better than the same mass of ice-cold water because
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Melting ice cools an orange drink better than the same mass of ice-cold water because ice absorbs latent heat during melting. Latent heat is the heat energy required to change the state of a substance without changing its temperature. When ice melts, it absorbs latent heat from the drink to overcome the intermolecular forces holding its particles together. This absorption of latent heat results in cooling of the drink. On the other hand, ice-cold water has already reached its melting point, and there is no further absorption of latent heat during cooling. Water has a higher specific heat than ice, which means it can absorb more heat energy without changing its temperature. However, this does not contribute to better cooling of the drink. Option (i) is incorrect because ice-cold water is actually at a lower temperature than melting ice. Option (iii) is also incorrect because while ice floats, it does not cool the air above the drink significantly. Option (v) is not entirely correct because while ice can make better thermal contact than water, it is the absorption of latent heat that makes it more effective in cooling the drink. Therefore, the correct answer is option (iv) ice absorbs latent heat during melting.
Tambaya 34 Rahoto
A body moving with uniform acceleration has two points(5, 15) and (20, 60) on the velocity-times graph of its motion. Calculate
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Since the body is moving with uniform acceleration, the graph of its velocity vs time will be a straight line. Let the initial velocity of the body be u, and the acceleration be a. From the two given points, we can determine two equations: 1. At t = 5 s, v = 15 m/s: 15 = u + 5a 2. At t = 20 s, v = 60 m/s: 60 = u + 20a Subtracting equation 1 from equation 2, we get: 45 = 15a a = 3 m/s^2 Therefore, the acceleration of the body is 3.00 ms^-2. is the correct answer.
Tambaya 35 Rahoto
The diagram above shows a current carrying wire between the poles of a magnet. In which direction would the wire tend to move?
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Tambaya 37 Rahoto
Which of the following is not a mechanical wave?
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Radio waves are not mechanical waves. Mechanical waves are waves that require a physical medium, such as a solid, liquid, or gas, to travel through. They transfer energy through the vibration of particles in the medium. On the other hand, radio waves are a type of electromagnetic wave, which means they do not require a physical medium to travel through. They can travel through a vacuum, such as space, and are created by the acceleration of electric charges. Radio waves are used for communication, including radio and television broadcasting, as well as cellular and satellite communication. Therefore, the answer is option (iii) radio waves.
Tambaya 38 Rahoto
A ball is projected horizontally from the top of a hill with a velocity of 20m-1. If it reaches the ground 4 seconds later, what is the height of the hill? (g = 10ms-2)
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We can solve this problem using the kinematic equations of motion. Since the ball is projected horizontally, its initial vertical velocity is zero. We can use the following equation to find the height of the hill: h = (1/2)gt^2 Where h is the height of the hill, g is the acceleration due to gravity, and t is the time taken for the ball to reach the ground. In this case, g = 10ms^-2 and t = 4s. Plugging these values into the equation, we get: h = (1/2)(10)(4)^2 = 80m Therefore, the height of the hill is 80 meters. This is option (C). To summarize, the ball is projected horizontally with a velocity of 20m/s, which means its initial vertical velocity is zero. Using the kinematic equation for displacement, we can find the height of the hill. The ball takes 4 seconds to reach the ground, so we substitute this value along with the acceleration due to gravity into the equation. The resulting answer is 80 meters, which is the height of the hill.
Tambaya 39 Rahoto
what part of the camera corresponds to the iris of the eye?
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The part of the camera that corresponds to the iris of the eye is the diaphragm. Just like the iris controls the amount of light entering the eye, the diaphragm controls the amount of light entering the camera. It is a circular opening in the camera lens that can be adjusted to vary the size of the aperture. When the diaphragm is wider, more light enters the camera, and when it is narrower, less light enters. This helps control the depth of field and the amount of light that hits the camera's sensor or film.
Tambaya 40 Rahoto
a tap supplies water at 26oC while another supplies water at 82oC. If a man wishes to bathe with water at 40oC the ratio of hot mass of hot water to that of cold water required is
Tambaya 41 Rahoto
Which of the following statements is not correct? Isotopes of an element have
Tambaya 42 Rahoto
What type of motion does the skin of a talking drum perform when its being struck with the drum stick?
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The skin of a talking drum performs a vibratory motion when it is being struck with the drum stick. When the drum is struck, the skin vibrates back and forth, producing sound waves that travel through the air. The skin moves in a series of up-and-down motions that are perpendicular to the surface of the drum. The shape of the drum, and the tension of the skin, determine the frequency of the vibration and the pitch of the resulting sound. The player can vary the pitch by squeezing the drum under their arm or by applying pressure to the skin with their non-dominant hand. Therefore, the correct option is vibratory.
Tambaya 43 Rahoto
The work function of a metal is 4.65ev and the metal is illuminated with a radiation of 6.86ev. What is the kinetic energy of the electrons ejected from the surface of the metal?
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The kinetic energy of the ejected electrons can be calculated using the equation: Kinetic energy of electron = Energy of incident radiation - Work function of the metal Substituting the given values, we have: Kinetic energy of electron = 6.86ev - 4.65ev Kinetic energy of electron = 2.21ev Therefore, the kinetic energy of the electrons ejected from the surface of the metal is 2.21ev. Hence, the correct option is (b) 2.21ev.
Tambaya 44 Rahoto
An orange fruit drops to the ground from the top of a tree 45m tall. How long does it takes to reach the ground? (g = 10ms-2)
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To calculate the time it takes for the orange fruit to hit the ground, we can use the following formula: d = 1/2 * g * t^2 where d is the distance (in meters) the orange falls, g is the acceleration due to gravity (in meters per second squared), and t is the time (in seconds) it takes for the orange to fall. In this case, the orange falls from a height of 45m, and g is 10m/s^2. We can plug these values into the formula and solve for t: 45 = 1/2 * 10 * t^2 Simplifying the equation: t^2 = 9 t = 3 Therefore, it takes 3 seconds for the orange fruit to hit the ground. This is the answer to the problem.
Tambaya 45 Rahoto
When a metal surface is irradiated, photoelectrons may be ejected from the metal. The kinetic energy of the ejected electrons depends on the
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The kinetic energy of the photoelectrons ejected from a metal surface when irradiated depends on the frequency of the radiation. This is known as the photoelectric effect. When a photon of light is absorbed by a metal surface, it can transfer its energy to an electron within the metal, causing the electron to be ejected from the surface. The amount of energy transferred to the electron depends on the frequency of the photon, which is related to its wavelength. Photons with higher frequencies (shorter wavelengths) have more energy than photons with lower frequencies (longer wavelengths), and thus can transfer more energy to the ejected electron, resulting in a higher kinetic energy. Therefore, the frequency of the radiation is the key factor that determines the kinetic energy of the ejected electrons.
Tambaya 46 Rahoto
An induction coil is generally used to
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An induction coil is generally used to produce a large output voltage. An induction coil is a type of transformer that uses a magnetic field to convert a low voltage (input voltage) to a high voltage (output voltage). The coil works on the principle of electromagnetic induction, where a changing magnetic field induces an electric current in a nearby conductor. In an induction coil, the primary coil is connected to a source of electrical energy, and the secondary coil is connected to the load. When an electric current flows through the primary coil, it creates a changing magnetic field that induces an electric current in the secondary coil, resulting in a large output voltage. The ratio of the number of turns in the primary coil to the number of turns in the secondary coil determines the voltage transformation ratio.
Tambaya 47 Rahoto
A house is supplied with a 240V a.c mains. To operate a door bell rated at 8v, a transformer is used if the number of turns in the primary coil of the transformer is 900. calculate the number of turns in the secondary coil of the transformer
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To calculate the number of turns in the secondary coil of the transformer, we need to use the formula: Np/Ns = Vp/Vs Where Np is the number of turns in the primary coil, Ns is the number of turns in the secondary coil, Vp is the voltage in the primary coil, and Vs is the voltage in the secondary coil. In this case, we know that the voltage in the primary coil is 240V and the voltage in the secondary coil is 8V. We also know that the number of turns in the primary coil is 900. Plugging these values into the formula, we get: 900/Ns = 240/8 Simplifying this equation, we get: 900/Ns = 30 To solve for Ns, we can multiply both sides by Ns: 900 = 30Ns Dividing both sides by 30, we get: Ns = 30 Therefore, the number of turns in the secondary coil of the transformer is 30. So, the correct option is "30".
Tambaya 48 Rahoto
A particle of charge q and mass m moving with a velocity v enters a uniform magnetic field b in the direction of the field. The force on the particle is
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A particle of charge q and mass m moving with a velocity v enters a uniform magnetic field B in the direction of the field. The force on the particle is given by the equation F = qVB sinθ, where θ is the angle between the velocity vector and the magnetic field vector. In this case, since the particle enters the field in the direction of the field, θ = 0, and sinθ = 0. Therefore, the force on the particle is zero (F = qVB sinθ = qVB(0) = 0). Option (d) is correct - the force on the particle is zero. Option (a) is incorrect because it represents the formula for the magnitude of the force on a charged particle moving perpendicular to the magnetic field. Option (b) is incorrect because it represents the formula for the acceleration of a charged particle in a magnetic field. Option (c) is incorrect because it represents the formula for the radius of the circular path followed by a charged particle moving perpendicular to the magnetic field. Therefore, the correct answer is - O.
Tambaya 49 Rahoto
In which of the following is a stationary wave produced? i. A vibrating turning fork held near the end of a resonance tube closed at one end. ii. A string tightly stretched between two points and plucked at its middle. iii. The tuning fork vibrating in air
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A stationary wave is a type of wave that appears to be standing still, even though it is made up of two waves traveling in opposite directions. Option i describes a resonance tube closed at one end and a vibrating tuning fork held near the end. In this situation, the sound wave produced by the tuning fork travels down the resonance tube and reflects off the closed end, creating a stationary wave. Option ii describes a string that is tightly stretched between two points and plucked at its middle. In this situation, the string vibrates back and forth, creating a wave that travels along the string and reflects back when it reaches the ends, creating a stationary wave. Option iii describes a tuning fork vibrating in air, which produces a sound wave that travels through the air as a longitudinal wave. This wave does not create a stationary wave since it does not reflect back and interfere with itself. Therefore, the correct answer is option i and ii only.
Tambaya 50 Rahoto
Which of the following remains unchanged as light travels from one medium to the other? i. speed. ii. wavelength. iii. frequency.
Tambaya 51 Rahoto
A boatman facing north wants to cross a flowing river to a point directly opposite its position at the other bank. If the river is flowing eastwards, in what direction should he row his boat?
Tambaya 52 Rahoto
Which of the following is stored by dry Leclanche cell?
Tambaya 53 Rahoto
Two mirrors are inclined as shown in the diagram. A ray of light RO strikes the arrangement at O and emerges along PO. The emergent ray has been deviated through
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Tambaya 54 Rahoto
The diagram represents a block-and-tackle pulley system on which an effort of 50N is just able to lift a load of weight W. If the efficiency of the machine is 40%. find the value of W
Tambaya 55 Rahoto
The diagram above represents six maximum and minimum thermometer. If the temperature of the surrounding falls, which of the following correctly states how the thermometer would respond to the change in temperature?
Tambaya 57 Rahoto
The product PV where P is pressure and V is volume has the same unit as
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The unit of pressure is force per unit area, while the unit of volume is length cubed. Therefore, the product of pressure and volume is given by force times length cubed, which is equivalent to work, the amount of energy transferred when a force of one newton is applied over a distance of one meter. Therefore, the product PV has the same unit as work, which is joule (J).
Tambaya 58 Rahoto
Which of the following is a fundamental quantity?
Bayanin Amsa
In physics, a fundamental quantity is a physical quantity that cannot be defined in terms of other physical quantities. These quantities are also known as base quantities or fundamental units. The seven fundamental quantities in the International System of Units (SI) are length, mass, time, electric current, temperature, amount of substance, and luminous intensity. Out of the options given, "length" is a fundamental quantity because it is one of the seven base quantities in the SI system. It is defined as the distance between two points in space, and it is measured in units such as meters, feet, or miles. The other options, such as "speed," "density," "impulse," and "energy," are all derived quantities, which means that they can be expressed in terms of one or more of the fundamental quantities. For example, speed is defined as the rate at which an object changes its position with respect to time, and it is derived from the fundamental quantities of length and time. Therefore, the correct option is "length," as it is a fundamental quantity.
Tambaya 59 Rahoto
At which of the following distances from the lens should a slide be placed on a slide projector if f is the focal length of the projector length?
Bayanin Amsa
Tambaya 60 Rahoto
(a)(i) By means of a labelled diagram, describe the mode uf operation of a modern X-ray tube.
(ii) State the energy transformation which takes place during the operation.
(b) Explain the terms hardness and intensity as applied to X-rays
(c)(i) State three uses of X-rays
(ii) State one hazard of over-exposure to X-rays in a radiological laboratory, indicating two safety precautions.
None
Bayanin Amsa
None
Tambaya 61 Rahoto
(a) Distinguish between temperature and heat. State the units in which they are measured
(ii) State two physical properties used for measuring temperature.
(b) (i) Describe, with the aid of a diagram, how the upper fixed point is determined for a mecury-in-glass thermometer. State one precaution to ensure accurate results
(ii) State one advantage which a constant-volume gas thermometer has over other thermometers and one reason why it is seldom used as an everyday laboratory instrument.
(c) Using the kinetic theory of matter explain why evaporation causes cooling.
Tambaya 62 Rahoto
(a) Explain the terms reactance and impedance in an a.c circuit.
(b) A source of e.m.f. 240 v and frequency 50 Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 10A, the potential difference across the resistor is 140V and that across the inductor is 50V. Draw the vector diagram of the potential difference across the inductor, the capacitor and the resistor.
Calculate the (i) potential difference across the capacitor; (ii) capacitance of the capacitor; (iii) inductance of the inductor.
None
Bayanin Amsa
None
Tambaya 63 Rahoto
(a) Explain the terms:- uniform acceleration and average speed.
(b) A body at rest is given an initial uniform acceleration of 8.0ms\(^{-2}\) for 30s after which the acceleration is reduced to 5.0ms\(^{-2}\) for the next 20s. The body maintains the speed attained for 60s after which it is brought to rest in 20s. Draw the velocity-time graph of the motion using the information given above.
(c) Using the graph, calculate the: (i) maximum speed attained during the motion; (ii) average retardation as the body is being brought to rest; (iii) total distance travelled during the first 50s; (iv) average speed during the same interval as in (ii).
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