Ana loda....
Latsa & Riƙe don Ja Shi Gabaɗaya |
|||
Danna nan don rufewa |
Tambaya 1 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 and at the same time undergoes an acceleration, its motion is said to be rectilinear. When an object moves with constant speed, it means that it covers the same distance in equal time intervals. On the other hand, acceleration is the rate of change of velocity with time. If an object undergoes acceleration, its velocity changes with time. Therefore, if a body moves with constant speed and undergoes an acceleration, it means that its direction of motion changes while it covers equal distances in equal time intervals. This type of motion is called rectilinear motion, where the object moves in a straight line, but its velocity changes due to the acceleration. In contrast, circular motion is when an object moves in a circular path with a constant speed, while oscillatory motion is when an object moves back and forth around a fixed point. Rotational motion is when an object rotates around an axis. None of these descriptions fit the scenario of a body moving with constant speed and undergoing acceleration, so the answer is rectilinear motion.
Tambaya 2 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 3 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 5 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 6 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 7 Rahoto
Gases conduct electricity under
Bayanin Amsa
Gases conduct electricity under low pressure and high voltage
Tambaya 8 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 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
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 11 Rahoto
In the molecular explanation, heat is transferred by the
Bayanin Amsa
- Conduction is explained in terms of the free electrons
- Convection is explained in terms of the movement of the fluid involved
- Radiation is explained in terms of invisible electromagnetic waves.
Tambaya 12 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 13 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 14 Rahoto
A mass of 0.5kg is whirled in a vertical circle of radius 2m at a steady rate of 2 rev/s. Calculate the centripetal force
Bayanin Amsa
The centripetal force is the force that acts towards the center and keeps an object moving in a circular path. To calculate the centripetal force, we can use the following formula: f = m * v^2 / r where: - f = centripetal force - m = mass of the object (0.5 kg) - v = velocity of the object (2 rev/s * 2 * pi m/rev = 12.57 m/s) - r = radius of the circle (2 m) Plugging in the values, we get: f = 0.5 kg * 12.57 m/s^2 / 2 m f = 31.43 N Rounding to the nearest whole number, the centripetal force is 31 N. So, the closest answer from the options is 160N.
Tambaya 15 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 16 Rahoto
A man on a bench will exert the greatest pressure on the bench when he
Bayanin Amsa
The man on the bench will exert the greatest pressure when he stands on the toes of one foot. This is because when he stands on one foot, all his weight is concentrated on a smaller surface area of the bench, resulting in more pressure. The pressure he exerts is calculated by dividing his weight by the surface area in contact with the bench. When he stands on one foot, the surface area is smaller, which means the pressure exerted is greater. In comparison, when he lies flat on his back or belly, or when he stands on both feet, his weight is distributed over a larger surface area, resulting in less pressure.
Tambaya 17 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 18 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 19 Rahoto
According to kinetic molecular model, in gases
Bayanin Amsa
In kinetic molecular model, gases are energised and thus moves freely, fast as they occupy specific space
Tambaya 20 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 21 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 22 Rahoto
In semi-conductor, the carriers of current at room temperature are
Bayanin Amsa
In a semiconductor, the carriers of current at room temperature are both electrons and holes. Semiconductors are materials with properties that are in between those of conductors (e.g. metals) and insulators (e.g. rubber). At room temperature, a semiconductor crystal contains both free electrons and positively charged vacancies called holes. When a voltage is applied across the semiconductor, the electrons move towards the positive end of the circuit and the holes move towards the negative end. This movement of charge carriers constitutes an electric current. In summary, both electrons and holes can carry current in a semiconductor at room temperature, making the correct answer.
Tambaya 23 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 24 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 25 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 26 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 27 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 28 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 29 Rahoto
The lower fixed part of a faulty thermometer reads 2°C while the upper fixed point is 100°C.
What is the true temperature when the thermometer reads 51°C?
Bayanin Amsa
Since the thermometer is faulty, it is not measuring the temperature accurately. To find the true temperature, we need to determine the extent of the error in the thermometer. We can do this by comparing the difference between the lower fixed point and the reading with the difference between the upper fixed point and the true temperature. Since the lower fixed point reads 2°C and the upper fixed point reads 100°C, and the thermometer reading is 51°C, we can calculate the error as follows: True temperature = (51°C - 2°C) / (51°C - 2°C) * (100°C - 51°C) + 51°C = 50°C So, the true temperature when the thermometer reads 51°C is 50°C, which is option B.
Tambaya 30 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 31 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 32 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 33 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 34 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 35 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 36 Rahoto
The diagram above represents the stress-strain graph of a loaded wire. Which of these statements is correct?
Bayanin Amsa
- I is the elastic limit
- the end of the constant part J is the yield point
- L is the break point.
Tambaya 37 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 38 Rahoto
A cone is in unstable equilibrium has its potential energy
Bayanin Amsa
In unstable equilibrium, potential energy decreases as the height decreases.
Tambaya 39 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 40 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 |
Za ka so ka ci gaba da wannan aikin?