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Ajụjụ 1 Ripọtì
Three resistors with resistance 200Ω, 500Ω and 1kΩ are connected in series. A 6v battery is connected to either end of the combination. Calculate the potential difference between the ends of 200Ω resistance.
Akọwa Nkọwa
To calculate the potential difference between the ends of the 200Ω resistance, we need to use Ohm's Law, which states that the potential difference (V) across a resistor is equal to the current (I) flowing through the resistor multiplied by the resistance (R) of the resistor. First, we need to find the total resistance of the series combination of resistors. We add up the individual resistances: Total resistance = 200Ω + 500Ω + 1kΩ = 1.7kΩ Next, we can use Ohm's Law to find the current flowing through the circuit. We know that the battery voltage is 6V, and the total resistance is 1.7kΩ: I = V / R = 6V / 1.7kΩ = 0.0035A Now we can use Ohm's Law again to find the potential difference across the 200Ω resistor: V = IR = 0.0035A * 200Ω = 0.7V Therefore, the potential difference between the ends of the 200Ω resistance is 0.7V. The correct answer is option B.
Ajụjụ 2 Ripọtì
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
Akọwa Nkọwa
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.
Ajụjụ 3 Ripọtì
Gases conduct electricity under
Akọwa Nkọwa
Gases conduct electricity under low pressure and high voltage
Ajụjụ 4 Ripọtì
Heat may be transferred by conduction, convention and radiation. By which of these methods does heat travel through vacuum?
Akọwa Nkọwa
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".
Ajụjụ 5 Ripọtì
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
Akọwa Nkọwa
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
Ajụjụ 6 Ripọtì
Which of the following bodies, each with centre of gravity G, lying on a horizontal table, is/are in unstable equilibrium?
Akọwa Nkọwa
- 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)
Ajụjụ 8 Ripọtì
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.
Akọwa Nkọwa
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 |
Ajụjụ 9 Ripọtì
Workdone on an object to bring it to a certain point in space is called
Akọwa Nkọwa
The work done on an object to bring it to a certain point in space is called "Potential Energy". Potential energy is a form of energy that an object possesses due to its position relative to other objects. When an object is lifted or moved to a higher point against gravity, work is done on it, and this work is stored as potential energy. The potential energy of an object is directly proportional to its height and mass. It can be converted into other forms of energy, such as kinetic energy, when the object is released or allowed to move freely. Therefore, potential energy is a type of stored energy that an object has due to its position, and it can be released to do work.
Ajụjụ 10 Ripọtì
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
Akọwa Nkọwa
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.
Ajụjụ 11 Ripọtì
The angular dispersion of a prism depends on
Akọwa Nkọwa
Dispersion is due to different refractive indices speeds and wavelengths.
Ajụjụ 13 Ripọtì
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?
Akọwa Nkọwa
| f1 | = | f(v - vo )v - vs | = | 1200(340 - 15)340 - 20 | = | 1.22KHz |
Ajụjụ 15 Ripọtì
Which of the following is consistent with Charles' law?
I
II
III
IV.
Akọwa Nkọwa
This is the correct graph. The graph is volume against 1/ temperature where temperature is in Celsius.
Ajụjụ 16 Ripọtì
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.
Akọwa Nkọwa
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.
Ajụjụ 17 Ripọtì
- angle of dip is zero at the magnetic equator
- angle of variation is the same as angle of declination.
Akọwa Nkọwa
- angle of dip is zero at the magnetic equator
- angle of variation is the same as angle of declination.
Ajụjụ 18 Ripọtì
The volume of a stone having an irregular shape can be determined using?
Akọwa Nkọwa
The volume of a stone with an irregular shape can be determined using a measuring cylinder. A measuring cylinder is a glass or plastic container with a narrow cylindrical shape and markings on the side to indicate the volume it contains. To determine the volume of an irregularly shaped stone, you would fill the measuring cylinder with water, carefully lower the stone into the water, and note the increase in the volume of the water. The difference in the volume of the water before and after the stone was added is equal to the volume of the stone. The meter rule, vernier calliper, and micrometer screw gauge are all measuring instruments, but they are not designed to measure the volume of irregularly shaped objects. The meter rule is a measuring tool used for measuring length. The vernier calliper is used for measuring the diameter of objects, and the micrometer screw gauge is used for precise measurements of small distances.
Ajụjụ 19 Ripọtì
The volume of 0.354g of helium at 273°C and 114cm of mercury pressure is 2667cm3 . Calculate the volume
Akọwa Nkọwa
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 |
Ajụjụ 20 Ripọtì
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
Akọwa Nkọwa
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.
Ajụjụ 21 Ripọtì
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
Akọwa Nkọwa
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.
Ajụjụ 22 Ripọtì
A train has an initial velocity of 44m/s and an acceleration of -4m/s2 . Calculate its velocity after 10 seconds
Akọwa Nkọwa
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.
Ajụjụ 23 Ripọtì
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?
Akọwa Nkọwa
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 |
Ajụjụ 24 Ripọtì
If the time of flight is 96seconds, calculate the horizontal range through the point of projection.
Akọwa Nkọwa
Time of flight, T = 96s
R = (Ucosθ) *time* T = 640 × 96 = 61,440m
Ajụjụ 25 Ripọtì
One newton × One meter equals?
Akọwa Nkọwa
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.
Ajụjụ 26 Ripọtì
In semi-conductor, the carriers of current at room temperature are
Akọwa Nkọwa
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.
Ajụjụ 27 Ripọtì
Efficiency of conduction in liquids and gases compared to solids is
Akọwa Nkọwa
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.
Ajụjụ 28 Ripọtì
Which of the following statement about the electromagnet shown above is correct?
Akọwa Nkọwa
A - B = S - N.
Also, starting end of the current is south while terminating end is North.
Ajụjụ 29 Ripọtì
When blue and green colours of light are mixed, the resultant colour is
Ajụjụ 30 Ripọtì
Any line or section taken through an advancing wave in which all the particles are in the same phase is called the
Akọwa Nkọwa
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.
Ajụjụ 31 Ripọtì
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.
Akọwa Nkọwa
From two parts 20m apart
a = 10m, x = 6m, A = 5
V = ω√A2−X2
= 5√102−62
= 40m/s
Ajụjụ 32 Ripọtì
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
Akọwa Nkọwa
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
Ajụjụ 33 Ripọtì
A body was slightly displaced from its equilibrium position. Which one of the following is a condition for its stable equilibrium
Akọwa Nkọwa
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.
Ajụjụ 34 Ripọtì
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.
Akọwa Nkọwa
Devices with different liquids
d1
h1
= d2
h2
1 × 20 = 0.8 × h
| h | = | 200.8 | = | 25cm |
Ajụjụ 35 Ripọtì
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
Akọwa Nkọwa
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.
Ajụjụ 36 Ripọtì
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
Akọwa Nkọwa
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.
Ajụjụ 37 Ripọtì
Neutrons were discovered by
Akọwa Nkọwa
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.
Ajụjụ 38 Ripọtì
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
Akọwa Nkọwa
R3.6=7525=13
3R = 3.6
R = 1.2Ω
Ajụjụ 40 Ripọtì
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
Akọwa Nkọwa
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.
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